Part C Lévy Processes and Finance

Part C Levy´ Processes and Finance

Matthias Winkel1 University of Oxford

HT 2007

1Departmental lecturer (Institute of Actuaries and Aon Lecturer in Statistics) at the Department of Statistics, University of Oxford

MS3 Levy´ Processes and Finance Matthias Winkel – 16 lectures HT 2007

Prerequisites Part A Probability is a prerequisite. BS3a/OBS3a Applied Probability or B10 Martin- gales and Financial Mathematics would be useful, but are by no means essential; some material from these courses will be reviewed without proof. Aims Lévy processes form a central class of stochastic processes, contain both Brownian motion and the Poisson process, and are prototypes of Markov processes and semimartingales. Like Brownian motion, they are used in a multitude of applications ranging from biology and physics to insurance and finance. Like the Poisson process, they allow to model abrupt moves by jumps, which is an important feature for many applications. In the last ten years Lévy processes have seen a hugely increased attention as is reflected on the academic side by a number of excellent graduate texts and on the industrial side realising that they provide versatile stochastic models of financial markets. This continues to stimulate further research in both theoretical and applied directions. This course will give a solid introduction to some of the theory of Lévy processes as needed for financial and other applications. Synopsis Review of (compound) Poisson processes, Brownian motion (informal), Markov property. Connection with random walks, [Donsker’s theorem], Poisson limit theorem. Spatial Poisson processes, construction of Lévy processes. Special cases of increasing Lévy processes (subordinators) and processes with only positive jumps. Subordination. Examples and applications. Financial models driven by Lévy processes. Stochastic volatility. Level passage problems. Applications: option pricing, insurance ruin, dams. Simulation: via increments, via simulation of jumps, via subordination. Applications: option pricing, branching processes. Reading • J.F.C. Kingman: Poisson processes. Oxford University Press (1993), Ch.1-5, 8 • A.E. Kyprianou: Introductory lectures on fluctuations of Lévy processes with Ap- plications. Springer (2006), Ch. 1-3, 8-9 • W. Schoutens: Lévy processes in finance: pricing financial derivatives. Wiley (2003) Further reading • J. Bertoin: Lévy processes. Cambridge University Press (1996), Sect. 0.1-0.6, I.1, III.1-2, VII.1 • K. Sato: Lévy processes and infinite divisibility.Cambridge University Press (1999), Ch. 1-2, 4, 6, 9

Lecture 1

Introduction

Reading: Kyprianou Chapter 1 Further reading: Sato Chapter 1, Schoutens Sections 5.1 and 5.3

In this lecture we give the general definition of a Lévy process, study some examples of Lévy processes and indicate some of their applications. By doing so, we will review some results from BS3a Applied Probability and B10 Martingales and Financial Mathematics.

1.1 Deﬁnition of L´evy processes

Stochastic processes are collections of random variables Xt, t ≥ 0 (meaning t ∈ [0, ∞) as opposed to n ≥ 0 by which means n ∈ N = {0, 1, 2,...}). For us, all Xt, t ≥ 0, take values in a common state space, which we will choose speciﬁcally as R (or [0, ∞) or Rd for some d ≥ 2). We can think of Xt as the position of a particle at time t, changing as t varies. It is natural to suppose that the particle moves continuously in the sense that t 7→ Xt is continuous (with probability 1), or that it has jumps for some t ≥ 0:

∆Xt = Xt+ − Xt = lim Xt+ε − lim Xt ε. − ε 0 ε 0 − ↓ ↓

We will usually suppose that these limits exist for all t ≥ 0 and that in fact Xt+ = Xt, i.e. that t 7→ Xt is right-continuous with left limits Xt for all t ≥ 0 almost surely. The − path t 7→ Xt can then be viewed as a random right-continuous function.

Definition 1 (Lévy process) A real-valued (or Rd-valued) stochastic process X = (Xt)t 0 is called a Lévy process if ≥ − − ≥ (i) the random variables Xt0 ,Xt1 Xt0 ,...,Xtn Xtn−1 are independent for all n 1 and 0 ≤ t0 < t1 <...

(ii) Xt+s − Xt has the same distribution as Xs for all s, t ≥ 0 (stationary increments),

(iii) the paths t 7→ Xt are right-continuous with left limits (with probability 1).

It is implicit in (ii) that P(X0 = 0) = 1 (choose s = 0). 1 2 Lecture Notes – MS3b L´evy Processes and Finance – Oxford HT 2008

Figure 1.1: Variance Gamma process and a L´evy process with no positive jumps

Here the independence of n random variables is understood in the following sense: ( ) Deﬁnition 2 (Independence) Let Y j be an Rdj -valued random variable for j = 1,...,n. The random variables Y (1),...,Y (n) are called independent if, for all (Borel ( ) measurable) C j ⊂ Rdj P(Y (1) ∈ C(1),...,Y (n) ∈ C(n))= P(Y (1) ∈ C(1)) . . . P(Y (n) ∈ C(n)). (1)

(j) (j1) (jn) An infinite collection (Y )j J is called independent if Y ,...,Y are independent for ∈ (1) (n) every finite subcollection. Infinite-dimensional random variables (Yi )i I1 ,..., (Yi )i In (1) (n) ∈ ⊂ ∈ are called independent if (Yi )i F1 ,..., (Yi )i Fn are independent for all finite Fj Ij. ∈ ∈ (j) (j) (j) × × (j) (j) It is sufficient to check (1) for rectangles of the form C =(a1 , b1 ] . . . (adj , bdj ].

1.2 First main example: Poisson process

Poisson processes are Lévy processes. We recall the definition as follows. An N(⊂ R)- valued stochastic process X =(Xt)t 0 is called a Poisson process with rate λ ∈ (0, ∞) if ≥ X satisfies (i)-(iii) and Poi P (λt)k λt ≥ ≥ (iv) (Xt = k)= k! e− , k 0, t 0 (Poisson distribution). The Poisson process is a continuous-time Markov chain. We will see that all Lévy processes have a Markov property. Also recall that Poisson processes have jumps of size 1 (spaced by independent exponential random variables Zn = Tn+1−Tn, n ≥ 0, with param- λs eter λ, i.e. with density λe− , s ≥ 0). In particular, {t ≥ 0 : ∆Xt =6 0} = {Tn, n ≥ 1}

and ∆XTn = 1 almost surely (short a.s., i.e. with probability 1). We can deﬁne more general L´evy processes by putting

Ct = Yk, t ≥ 0, =1 Xk for a Poisson process (Xt)t 0 and independent identically distributed Yk, k ≥ 1. Such ≥ processes are called compound Poisson processes. The term “compound” stems from the ◦ representation Ct = S Xt = SXt for the random walk Sn = Y1 + . . . + Yn. You may think of Xt as the number of claims up to time t and of Yk as the size of the kth claim. Recall (from BS3a) that its moment generating function, if it exists, is given by

γY1 E(exp{γCt}) = exp{−λt(E(e − 1)}. This will be an important building block of a general L´evy process. Lecture 1: Introduction 3

Figure 1.2: Poisson process and Brownian motion

1.3 Second main example: Brownian motion

Brownian motion is a Lévy process. We recall (from B10b) the definition as follows. An R-valued stochastic process X =(Xt)t 0 is called Brownian motion if X satisfies (i)-(ii) ≥ and BM (iii) the paths t 7→ Xt are continuous almost surely,

2 (iv)BM P(X ≤ x)= x 1 exp − y dy, x ∈ R, t> 0. (Normal distribution). t √2πt 2t −∞ The paths of BrownianR motion aren continuous,o but turn out to be nowhere diﬀerentiable (we will not prove this). They exhibit erratic movements at all scales. This makes Brownian√ motion an appealing model for stock prices. Brownian motion has the scaling property ( cXt/c)t 0 ∼ X where “∼” means “has the same distribution as”. ≥ Brownian motion will be the other important building block of a general L´evy process. The canonical space for Brownian paths is the space C([0, ∞), R) of continuous real- valued functions f : [0, ∞) → R which can be equipped with the topology of locally uniform convergence, induced by the metric

k d(f,g)= 2− min{dk(f,g), 1}, where dk(f,g) = sup |f(x) − g(x)|. [0 ] k 1 x ,k X≥ ∈ This metric topology is complete (Cauchy sequences converge) and separable (has a countable dense subset), two attributes important for the existence and properties of limits. The bigger space D([0, ∞), R) of right-continuous real-valued functions with left limits can also be equipped with the topology of locally uniform convergence. The space is still complete, but not separable. There is a weaker metric topology, called Skorohod’s topology, that is complete and separable. In the present course we will not develop this and only occasionally use the familiar uniform convergence for (right-continuous) functions f, fn : [0,k] → R, n ≥ 1:

sup |fn(x) − f(x)|→ 0, as n →∞, x [0,k] ∈ which for stochastic processes X,X(n), n ≥ 1, with time range t ∈ [0, T ] takes the form | (n) − |→ →∞ sup Xt Xt 0, as n , t [0,T ] ∈ and will be as a convergence in probability or as almost sure convergence (from BS3a or 2 2 2 B10a) or as L -convergence, where Zn → Z in the L -sense means E(|Zn − Z| ) → 0. 4 Lecture Notes – MS3b L´evy Processes and Finance – Oxford HT 2008

1.4 Markov property

The Markov property is a consequence of the independent increments property (and the stationary increments property): Proposition 3 (Markov property) Let X be a Lévy process and t ≥ 0 a fixed time, then the pre-t process (Xr)r t is independent of the post-t process (Xt+s − Xt)s 0, and the ≤ ≥ post-t process has the same distribution as X. − Proof: By Definition 2, we need to check the independence of (Xr1 ,...,Xrn )and (Xt+s1 − Xt,...,Xt+sm Xt). By property (i) of the Lévy process, we have that increments over disjoint time intervals are independent, in particular the increments − − − − − Xr1 ,Xr2 Xr1 ,...,Xrn Xrn−1 ,Xt+s1 Xt,Xt+s2 Xt+s1 ,...,Xt+sm Xt+sm−1 . Since functions (here linear transformations from increments to marginals) of independent random variables are independent, the proof of independence is complete. Identical distribution follows first on the level of single increments from (ii), then by (i) and linear transformation also for finite-dimensional marginal distributions. 2

1.5 Some applications

Example 4 (Insurance ruin) A compound Poisson process (Zt)t 0 with positive jump ≥ sizes Ak, k ≥ 1, can be interpreted as a claim process recording the total claim amount incurred before time t. If there is linear premium income at rate r > 0, then also the gain process rt − Zt, t ≥ 0, is a L´evy process. For an initial reserve of u> 0, the reserve process u + rt − Zt is a shifted L´evy process starting from a non-zero initial value u.

Example 5 (Financial stock prices) Brownian motion (Bt)t 0 or linear Brownian mo- ≥ tion σBt + µt, t ≥ 0, was the first model of stock prices, introduced by Bachelier in 1900. Black, Scholes and Merton studied geometric Brownian motion exp(σBt + µt) in 1973, which is not itself a Lévy process but can be studied with similar methods. The Economics Nobel Prize 1997 was awarded for their work. Several deficiencies of the Black-Scholes model have been identified, e.g. the Gaussian density decreases too quickly, no variation of the volatility σ over time, no macroscopic jumps in the price processes. These deficiencies can be addressed by models based on Lévy processes. The Variance gamma model is a time-changed Brownian motion BTs by an independent increasing jump process, a ∼ so-called Gamma Lévy process with Ts Gamma(αs,β). The process BTs is then also a Lévy process itself. Example 6 (Population models) Branching processes are generalisations of birth- and-death processes (see BS3a) where each individual in a population dies after an exponentially distributed lifetime with parameter µ, but gives birth not to single children, but to twins, triplets, quadruplet etc. To simplify, it is assumed that children are only born at the end of a lifetime. The numbers of children are independent and identically distributed according to an offspring distribution q on {0, 2, 3,...}. The population size process (Zt)t 0 can jump downwards by 1 or upwards by an integer. It is not a Lévy ≥ process but is closely related to Lévy processes and can be studied with similar methods. There are also analogues of processes in [0, ∞), so-called continuous-state branching processes that are useful large-population approximations. Lecture 2

L´evy processes and random walks

Reading: Kingman Section 1.1, Grimmett and Stirzaker Section 3.5(4) Further reading: Sato Section 7, Durrett Sections 2.8 and 7.6, Kallenberg Chapter 15 Lévy processes are the continuous-time analogues of random walks. In this lecture we examine this analogy and indicate connections via scaling limits and other limiting results. We begin with a first look at infinite divisibility.

2.1 Increments of random walks and L´evy processes

Recall that a random walk is a stochastic process in discrete time n

S0 =0, Sn = Aj, n ≥ 1, j=1 X d for a family (Aj)j 1 of independent and identically distributed real-valued (or R -valued) ≥ random variables. Clearly, random walks have stationary and independent increments. Specifically, the Aj, j ≥ 1, themselves are the increments over single time units. We refer to Sn+m − Sn as an increment over m time units, m ≥ 1. While every distribution may be chosen for Aj, increments over m time units are sums of m independent and identically distributed random variables, and not every distribution has this property. This is not a deep observation, but it becomes important when moving to Lévy processes. In fact, the increment distribution of Lévy processes is restricted: any increment Xt+s − Xt, or Xs for simplicity, can be decomposed, for every m ≥ 1, m

Xs = (Xjs/m − X(j 1)s/m) − j=1 X into a sum of m independent and identically distributed random variables. Definition 7 (Infinite divisibility) A random variable Y is said to have an infinitely divisible distribution if for every m ≥ 1, we can write ∼ (m) (m) Y Y1 + . . . + Ym (m) (m) for some independent and identically distributed random variables Y1 ,...,Ym .

(m) We stress that the distribution of Yj may vary as m varies, but not as j varies. 5 6 Lecture Notes – MS3b L´evy Processes and Finance – Oxford HT 2008

The argument just before the definition shows that increments of Lévy processes are infinitely divisible. Many known distributions are infinitely divisible, some are not. Example 8 The Normal, Poisson, Gamma and geometric distributions are infinitely divisible. This often follows from the closure under convolutions of the type 2 2 2 2 Y1 ∼ Normal(µ, σ ),Y2 ∼ Normal(ν, τ ) ⇒ Y1 + Y2 ∼ Normal(µ + ν, σ + τ ) for independent Y1 and Y2 since this implies by induction that for independent (m) (m) ∼ 2 ⇒ (m) (m) ∼ 2 Y1 ,...,Ym Normal(µ/m, σ /m) Y1 + . . . + Ym Normal(µ, σ ). The analogous arguments (and calculations, if necessary) for the other distributions are left as an exercise. The geometric(p) distribution here is P(X = n)= pn(1 − p), n ≥ 0. Example 9 The Bernoulli(p) distribution, for p ∈ (0, 1), is not infinitely divisible. As- sume that you can represent a Bernoulli(p) random variable X as Y1 +Y2 for independent identically distributed Y1 and Y2. Then

P(Y1 > 1/2) > 0 ⇒ 0= P(X > 1) ≥ P(Y1 > 1/2,Y2 > 1/2) > 0 is a contradiction, so we must have P(Y1 > 1/2) = 0, but then √ P(Y1 > 1/2)=0 ⇒ p = P(X =1)= P(Y1 =1/2)P(Y2 =1/2) ⇒ P(Y1 =1/2) = p. Similarly, P(Y1 < 0) > 0 ⇒ 0= P(X < 0) ≥ P(Y1 < 0,Y2 < 0) > 0 is a contradiction, so we must have P(Y1 < 0) = 0 and then

1 − p = P(X =0)= P(Y1 =0,Y2 = 0) ⇒ P(Y1 =0)= 1 − p> 0. √ √ This is impossible for several reasons. Clearly, p + 1 − p> 1, butp also

0= P(X =1/2) ≥ P(Y1 = 0)P(Y2 =1/2) > 0.

2.2 Central Limit Theorem and Donsker’s theorem

2 Theorem 10 (Central Limit Theorem) Let (Sn)n 0 be a random walk with E(S1 )= 2 ≥ E(A1) < ∞. Then, as n →∞,

S − E(S ) S − nE(A1) n n = n → Normal(0, 1) in distribution. Var(Sn) nVar(A1) This resultp as a result forp one time n →∞ can be extended to a convergence of processes, a convergence of the discrete-time process (Sn)n 0 to a (continuous-time) Brownian ≥ motion, by scaling of both space and time. The processes

S[ ] − [nt]E(A1) nt , t ≥ 0, nVar(A1) where [nt] ∈ Z with [nt] ≤ nt < [pnt]+1 denotes the integer part of nt, are scaled versions of the random walk (Sn)n 0, now performing n steps per time unit (holding time 1/n), ≥ centred and each only a multiple 1/ nVar(A1) of the original size. If E(A1) = 0, you may think that you look at (Sn)n 0 from further and further away, but note that space ≥ p and time are scaled diﬀerently, in fact so as to yield a non-trivial limit. Lecture 2: L´evy processes and random walks 7

Figure 2.1: Random walk converging to Brownian motion

2 2 Theorem 11 (Donsker) Let (Sn)n 0 be a random walk with E(S1 ) = E(A1) < ∞. ≥ Then, as n →∞,

S[nt] − [nt]E(A1) → Bt locally uniformly in t ≥ 0, nVar(A1)

“in distribution”, forp a Brownian motion (Bt)t 0. ≥ Proof: [only for A1 ∼ Normal(0, 1)] This proof is a coupling proof. We are not going to work directly with the original random walk (Sn)n 0, but start from Brownian motion ≥ (Bt)t 0 and deﬁne a family of embedded random walks ≥ (n) ≥ ≥ Sk := Bk/n, k 0, n 1.

Then note using in particular E(A1) = 0 and Var(A1) = 1 that

(n) S1 − E(A1) S1 ∼ Normal(0, 1/n) ∼ , nVar(A1) and indeed p − E (n) ∼ S[nt] [nt] (A1) S[nt] . t 0 nVar(A1) ≥ !t 0 ≥ To show convergence in distribution for thep processes on the right-hand side, it suﬃces to establish convergence in distribution for the processes on the left-hand side, as n →∞. To show locally uniform convergence we take an arbitrary T ≥ 0 and show uniform convergence on [0, T ]. Since (Bt)0 t T is uniformly continuous (being continuous on a ≤ ≤ compact interval), we get a.s.

sup S[nt] − Bt ≤ sup |Bs − Bt|→ 0 0 t T 0 s t T : s t 1/n ≤ ≤ ≤ ≤ ≤ | − |≤ as n →∞. This establishes a.s. convergence, which “implies” convergence in distribution for the embedded random walks and for the original scaled random walk. This completes the proof for A1 ∼ Normal(0, 1). 2 Note that the almost sure convergence only holds for the embedded random walks (n) ≥ (Sk )k 0, n 1. Since the identity in distribution with the rescaled original random ≥ walk only holds for ﬁxed n ≥ 1, not jointly, we cannot deduce almost sure convergence in the statement of the theorem. Indeed, it can be shown that almost sure convergence will fail. The proof for general increment distribution is much harder and will not be given in this course. If time permits, we will give a similar coupling proof for another important special case where P(A1 =1)= P(A1 = −1)=1/2, the simple symmetric random walk. 8 Lecture Notes – MS3b L´evy Processes and Finance – Oxford HT 2008

2.3 Poisson limit theorem

The Central Limit Theorem for Bernoulli random variables A1,...,An says that for large n, the number of 1s in the sequence is well-approximated by a Normal random variable. In practice, the approximation is good if p is not too small. If p is small, the Bernoulli random variables count rare events, and a diﬀerent limit theorem is relevant:

Theorem 12 (Poisson limit theorem) Let Wn be binomially distributed with parameters n and pn = λ/n (or if npn → λ, as n →∞). Then we have

Wn → P oi(λ), in distribution, as n →∞. Proof: Just calculate that, as n →∞, − − k − npn n k n k n k n(n 1) . . . (n k + 1) (npn) 1 n λ λ p (1 − p ) − = → e− . k n n k! nk − k k! (1 pn) 2 (n) (n) (n) ≥ Theorem 13 Suppose that Sk = A1 + . . . + Ak , k 0, is the sum of independent Bernoulli(pn) random variables for all n ≥ 1, and that npn → λ ∈ (0, ∞). Then (n) → ≥ S[nt] Nt “in the Skorohod sense” as functions of t 0,

“in distribution” as n →∞, for a Poisson process (Nt)t 0 with rate λ. ≥ ( ) ( ) The proof of so-called ﬁnite-dimensional convergence for vectors (S n ,...,S n ) is [nt1] [ntm] (n) not very hard but not included here. One can also show that the jump times (Tm )m 1 (n) ≥ of (S[ ])t 0 converge to the jump times of a Poisson process. E.g. nt ≥ [nt] (n) [nt] [nt]pn P(T1 > t)=(1 − p ) = 1 − → exp{−λt}, n [nt] since [nt]/n → t (since (nt − 1)/n → t and nt/n = t) and so [nt]pn → tλ. The general statement is hard to make precise and prove, certainly beyond the scope of this course.

2.4 Generalisations

Infinitely divisible distributions and Lévy processes are precisely the classes of limits that arise for random walks as in Theorems 10 and 12 (respectively 11 and 13) with different 1/α step distributions. Stable Lévy processes are ones with a scaling property (c Xt/c)t 0 ∼ ≥ X for some α ∈ R. These exist, in fact, for α ∈ (0, 2]. Theorem 10 (and 11) for suitable 2 distributions of A1 (depending on α and where E(A1) = ∞ in particular) then yield convergence in distribution

S − nE(A1) S n → stable(α) for α ≥ 1, or n → stable(α) for α ≤ 1. n1/α n1/α Example 14 (Brownian ladder times) For a Brownian motion B and a level r > 0, the distribution of Tr = inf{t ≥ 0 : Bt >r} is 1/2-stable, see later in the course. Example 15 (Cauchy process) The Cauchy distribution with density a/(π(x2 + a2)), x ∈ R, for some parameter c ∈ R is 1-stable, see later in the course. Lecture 3

Spatial Poisson processes

Reading: Kingman 1.1 and 2.1, Grimmett and Stirzaker 6.13, Kyprianou Section 2.2 Further reading: Sato Section 19

We will soon construct the most general nonnegative Lévy process (and then general real-valued ones). Even though we will not prove that they are the most general, we have already seen that only infinitely divisible distributions are admissible as increment distributions, so we know that there are restrictions; the part missing in our discussion will be to show that a given distribution is infinitely divisible only if there exists a Lévy process X of the type that we will construct such that X1 has the given distribution. Today we prepare the construction by looking at spatial Poisson processes, objects of interest in their own right.

3.1 Motivation from the study of L´evy processes

Brownian motion (Bt)t 0 has continuous sample paths. It turns out that (σBt + µt)t 0 ≥ ≥ for σ ≥ 0 and µ ∈ R is the only continuous Lévy process. To describe the full class of Lévy processes (Xt)t 0, it is vital to study the process (∆Xt)t 0 of jumps. ≥ ≥ Take e.g. the Variance Gamma process. In Assignment 1.2.(b), we introduce this process as Xt = Gt − Ht, t ≥ 0, for two independent Gamma Lévy processes G and H. But how do Gamma Lévy processes evolve? We could simulate discretisations (and will do!) and get some feeling for them, but we also want to understand them mathematically. Do they really exist? We have not shown this. Are they compound Poisson processes? Let us look at their moment generating function (cf. Assignment 2.4.):

αt β ∞ γx 1 βx E(exp{γGt})= = exp αt (e − 1) e− dx . β − γ 0 x Z This is almost of the form of a compound Poisson process of rate λ with non-negative ≥ jump sizes Yj, j 1, that have a probability density function h(x)= hY1 (x), x> 0:

∞ γx E(exp{γCt}) = exp{λt (e − 1)h(x)dx} 0 Z To match the two expressions, however, we would have to put

(0) α βx λh(x)= λ0h (x)= e− , x> 0, x 9 10 Lecture Notes – MS3b L´evy Processes and Finance – Oxford HT 2008

(0) α βx and h cannot be a probability density function, because x e− is not integrable at x ↓ 0. What we can do is e.g. truncate at ε> 0 and specify

(ε) α βx (ε) λ h (x)= e− , x>ε, h (x)=0, x ≤ ε. ε x ( ) ε ∞ α βx In order for h to be a probability density, we just put λε = ε x e− dx, and notice that λ → ∞ as ε ↓ 0. But λ is the rate of the Poisson process driving the compound ε ε R Poisson process, so jumps are more and more frequent as ε ↓ 0. On the other hand, the average jump size, the mean of the distribution with density h(ε) tends to zero, so most of these jumps are very small. In fact, we will see that

Gt = ∆Gs, s t X≤ as an absolutely convergent series of inﬁnitely (but clearly countably) many positive jump α βx sizes, where (∆Gs)s 0 is a Poisson point process with intensity g(x)= e− , x> 0, the ≥ x collection of random variables

N((a, b] × (c,d]) = #{t ∈ (a, b] : ∆Gt ∈ (c,d]}, 0 ≤ a < b, 0

a Poisson counting measure (evaluated on rectangles) with intensity function λ(t, x) = g(x), x > 0, t ≥ 0; the random countable set {(t, ∆Gt) : t ≥ 0 and ∆Ct =06 } a spatial Poisson process with intensity λ(t, x). Let us now formally introduce these notions.

3.2 Poisson counting measures

The essence of one-dimensional Poisson processes (Nt)t 0 is the set of arrival (“event”) ≥ times Π = {T1, T2, T3,...}, which is a random countable set. The increment N((s, t]) := Nt − Ns counts the number of points in Π ∩ (s, t]. We can generalise this concept to counting measures of random countable subsets on other spaces, say Rd. Saying directly what exactly (the distribution of) random countable sets is, is quite diﬃcult in general. Random counting measures are a way to describe the random countable sets implicitly.

Deﬁnition 16 (Spatial Poisson process) A random countable subset Π ⊂ Rd is called a spatial Poisson process with (constant) intensity λ if the random variables N(A) = #Π ∩ A, A ⊂ Rd (Borel measurable, always, for the whole course, but we stop saying this all the time now), satisfy

d (a) for all n ≥ 1 and disjoint A1,...,An ⊂ R , the random variables N(A1),...,N(An) are independent, hom(b) N(A) ∼ Poi(λ|A|), where |A| denotes the volume (Lebesgue measure) of A.

Here, we use the convention that X ∼ Poi(0) means P(X =0)=1 and X ∼ Poi(∞) means P(X = ∞) = 1. This is consistent with E(X) = λ for X ∼ Poi(λ), λ ∈ (0, ∞). This convention captures that Π does not have points in a given set of zero volume a.s., and it has infinitely many points in given sets of infinite volume a.s. In fact, the definition fully specifies the joint distributions of the random set function d d N on subsets of R , since for any non-disjoint B1,...,Bm ⊂ R we can consider all Lecture 3: Spatial Poisson processes 11

∩ ∩ intersections of the form Ak = B1∗ . . . Bm∗ , where each Bj∗ is either Bj∗ = Bj or Bj∗ = c Rd \ m Bj = Bj. They form n = 2 disjoint sets A1,...,An to which (a) of the definition applies. (N(B1),...,N(Bm)) is a just a linear transformation of (N(A1),...,N(An)). Grimmett and Stirzaker collect a long list of applications including modelling stars in a galaxy, galaxies in the universe, weeds in the lawn, the incidence of thunderstorms and tornadoes. Sometimes the process in Definition 16 is not a perfect description of such a system, but useful as a first step. A second step is the following generalisation: Definition 16 (Spatial Poisson process, continued) A random countable subset Π ⊂ D ⊂ Rd is called a spatial Poisson process with (locally integrable) intensity function λ : D → [0, ∞), if N(A) = #Π ∩ A, A ⊂ D, satisfy

(a) for all n ≥ 1 and disjoint A1,...,An ⊂ D, the random variables N(A1),...,N(An) are independent, inhom ∼ (b) N(A) Poi A λ(x)dx . Definition 17 (PoissonR counting measure) A set function A 7→ N(A) that satisfies (a) and inhom(b) is referred to as a Poisson counting measure with intensity function λ(x). (j) (j) × × (j) (j) It is sufficient to check (a) and (b) for rectangles Aj =(a1 , b1 ] . . . (ad , bd ]. The set function Λ(A) = A λ(x)dx is called the intensity measure of Π. Definitions 16 and 17 can be extended to measures that are not integrals of intensity functions. R Only if Λ({x}) > 0, we would require P(N({x}) ≥ 2) > 0 and this is incompatible with N({x}) = #Π ∩{x} for a random countable set Π, so we prohibit such “atoms” of Λ.

Example 18 (Compound Poisson process) Let (Ct)t 0 be a compound Poisson pro- ≥ cess with independent jump sizes Yj, j ≥ 1 with common probability density h(x), x> 0, at the times of a Poisson process (Xt)t 0 with rate λ> 0. Let us show that ≥

N((a, b] × (c,d]) = #{t ∈ (a, b] : ∆Ct ∈ (c,d]}

deﬁnes a Poisson counting measure. First note N((a, b] × (0, ∞)) = Xb − Xa. Now recall Thinning property of Poisson processes: If each point of a Poisson process (Xt)t 0 of rate λ is of type 1 with probability p and of type 2 with prob- ≥ ability 1 − p, independently of one another, then the processes X(1) and X(2) counting points of type 1 and 2, respectively, are independent Poisson processes with rates pλ and (1 − p)λ, respectively.

Consider the thinning mechanism, where the jth jump is of type 1 if Yj ∈ (c,d]. Then, the process counting jumps in (c,d] is a Poisson process with rate λP(Y1 ∈ (c,d]), and so × (1) − (1) ∼ − P ∈ N((a, b] (c,d]) = Xb Xa Poi((b a)λ (Y1 (c,d])).

We identify the intensity measure Λ((a, b] × (c,d])=(b − a)λP(Y1 ∈ (c,d]). For the independence of counts in disjoint rectangles A1,...,An, we cut them into smaller rectangles Bi =(ai, bi]×(ci,di], 1 ≤ i ≤ m such that for any two Bi and Bj either (ci,di]=(cj,dj] or (ci,di] ∩ (cj,dj] = ∅. Denote by k the number of diﬀerent intervals (ci,di], w.l.o.g. (ci,di] for 1 ≤ i ≤ k. Now a straightforward generalisation of the thinning (i) property to k types splits (Xt)t 0 into k independent Poisson processes X with rates ≥ λP(Y1 ∈ (ci,di]), 1 ≤ i ≤ k. Now N(B1),...,N(Bm) are independent as increments of independent Poisson processes or of the same Poisson process over disjoint time intervals. 12 Lecture Notes – MS3b L´evy Processes and Finance – Oxford HT 2008

3.3 Poisson point processes

In Example 18, the intensity measure is of the product form Λ((a, b] × (c,d]) = (b − a)ν((c,d]) for a measure ν on D0 = (0, ∞). Take D = [0, ∞) × D0 in Deﬁnition 16. This means, that the spatial Poisson process is homogeneous in the ﬁrst component, the time component, like the Poisson process.

Proposition 19 If Λ((a, b] × A0)=(b − a) g(x)dx for a locally integrable function g A0 on D0 (or =(b − a)ν(A0) for a locally ﬁnite measure ν on D0), then no two points of Π share the same ﬁrst coordinate. R

Proof: If ν is finite, this is clear, since then Xt = N([0, t] × D0), t ≥ 0, is a Poisson process with rate ν(D0). Let us restrict attention to D0 = R∗ = R \{0} for simplicity – this is the most relevant case for us. The local integrability condition means that we can find intervals (In)n 1 such that 1 In = D0 and ν(In) < ∞, n ≥ 1. Then the ≥ n × ≥ ≥ (n) × independence of N((tj 1, tj] In), j =1,...,m, n 1, implies that Xt = N([0, t] In), − S t ≥ 0, are independent Poisson processes with rates ν(In), n ≥ 1. Therefore any two of (n) ≥ ≥ the jump times (Tj , j 1, n 1) are jointly continuously distributed and take different values almost surely: x P (n) (m) ∞ 6 (Tj = Ti )= fT (n) (x)fT (m) (y)dydx =0 forall n = m. 0 j i Z Zx (n) − (m) [Alternatively, show that Tj Ti has a continuous distribution and hence does not take a fixed value 0 almost surely]. Finally, there are only countably many pairs of jump times, so almost surely no two jump times coincide. 2 Let Π be a spatial Poisson process with intensity measure Λ((a, b] × (c,d]) = (b − d 0 − a) c g(x)dx for a locally integrable function g on D (or = (b a)ν((c,d]) for a locally finite measure ν on D0), then the process (∆t)t 0 given by R ≥ ∆t = 0 if Π ∩{t} × D = ∅, ∆t = x if Π ∩{t} = {(t, x)} is a Poisson point process in D0 ∪{0} with intensity function g on D0 in the sense of the following definition. d 1 Definition 20 (Poisson point process) Let g be locally integrable on D0 ⊂ R − \{0} (or ν locally finite). A process (∆t)t 0 in D0 ∪{0} such that ≥ N((a, b] × A0) = #{t ∈ (a, b] : ∆t ∈ A0}, 0 ≤ a < b, A0 ⊂ D0 (measurable), is a Poisson counting measure with intensity Λ((a, b] × A0)=(b − a) g(x)dx (or A0 Λ((a, b] × A0)=(b − a)ν(A0)), is called a Poisson point process with intensity g (or R intensity measure ν).

Note that for every Poisson point process, the set Π = {(t, ∆t) : t ≥ 0, ∆t =6 0} is a spatial Poisson process. Poisson random measure and Poisson point process are representations of this spatial Poisson process. Poisson point processes as we have deﬁned them always have a time coordinate and are homogeneous in time, but not in their spatial coordinates. In the next lecture we will see how one can do computations with Poisson point processes, notably relating to ∆t. P Lecture 4

Spatial Poisson processes II

Reading: Kingman Sections 2.2, 2.5, 3.1; Further reading: Williams Chapters 9 and 10

In this lecture, we construct spatial Poisson processes and study sums s t f(∆s) over ≤ Poisson point processes (∆t)t 0. We will identify s t ∆s as L´evy process next lecture. ≥ ≤ P P 4.1 Series and increasing limits of random variables

Recall that for two independent Poisson random variables X ∼ Poi(λ) and Y ∼ Poi(µ) we have X + Y ∼ Poi(λ + µ). Much more is true. A simple induction shows that

Xj ∼ Poi(µj), 1 ≤ j ≤ m, independent ⇒ X1 + . . . + Xm ∼ Poi(µ1 + . . . + µm). ∞ What about countably inﬁnite families with µ = m 1 µm < ? Here is a general result, a bit stronger than the convergence theorem for mome≥ nt generating functions. P Lemma 21 Let (Zm)m 1 be an increasing sequence of [0, ∞)-valued random variables. ≥ Then Z = limm Zm exists a.s. as a [0, ∞]-valued random variable. In particular, →∞ E(eγZm ) → E(eγZ )= M(γ) for all γ =06 . We have

P(Z < ∞)=1 ⇐⇒ lim M(γ)=1 γ 0 ↑ and P(Z = ∞)=1 ⇐⇒ M(γ)=0 for all (one) γ < 0.

Proof: Limits of increasing sequences exist in [0, ∞]. Hence, if a random sequence (Zm)m 1 is increasing a.s., its limit Z exists in [0, ∞] a.s. Therefore, we also have γZm ≥ γZ e → e ∈ [0, ∞] with the conventions e−∞ = 0 and e∞ = ∞. Then (by monotone convergence) E(eγZm ) → E(eγZ ). If γ < 0, then eγZ = 0 ⇐⇒ Z = ∞, but E(eγZ ) is a mean (weighted average) of nonnegative numbers (write out the deﬁnition in the discrete case), so P(Z = ∞)=1if γZ γZ γZ and only if E(e ) = 0. As γ ↑ 0, we get e− ↑ 1 if Z < ∞ and e− =0 → 0 if Z = ∞, so (by monotone convergence)

γZ E(e ) ↑ E(1 Z< )= P(Z < ∞) { ∞} and the result follows. 2 13 14 Lecture Notes – MS3b L´evy Processes and Finance – Oxford HT 2008

Example 22 For independent Xj ∼ Poi(µj) and Zm = X1 + . . . + Xm, the random variable Z = limm Zm exists in [0, ∞] a.s. Now →∞ γ γ γZm γ Zm (e 1)(µ1+...+µm) (1 e )µ E(e )= E((e ) )= e − → e− − ∞ shows that the limit is Poi(µ) if µ = m µm < . We do not need the lemma for this, since we can even directly identify the→∞ limiting moment generating function. If µ = ∞, the limit of the moment generatingP function vanishes, and by the lemma, we obtain P(Z = ∞) = 1. So we still get S ∼ Poi(µ) within the extended range 0 ≤ µ ≤∞.

4.2 Construction of spatial Poisson processes

The examples of compound Poisson processes are the key to constructing spatial Poisson processes with finite intensity measure. Infinite intensity measures can be decomposed. Theorem 23 (Construction) Let Λ be an intensity measure on D ⊂ Rd and suppose that there is a partition (In)n 1 of D into regions with Λ(In) < ∞. Consider independently ≥ ∩· ∩ ∼ (n) (n) ∼ Λ(In ) P (n) ∈ Λ(In A) Nn Poi(Λ(In)),Y1 ,Y2 ,... , i.e. (Yj A)= Λ(In) Λ(In) { (n) ≤ ≤ } and define Πn = Yj : 1 j Nn . Then Π = n 1 Πn is a spatial Poisson process with intensity measure Λ. ≥ S Proof: First fix n and show that Πn is a spatial Poisson process on In Thinning property of Poisson variables: Consider a sequence of independent Bernoulli(p) random variables (Bj)j 1 and independent X ∼ Poi(λ). ≥ Then the following two random variables are independent:

X X

X1 = Bj ∼ Poi(pλ) and X2 = (1 − Bj) ∼ Poi((1 − p)λ). j=1 j=1 X X To prove this, calculate the joint probability generating function

∞ X1 X2 B1+...+Bn n B1 ... Bn E(r s ) = P(X = n)E(r s − − − ) n=0 X n ∞ n λ λ n k n k k n k = e− p (1 − p) − r s − n! k n=0 =0 X Xk ∞ n λ λ n λp(1 r) λ(1 p)(1 s) = e− (pr + (1 − p)s) = e− − e− − − , n! n=0 X so the probability generating function factorises giving independence and we recognise the Poisson distributions as claimed. ⊂ ∼ For A In, consider X = Nn and the thinning mechanism, where Bj = 1 Y (n) A { j ∈ } P (n) ∈ Bernoulli( (Yj A)), then we get property (b): P (n) ∈ Nn(A)= X1 is Poisson distributed with parameter (Yj A)Λ(In)=Λ(A). Lecture 4: Spatial Poisson processes II 15

For property (a), disjoint sets A1,...,Am ⊂ In, we apply the analogous thinning (n) ∈ \ ∪ ∪ property for m + 1 types Yj Ai i = 0,...,m, where A0 = In (A1 . . . Am) to deduce the independence of Nn(A1),...,Nn(Am). Thus, Πn is a spatial Poisson process. ∩ Now for N(A)= n 1 Nn(A In), we add up inﬁnitely many Poisson variables and, ≥ ∩ by Example 22, obtain a Poi(µ) variable, where µ = n 1 Λ(A In)=Λ(A), i.e. property P ≥ (b). Property (a) also holds, since Nn(Aj ∩ In), n ≥ 1, j =1,...,m, are all independent, P and N(A1),...,N(Am) are independent as functions of independent random variables. 2

4.3 Sums over Poisson point processes

Recall that a Poisson point process (∆t)t 0 with intensity function g : D0 → [0, ∞) – ≥ focus on D0 = (0, ∞) ﬁrst but this can then be generalised – is a process such that

d N((a, b] × (c,d]) = #{a < t ≤ b : ∆t ∈ (c,d]} ∼ Poi (b − a) g(x)dx , Zc 0 ≤ a < b, (c,d] ⊂ D0, deﬁnes a Poisson counting measure on D = [0, ∞) × D0. This means that Π= {(t, ∆t) : t ≥ 0 and ∆t =06 } is a spatial Poisson process. Thinking of ∆s as a jump size at time s, let us study

Xt = 0 s t ∆s, the process performing all these jumps. Note that this is the situation for compound≤ ≤ Poisson processes X; in Example 18, g : (0, ∞) → [0, ∞) is integrable. P

Theorem 24 (Exponential formula) Let (∆t)t 0 be a Poisson point process with lo- ≥ cally integrable intensity function g : (0, ∞) → [0, ∞). Then for all γ ∈ R

∞ γx E exp γ ∆s = exp t (e − 1)g(x)dx . 0 ( 0 s t )! X≤ ≤ Z Proof: Local integrability of g on (0, ∞) means in particular that g is integrable on n n+1 In = (2 , 2 ], n ∈ Z. The properties of the associated Poisson counting measure N immediately imply that the random counting measures Nn counting all points in In, n ∈ Z, deﬁned by

Nn((a, b] × (c,d]) = {a < t ≤ b : ∆t ∈ (c,d] ∩ In}, 0 ≤ a < b, (c,d] ⊂ (0, ∞), are independent. Furthermore, Nn is the Poisson counting measure of jumps of a com- − d − P (n) ∈ ≤ pound Poisson process with (b a) c g(x)dx =(b a)λn (Y1 (c,d]) for 0 a < b and 1 (c,d] ⊂ In (cf. Example 18), so λn = g(x)dx and (if λn > 0) jump density hn = λ− g R In n on I , zero elsewhere. Therefore, we obtain n R

( ) ( ) ∆ if ∆ ∈ I E exp γ ∆ n = exp t (eγx − 1)g(x)dx , where ∆ n = s s n s s 0 otherwise ( 0 s t )! In X≤ ≤ Z 16 Lecture Notes – MS3b L´evy Processes and Finance – Oxford HT 2008

Now we have m (n) ↑ →∞ Zm = ∆s ∆s as m , n= m 0 s t 0 s t X− X≤ ≤ X≤ ≤ and (cf. Lemma 21 about ﬁnite or inﬁnite limits), the associated moment generating functions (products of individual moment generating functions) converge as required: m 2n+1 ∞ exp t (eγx − 1)g(x)dx → exp t (eγx − 1)g(x)dx . 2n 0 n= m ( Z ) Z Y− 2

4.4 Martingales (from B10a)

A discrete-time stochastic process (Mn)n 0 in R is called a martingale if for all n ≥ 0 ≥ E(Mn+1|M0,...,Mn)= Mn, i.e. if E(Mn+1|M0 = x0,...,Mn = xn)= xn for all xj. This is the principle of a fair game. What can I expect from the future if my current state is Mn = xn? No gain and no loss, on average, whatever the past. The following important rules for conditional expectations are crucial to establish the martingale property • If X and Y are independent, then E(X|Y )= E(X). • If X = f(Y ), then E(X|Y )= E(f(Y )|Y )= f(Y ) for functions f : R → R for which the conditional expectations exist.

• Conditional expectation is linear E(αX1 + X2|Y )= αE(X1|Y )+ E(X2|Y ). • More generally: E(g(Y )X|Y )= g(Y )E(X|Y ) for functions g : R → R for which the conditional expectations exist. These are all not hard to prove for discrete random variables. The full statements (continuous analogues) are harder. Martingales in continuous time can also be deﬁned, but (formally) the conditioning needs to be placed on a more abstract footing. Denote by Fs the “information available up to time s ≥ 0”, for us just the process (Mr)r s up to time ≤ s – this is often written Fs = σ(Mr,r ≤ s). Then the four bullet point rules still hold for Y =(Mr)r s or for Y replaced by Fs. ≤ We call (Mt)t 0 a martingale if for all s ≤ t ≥ E(Mt|Fs)= Ms.

Example 25 Let (Ns)s 0 be a Poisson process with rate λ. Then Ms = Ns − λs is a ≥ martingale: by the ﬁrst three bullet points and by the Markov property (Proposition 3)

E(Nt − λt|Fs)= E(Ns +(Nt − Ns) − λt|Fs)= Ns +(t − s)λ − λt = Ns − λs. γ Also Es = exp{γNs − λs(e − 1)} is a martingale since by the ﬁrst and last bullet points above, and by the Markov property γ E(Et|Fs) = E(exp{γNs + γ(Nt − Ns) − λt(e − 1)}|Fs) γ = exp{γNs − λt(e − 1)}E(exp{γ(Nt − Ns)}) γ γ = exp{γNs − λt(e − 1)} exp{−λ(t − s)(e − 1)} = Es. We will review relevant martingale theory when this becomes relevant. Lecture 5

The characteristics of subordinators

Reading: Kingman Section 8.4 We have done the leg-work. We can now harvest the fruit of our efforts and proceed to a number of important consequences. Our programme for the next couple of lectures is: • We construct Lévy processes from their jumps, first the most general increasing Lévy process. As linear combinations of independent Lévy processes are Lévy processes (Assignment A.1.2.(a)), we can then construct Lévy processes such as Variance Gamma processes of the form Zt = Xt − Yt for two increasing X and Y .

• We have seen martingales associated with Nt and exp{Nt} for a Poisson process N. Similar martingales exist for all Lévy processes (cf. Assignment A.2.3.). Martin- gales are important for finance applications, since they are the basis of arbitrage-free models (more precisely, we need equivalent martingale measures, but we will assume here a “risk-free” measure directly to avoid technicalities). • Our rather restrictive first range of examples of Lévy processes was obtained from known infinitely divisible distributions. We can now model using the intensity function of the Poisson point process of jumps to get a wider range of examples. • We can simulate these Lévy processes, either by approximating random walks based on the increment distribution, or by constructing the associated Poisson point process of jumps, as we have seen, from a collection of independent random variables.

5.1 Subordinators and the L´evy-Khintchine formula

We will call (weakly) increasing L´evy processes “subordinators”. Recall “ν(dx) ˆ=g(x)dx”.

Theorem 26 (Construction) Let a ≥ 0, and let (∆t)t 0 be a Poisson point process ≥ with intensity measure ν on (0, ∞) such that

(1 ∧ x)ν(dx) < ∞, (0, ) Z ∞ then the process Xt = at + s t ∆s is a subordinator with moment generating function ≤ E(exp{γXt}) = exp{tΨ(γ)}, where P Ψ(γ)= aγ + (eγx − 1)ν(dx). (0, ) Z ∞

17 18 Lecture Notes – MS3b L´evy Processes and Finance – Oxford HT 2008

Proof: Clearly (at)t 0 is a deterministic subordinator and we may assume a = 0 in ≥ the sequel. Now the Exponential formula gives the moment generating function of Xt = ∞ s t ∆s. We can now use Lemma 21 to check whether Xt < for t> 0: ≤

P ∞ γx P(Xt < ∞)=1 ⇐⇒ E (exp {γXt}) = exp t (e − 1)ν(dx) → 1 as γ ↑ 0. 0 Z This happens, by monotone convergence, if and only if for some (equivalently all) γ < 0

∞ ∞ (1 − eγx)ν(dx) < ∞ ⇐⇒ (1 ∧ x)ν(dx) < ∞. 0 0 Z Z It remains to check that (Xt)t 0 is a L´evy process. Fix 0 ≤ t0 < t1 <...

N((a, b] × (c,d]) = {a ≤ t < b : ∆t ∈ (c,d]}, 0 ≤ a < b, 0

and so are the sums tj− s

Note also that, due to the Exponential formula, P(Xt < ∞) > 0 already implies P(Xt < ∞) = 1. We shall now state but not prove the Lévy-Khintchine formula for nonnegative random variables. Theorem 27 (Lévy-Khintchine) A nonnegative random variable Y has an infinitely divisible distribution if and only if there is a pair (a, ν) such that for all γ ≤ 0

E(exp{γY }) = exp aγ + (eγx − 1)ν(dx) , (1) (0, ) Z ∞ ≥ ∧ ∞ where a 0 and ν is such that (0, )(1 x)ν(dx) < . ∞ Corollary 28 Given a nonnegativeR random variable Y with infinitely divisible distribution, there exists a subordinator (Xt)t 0 with X1 ∼ Y . ≥ Proof: Let Y have an infinitely divisible distribution. By the Lévy-Khintchine theorem, its moment generating function is of the form (1) for parameters (a, ν). Theorem 26 constructs a subordinator (Xt)t 0 with X1 ∼ Y . 2 ≥ This means that the class of subordinators can be parameterised by two parameters, the nonnegative “drift parameter” a ≥ 0, and the “Lévy measure” ν, or its density, the “Lévy density” g : (0, ∞) → [0, ∞). The parameters (a, ν) are referred to as the “Lévy- Khintchine characteristics” of the subordinator (or of the infinitely divisible distribution). Using the Uniqueness theorem for moment generating functions, it can be shown that a and ν are unique, i.e. that no two sets of characteristics refer to the same distribution. Lecture 5: The characteristics of subordinators 19

5.2 Examples

Example 29 (Gamma process) The Gamma process, where Xt ∼ Gamma(αt, β), is an increasing L´evy process. In Assignment A.2.4. we showed that αt β ∞ γx 1 βx E (exp {γXt})= = exp t (e − 1)αx− e− dx , γ<β. β − γ 0 Z 1 βx We read oﬀ the characteristics a = 0 and g(x)= αx− e− , x> 0.

Example 30 (Poisson process) The Poisson process, where Xt ∼ Poi(λt), has γ E (exp {γXt}) = exp {tλ(e − 1)}

This corresponds to characteristics a = 0 and ν = λδ1, where δ1 is the discrete unit point mass in (jump size) 1. Example 31 (Increasing compound Poisson process) The compound Poisson process Ct = Y1 + . . . + YXt , for a Poisson process X and independent identically distributed nonnegative Y1,Y2,... with probability density function h(x), x> 0, satisﬁes

∞ γx E (exp {γCt}) = exp t (e − 1)λh(x)dx , 0 Z and we read off characteristics a = 0 and g(x) = λh(x), x > 0. We can add a drift and consider Ct = at + Ct for some a> 0 to get a compound Poisson process with drift. Example 32 (Stable subordinator) The stable subordinator is best defined in terms e α 1 of its Lévy-Khintchinee characteristicse a = 0 and g(x)= x− − . This gives for γ ≤ 0 − ∞ γx α 1 Γ(1 α) α E (exp {γXt}) = exp t (e − 1)x− − dx = exp t (−γ) . 0 α Z 1/α 1/α Note that E(exp{γc Xt/c})= E(exp{γXt}), so that (c Xt/c)t 0 ∼ X. More generally, ≥ α 1 we can also consider e.g. tempered stable processes with g(x)= x− − exp{−ρx}, ρ> 0.

Figure 5.1: Examples: Poisson process, Gamma process, stable subordinator

5.3 Aside: nonnegative L´evy processes

It may seem obvious that a nonnegative Lévy process, i.e. one where Xt ≥ 0 a.s. for all t ≥ 0, is automatically increasing, since every increment Xs+t − Xs has the same distribution Xt and is hence also nonnegative. Let us be careful, however, and remember that there is a difference between something never happening at a fixed time and something never happening at any time. We have e.g. for a (one-dimensional) Poisson process (Nt)t 0 ≥

P(∆Nt =0)=6 P(Tn = t)=0 forall t ≥ 0, but P(∃t : ∆Nt =0)=16 . n 1 X≥ 20 Lecture Notes – MS3b L´evy Processes and Finance – Oxford HT 2008

Here we can argue that if f(t) < f(s) for some s < t and a right-continuous function, then there are also two rational numbers s0 < t0 for which f(t0) < f(s0), so P ∃ ∈ ∞ − ⇒ P ∃ ∈ ∞ ∩ Q − ( s, t (0, ),s 0 ( s0, t0 (0, ) : Xt0 Xs0 < 0) > 0 P ≤ P However, the latter can be bounded above (by subadditivity ( n An) n (An)) P ∃ ∈ ∞ ∩ Q − ≤ P ( s0, t0 (0, ) : Xt0 Xs0 < 0) (SXt0 s0 < 0)=0P . − s0,t0 (0, ) Q ∈X∞ ∩ Another instance of such delicate argument is the following: if Xt ≥ 0 a.s. for one t > 0 and a subordinator X, then Xt ≥ 0 a.s. for all t ≥ 0. It is true, but to say if P(Xs < 0) > 0 for some s < t then P(Xt < 0) > 0 may not be all that obvious. It is, however, easily justiﬁed for s = t/m, since then P(Xt < 0) ≥ P(Xtj/m − Xt(j 1)/m < − 0 for all j =1,...,m) > 0. We have to apply a similar argument to get P(Xtq < 0)=0 for all rational q > 0. Then we use again right-continuity to see that a function that is nonnegative at all rationals cannot take a negative value at an irrational either, so we get

P(∃s ∈ [0, ∞) : Xs < 0) = P(∃s ∈ [0, ∞) ∩ Q : Xs < 0) ≤ P(Xs < 0)=0. s [0, ) Q ∈ X∞ ∩ 5.4 Applications

Subordinators have found a huge range of applications, but are not directly models for a lot of real world phenomena. We can now construct more general Lévy processes of the form Zt = Xt − Yt for two subordinators X and Y . Let us here indicate some subordinators as they are used/arise in connection with other Lévy processes. Example 33 (Subordination) For a Lévy process X and an independent subordinator ≥ T , the process Ys = XTs , s 0, is also a Lévy process (we study this later in the course). − The rough argument is that (XTs+u XTs )u 0 is independent of (Xr)r Ts and distributed ≥− ≤ as X, by the Markov property. Hence XTs+r XTs is independent of XTs and distributed as XTr . A rigorous argument can be based on calculations of joint moment generating functions. Hence, subordinators are a useful tool to construct Lévy processes, e.g. from Brownian motion X. Many models of financial markets are of this type. The operation

Ys = XTs is called subordination – this is where subordinators got their name from.

Example 34 (Level passage) Let Zt = at − Xt where a = E(X1). It can be shown that τs = inf{t ≥ 0 : Zt > s} < ∞ a.s. for all s ≥ 0 (from the analogous random walk result). It turns out (cf. later in the course) that (τs)s 0 is a subordinator. ≥ Example 35 (Level set) Look at the zero set Z = {t ≥ 0 : Bt = 0} for Brownian motion (or indeed any other centred Lévy process) B. Z is unbounded since B crosses zero at arbitrarily large times so as to pass beyond all s and −s. Recall that (tB1/t)t 0 ≥ is also a Brownian motion. Therefore, Z also has an accumulation point at t = 0, i.e. crosses zero infinitely often at arbitrarily small times. In fact, it can be shown that Z { ≥ }cl is the closed√ range Xr,r 0 of a subordinator (Xr)r 0. The Brownian scaling cl ≥ property ( cBt/c)t 0 ∼ B shows that {Xr/c,r ≥ 0} ∼ Z, and so X must have a scaling ≥ property. In fact, X is a stable subordinator of index 1/2. Similar results, with different subordinators, hold not just for all Lévy processes but even for most Markov processes. Lecture 6

L´evy processes with no negative jumps

Reading: Kyprianou 2.1, 2.4, 2.6, Schoutens 2.2; Further reading: Williams 10-11 Subordinators X are processes with no negative jumps. We get processes that can de- crease by adding a negative drift at for a< 0. Also, Brownian motion B has no negative jumps. A guess might be that Xt + at + σBt is the most general L´evy process with no negative jumps, but this is false. It turns out that even a non-summable amount of positive jumps can be incorporated, but we will have to look at this carefully.

6.1 Bounded and unbounded variation

The (total) variation of a right-continuous function f : [0, t] → R with left limits is n

||f||TV := sup |f(tj) − f(tj 1)| :0= t0 < t1 <...

21 22 Lecture Notes – MS3b Lévy Processes and Finance – Oxford HT 2008 but then assuming finite total variation with positive probability, the uniform continuity of the Brownian path implies 2n 2n 2 | − − − | ≤ | − − − | | − − − |→ Btj2 n Bt(j 1)2 n sup Btj2 n Bt(j 1)2 n Btj2 n Bt(j 1)2 n 0 − j=1,...,2n − − j=1 j=1 X X with positive probability, but this is incompatible with convergence to t, so the assumption of finite total variation must have been wrong. Here is how jumps influence total variation:

Proposition 36 Let f be a right-continuous function with left limits and jumps (∆fs)0 s t. Then ≤ ≥

||f||TV ≥ |∆fs| 0 s t X≤ ≤ ∈ N Proof: Enumerate the jumps in decreasing order of size by (Tn, ∆fTn )n 0. Fix N ≥ and δ > 0. Choose ε > 0 so small that [Tn − ε, Tn] is a disjoint union and such that |f(Tn − ε) − f(Tn−)| < δ/N. Then for {Tn − ε, Tn : n =1,...,N} = {t1,...,t2N+1} such S that 0 = t0 < t1 <...

2N+2 N

|f(tj) − f(tj 1)| ≥ ∆f(Tn) − δ. − j=1 n=1 X X Since N and δ were arbitrary, this completes the proof, whether the right-hand side is ﬁnite or inﬁnite. 2

6.2 Martingales (from B10a)

Three martingale theorems are of central importance. We will require in this lecture just the maximal inequality, but we formulate all three here for easier reference. They all come in several different forms. We present the L2-versions as they are most easily formulated and will suffice for us. A stopping time is a random time T such that for every s ≥ 0 the information Fs allows to decide whether T ≤ s. More formally, if the event {T ≤ s} can be expressed in terms of (Mr,r ≤ s) (is measurable with respect to Fs). The prime example of a stopping time is the first entrance time TA = inf{t ≥ 0 : Mt ∈ A}. Note that

{T ≤ s} = {Mr 6∈ A for all r ≤ s} (and at least for closed sets A we can drop the irrational r ≤ s and see measurability, then approximate open sets.)

Theorem 37 (Optional stopping) Let (Mt)t 0 be a martingale and T a stopping time. E 2 ∞ E E ≥ If supt 0 (Mt ) < , then (MT )= (M0). ≥ E 2 Theorem 38 (Convergence) Let (Mt)t 0 be a martingale such that supt 0 (Mt ) < ≥ ≥ ∞, then Mt → M almost surely. ∞ E { 2 Theorem 39 (Maximal inequality) Let (Mt)t 0 be a martingale. Then (sup Ms : ≤ ≤ } ≤ E 2 ≥ 0 s t ) 4 (Mt ). Lecture 6: L´evy processes with no negative jumps 23

6.3 Compensation

Let g : (0, ∞) → [0, ∞) be the intensity function of a Poisson point process (∆t)t 0. If { ≤ ≤ } ∼ ≥ ∞ g is not integrable at inﬁnity, then # 0 s t : ∆s > 1 Poi( 1∞ g(x)dx) = Poi( ), and it is impossible for a right-continuous function with left limits to have accumulation R points in the set of such jumps (lower and upper points of a sequence of jumps will then have diﬀerent limit points). If however g is not integrable at zero, we have to investigate this further.

Proposition 40 Let (∆t)t 0 be a Poisson point process with intensity measure ν on ≥ (0, ∞). ∞ (i) If 0∞ xν(dx) < , then

R ∞ E ∆s = t xν(dx). 0 s t ! X≤ Z 2 ∞ (ii) If 0∞ x ν(dx) < , then

R ∞ 2 Var ∆s = t x ν(dx). 0 s t ! X≤ Z Proof: These are the two leading terms in the expansion with respect to γ of the Expo- nential formula: the ﬁrst moment can always be obtained from the moment generating ∂ | =0 function by taking ∂γ γ , here

∂ ∞ ∞ ∞ exp t (eγx − 1)ν(dx) = t xeγxν(dx) = t ν(dx), ∂γ 0 0 0 Z γ=0 Z γ=0 Z and the second moment follows from the second derivative in the same way. 2

Consider compound Poisson processes, with a drift that turns them into martingales 1 ε − Zt = ∆s1 ε<∆s 1 t xν(dx) (1) { ≤ } s t ε X≤ Z We have deliberately excluded jumps in (1, ∞). These are easier to handle separately. ε ↓ What integrability condition on ν do we need for Zt to converge as ε 0?

Lemma 41 Let (∆t)t 0 be a Poisson point process with intensity measure ν on (0, 1). 1 ε ≥ ε 2 2 ∞ With Z defined in (1), Zt converges in L if 0 x ν(dx) < . Proof: We only do this for ν(dx)= g(x)dx. NoteR that for 0 <δ<ε< 1, by Proposition 40(ii) applied to gδ,ε(x)= g(x)1 δ x<ε , { ≤ } ε E | ε − δ|2 2 ( Zt Zt )= t x g(x)dx Zδ ε ↓ 2 E − 2 so that (Zt )0<ε<1 is a Cauchy family as ε 0, for the L -distance d(X,Y )= ((X Y ) ). By completeness of L2-space, there is a limiting random variable Z as required. 2 t p We can slightly tune this argument to establish a general existence theorem: Theorem 42 (Existence) There exists a Lévy process whose jumps form a Poisson ∞ ∧ 2 ∞ point process with intensity measure ν on (0, ) if and only if (0, )(1 x )ν(dx) < . ∞ R 24 Lecture Notes – MS3b Lévy Processes and Finance – Oxford HT 2008

Proof: The “only if” statement is a consequence of a L´evy-Khintchine type characteri- sation of inﬁnitely divisible distributions on R, cf. Theorem 44, which we will not prove. Let us prove the “if” part in the case where ν(dx)= g(x)dx. E ε − δ ε − δ By Proposition 40(i), (Zt Zt ) = 0. By Assignment A.2.3.(c), the procecss Zt Zt is a martingale, and the maximal inequality (Theorem 39) shows that

ε E ε − δ ≤ E | ε − δ|2 2 sup Zt Zt 4 ( Zt Zt )=4t x g(x)dx 0 s t δ ≤ ≤ Z

ε ≤ ≤ ↓ 2 so that (Zs , 0 s t)0<ε<1 is a Cauchy family as ε 0, for the uniform L -distance E | − |2 2 d[0,t](X,Y )= (sup0 s t Xs Ys ). By completeness of L -space, there is a limiting ≤ ≤ (1) 2 ε process (Zs )0 s t, which as the uniform limit (in L ) of (Zs )0 s t is right-continuous p≤ ≤ ≤ ≤ with left limits. Also consider the independent compound Poisson process

(2) (1) (2) Zt = ∆s1 ∆s>1 and set Z = Z + Z . { } s t X≤

It is not hard to show that Z is a Lévy process that incorporates all jumps (∆s)0 s t. 2 ≤ ≤ 5/2 Example 43 Let us look at a Lévy density g(x) = |x|− , x ∈ [−3, 0). Then the 3 compensating drifts ε xg(x)dx take values 0.845, 2.496, 5.170 and 18.845 for ε = 1, ε = 0.3, ε = 0.1 and ε = 0.01. In the simulation, you see that the slope increases (to R infinity, actually as ε ↓ 0), but the picture begins to stabilise and converge to a limit.

Figure 6.1: Approximation of a L´evy process with no positive jumps – compensating drift Lecture 7

General L´evy processes and simulation

Reading: Schoutens Sections 8.1, 8.2, 8.4

For processes with no negative jumps, we compensated jumps by a linear drift and incorporated more and more smaller jumps while letting the slope of the linear drift tend to negative infinity. We will now construct the most general real-valued Lévy process as the difference of two such processes (and a Brownian motion). For explicit marginal distributions, we can simulate Lévy processes by approximating random walks. In practice, we often only have explicit characteristics (drift coefficient, Brownian coefficient and Lévy measure). We will also simulate Lévy processes based on the characteristics.

7.1 Construction of L´evy processes

The analogue of Theorem 27 for real-valued random variables is as follows.

Theorem 44 (Lévy-Khintchine) A real-valued random variable X has an infinitely divisible distribution if there are parameters a ∈ R, σ2 ≥ 0 and a measure ν on R \{0} 2 iλX ψ(λ) with ∞ (1 ∧ x )ν(dx) < ∞ such that E(e )= e− , where −∞ R 1 2 2 ∞ iλx ψ(λ)= −iaλ + σ λ − (e − 1 − iλx1 x 1 )ν(dx), λ ∈ R. 2 {| |≤ } Z−∞ Lévy processes are parameterised by their Lévy-Khintchine characteristics (a, σ2, ν), where we call a the drift coefficient, σ2 the Brownian coefficient and ν the Lévy measure or jump measure. ν(dx) will often be of the form g(x)dx, and we then refer to g as the Lévy density or jump density.

2 Theorem 45 (Existence) Let (a, σ , ν) be L´evy-Khintchine characteristics, (Bt)t 0 a ≥ standard Brownian motion and (∆t)t 0 an independent Poisson point process of jumps ≥ with intensity measure ν. Then there is a L´evy process

Zt = at + σBt + Mt + Ct, where Ct = ∆s1 ∆s >1 , {| | } s t X≤ 25 26 Lecture Notes – MS3b L´evy Processes and Finance – Oxford HT 2008 is a compound Poisson process (of big jumps) and

− Mt = lim ∆s1 ε< ∆s 1 t xν(dx) ε 0 { | |≤ } : 1 ↓ s t x R ε< x ! X≤ Z{ ∈ | |≤ } is a martingale (of small jumps – compensated by a linear drift).

Proof: The construction of Mt = Pt − Nt can be made from two independent processes Pt and Nt with no negative jumps as in Theorem 42. Nt will be built from a Poisson point process with intensity measure ν((c,d]) = ν([−d, −c)), 0 1 Z| | The last formula is checked in analogy with the moment generating function computation of Assignment A.1.3 (in general, the moment generating function will not be well-defined for this component). For the others, now “replace” γ by iλ. A formal justification can be obtained by analytic continuation, since the moment generating functions of these components are entire functions of γ as a complex parameter. Now the characteristic function of Z1 is the product of characteristic functions of the independent components, and this yields the formula required. 2 We stress in particular, that every Lévy process is the difference of two processes with only positive jumps. In general, these processes are not subordinators, but of the form in Theorem 42 plus a Brownian motion component. They can then both take positive and negative values. Example 46 (Variance Gamma process) We introduced the Variance Gamma process as difference X = G − H of two independent Gamma subordinators G and H. We can generalise the setting of Exercise A.1.2.(b) and allow G1 ∼ Gamma(α+, β+) and H1 ∼ Gamma(α , β ). The moment generating function of the Variance Gamma process − − is α+t α−t + γXt γGt γHt β β E(e ) = E(e )E(e− )= − β+ − γ β + γ − ∞ ∞ γx 1 β+x γy 1 β−y = exp t (e − 1)α+x− e− dx exp t (e− − 1)α y− e− dy 0 0 − Z 0Z ∞ γx 1 β+ x γx 1 β− x = exp t (e − 1)α+|x|− e− | |dx + t (e − 1)α |x|− e− | |dx . 0 − Z Z−∞ Lecture 7: General Lévy processes and simulation 27 and this is in Lévy-Khintchine form with ν(dx)= g(x)dx with

1 β+ x α+|x|− e− | | x> 0 g(x)= 1 β− x α |x|− e− | | x< 0 − The process (∆Xt)t 0 is a Poisson point process with intensity function g. ≥ Example 47 (CGMY process) Theorem 45 encourages to specify L´evy processes by their characteristics. As a natural generalisation of the Variance Gamma process, Carr, Geman, Madan and Yor (CGMY) suggested the following for ﬁnancial price processes

Y 1 C+ exp{−G|x|}|x|− − x> 0 g(x)= Y 1 C exp{−M|x|}|x|− − x< 0 − for parameters C > 0, G > 0, M > 0, Y ∈ [0, 2). While the Lévy density is a nice ± function, the probability density function of an associated Lévy process Xt is not available in closed form, in general. The CGMY model contains the Gamma model for Y = 0. When this model is fitted to financial data, there is usually significant evidence against Y = 0, so the CGMY model is more appropriate than the Variance Gamma model. We can construct Lévy processes from their Lévy density and will also simulate from Lévy densities. Note that this way of modelling is easier than searching directly for infinitely divisible probability density functions.

7.2 Simulation via embedded random walks

“Simulation” usually refers to the realisation of a random variable using a computer. Most mathematical and statistical packages provide functions, procedures or commands for the generation of sequences of pseudo-random numbers that, while not random, show features of independent and identically distributed random variables that are adequate for most purposes. We will not go into the details of the generation of such sequences, but assume that we a sequence (Uk)k 1 of independent Unif(0, 1) random variables. ≥ If the increment distribution is explicitly known, we simulate via time discretisation.

Method 1 (Time discretisation) Let (Xt)t 0 be a L´evy process so that Xt has prob- ≥ x ability density function ft. Fix a time lag δ > 0. Denote Ft(x) = ft(y)dy and 1 −∞ F − (u) = inf{x ∈ R : F (x) >u}. Then the process t t R n (1,δ) 1 Xt = S[t/δ], where Sn = Yk and Yk = Fδ− (Uk), =1 Xk is called the time discretisation of X with time lag δ. 1 One usually requires numerical approximation for Ft− , even if ft is available in closed form. That the approximations converge, is shown in the following proposition. ↓ (1,δ) → Proposition 48 As δ 0, we have Xt Xt in distribution. Proof: We can employ a coupling proof: t is a.s. not a jump time of X, so we have → (1,δ) ∼ 2 X[t/δ]δ Xt a.s., and so convergence in distribution for Xt X[t/δ]δ. 28 Lecture Notes – MS3b L´evy Processes and Finance – Oxford HT 2008 0 1 2 3 4 5 6 7 0.0 0.2 0.4 0.6 0.8 1.0

0 2 4 6 8 10 0 2 4 6 8 10

time time Gamma process with shape parameter 0.1 and scale parameter 0.1 Gamma process with shape parameter 1 and scale parameter 1 0 2 4 6 8 0 2 4 6 8 10

0 2 4 6 8 10 0 2 4 6 8 10

time time Gamma process with shape parameter 10 and scale parameter 10 Gamma process with shape parameter 100 and scale parameter 100

Figure 7.1: Simulation of Gamma processes from random walks with Gamma increments

Example 49 (Gamma processes) For Gamma processes, Ft is an incomplete Gamma 1 function, which has no closed-form expression, and Ft− is also not explicit, but numerical evaluations have been implemented in many statistical packages. There are also Gamma generators based on more uniform random variables. We display a range of parameter 1 choices. Since for a Gamma(1, 1) process X, the process (β− Xαt)t 0 is Gamma(α, β): ≥

αt αt 1 1 β E(exp{γβ− X })= = , αt 1 − γβ 1 β − γ − we chose α = β (keeping mean 1 and comparable spatial scale) but a range of parameters α ∈ {0.1, 1, 10, 100} on a ﬁxed time interval [0, 10]. We “see” convergence to a linear drift as α →∞ (for ﬁxed t this is due to the laws of large numbers). 0 2 4 6 8 10 0 2 4 6 8 10 0 2 4 6 8 10

0 2 4 6 8 10 0 2 4 6 8 10 0 2 4 6 8 10

time33 time22 time1 Approximation of Gamma(1,1) process, delta=1 Approximation of Gamma(1,1) process, delta=0.1 Approximation of Gamma(1,1) process, delta=0.01

Figure 7.2: Random walk approximation to a L´evy process, as in Proposition 48 Lecture 7: General L´evy processes and simulation 29

Example 50 (Variance Gamma processes) We represent the Variance Gamma process as the difference of two independent Gamma processes and focus on the symmetric case, so achieve mean 0 and fix variance 1 by putting β = α2/2; we consider α ∈{1, 10, 100, 1000}. We “see” convergence to Brownian motion as α →∞ (for fixed t due to the Central Limit Theorem). 0 1 2 3 4 5 −2.0 −1.5 −1.0 −0.5 0.0

0 2 4 6 8 10 0 2 4 6 8 10

time time Variance Gamma process with shape parameter 0.5 and scale parameter 1 Variance Gamma process with shape parameter 50 and scale parameter 10 0.0 0.5 1.0 1.5 2.0 −0.5 0.0 0.5 1.0 1.5 2.0

0 2 4 6 8 10 0 2 4 6 8 10

time time Variance Gamma process with shape parameter 5000 and scale parameter 100 Variance Gamma process with shape parameter 5e+05 and scale parameter 1000

Figure 7.3: Simulation of Variance Gamma processes as diﬀerences of random walks

7.3 R code – not examinable

The following code is posted on the course website as gammavgamma.R. psum <- function(vector){ b=vector; b[1]=vector[1]; for (j in 2:length(vector)) b[j]=b[j-1]+vector[j]; b} gammarw <- function(a,p){ unif=runif(10*p,0,1) pos=qgamma(unif,a/p,a); space=psum(pos); time=(1/p)*1:(10*p); plot(time,space, pch=".", sub=paste("Gamma process with shape parameter",a,"and scale parameter",a))} 30 Lecture Notes – MS3b L´evy Processes and Finance – Oxford HT 2008

vgammarw <- function(a,p){ unifpos=runif(10*p,0,1) unifneg=runif(10*p,0,1) pos=qgamma(unifpos,a*a/(2*p),a); neg=qgamma(unifneg,a*a/(2*p),a); space=psum(pos-neg); time=(1/p)*1:(10*p); plot(time,space, pch=".", sub=paste("Variance Gamma process with shape parameter",a*a/2, "and scale parameter",a))}

Now you can try various values of parameters a> 0 and steps per time unit p =1/δ in gammarw(a,p), e.g. gammarw(10,100) vgammarw(10,1000) Lecture 8

Simulation II

Reading: Ross 11.3, Schoutens Sections 8.1, 8.2, 8.4

In practice, the increment distribution is often not known, but the L´evy characteristics are, so we have to simulate Poisson point processes of jumps, by “throwing away the small jumps” and then analyse (and correct) the error committed.

8.1 Simulation via truncated Poisson point processes

Example 51 (Compound Poisson process) Let (Xt)t 0 be a compound Poisson pro- ≥ cess with L´evy density g(x) = λh(x), where h is a probability density function. Denote x 1 1 H(x) = h(y)dy and H− (u) = inf{x ∈ R : H(x) > u}. Let Yk = H− (U2k) and −∞1 Zk = −λ− ln(U2k 1), k ≥ 1. Then the process R − n n (2) { ≥ ≤ } Xt = SNt , where Sn = Yk, Tn = Zk, Nt = # n 1 : Tn t , =1 =1 Xk Xk has the same distribution as X.

Method 2 (Throwing away the small jumps) Let (Xt)t 0 be a L´evy process with ≥ characteristics (a, 0,g), where g is not the multiple of a probability density function. Fix a jump size threshold ε> 0 so that λε = x R: x >ε g(x)dx > 0, and write { ∈ | | } R g(x)= λεhε(x), |x| >ε, hε(x)=0, |x| ≤ ε,

x 1 for a probability density function hε. Denote H(x)= h(y)dy and H− (u) = inf{x ∈ 1 1 −∞ R : H(x) >u}. Let Yk = H− (U2k) and Zk = −λ− ln(U2k 1), k ≥ 1. Then the process R − n n (2,ε) − { ≥ ≤ } Xt = SNt bεt, where Sn = Yk, Tn = Zk, Nt = # n 1 : Tn t , =1 =1 Xk Xk − and bε = a x R:ε< x 1 xg(x)dx, is called the process with small jumps thrown away. { ∈ | |≤ } R 2 (1,δ) (2,ε) For characteristics (a, σ ,g) we can now simulate Lt = σBt + Xt by σBt + Xt . The following proposition says that such approximations converge as ε ↓ 0 (and δ ↓ 0). This is illustrated in Figure 6.3. 31 32 Lecture Notes – MS3b L´evy Processes and Finance – Oxford HT 2008

↓ (2,ε) → Proposition 52 As ε 0, we have Xt Xt in distribution.

Proof: For a process with no negative jumps and characteristics (0, 0,g), this is a consequence of the stronger Lemma 41, which gives a coupling for which convergence holds in the L2 sense. For a general L´evy process with characteristics (a, 0,g) that argument can be adapted, or we write Xt = at + Pt − Nt and deduce the result:

E { (2,ε)} iatE { (2,ε)} E {− (2,ε)} (exp iλXt ) = e (exp iλPt ) (exp iλNt ) iat iλXt → e E(exp{iλPt})E(exp{−iλNt})= E(e ). 2

Example 53 (Symmetric stable processes) Symmetric stable processes X are L´evy α 1 processes with characteristics (0, 0,g), where g(x) = c|x|− − , x ∈ R \ {0} for some α ∈ (0, 2) (cf. Assignment 3.2.). We decompose X = P − N for two independent processes with no negative jumps and simulate P and N. By doing this, we have

∞ c α α 1 1/α λ = g(x)dx = ε− , H (x)=1 − (ε/x) and H− (u)= ε(1 − u)− . ε α ε ε Zε For the simulation we choose ε = 0.01. We compare α ∈ {0.5, 1, 1.5, 1.8}. All processes are centred with inﬁnite variance. Big jumps dominate the plots for small α. Recall that α 2 iλXt b λ bλ E(e )= e− | | → e− as α ↑ 2, and we get, in fact, convergence to Brownian motion, the stable process of index 2. −40 −20 0 20 −600 −400 −200 0

0 2 4 6 8 10 0 2 4 6 8 10

time time Stable process with index 0.5 and cplus= 1 and cminus= 1 Stable process with index 1 and cplus= 1 and cminus= 1 −2 0 2 4 6 8 −2 0 2 4 6 8 10 12 0 2 4 6 8 10 0 2 4 6 8 10

time time Stable process with index 1.5 and cplus= 1 and cminus= 1 Stable process with index 1.8 and cplus= 1 and cminus= 1

Figure 8.1: Simulation of symmetric stable processes from their jumps Lecture 8: Simulation II 33 0 10000 20000 30000 0 20 40 60 80

0 2 4 6 8 10 0 2 4 6 8 10

time time Stable process with index 0.5 and cplus= 1 and cminus= 0 Stable process with index 0.8 and cplus= 1 and cminus= 0 0 2 4 6 8 10 −10 −5 0 5 10 15 20

0 2 4 6 8 10 0 2 4 6 8 10

time time Stable process with index 1.5 and cplus= 1 and cminus= 0 Stable process with index 1.8 and cplus= 1 and cminus= 0

Figure 8.2: Simulation of stable processes with no negative jumps

Example 54 (Stable processes with no negative jumps) For stable processes with α 1 no negative jumps, we have g(x) = c+x− − , x > 0. The subordinator case α ∈ (0, 1) ∈ was discussed in Assignment A.3.1. – Xt = s t ∆s. The case α [1, 2), where com- ≤ pensation is required, is such that E(X1) = 0, i.e. a = − ∞ xg(x)dx. We choose ǫ =0.1 P 1 for α ∈{0.5, 0.8} and ε =0.01 for α ∈{1.5, 1.8}. R Strictly speaking, we take as triplet (a, σ2,g) in Theorem 45 g as given, but for ∈ 1 α (0, 1) we take a = 0 xg(x)dx so that

R ∞ E(eiλXt ) = exp t (eiλx − 1)g(x)dx 0 Z ∞ iλx = exp −t −iλa − (e − 1 − iλx1 x 1 )g(x)dx , 0 {| |≤ } Z since compensation of small jumps is not needed and we obtain a subordinator if we do ∈ − not compensate, and for α (1, 2), we take a = 1∞ xg(x)dx so that

∞ R E(eiλXt ) = exp t (eiλx − 1 − iλx)g(x)dx 0 Z ∞ iλx = exp −t −iλa − (e − 1 − iλx1 x 1 )g(x)dx , 0 {| |≤ } Z since we can compensate all jump and achieve E(Xt) = 0. Only with these choices we obtain the Lévy processes with no negative jumps that satisfy the scaling property. This discussion shows that the representation in Theorem 45 is artificial and representations with different compensating drifts are often more natural. 34 Lecture Notes – MS3b Lévy Processes and Finance – Oxford HT 2008

8.2 Generating speciﬁc distributions

In this course, we will not go into the computational details of simulations. However, we do point out some principles here that lead to improved simulations, and we discuss some of the resulting modiﬁcations to the methods presented. Often, it is not eﬃcient to compute the inverse cumulative distribution function. For a number of standard distributions, other methods have been developed. We will here look at standard Normal generators. The Gamma distribution is discussed in Assignment A.4.2.

Example 55 (Box-Muller generator) Consider the following procedure

1. Generate two independent random numbers U ∼ Unif(0, 1) and V ∼ Unif(0, 1).

2. Set X = −2ln(U) cos(2πV ) and Y = −2 ln(U) sin(2πV ).

3. Return thep pair (X,Y ). p

The claim is the X and Y are independent standard Normal random variables. The proof is an exercise on the transformation formula. First, the transformation is clearly bijective from (0, 1)2 to R2. The inverse transformation can be worked out from X2+Y 2 = −2 ln(U) and Y/X = tan(2πV ) as

1 (X2+Y 2)/2 1 (U, V )= T − (X,Y )=(e− , (2π)− arctan(Y/X))

(with an appropriate choice of the branch of arctan, which is not relevant here). The Jacobian of the inverse transformation is − (x2+y2)/2 − (x2+y2)/2 xe− ye− ⇒| | 1 (x2+y2)/2 J = − 1 y 1 1 1 1 det(J) = e− 2 2 1+ 2 2 2 1+ 2 2 2π π x y /x π x y /x and so, as required,

1 1 (x2+y2)/2 f (x, y)= f (T − (x, y))| det(J)| = e− . X,Y U,V 2π For a more eﬃcient generation of standard Normal random variables, it turns out useful to ﬁrst generate uniform random variables on on the disk of radius 1:

Example 56 (Uniform distribution on the disk) For U1 ∼ Unif(0, 1) and U2 ∼ Unif(0, 1) independent, we have that (V1,V2) = (2U1 − 1, 2U2 − 1) is uniformly distributed on the square (−1, 1)2 centered at (0, 0), which contains the disk D = {(x, y) ∈ 2 2 2 2 R : x +y < 1}, and we have, in particular P((V1,V2) ∈ D)= π/4. Now, for all A ⊂ R , we have area(A ∩ D) P((V1,V2) ∈ A|(V1,V2) ∈ D)= , π so the conditional distribution of (V1,V2) given (V1,V2) ∈ D is uniform on D. By the following lemma, this conditioning can be turned into an algorithm by repeated trials:

1. Generate two independent random numbers U1 ∼ Unif(0, 1) and U2 ∼ Unif(0, 1). Lecture 8: Simulation II 35

2. Set (V1,V2)=(2U1 − 1, 2U2 − 1).

3. If (V1,V2) ∈ D, go to 4., else go to 1.

4. Return the numbers (V1,V2).

The pair of numbers returned will be uniformly distributed on the disk D.

Lemma 57 (Conditioning by repeated trials) Let X,X1,X2,... be independent and identically distributed d-dimensional random vectors. Also let A ⊂ Rd such that p = P(X ∈ A) > 0. Denote N = inf{n ≥ 1 : Xi ∈ A}. Then N ∼ geom(p) is independent of XN , and XN has as its (unconditional) distribution the conditional distribution of X given X ∈ A, i.e.

d P(XN ∈ B)= P(X ∈ B|X ∈ A) for all B ⊂ R .

Proof: We calculate the joint distribution

P(N = n, Xn ∈ B) = P(X1 6∈ A,...,Xn 1 6∈ A, Xn ∈ A ∩ B) − n 1 n 1 = (1 − p) − P(Xn ∈ A ∩ B)=(1 − p) − pP(Xn ∈ B|Xn ∈ A).

We now get the following modiﬁcation of the Normal generator:

Example 58 (Polar method) The following is a more eﬃcient method to generate two independent standard Normal random variables:

1. Generate two independent random numbers U1 ∼ Unif(0, 1) and U2 ∼ Unif(0, 1).

2 2 2. Set (V1,V2)=(2U1 − 1, 2U2 − 1) and S = V1 + V2 .

3. If S ≤ 1, go to 4., else go to 1.

4. Set P = −2(ln(S))/S p 5. Return the pair (X,Y )=(PV1,PV2).

The gain in eﬃciency mainly stems from the fact that no sine and cosine need to be com- puted. The method works because in polar coordinates (V1,V2)=(R cos(Θ), R sin(Θ), we have independent S = R2 ∼ Unif(0, 1) and Θ ∼ Unif(0, 2π) (as is easily checked), so we can choose U = S and 2πV = Θ in the Box-Muller generator. 36 Lecture Notes – MS3b L´evy Processes and Finance – Oxford HT 2008

8.3 R code – not examinable

The following code is posted on the course website as stable.R. stableonesided <- function(a,c,eps,p){ f=c*eps^(-a)/a; n=rpois(1,10*f); t=runif(n,0,10); y=(eps^(-a)-a*f*runif(n,0,1)/c)^(-1/a); ytemp=1:n;res=(1:(10*p))/100;{ for (k in 1:(10*p)){{for (j in 1:n){ if(t[j]<=k/p)ytemp[j]<-y[j] else ytemp[j]<-0}}; res[k]<-sum(ytemp)}}; res} stable <- function(a,cp,cn,eps,p){ pos=stableonesided(a,cp,eps,p); neg=stableonesided(a,cn,eps,p); space=pos-neg;time=(1/p)*1:(10*p); plot(time,space, pch=".", sub=paste("Stable process with index",a,"and cplus=",cp,"and cminus=",cn))} stableonesidedcomp <- function(a,c,eps,p){ f=(c*eps^(-a))/a; n=rpois(1,10*f); t=runif(n,0,10); y=(eps^(-a)-a*f*runif(n,0,1)/c)^(-1/a); ytemp=1:n; res=(1:(10*p))/100;{ for (k in 1:(10*p)){{if (n!=0)for (j in 1:n){ if(t[j]<=k/p)ytemp[j]<-y[j] else ytemp[j]<-0}};{ if (n!=0)res[k]<-sum(ytemp)-(c*k/(p*(a-1)))*(eps^(1-a)) else res[k]<--c*k/(p*(a-1))*(eps^(1-a))}}}; res} Lecture 9

Simulation III

Reading: Ross 11.3, Schoutens Sections 8.1, 8.2, 8.4; Further reading: Kyprianou Section 3.3

9.1 Applications of the rejection method

Lemma 57 can be used in a variety of ways. A widely applicable simulation method is the rejection method. Suppose you have an explicit probability density function f, but the inverse distribution function is not explicit. If h ≥ cf, for some c < 1 is a probability density function whose inverse distribution function is explicit (e.g. uniform or exponential) or from which we can simulate by other means, then the procedure 1. Generate a random variable X with density h.

2. Generate an independent uniform variable U.

3. If Uh(X) ≤ cf(X), go to 4., else go to 1.

4. Return X.

Proposition 59 The procedure returns a random variable with density f.

Proof: Denote p = P(Uh(X) ≤ cf(X)). By Lemma 57 (applied to the vector (X, U)), the procedure returns a random variable with distribution P(X ≤ x, Uh(X) ≤ cf(X)) P(X ≤ x|Uh(X) ≤ cf(X)) = p 1 x c x = h(z)P(U ≤ cf(z)/h(z))dz = f(z)dz, p p Z−∞ Z−∞ and letting x →∞ shows c = p. 2

Example 60 (Gamma distribution) Note that the Gamma density for α> 1 satisﬁes

α 1 1 α α 1 βx β − βx β x − e− ≤ βe− Γ(α) Γ(α)

βx α 1 so we can apply the procedure with f(x)= βe− and c = Γ(α)/β − . 37 38 Lecture Notes – MS3b L´evy Processes and Finance – Oxford HT 2008

It is important that c is not too small, since otherwise lots of iterations are needed until the random variable is returned. The number of iterations is geometrically distributed (ﬁrst success in a sequence of independent Bernoulli trials) with success parameter c, so on average 1/c trials are required. For the simulation via Poisson point processes (with truncation at ε, say), we can use properties of Poisson point processes to simulate separately jumps of sizes in intervals In, n = 1,...,n0 and can choose intervals In so that the intensity function g is almost constant.

Example 61 (Distribution on In) Suppose that we simulate a Poisson point process on a bounded spatial interval In = (a, b], with some intensity function g : In → [0, ∞). Then we can take uniform

b g(x)dx − a h(x)=1/(b a) and c = − { ≤ }. (b a) maxR g(x) : a < x b b and simulate Exp( a g(x)dx)-spaced times Tn and spatial coordinates ∆Tn by the rejection method with h and c as given. R 9.2 “Errors increase in sums of approximated terms.”

Methods 1 and 2 are based on sums of many, mostly small, independent identically distributed random variables. As δ ↓ 0 or ε ↓ 0, these are more and more smaller and smaller random variables. If each is aﬀected by a small error, then adding up these errors makes the approximations worse whereas the precision should increase. For Method 1, this can often be prevented by suitable conditioning, e.g. on the terminal value:

Example 62 (Poisson process) A Poisson process with intensity λ on the time interval [0, 1] can be generated as follows:

1. Generate a Poisson random variable N with parameter λ.

2. Given N = n, generate n independent Unif(0, 1) random variables U1,...,Un.

3. Return Xt = #{1 ≤ i ≤ N : Ui ≤ t}.

Clearly, this process is a Poisson process, since Xs (and indeed (Xt1 ,...,Xtn tn−1 )) is − obtained from X1 ∼ Poi(λ) by thinning as in Section 4.2.

This is not of much practical use since we would usually simulate a Poisson random variable by evaluating a unit rate Poisson process (simulated from standard exponential interarrival times) at λ. In the case of Brownian motion (and the Gamma process, see Assignment A.4.3.), however, such conditioning is very useful and can then be iterated, e.g. in a dyadic scheme:

Example 63 (Brownian motion) Consider the following method to generate Brown- ian motion on the time interval [0, 1]. Lecture 9: Simulation III 39

1. Set X0 = 0 and generate X1 ∼ Normal(0, 1) hence specifying Xk2−n for n = 0, k =0,..., 2n.

n 2. For k =1,..., 2 , conditionally given X(k 1)2−n = x and Xk2−n = z, generate − x + z 2 − − ∼ n X(2k 1)2 n 1 Normal , 2− − − 2 3. If the required precision has been reached, stop, else increase n by 1 and go back to 2. This process is Brownian motion, since the following lemma shows that Brownian motion has these conditional distributions. Speciﬁcally, the n = 0, k = 1 case of 2. is obtained n directly for s = 1/2, t = 1. For n ≥ 1, k =1,..., 2 , note that X(2k 1)2−n−1 − X(k 1)2−n − − is independent of X(k 1)2−n and so, we are really saying that for Brownian motion − − z x n 2 X(2k 1)2−n−1 − X(k 1)2−n ∼ Normal , 2− − , − − 2 − − − − conditionally given Zk2 n Z(k 1)2 n = z x, which is equivalent to the speciﬁcation in 2. −

n A further advantage of this method is that δ =2− can be decreased without having to start afresh. Previous less precise simulations can be reﬁned.

Lemma 64 Let (Xt)t 0 be Brownian motion and 0

Proof: Note that Xs ∼ Normal(0,s) and Xt − Xs ∼ Normal(0, t − s) are independent. By the transformation formula (Xs,Xt) has joint density 1 x2 (z − x)2 f (x, z)= exp − − , Xs,Xt − 2s 2(t − s) 2π s(t s) and so the conditional density is p f (x, z) 1 x2 (z − x)2 z2 Xs,Xt √ − − fXs Xt=z(x) = = exp + | f (z) − 2s 2(t − s) 2t Xt 2π s(t s)/t 1 (x − zs/t)2 = √ expp − . − 2s(t − s)/t 2π s(t s)/t 2 p

For Method 2, we can achieve similar improvements by simulating (∆t)t 0 in stages. ≥ Choose a strictly decreasing sequence ∞ = a0 > a1 > a2 >...> 0 of jump size thresholds with an ↓ 0 as n →∞ (k) ( k) − ≥ ≥ ∆t = ∆t1 ak ∆t ak− , k 1, t 0. { ≤ 1} {− ≥ − 1} Simulate the Poisson counting processes N (k) associated with ∆(k) as in Example 62 and otherwise construct ak−1 (k) (k) − Zt = ∆s t x1 0

9.3 Approximation of small jumps by Brownian motion

Theorem 65 (Asmussen-Rosinski) Let (Xt)t 0 be a Lévy process with characteristics ≥ (a, 0,g). Denote ε σ2(ε)= x2g(x)dx ε Z− If σ(ε)/ε →∞ as ε ↓ 0, then (2 ) X − X ,ε t t → B in distribution as ε ↓ 0 σ(ε) t for an independent Brownian motion (Bt)t 0 ≥ If σ(ε)/ε →∞, it is well-justified to adjust Method 2 to set (2+,ε) (2,ε) Xt = Xt + σ(ε)Bt for an independent Brownian motion. In other words, we may approximate the small jumps by an independent Brownian motion. Example 66 (CGMY process) The CGMY process is a popular process in Mathe- matical Finance. It is defined via its characteristics (0, 0,g), where Y 1 Y 1 g(x)= C exp{−G|x|}|x|− − , x< 0, g(x)= C exp{−M|x|}|x|− − , x> 0. for some C ≥ 0, G > 0, M > 0 and Y < 2. Let (Xt)t 0 be a CGMY process. We ≥ calculate ε ε 2 2 ≤ | |1 Y 2C 2 Y σ (ε)= x g(x)dx C x − dx = − ε − ε ε 2 Y Z− Z− and for every given δ > 0 and all ε> 0 small enough, the same quantity with C replaced by C − δ is a lower bound, so that

σ(ε) ∼ 2C Y/2 →∞ ⇐⇒ − ε− Y > 0 ε r2 Y Hence an approximation of the small jumps of size (−ε,ε) thrown away by a Brownian motion σ(ε)Bt is appropriate if and only if Y > 0. In fact, for Y < 0, the process has ﬁnite jump intensity, so all jumps can be simulated. Therefore, only the case Y = 0 is problematic. This is the Variance Gamma process (and its asymmetric companions). Whether or not we can approximate small jumps by a Brownian motion, we have to decide what value of ε to choose. By the independence properties of Poisson point − 2,ε processes, the remainder term that Xt Xt is a (zero mean, for ε < 1) L´evy process with intensity function g on [−ε,ε] and variance 2 − (2,ε) σ (ε) = Var(Xt Xt ) ↓ (2,ε) (let δ 0 in the proof of Lemma 41). We can choose ε so that the accuracy of Xt is within an agreed deviation h, i.e. e.g. 2σ(ε) = h. In the setting of Theorem 65, this (2,ε) means that a deviation of Xt from Xt by more than h would happen with probability about 0.05. Lecture 9: Simulation III 41

9.4 Appendix: Consolidation on Poisson point processes

This section and and the next should not be necessary this year, because the relevant material has been included in earlier lectures. They may still be useful as a reminder of key concepts. We can consider Poisson point processes (∆t)t 0 in very general spaces, e.g. (topo- ≥ logical) spaces (E, O) where we have a collection/notion of open sets O ∈ O (and an associated Borel σ-algebra B = σ(O), the smallest σ-algebra that contains all open sets, for which we also require that {x} ∈ B for all x ∈ E and G = {(x, x) : x ∈ E}∈B⊗B). We just require that (∆t)t 0 is a family of ∆t ∈ E ∪{0} such that there is an intensity ≥ (Borel) measure ν on E with ν({x})=0 for all x ∈ E,

(a) for disjoint A1 =(a1, b1] × O1,...,An =(an, bn] × On, Oi ∈ O, the counts

N(Ai)= N((a1, b1] × Oi) = #{t ∈ (ai, bi] : ∆t ∈ Oi}, i =1,...,n

are independent random variables and inhom (b) N(Ai) ∼ Poi((bi − ai)ν(Oi)).

For us, E = R \{0} is the space of jump sizes, and ν(O ) = ν((c ,d )) = di g(x)dx i i i ci for an intensity function g : R \{0} → [0, ∞). Property (a) for all open sets is then R equivalent to property (a) for all measurable sets or all half-open intervals or all closed intervals etc. (all that matters is that the collection generates the Borel σ-algebra). It is an immediate consequence of the deﬁnition (and this discussion) that for (measurable) disjoint B1, B2,... ⊂ R \{0}, the “restricted” processes

(i) ≥ ∆t = ∆t1 ∆t Bi , t 0, { ∈ }

are also Poisson point processes with the restriction of g to Bi as intensity function, and they are independent. We used this fact crucially and repeatedly in two forms. Firstly, for B1 = (0, ∞) and B2 = (−∞, 0) (and B3 = B4 = . . . = ∅), we consider Poisson point processes of positive points (jump sizes) and of negative points (jump sizes). We constructed from them independent L´evy processes. Secondly, for a sequence ∞ = a0 > a1 > a2 >..., we considered Bi = [ai, ai 1), i ≥ 1, so as to simulate separately − independent L´evy processes (in fact compound Poisson processes with linear drift) with jump sizes only in Bi.

9.5 Appendix: Consolidation on the compensation of jumps

The general L´evy process requires compensation of small jumps in its approximation by processes with no jumps in (−ε,ε), as ε ↓ 0. This is reﬂected in its characteristic function of the form

iλXt tψ(λ) 1 2 2 ∞ iλx E(e )= e− , ψ(λ)= −ia1λ+ σ λ − (e −1−iλx1 x 1 )ν(dx), λ ∈ R, (1) 2 {| |≤ } Z−∞ 42 Lecture Notes – MS3b L´evy Processes and Finance – Oxford HT 2008

2 2 where usually ν(dx) = g(x)dx. This is a parametrisation by (a1, σ , ν) or (a1, σ ,g), 2 where we require the (weak) integrability condition ∞ (1 ∧ x )ν(dx) < ∞. The ﬁrst class of L´evy processes that we constructed−∞ systematically were subordina- R tors, where no compensation was necessary. We parametrised them by parameters a2 ≥ 0 and g : (0, ∞) → [0, ∞) (or ν measure on (0, ∞)) so that the moment generating function is of the form

γX tΨ(γ) ∞ γx E(e t )= e , Ψ(γ)= a2γ + (e − 1)ν(dx), γ ≤ 0. (2) 0 Z ∧ ∞ We required the stronger integrability condition 0∞(1 x)ν(dx) < . Similarly, for diﬀerences of subordinators, we have a characteristic function R

iλXt tψ(λ) 1 2 2 ∞ iλx E(e )= e− , ψ(λ)= −ia2λ + σ λ − (e − 1)ν(dx), (3) 2 Z−∞ under the stronger integrability condition ∞ (1 ∧|x|)ν(dx) < ∞. Compensation in (1) −∞ is only done for small jumps. This is, because, in general the indicator 1 x 1 cannot be 2 R {| |≤ } omitted. However, if ∞ (x ∧ x )ν(dx) < ∞, then we can also represent −∞ R iλXt tψ(λ) 1 2 2 ∞ iλx E(e )= e− , ψ(λ)= −ia3λ + σ λ − (e − 1 − iλx)ν(dx). (4) 2 Z−∞ Equations (1), (3) and (4) are compatible whenever any two integrability conditions are fulﬁlled, since the linear (in λ) terms under the integral can be added to a1 to give

a2 = a1 + xν(dx) and a3 = a1 − xν(dx). x 1 x >1 Z{| |≤ } Z{| | }

Note that then (by differentiation at λ = 0), we get a3 = E(X1). If a3 = 0, then (Xt)t 0 ≥ is a martingale. On the other hand, for processes with finite jump intensity, i.e. under the even stronger integrability condition ∞ g(x)dx < ∞, we get a1 as the slope of the −∞ paths of X between the jumps. Both a1 and a3 are therefore natural parameterisations, R but not available, in general. a2 is available in general, but does not have such a natural interpretation. We use characteristic functions for similar reasons: in general, moment generating functions do not exist. If they do, i.e. under a strong integrability condition 1 γx ∞ γx ∞ 1∞ e ν(dx) < for some γ > 0 or − e ν(dx) < for some γ < 0, we get −∞ R R γXt tΨ(γ) 1 2 2 ∞ γx E(e )= e , Ψ(γ)= a1γ + σ γ + (e − 1 − γx1 x 1 )g(x)dx. (5) 2 {| |≤ } Z−∞ Moment generating functions are always defined on an interval I possibly including end points γ ∈ [−∞, 0] and/or γ+ ∈ [0, ∞], we always have 0 ∈ I, but maybe γ = γ+ = 0. − − Xt If 1 ∈ I and a1 is such that Ψ(1) = 0, then (e )t 0 is a martingale. ≥ Lecture 10

L´evy markets and incompleteness

Reading: Schoutens Chapters 3 and 6

10.1 Arbitrage-free pricing (from B10b)

By Donsker’s Theorem, Brownian motion is the scaling limit of most random walks and in particular of the simple symmetric random walk Rn = X1 + . . . + Xn where X1,X2,... are independent with P(Xi =1)= P(Xi = −1)=1/2.

R[nt]/√n Bt Corollary 67 For simple symmetric random walk (Rn)n 0, we have e → e , ≥ geometric Brownian motion, in distribution as n →∞.

Proof: First note that E(X1) = 0 and Var(X1) =1. Now for all x> 0 √ R[nt]/√n Bt P(e ≤ x)= P(R[nt]/ n ≤ ln(x)) → P(Bt ≤ ln(x)) = P(e ≤ x), by the Central Limit Theorem. 2 This was convergence of for ﬁxed t. Stronger convergence, locally uniformly in t can also be shown. Note that (Rn)n 0 is a martingale, and so is (Bt)t 0. However, ≥ ≥ n Rn 1 1 1 E(e )= e− + e →∞ 2 2 Proposition 68 For non-symmetric simple random walk (Rn)n 0 with P(Xi = 1) = p, Rn ≥ the process (e )n 0 is a martingale if and only if p =1/(1 + e). ≥ Proof: By the fourth and ﬁrst rules for conditional expectations, we have

E(eRn+1 |eR0 ,...,eRn )= E(eRn eXn+1 |eR0 ,...,eRn )= eRn E(eXn+1 )

Rn and so, (e )n 0 is a martingale if and only if ≥ Xn+1 1 2 1= E(e )= pe + (1 − p)e− ⇐⇒ p(e − 1/e)=1 − 1/e ⇐⇒ p =1/(e + 1).

The argument works just assuming that Rn = X1 + . . . + Xn, n ≥ 0, satisﬁes P(|Xn+1| = 1|R0 = r0,...,Rn = rn) = 1. Among all joint distributions, the non-symmetric exponen- tiated random walk with p =1/(1 + e) is the only martingale. The concept of arbitrage-free pricing in binary models is leaving aside any randomness. We will approach the Black-Scholes model from discrete models. 43 44 Lecture Notes – MS3b L´evy Processes and Finance – Oxford HT 2008

Suppose we have a risky asset with random price process S per unit and a risk-free asset with deterministic value function A per unit. Consider portfolios (U, V ) of U units of the risky asset and V units of the risk-free asset. We allow that (Ut,Vt) depends on the performance of (Ss)0 s 0) > 0. We will be interested in models where no arbitrage opportunities exist.

Example 69 (One-period model) There are two scenarios “up” and “down” (to which we may later assign probabilities p ∈ (0, 1) and 1 − p). The model consists of (S0,S1) only, where S0 changes to S1(up) or S1(down)

W1(up) = U0S1(up) + V0A1 or W1(down) = U0S1(down) + V0A1. (1)

It is easily seen that arbitrage opportunities occur if and only if A1/A0 ≥ S1(up)/S0 or S1(down)/S0 ≥ A1/A0, i.e. if one asset is uniformly preferable to the other.

A derivative security (or contingent claim) with maturity t is a contract that provides the owner with a payoﬀ Wt dependent on the performance of (Ss)0 s t. If there is a ≤ ≤ self-ﬁnancing portfolio process (U, V ) with value Wt at time t, then such a portfolio process is called a hedging portfolio process replicating the contingent claim. The value W0 = U0S0 + V0A0 of the hedging portfolio at time 0 is called the arbitrage-free price of the derivative security. It is easily seen that there would be an arbitrage opportunity, if the derivative security was available to buy and sell at any other price (as an additional asset in the model). In general, not all contingent claims can be hedged.

Example 69 (One-period model, continued) Consider any contingent claim, i.e. a payoﬀ of W1(up) or W1(down) according to whether scenario “up” or “down” take place. Equations (1) can now be used to set up a hedging portfolio (U0,V0) and calculate the unique arbitrage-free price. Note that the arbitrage-free price is independent of probabilities p ∈ (0, 1) and 1 − p that we may assign to the two scenarios as part of our model speciﬁcation. Because of the linearity of (1), there is a unique q ∈ (0, 1) such that for all contingent claims W1 : {up, down}→ R

W0 = qW1(up) + (1 − q)W1(down).

If we refer to q and 1− q as probabilities of “up” and “down”, then W0 is the expectation 1 of W1 under this distribution. If A0 = A1 = 1, S0 = 1, S1(up) = e, S1(down) = e− , then we identify (for W1 = S1 and hence (U0,V0)=(1, 0)) that q =1/(1 + e).

The property that every contingent claim can be hedged by a self-ﬁnancing portfolio process is called completeness of the market model. Lecture 10: L´evy markets and incompleteness 45

Example 70 (n-period model) Each of n periods has two scenarios, “up” and “down”, Rk as is the case for the model Sk = e for a simple random walk (Rk)0 k n. Then there n ≤ ≤ are 2 different combinations of “up” (Xk = 1) and “down” (Xk = −1). A contingent claim at time n is now any function Wn assigning a payoff to each of these combinations. By a backward recursion using the one-period model as induction step, we can work out hedging portfolios (Uk,Vk) and the value Wk of the derivative security at times k = n − 1, n − 2,..., 0, where in each case, (Uk,Vk) and Wk will depend on previous k “up”s and “down”s X1,...,Xk, so this is a specification of 2 values each. W0 will be the unique arbitrage-free price for the derivative security. The induction also shows that, for A0 = A1 = . . . = An = 1, it can be worked out as

W0 = E(Wn(X1,...,Xn)), where Xk independent with P(Xk =1)=1/(1 + e), and that (Wk)0 k n is a martingale with Wk = E(Wn|X1,...,Xk), e.g. (Sk)0 k n. If k ≤ δk≤ ≤ ≤ Ak = (1+ i) = e , we get an arbitrage-free model if and only if −1 <δ< 1, and then δn 1+δ 2 W0 = e− E(Wn), where Xk independent with P(Xk =1)=(e − 1)/(e − 1), δk where now (e− Wk)0 k n is a martingale. In particular, the n-period model is complete. ≤ ≤ { − 1 2 } Example 71 (Black-Scholes model) Let St = S0 exp σBt +(µ 2 σ )t for a Brow- 2 δt nian motion (Bt)t 0 and two parameters µ ∈ R and σ > 0. Also put At = e . It ≥ can be shown that also in this model, every contingent claim can be hedged, i.e. the Black-Scholes model is complete. Moreover, the pricing of contingent claims Wt can be carried out using the risk-neutral process { − 1 2 } ≥ Rt = S0 exp σBt +(δ 2 σ ))t , t 0, δt where the drift parameter is δ, not µ. The discounted process Mt = e− Rt is a martingale and has analogous uniqueness properties to the martingale for the n-period model, but they are much more complicated to formulate here. The arbitrage-free price of Wt = G((Ss)0 s t) is now (for all µ ∈ R) ≤ ≤ δt W0 = e− E(G((Rs)0 s t)). ≤ ≤ + Examples are G((Ss)0 s t)=(St − K) for the European call option and G((Ss)0 s t)= + ≤ ≤ ≤ ≤ (K − St) for the European put option. We will also consider path-dependent options − + such as Up-and-out barrier options with payoﬀ (St K) 1 St

10.2 Introduction to L´evy markets

The Black-Scholes model is widely used for option pricing in the finance industry, largely because many options can be priced explicitly and there are computationally efficient methods also for more complicated derivatives, that can be carried out frequently and for high numbers of options. However, its model fit is poor and any price that is obtained from the Black-Scholes model must be adjusted to be realistic. There are several models based on Lévy processes that offer better model fit, but the Black-Scholes methods for option pricing do not transfer one-to-one. Lévy processes give a very wide modelling freedom. For practical applications it is useful to work with parametric subfamilies. Several such families have been suggested. 46 Lecture Notes – MS3b Lévy Processes and Finance – Oxford HT 2008

Example 72 (CGMY process) Carr, Geman, Madan and Yor proposed a model with four parameters (the ﬁrst letters of their names. It is deﬁned via its intensity

Y 1 Y 1 g(x)= C exp{−G|x|}|x|− − , x< 0, g(x)= C exp{−M|x|}|x|− − , x> 0.

Let (Xt)t 0 be a CGMY process. If M > 1, then a risk-neutral price process can be ≥ δt modelled as Rt = R0 exp{Xt − t(φ(1) − δ)}. Then the discounted process (e− Rt)t 0 ≥ is a martingale, and it can be shown that arbitrage-free prices for contingent claims Wt = G((Rs)0 s t) can be calculated as ≤ ≤ δt W0 = e− E(G((Rs)0 s t)). ≤ ≤ It can also be shown, however, that this is not the only way to obtain arbitrage-free prices, and other prices do not necessarily lead to arbitrage opportunities. Also, not every contingent claim can be hedged, the model is not complete.

10.3 Incomplete discrete ﬁnancial markets

Essentially, arbitrage-free discrete models are complete only if the number of possible (1) (m) scenarios ω0,...,ωm (for one period) is the same as the number of assets S1 ,...,S1 : Ω = {ω0,...,ωm} → R in the model, since this leads to a system of linear equations to (1) (m) relate a contingent claim W1 : {ω0,...,ωm}→ R to a portfolio (U ,...,U ,V )

m (i) (i) V0A1 + U0 S1 (ωj)= W1(ωj), j =0,...,m i=1 X (1) (m) that can usually be uniquely solved for (U0 ,...,U0 ,V0), and we can read oﬀ

m (i) (i) W0 = V0A0 + U0 S0 . i=1 X If the number of possible scenarios is higher, then the system does not have a solution, in general (and hedging portfolios will not exist, in general). If the number of possible scenarios is lower, there will usually be inﬁnitely many solutions. If the system has no solution in general, the model is incomplete, but this does not mean that there is no price. It means that there is not a unique price. We can, in general, get some lower and upper bounds for the price imposed by no-arbitrage. One way of approaching this is to add a derivative security to the market as a further asset with an initial price that keeps the no arbitrage property for the extended model. One can, in fact, add more and more assets until the model is complete. Then there exist (j) unique probabilities qj = P(ωj), 0 ≤ j ≤ m, that make all discounted assets (A0/A1)S , 1 ≤ j ≤ m (including the ones added to complete the market) martingales.

Example 73 (Ternary model) Suppose there are three scenarios, but only two assets. The model with S0 =1= A0 < A1 = 2, and 1 = S1(ω0) < 2 = S1(ω1) < 3 = S1(ω2) is easily seen to be arbitrage-free since S1(ω0) < A1 < S1(ω2). The contingent claim 0= W1(ω0)= W1(ω1) < W1(ω2) = 1 can be hedged if and only if

2V0 + U0 =0, 2V0 +2U0 =0, 2V0 +3U0 =1, Lecture 10: L´evy markets and incompleteness 47

but the ﬁrst two equations already imply U0 = V0 = 0 and then the third equation is false. Therefore the contingent claim W1 cannot be hedged. The model is not complete. For the model (W, S, A) to be arbitrage-free we clearly require W0 > 0 since otherwise we could make arbitrage with a portfolio (1, 0, 0), just “buying” the security. Its cost at time zero is nonpositive and its value at time one is nonnegative and positive for scenario ω2. Now note that (S, A) is arbitrage-free, so any arbitrage portfolio must be of the form (−1/W0, U0,V0) with zero value −1+ U0 + V0 = 0 at time 0 and values at time 1

U0 +2V0 =2 − U0, 2U0 +2V0 =2 > 0, −1/W0 +3U0 +2V0 = −1/W0 +2+ U0, so that we need 2 ≥ U0 ≥ 1/W0 − 2, and this is possible if and only if W0 ≥ 1/4. Therefore, the range of arbitrage-free prices is W0 ∈ (0, 1/4). We can get these prices as expectations under martingale probabilities: 1 1 3 1 1 1= E (S1)= q0 + q1 + q2, W0 = E (W1)= q2, q0 + q1 + q2 =1. 2 q 2 2 2 q 2

This is a linear system for (q0, q1, q2) that we solve to get

q0 = q2 =2W0, q1 =1 − 4W0 and this specifies a probability distributon on all three scenarios iff W0 ∈ (0, 1/4). Since we can express every contingent claim as a linear combination of A1,S1, W1, we can now 1 price every contingent claim X1 under the martingale probabities as X0 = 2 Eq(X1). 48 Lecture Notes – MS3b Lévy Processes and Finance – Oxford HT 2008 Lecture 11

L´evy markets and time-changes

11.1 Incompleteness and martingale probabilities in L´evy markets

By a L´evy market we will understand a model (S, A) of a risky asset St = exp{Xt} for a L´evy process X = (Xt)t 0 and a deterministic risk-free bank account process, usually δt ≥ At = e , t ≥ 0. We exclude deterministic Xt = µt in the sequel.

Theorem 74 (No arbitrage) A L´evy market allows arbitrage if and only if either Xt − δt is a subordinator or δt − Xt is a subordinator.

Proof: We only prove that these cases lead to arbitrage opportunities. If Xt − δt is a subordinator, then the portfolio (1, −1) is an arbitrage portfolio. 2 The other direction of proof is difficult, since we would need technical definitions of admissible portfolio processes and related quantities. No arbitrage is closely related (almost equivalent) to the existence of martingale probabilities. Formally, an equivalent martingale measure Q is a probability measure which has the same sets of zero probability as P, i.e. under which the same things are possi- δt ble/impossible as under P, and under which (e− St)t 0 is a martingale. For simplicity, ≥ we will not bother about this passage to a so-called risk-neutral world that is different δt from the physical world. Instead, we will consider models where (e− St)t 0 is already δt ≥ a martingale. Prices of the form W0 = e− EQ(Wt) are then arbitrage-free prices. The range of arbitrage-free prices is δt e− EQ(Wt) : Q martingale measure equivalent to P . The proof of incompleteness is also difficult, but the result is not hard to state:

Theorem 75 (Completeness) A Lévy market is complete if and only if (Xt)t 0 is ≥ either a multiple of Brownian motion with drift, Xt = µt + σBt or a multiple of the Poisson process with drift, Xt = at + bNt (with (a − δ)b< 0 to get no arbitrage). Completeness is closely related (almost equivalent) to the uniqueness of martingale probabilities. In an incomplete market, there are infinitely many choices for these martingale probabilities. This raises the question of how to make the right choice. While we can determine an arbitrage-free system of prices for all contingent claims, we cannot hedge the contingent claim, and this presents a risk to someone selling e.g. options. 49 50 Lecture Notes – MS3b Lévy Processes and Finance – Oxford HT 2008

11.2 Option pricing by simulation

If we are given a risk-neutral price process (martingale) (St)t 0, we can price contingent ≥ claims G((Ss)0 s t) as expectations ≤ ≤ δt P = e− E(G((Ss)0 s t)). ≤ ≤ Often, such expectations are diﬃcult to work out theoretically or numerically, particularly for path-dependent options such as barrier options. Monte-Carlo simulation always works, by the strong law of large numbers:

n 1 (k) → E G((Ss )0 s t) (G((Ss)0 s t) almost surely, ≤ ≤ ≤ ≤ n =1 Xk (k) as n → ∞, where (Ss )0 s t are independent copies of (Ss)0 s t. By simulating these ≤ ≤ ≤ ≤ copies, we can approximate the expectation on the right to get the price of the option.

Figure 11.1: Option pricing by simulation

11.3 Time changes

Lévy markets are one way to address shortcomings of the Black-Scholes model. Particu- larly quantities such as one-day return distributions can be fitted well. Other possibilities include modifications to the Black-Scholes model, where the speed of the market is modelled separately. The rationale behind this is to capture days with increased activity (and hence larger price movements) by notions of operational versus real time. In operational time, the price process follows a Brownian motion, but in real time, a busy day corresponds to several days in operational time, while a quiet day corresponds to a fraction of a day in operational time. The passage from operational to real time is naturally modelled by a time-change y 7→ τy, which we will eventually model by a stochastic process built from a Poisson point process. The price process is then (Bτy )y 0. This stochastic process cannot be observed ≥ directly in practice, but approximations of quadratic variation permit to estimate the time change. The most elementary time change is for τy = f(y), a deterministic continuous strictly increasing function f : [0, ∞) → [0, ∞) with f(0) = 0 and f(∞)= ∞. In this case, the Lecture 11: Lévy markets and time-changes 51

time-changed process Zy = Xf(y), y ≥ 0, visits the same states as X in the same order as X, performing the same jumps as X, but travelling at a different speed. Specifically, if f(y) << y, then, by time y, the process X will have gone to Xy, but Z only to Zy = Xf(y). We say that Z has evolved more slowly than X, and faster if instead f(y) >> y. If f is differentiable, we can more appropriately make local speed statements according to whether f ′(y) < 1 or f ′(y) > 1. Note, however, that “speed” really is “relative speed” when comparing X and Z, since X is not “travelling at unit speed” in a sense of rate of spatial displacement; jumps and particularly unbounded variation make such notions useless. We easily calculate

iλZy iλX f(t)ψ(λ) iλXt tψ(λ) E(e )= E(e f(y) )= e− , if E(e )= e− . and see that Z is a stochastic process with independent increments and right-continuous paths with left limits, but will only have stationary increments if f(y)= cy for all y ≥ 0 and some c ∈ (0, ∞).

Example 76 (Foreign exchange rates) Suppose that the EUR/USD-exchange rate today is S0 and you wish to model the exchange rate (St)t 0 over the next couple of days. ≥ As a ﬁrst model you might think of

2 St = S0 exp{σBt − tσ /2}, where B is a standard Brownian motion σ is a volatility parameter that measures the magnitude of variation. This magnitude is related to the amount of activity on the exchange markets and will be much higher for the EUR/USD-exchange rate than e.g. for the EUR/DKK-exchange rate (DansKe Kroner, Danish crowns are not traded so frequently in such high volumes. Also, DKK is closely aligned with EUR due to strong economic ties between Denmark and the Euro countries). However, in practice, trading activity is not constant during the day. When stock markets in the relevant countries are open, activity is much higher than when they are all closed and a periodic function f ′ : [0, ∞) → [0, ∞) can explain a good deal of this variability and provide a better model

2 { − } { e − } St = S0 exp σBf(y) f(y)σ /2 = S0 exp Bf(y) f(y)/2 ,

≥ 2 where Bs = σBs/σ2 , s 0, is also a standard Brownian motione ande f(y)= f(y)σ makes the parameter σ redundant – the ﬂexibility for f retains all modelling freedom. e e If we weaken the requirement of strict monotonicitye to weak monotonicity and f(y)= c, y ∈ [l,r), is constant on an interval, then Zy = Xc, y ∈ [l,r), during this interval. For a ﬁnancial market model this can be interpreted as time intervals with no market activity, when the price will not change.

If we weaken the continuity of f to allow (upward) jumps, then Zy = Xf(y), y ≥ 0, does not evaluate X everywhere. Speciﬁcally, if ∆f(y) > 0 is the only jump of f, then Z will visit the same points as X in the same order until Xf(y ) and then skip over − − (Xf(y )+s)0 s<∆f(y) to directly jump to Xf(y). In general, this is the behaviour at every jump− of f. ≤ 52 Lecture Notes – MS3b L´evy Processes and Finance – Oxford HT 2008

Figure 11.2: piecewise constant volatility

11.4 Quadratic variation of time-changed Brownian motion

In Section 6.1 we studied quadratic variation of Brownian motion in order to show that Brownian motion has inﬁnite total variation (and is therefore not the diﬀerence of two increasing processes). Let us here look at quadratic variation of time-changed Brownian motion Zy = Bf(y) for an increasing function f : [0, ∞) → [0, ∞):

[2nt] (n) (n) 2 − − − − [Z]t = p lim [Z]t , where [Z]t = (Zj2 n Z(j 1)2 n ) n − →∞ j=1 X and p − lim denotes a limit of random variables in probability. One may expect that [B]t = t implies that [Z]y = f(y), and this is true under suitable assumptions.

Proposition 77 Let B be Brownian motion and f : [0, ∞) → [0, ∞) continuous and increasing with f(0) = 0. Then [Z]y = f(y) for all y ≥ 0.

Proof: The proof (for Z) is the same as for Brownian motion (B) itself, see Section 6.1 and Assignment 6. 2

Quadratic variation is accumulated locally. Under the continuity assumption of Brow- nian motion and its time change, it is the wiggly local behaviour of Brownian motion that generates quadratic variation. In Section 6.1 we showed that under the no-jumps assumption, positive quadratic variation implies inﬁnite total variation. Hence, still under the no-jumps assumption, ﬁnite total variation implies zero quadratic variation. But what is the impact of jumps on quadratic variation? It can be shown as in Proposition 3 that

2 [f]y ≥ |∆fs| . s y X≤ Example 78 Consider a piecewise linear function f : [0, ∞) → [0, ∞) with slope 0.1 and jumps ∆f2k 1 =1.8, k ≥ 1. Then f(2k)=2k, but by Proposition 77 − 2 2 [f]2k ≥ |∆fs| = k(1.8) =3.24k s 2k X≤ Lecture 11: L´evy markets and time-changes 53 and, in fact, this is an equality, since

(n) n 2 n+1 − n 2 → 2 [f]2k = k(1.8+2− 0.1) + (2 1)k(2− 0.1) k(1.8) .

Now deﬁne Zy = Bf(y) and note that

k 2n 1 (n) − 2 − − − [Z]2k = (B(2i 2)+j2 n0.1 B(2i 2)+(j 1)2 n0.1) − − − i=1 j=1 X X 2 +(B2i 0.1 − B(2i 2)+0.1 2−n ) − − − 2n 2 − − − + (B2i (j 1)2 n0.1 B2i j2 n0.1) − − − j=1 ! X k 2 → 2k0.1+ (B2i 0.1 − B(2i 2)+0.1) , − − i=1 X →∞ | |2 as n , which is actually 0.1(2k)+ s 2k ∆Zs . Note that this is a random quantity. ≤ In general, quadratic variation consistsP of a continuous part due to Brownian ﬂuctu- ations and the sum of squared jump sizes. 54 Lecture Notes – MS3b L´evy Processes and Finance – Oxford HT 2008 Lecture 12

Subordination and stochastic volatility

≥ Subordination is the operation Xτy , y 0, for a L´evy (or more general Markov) process (Xt)t 0 and a subordinator (τy)y 0. One distinguishes subordination in the sense of ≥ ≥ Bochner, where X and τ are independent and subordination in the wide sense where τy is a stopping time for all y ≥ 0. These are both special cases of the more general concept of time change, where (τy)y 0 does not have to be a subordinator. ≥ 12.1 Bochner’s subordination

Theorem 79 (Bochner) Let (Xt)t 0 be a L´evy process and (τy)y 0 an independent sub- ≥ ≥ ≥ ordinator. Then the process Zy = Xτy , y 0, is a L´evy process, and we have

iλZy yΦ(ψ(λ)) iλXt tψ(λ) qτy yΦ(q) E(e )= e− , where E(e )= e− and E(e− )= e− .

Proof: First calculate by conditioning on τy (assuming that τy is continuous with probability density function fτy )

E { } E { } ∞ E { } (exp iλZy ) = (exp γXτy )= fτy (t) (exp iλXt )dt 0 Z ∞ {− } yΦ( ψ(λ)) = fτy (t) exp tψ(λ) dt = e− − . 0 Z Now, for r, s ≥ 0,

E(exp{iλZy + iµ(Zy+x − Zy)}) ∞ ∞ E { − } = fτy,τy+x τy (t, u) (exp iλXt + iµ(Xt+u Xt) )dtdu 0 0 − Z Z ∞ ∞ tψ(λ) uψ(µ) yΦ(ψ(λ)) xΦ(ψ(µ)) = fτy (t)fτx (u)e− e− dtdu = e− e− , 0 0 Z Z so that we deduce that Zy and Zy+x − Zy are independent, and that Zy+x − Zy ∼ Zx. For the right-continuity of paths, note that

lim Zy+ε = lim Xτy+ε = Xτy = Zy, ε 0 ε 0 ↓ ↓ 55 56 Lecture Notes – MS3b L´evy Processes and Finance – Oxford HT 2008

↓ → since τy + δ := τy+ε τy and therefore Xτy+δ Xτy . For left limits, the same argument applies. 2

− − 6 6 Note that ∆Zy = Zy Zy = Xτy Xτy = can be non-zero if either ∆τy = 0, or 6 − − ∆Xτy = 0, so Z inherits jumps from τ and from X. We have, with probability 1 for all y ≥ 0 that 6 − (∆X)τy if (∆X)τy = 0, ∆Zy = Xτy Xτy− = − X − X − if ∆τ =6 0. τy τy y 6 6 ≥ Note that we claim that Xτy− = Xτy− , i.e. (∆X)τy− = 0 if ∆τy = 0, for all y 0 − with probability 1. This is due to the fact that the countable set of times {τy , τy : y ≥ − 0 and ∆τy =06 } is a.s. disjoint from {t ≥ 0 : ∆Xt =06 }.

Note also that Xτy = Xτy− is possible with positive probability, certainly in the case of a compound Poisson process X. Heuristically, if Xt has density ft and τ L´evy density gτ , then Z will have L´evy density

∞ g(z) := ft(z)gτ (t)dt, z ∈ R, (1) 0 Z − ∼ since every jump of τ of size ∆τy = t leads to a jump Xτy Xτy Xt, and the total − intensity of jumps of size z receives contributions from τ-jumps of all sizes t ∈ (0, ∞). We can make this precise as follows:

Proposition 80 Let X be a Lévy process with probability density function ft of Xt, ≥ t 0, τ a subordinator with Lévy-Khintchine characteristics (0,gτ ), then Zy = Xτy has Lévy-Khintchine characteristics (0, 0,g), where g is given by (1).

Proof: Consider a Poisson point process (∆y)y 0 with intensity function g, then, by the ≥ Exponential Formula

E exp iλ ∆s1 ∆s >ε {| | } ( s y )! X≤ ∞ iλz = exp y (e − 1)g(z)1 z >ε dz {| | } Z−∞ ∞ ∞ iλz = exp y (e − 1)ft(z)1 z >ε dzgτ (t)dt 0 {| | } Z Z−∞ ∞ tψ(λ) → exp −y (1 − e− )gτ (t)dt = exp {−yΦ(ψ(λ))} , 0 Z as ε ↓ 0, and this is the same distribution as we established for Zy. 2

Note that we had to prove convergence in distribution as ε ↓ 0, since we have not studied integrability conditions for g. This is done on Assignment sheet 6.

Corollary 81 If X is Brownian motion and τ has characteristics (b, gτ ), then Zy = Xτy has characteristics (0,b,g). Lecture 12: Subordination and stochastic volatility 57

Proof: Denote Φ0(q)= Φ(q) − bq. Then the calculation in the proof of the proposition does not yield a characteristic exponent Φ(ψ(λ)), but Φ0(ψ(λ)). Note that Φ(ψ(λ)) = 1 2 1 2 Φ( 2 λ )= 2 bλ + Φ0(ψ(λ)), so we consider

bBt + ∆s s t X≤ for an independent Brownian motion B, which has characteristic exponent as required. 2

Example 82 If we define the Variance Gamma process by subordination of Brownian 1 θy motion B by a Gamma(α, θ) subordinator τ with Lévy density gτ (y) = αy− e− , then we obtain a Lévy density

∞ ∞ √1 z2/(2t) 1 θt g(z)= ft(z)gτ (t)dt = e− αt− e− dt 0 0 2πt Z Z and we can calculate this integral to get

1 √2θ z g(z)= α|z|− e− | | as Lévy density. The Variance Gamma process as subordinated Brownian motion has an interesting interpretation when modelling financial price processes. In fact, the stock price is then considered to evolve according to the Black-Scholes model, but not in real time, but in operational time τy, y ≥ 0. Time evolves as a Gamma process, with infinitely many jumps in any bounded interval. Note that all Lévy processes that we can construct as subordinated Brownian motions are symmetric. However, not all symmetric Lévy processes are subordinated Brownian motions,

12.2 Ornstein-Uhlenbeck processes

Example 83 (Gamma-OU process) Let (Nt)t 0 be a Poisson process with intensity ≥ aλ and jump times (Tk)k 1, (Xn)n 1 a sequence of independent Gamma(1, b) random ≥ ≥ variables, Y0 ∼ Gamma(a, b), consider the stochastic process

Nt λt λ(t Tk ) Yt = Y0e− + Xne− − =1 Xk We use this model for the speed of the market and think of an initial speed of Y0 which slows down exponentially, but at times of a Poisson process, events occur that make the speed jump up at times Tk, k ≥ 0. Each of these also slow down exponentially. In fact, there is a strong equilibrium in that

Nt −λt qYt qe Y0 λ(t Tk) E(e− ) = E(e− )E exp −q Xke− − ( k=1 )! a X n ∞ n t b (λat) λat 1 b = λt e− λs ds b + qe n! 0 t b + qe − n=0 − X Z b a b + qe λt a b a = − = , b + qe λt b + q b + q − 58 Lecture Notes – MS3b L´evy Processes and Finance – Oxford HT 2008

so Yt has the same distribution as Y0. In fact, (Yt)t 0 is a stationary Markov process. ≥ The process Y is called a Gamma-OU process, since it has the Gamma distribution as its stationary distribution. The time change process y τy = Ysds 0 Z associated with speed Y is called integrated Ornstein-Uhlenbeck process. Note that the stationarity of Y implies that τ has stationary increments, but note that τ does not have independent increments. The associated stochastic volatility model is now the time-changed Brownian motion (Bτy )y 0. ≥ In general, we can deﬁne Ornstein-Uhlenbeck processes associated with any subordinator Z or rather its Poisson point process (∆Zt)t 0 of jumps as ≥ λt λ(t s) Yt = Y0e− + ∆Zse− − , s t X≤ for any initial distribution for Y0, but a stationary distribution can also be found.

We can always associate a stochastic volatility model (Bτy )y 0 using the integrated ≥ volatility τy = y Ysds as time change. Note that, by the discussion of the last section, we can actually infer the time change from sums of squared increments for a small time n R lag 2− , even though the actual time change is not observed. In practice, the so-called market microstructure (piecewise constant prices) destroys model ﬁt for small times, so n we need to choose a moderately small 2− . In practice, 5 minutes is a good choice.

12.3 Simulation by subordination

Note that we can simulate subordinators using simulation Method 1 (Time discretisation) or Method 2 (Throwing away the small jumps). The latter consisted essentially in simulating the Poisson point process of jumps of the subordinator. Clearly, we can apply this method also to simulate an Ornstein-Uhlenbeck process.

Method 3 (Subordination) Let (τy)y 0 be an increasing process that we can simulate, ≥ and let (Xt)t 0 be a L´evy process with cumulative distribution function Ft of Xt. Fix a ≥ time lag δ > 0. Then the process

n (3,δ) 1 Z = S[ ], where S = A and A = F − (U ) y y/δ n k k τkδ τ(k−1)δ k =1 − Xk is the time discretisation of the subordinated process Zy = Xτy .

Example 84 We can use Method 3 to simulate the Variance Gamma process, since we can simulate the Gamma process τ and we can simulate the Ak. Actually, we can use the Box-Muller method to generate standard Normal random variables Nk and then use

Ak ∼ τkδ − τ(k 1)δNk, k ≥ 1, − ≥ p instead of Ak, k 1. e Lecture 13

Level passage problems

Reading: Kyprianou Sections 3.1 and 3.3

13.1 The strong Markov property

Recall that a stopping time is a random time T ∈ [0, ∞] such that for every s ≥ 0 the information Fs up to time s allows to decide whether T ≤ s. More formally, if the event {T ≤ s} can be expressed in terms of (Xr,r ≤ s) (is measurable with respect to Fs = σ(Xr,r ≤ s)). The prime example of a stopping time is the ﬁrst entrance time TI = inf{t ≥ 0 : Xt ∈ I} into a set I ⊂ R. Note that

{T ≤ s} = {there is r ≤ s such that Xr ∈ I}

(for open sets I we can drop the irrational r ≤ s to show measurability.).

We also denote the information FT up to time T . More formally,

FT = {A ∈F : A ∩{T ≤ s}∈Fs for all s ≥ 0}, i.e. FT contains those events that, if T ≤ s, can be expressed in terms of (Xr,r ≤ s), for all s ≥ 0. Recall the simple Markov property which we can now state as follows. For a L´evy process (Xt)t 0 and a ﬁxed time t, the post-t process (Xt+s −Xt)s 0 has the same distribution ≥ ≥ as X and is independent of the pre-t information Ft.

Theorem 85 (Strong Markov property) Let (Xt)t 0 be a L´evy process and T a stop- ≥ ping time. Then given T < ∞, the post-T process (XT +s − XT )s 0 has the same distri- ≥ bution as X and is independent of the pre-T information FT .

Proof: Let 0

P ∞ − ≤ − ≤ (A,T < ,XT +s1 XT c1,...,XT +sm XT cm) P ∞ P ≤ ≤ = (A,T < ) (Xs1 c1,...,Xsm cm). 59 60 Lecture Notes – MS3b L´evy Processes and Finance – Oxford HT 2008

n n First deﬁne stopping times Tn = 2− ([2 T ]+ 1), n ≥ 1, that only take countably many values. These are the next dyadic rationals after time T . Note that Tn ↓ T as n → ∞. n Now note that A ∩{Tn = k2− }∈Fk2−n and the simple Markov property yields P ∞ − ≤ − ≤ (A, Tn < ,XTn+s1 XTn c1,...,XTn+sm XTn cm) ∞ n P −n − −n ≤ −n − −n ≤ = (A, Tn = k2− ,Xk2 +s1 Xk2 c1,...,Xk2 +sm Xk2 cm) =0 Xk ∞ P n P ≤ ≤ = (A, Tn = k2− ) (Xs1 c1,...,Xsm cm) k=0 XP ∞ P ≤ ≤ = (A, Tn < ) (Xs1 c1,...,Xsm cm). → → ∞ Now the right-continuity of sample paths ensures XTn+sj XT +sj as n and we conclude P ∞ − ≤ − ≤ (A,T < ,XT +s1 XT c1,...,XT +sm XT cm) P ∞ − ≤ − ≤ = lim (A, Tn < ,XTn+s XTn c1,...,XTn+sm XTn cm) n 1 →∞ P ∞ P ≤ ≤ = lim (A, Tn < ) (Xs c1,...,Xsm cm) n 1 P→∞ ∞ P ≤ ≤ = (A,T < ) (Xs1 c1,...,Xsm cm), P − for all (c1,...,cm) such that (XT +sj XT = cj) = 0, j = 1,...,m. Finally note that (XT +s − XT )s 0 clearly has right-continuous paths with left limits. 2 ≥ 13.2 The supremum process

Let X =(Xt)t 0 be a L´evy process. We denote its supremum process by ≥

Xt = sup Xs, t ≥ 0. 0 s t ≤ ≤

We are interested in the joint distribution of (Xt, Xt), e.g. for the payoﬀ of a barrier or lookback option. Moment generating functions are easier to calculate and can be numerically inverted. We can also take such a transform over the time variable, e.g.

∞ qt γXt 1 γX q 7→ e− E(e )dt = uniquely identiﬁes E(e t ), 0 q − Ψ(γ) Z qt γXt γXτ ∼ and the distribution of Xt. But q 0∞ e− E(e )dt = E(e ) for τ Exp(q).

Proposition 86 (Independence)R Let X be a L´evy process, τ ∼ Exp(q) an independent random time. Then Xτ is independent of Xτ − Xτ .

Proof: We only prove the case where G1 = inf{t> 0 : Xt > 0} satisﬁes P(G1 > 0) = 1. { } ≥ In this case we can deﬁne successive record times Gn = inf t>Gn 1 : Xt > XGn−1 , n − 2, and also set G0 = 0. Note that, by the strong Markov property at the stopping times − Gn we have that XGn > XGn−1 (otherwise the post-Gn 1 process Xt = XGn−1+t XGn−1 − would have the property G1 = 0, but the strong Markov property yields P(G1 > 0) = e e e Lecture 13: Level passage problems 61

P ∈{ ≥ } (G1 > 0) = 1). So X can only reach new records by upward jumps, Xτ XGn , n 0 and more speciﬁcally, we will have Xτ = Gn if and only if Gn ≤ τ

βXτ +γ(Xτ Xτ ) ∞ qt βXt+γ(Xt Xt) E(e − ) = qe− E(e − )dt 0 Z ∞ Gn+1 qt βXt+γ(Xt Xt) = qE e− e − dt n=0 Gn ! X Z Gn+1 Gn ∞ − βXG qGn qs γ(XG s XG ) = q E e n e− e− e− n+ − n ds 0 n=0 Z X e G ∞ 1 e qGn+βXG qs γXs = q E e− n E e− − ds 0 n=0 Z ! X where we applied the strong Markov property at Gn to split the expectation in the last row Gn+1 Gn qs γ(XG +s XG ) – note that 0 − e− − n − n ds is a function of the post-Gn process, whereas qGn+βXG e− n is a function of the pre-G process, and the expectation of the product of R n independent random variables is the product of their expectations. This completes the proof, since the last row is a product of a function of β and a function of γ, which is enough to conclude. More explicitly, we can put β = 0, γ = 0 and β = γ = 0, respectively, to see that indeed the required identity holds:

βXτ +γ(Xτ Xτ ) βXτ γ(X τ Xτ ) E(e − )= E(e )E(e − ). 2

13.3 L´evy processes with no positive jumps

Consider stopping times Tx = inf{t ≥ 0 : Xt ∈ (x, ∞)}, so-called ﬁrst passage times. For ∞ L´evy processes with no positive jumps, we must have XTx = x, provided that Tx < . This observation allows to calculate the moment generating function of Tx. To prepare this result, recall that the distribution of Xt has moment generating function 0 γXt tΨ(γ) 1 2 2 γx E(e )= e , Ψ(γ)= a1γ + σ γ + (e − 1 − γx1 x 1 )g(x)dx. 2 {| |≤ } Z−∞ 2 Let us exclude the case where −X is a subordinator, i.e. where σ = 0 and a1 − 0 ≤ ∞ 1 xg(x)dx 0, since in that case Tx = . Then note that − 0 R 2 2 γx Ψ′′(γ)= σ + x e g(x)dx > 0, Z−∞ so that Ψ is convex and hence has at most two zeros, one of which is Ψ(0) = 0. There is a second zero γ0 > 0 if and only if Ψ′(0) = E(X1) < 0, since we excluded the case where −X is a subordinator, and P(Xt > 0) > 0 implies that Ψ(∞)= ∞.

Theorem 87 (Level passage) Let (Xt)t 0 be a Lévy process with no positive jumps ≥ and Tx the first passage time across level x. Then Φ( ) E qTx x q (e− 1 Tx< )= e− , { ∞} where Φ(q) is the largest γ for which Ψ(γ)= q. 62 Lecture Notes – MS3b Lévy Processes and Finance – Oxford HT 2008

Proof: We only prove this for the case where P(Tx < ∞) = 1, i.e. E(X1) ≥ 0 and γ0 = 0. γXt tΨ(γ) By Exercise A.2.3.(a) the processes Mt = e − are martingales. We will apply the E 2 t(Ψ(2γ) 2Ψ(γ)) Optional stopping theorem to Tx. Note that (Mt ) = e − is not such that E 2 ∞ supt 0 (Mt ) < . However, if we put ≥ (u) ≤ (u) ≥ Mt = Mt if t u and Mt = Mu if t u, u E (u) 2 ∞ ∧ then (Mt )t 0 is a martingale which satisﬁes supt 0 ((Mt ) ) < . Also, Tx u is a ≥ ≥ stopping time, so that for γ ≥ γ0 = 0 (so that Ψ(γ) ≥ 0) ( ) Ψ( ) E u E → E E γx γ Tx →∞ 1= (MTx u)= (MTx u) (MTx )= (e − ), as u , ∧ ∧ ≤ { − } ≤ { } by dominated convergence (MTx u exp γx Ψ(γ)Tx exp γx ). We now conclude ∧ that

Ψ(γ)Tx γx E(e− )= e− which for q = Ψ(γ) and Φ(q) the unique γ ≥ γ0 = 0 with Ψ(γ)= q. 2 Corollary 88 Let X be a L´evy process with no positive jumps and τ ∼ Exp(q) independent. Then Xτ ∼ Exp(Φ(q)). Φ( ) P P ≤ P ≥ E qTx q x 2 Proof: (Xτ > x)= (Tx τ)= 0∞ (τ t)fTx (t)dt = (e− )= e− . If we combine this with the IndependenceR Theorem of the previous section we obtain. Corollary 89 Let X be a L´evy process with no positive jumps and τ ∼ Exp(q) independent. Then − β(Xτ Xτ ) q(Φ(q) β) E(e− − )= Φ(q)(q − Ψ(β)) Proof: Note that we have from the Independence Theorem that

βXτ β(Xτ Xτ ) βXτ ∞ qt βXt q E(e )E(e− − )= E(e )= qe− E(e )dt = 0 q − Ψ(β) Z and from the preceding corollary − βXτ Φ(q) β(Xτ Xτ ) q Φ(q) β E(e )= and so E(e− − )= . Φ(q) − β q − Ψ(β) Φ(q) 2

13.4 Application: insurance ruin

Proposition 86 splits the Lévy process at its supremum into two increments. If you turn the picture of a Lévy process by 180◦, this split occurs at the infimum, and it can be ∼ − E βXτ shown (Exercise A.7.1) that Xτ Xτ Xτ . Therefore, Corollary 89 gives (e ), also for q ↓ 0 if E(X1) > 0, since then

q(Φ(q) − β) βE(X1) E(eβX∞ ) = lim = q 0 − ↓ Φ(q)(q Ψ(β)) Ψ(β) since Φ′(0) = 1/Ψ′(0) = 1/E(X1) and note that for an insurance reserve process Rt = u + Xt, the probability of ruin is r(u) = P(X < −u), the distribution funcion of X which is uniquely identiﬁed by E(eβX∞ ). ∞ ∞ Lecture 14

Ladder times and storage models

Reading: Kyprianou Sections 1.3.2 and 3.3

14.1 Case 1: No positive jumps

In Theorem 87 we derived the moment generating function of Tx = inf{t ≥ 0 : Xt > x} for any L´evy process with no positive jumps. We also indicated the complication that Tx = ∞ is a possibility, in general. Let us study this is in more detail in our standard setting 0 γXt tΨ(γ) 1 2 2 γx E(e )= e , Ψ(γ)= a1γ + σ γ + (e − 1 − γx1 x 1 )g(x)dx. 2 {| |≤ } Z−∞ The important quantity is 0 ∂ γXt E(X1)= E(e ) =Ψ′(0) = a1 − xg(x)dx. ∂γ γ=0 1 Z−

The formula that we derived was Φ( ) E qTx x q (e− 1 Tx< )= e− { ∞} where for q > 0, Φ(q) > 0 is unique with Ψ(Φ(q)) = q. Letting q ↓ 0, we see that Φ(0+) P ∞ E qTx x (Tx < ) = lim (e− 1 Tx< )= e− . q 0 { ∞} ↓ Here the convexity of Ψ that we derived last time implies that Φ(0+) = 0 if and only if E(X1)= φ′(0) ≥ 0. Therefore, P(Tx < ∞) = 1 if and only if E(X1) ≥ 0. Part of this could also be deduced by the strong (or weak) law of large numbers. Applied to increments Yk = Xkδ − X(k 1)δ it implies that − n Xnδ 1 1 1 1 = Yk → E(Y1)= E(Xδ)= E(X1), nδ δ n =1 δ δ Xk almost surely (or in probability) as n → ∞. We can slightly improve this result to a convergence as t →∞ as follows

γX /t tφ(γ/t) γφ′(0) γE(X ) Xt E(e t )= e → e = e 1 ⇒ → E(X1), t 63 64 Lecture Notes – MS3b L´evy Processes and Finance – Oxford HT 2008

in probability. We used here that Zt → a in distribution implies Zt → a in probability, which holds only because a is a constant, not a random variable. Note that indeed for all ε> 0, as t →∞,

P(|Zt − a| >ε) ≤ P(Zt ≤ a − ε)+1 − P(Zt ≤ a + ε) → 0+1 − 1=0.

From this, we easily deduce that Xt → ±∞ (in probability) if E(Xt) =6 0, but the case E(Xt) = 0 is not so clear from this method. In fact, it can be shown that all convergences hold in the almost sure sense, here. By an application of the Strong Markov property we can show the following.

Proposition 90 The process (Tx)x 0 is a subordinator. ≥

Proof: Let us here just prove that Tx+y − Tx is independent of Tx and has the same distribution as Ty. The remainder is left as an exercise.

Note ﬁrst that XTx = x, since there are no positive jumps. The Strong Markov − F property at Tx can therefore be stated as X =(XTx+s x)s 0 is independent of Tx and ≥ has the same distribution as X. Now note that e { ≥ } { ≥ } Tx + Ty = Tx + inf s 0 : Xs >y = Tx + inf s 0 : XTx+s > x + y

= inf{t ≥ 0 : Xt > x + y} = Tx+y e e so that Tx+y − Tx = Ty, and we obtain

P(Tx ≤ s, Tex+y − Tx ≤ t)= P(Tx ≤ s, Ty ≤ t)= P(Tx ≤ s)P(Ty ≤ t), { ≤ }∈F { ≤ }∩{ ≤ } { ≤ ∧ }∈F ≥ since Tx s Tx . Formally, Tx s Txe r = Tx s r r for all r 0 since Tx is a stopping time. 2

We can understand what the jumps of (Tx)x 0 are: in fact, X can be split into its ≥ supremum process Xt = sup0 s t Xs and the bits of path below the supremum. Roughly, the times ≤ ≤

{Tx, x ≥ 0} = {t ≥ 0 : Xt = Xt} are the times when the supremum increases. Tx − Tx > 0 if the supremum process − remains constant at height x for an amount of time Tx − Tx . The process (Tx)x 0 is − ≥ called “ladder time process”. The process (XTx )x 0 is called ladder height process. In ≥ this case, XTx = x is not very illuminating. Note that (Tx,XTx )x 0 is a bivariate L´evy ≥ process.

Example 91 (Storage models) Consider a L´evy process of bounded variation, repre- sented as At − Bt for two subordinators A and B. We interpret At as the amount of work arriving in [0, t] and Bt as the amount of work that can potentially exit from the system. Let us focus on the case where A is a compound Poisson process and Bt = t for a continuously working processor. The quantity of interest is the amount Wt of work waiting to be carried out and requiring storage, where W0 = w ≥ 0 is an initial amount of work stored. Lecture 14: Ladder times and storage models 65

Note that Wt =6 w + At − Bt, in general, since w + At − Bt can become negative, whereas Wt ≥ 0. In fact, we can describe as follows: if the storage is empty, then no work exits from the system. Can we see from At − Bt when the storage will be empty? We can express the first time it becomes empty and the first time it is refilled thereafter as

L1 = inf{t ≥ 0 : w + At − Bt =0} and R1 = inf{t ≥ L1 : ∆At > 0}.

On [L1, R1], Xt = Bt −At increases linearly at unit rate from w to w +(R1 −L1), whereas W remains constant equal to zero. In fact,

t t − − ∨ − ≤ ≤ 1 Wt = w Xt + 1ds = w Xt + 1 Xs=Xs w ds =(w Xt) Xt, 0 t R . L1 t 0 { ≥ } Z ∧ Z

An induction now shows that the system is idle if and only if Xt = Xt ≥ w, so that Wt = Xt +(w ∨ Xt) for all t ≥ 0. + In this context, (Xt −w) is the amount of time the system was idle before time t, and Tx = inf{t ≥ 0 : Xt > x} is the time by which the system has accumulated time x − w in the idle state, x ≥ w, and we see that (x − w)/Tx ∼ x/Tx → 1/E(T1)=1/Φ′(0) = φ′(0) = E(X1)=1 − E(A1) in probability, if E(A1) ≤ 1.

Example 92 (Dams) Suppose that the storage model refers more particularly to a dam that releases a steady stream of water at a constant intensity a2. Water is added according to a subordinator (At)t 0. The dam will have a maximal capacity of, say, M > 0. Given ≥ an initial water level of w ≥ 0, the water level at time t is, as before

Wt =(w ∨ Xt) − Xt, where Xt = a2t − At.

The time F = inf{t ≥ 0 : Wt > M}, the ﬁrst time when the dam overﬂows, is a quantity of interest. We do not pursue this any further theoretically, but note that this can be simulated, since we can simulate X and hence W .

14.2 Case 2: Union of intervals as ladder time set

Proposition 93 If Xt = a2t − Ct for a compound Poisson process (Ct)t 0 and a drift ≥ a2 ≥ 0 ∨ E(C1), then the ladder time set is a collection of intervals. More precisely,

{t ≥ 0 : Xt = Xt} is the range {σy,y ≥ 0} of a compound Poisson subordinator with positive drift coeﬃcient.

Proof: Deﬁne L0 = 0 and then for n ≥ 0 stopping times

Rn = inf{t ≥ Ln : ∆Ct > 0}, Ln+1 = inf{t ≥ Rn : Xt = Xt}.

The strong Markov property at these stopping times show that (Rn −Ln)n 0 is a sequence ≥ of Exp(λ) random variables where λ = 0∞ g(x)dx is the intensity of positive jumps, and (Ln − Rn 1)n 1 is a sequence of independent identically distributed random variables. − ≥ R Now deﬁne Tn = R0 − L0 + . . . + Rn 1 − Ln 1 and (σy)y 0 to be the compount Poisson − − ≥ process with unit drift, jump times Tn, n ≥ 1, and jump heights Ln − Rn 1, n ≥ 1. 2 − 66 Lecture Notes – MS3b L´evy Processes and Finance – Oxford HT 2008

The ladder height process (Xσy )y 0 will then also have a positive drift coefficient and ≥ share some jump times with σ (whenever X jumps from below Xt above Xt , but have − − extra jump times when X jumps from Xt upwards, some jump times of (σy)y 0 are not − ≥ jump times of (Xσy )y 0 – if X reaches Xt again without a jump.) ≥ − Example 94 (Storage models) In the context of the previous examples, consider more general subordinators Bt with unit drift. Interpret jumps of B as unfinished work potentially exiting to be carried out elsewhere. We should be explicit and make a convention that if the current storage amount Wt is not sufficient for a jump of B, then all remaining work exits. With this convention, the amount Wt waiting to be carried out is still

t − ≥ Wt = w Xt + 1 Xs=Xs w ds t 0, 0 { ≥ } Z but note that the latter integral cannot be expressed in terms of Xt so easily, but σy is still the time by which the system has accumulated time y − w in the idle state, for y ≥ w, so It = inf{y ≥ 0 : σy > t} is the amount of idle time before t.

14.3 Case 3: Discrete ladder time set

If Xt = a2t − Ct for a compound Poisson process (or indeed bounded variation pure jump process) (Ct)t 0 and a drift a2 < 0, then the ladder time set is discrete. We can still think ≥ of {t ≥ 0 : Xt = Xt} as the range {σy,y ≥ 0} of a compound Poisson subordinator with zero drift coefficient. More naturally, we would define successive ladder times G0 = 0 and Gn+1 = inf{t>Gn : Xt = Xt}. By the strong Markov property, Gn+1 − Gn, n ≥ 0, is a sequence of independent and identically distributed random variables, and for any intensity λ> 0, we can specify (σy)y 0 to be a compound Poisson process with rate λ> 0 ≥ and jump sizes Gn+1 − Gn, n ≥ 0. Note that (σy)y 0 is not unique since we have to choose λ. In fact, once a choice has ≥ been made and q > 0, we have {σy : y ≥ 0} = {σqy,y ≥ 0}, not just here, but also in Cases 1 and 2. In Cases 1 and 2, however, we identified a natural choice (of q) in each case.

14.4 Case 4: non-discrete ladder time set and positive jumps

The general case is much harder. It turns out that we can still express

{t ≥ 0 : Xt = Xt} = {σy : y ≥ 0} for a subordinator (σy)y 0, but, as in Case 3, there is no natural way to choose this process. ≥ It can be shown that the bivariate process (σy,Xσy )y 0 is a bivariate subordinator in this ≥ general setting, called the ladder process. There are descriptions of its distribution and relations between these processes of increasing ladder events and analogous processes of decreasing ladder events. Lecture 15

Branching processes

Reading: Kyprianou Section 1.3.4

15.1 Galton-Watson processes

Let ξ =(ξk)k 0 be (the probability mass function of) an oﬀspring distribution. Consider ≥ a population model where each individual gives birth to independent and identically distributed numbers of children, starting from Z0 = 1 individual, the common ances- tor. Then the (n + 1)st generation Zn+1 consists of the sum of numbers of children

Nn,1,...,Nn,Zn of the nth generation:

Zn+1 = Nn,i, where Nn,i ∼ ξ independent, i ≥ 1, n ≥ 0. i=1 X k Proposition 95 Let ξ be an oﬀspring distribution, g(s)= k 0 ξks its generating function, then ≥ P Z1 Z2 Zn (n) E(s )= g(s), E(s )= g(g(s)),..., E(s )= g◦ (s),

(0) (n+1) (n) where g◦ (s)= s, g◦ (s)= g◦ (g(s)), n ≥ 0. Proof: The result is clearly true for n = 0 and n = 1. Now note that

P P Zn ∞ j Xn+1 Nn,i Nn,i E(s ) = E s i=1 = P(Zn = j)E s i=1 j=0 X ∞ j Zn (n) = P(Zn = j)(g(s)) = E((g(s)) )= g◦ (g(s)). j=0 X 2

Proposition 96 (Zn)n 0 is a Markov chain whose transition probabilities are given by ≥

pij = P(N1 + . . . + Ni = j), where N1,...,Ni ∼ ξ independent. (1) (2) In particular, if (Zn )n 0 and (Zn )n 0 are two independent Markov chains with tran- ≥ ≥ sition probabilities (pij)i,j 0 starting from population sizes k and l, respectively, then (1) (2) ≥ Zn + Zn , n ≥ 0, is also a Markov chain with transition probabilities (pij)i,j 0 starting ≥ from k + l. 67 68 Lecture Notes – MS3b L´evy Processes and Finance – Oxford HT 2008

Proof: Just note that the independence of (Nn,i)i 1 and (Nk,i)0 k n 1,i 1 implies that ≥ ≤ ≤ − ≥ P | P | (Zn+1 =j Z0 =i0,...,Zn 1 =in 1,Zn =in) = (Nn,1 + ...,Nn,in =j Z0 =i0,...,Zn =in) − − P = (Nn,1 + ...,Nn,in = j)= pinj, as required. For the second assertion note that P (1) (2) | (1) (2) b(i1,i2),j := (Zn+1 + Zn+1 = j Zn = i1,Zn = i2) (1) (1) (2) (2) P + = (Nn,1 + . . . + Nn,i1 + Nn,1 + . . . + Nn,i2 = j)= pi1 i2,j only depends on i1 + i2 (not i1 or i2 separately) and is of the form required to conclude that

i (1) (2) P(Z = i1,Z = i − i1)b( ) (1) (2) (1) (2) i1=0 n n i1,i i1 ,j P(Z +1 + Z +1 = j|Z + Z = i)= − = p . n n n n P (1) (2) ij P (Zn + Zn = i) 2

The second part of the proposition is called the branching property and expresses the property that the families of individuals in the same generation evolve completely independently of one another.

15.2 Continuous-time Galton-Watson processes

We can also model lifetimes of individuals by independent exponentially distributed random variables with parameter λ > 0. We assume that births happen at the end of a lifetime. This breaks the generations. Since continuously distributed random variables are almost surely distinct, we will observe one death at a time, each leading to a jump of size k − 1 with probability ξk, k ≥ 0. It is customary to only consider oﬀspring distributions with ξ1 = 0, so that there is indeed a jump at every death time. Note that at any given time, if j individuals are present in the population, the next death occurs at a time

H = min{L1,...,Lj} ∼ Exp(jλ), where L1,...,Lj ∼ Exp(λ).

From these observations, one can construct (and simulate!) the associated population size process (Yt)t 0 by induction on the jump times. ≥ (1) (2) Proposition 97 (Yt)t 0 is a Markov process. If Y and Y are independent Markov ≥ processes with these transition probabilities starting from k and l, then Y (1) + Y (2) is also a Markov process with the same transition probabilities starting from k + l.

Proof: Based on BS3a Applied Probability, the proof is not diﬃcult. We skip it here. 2

(Yt)t 0 is called a continuous-time Galton-Watson process. In fact, these are the only ≥ Markov processes with the branching property (i.e. satisfying the second statement of the proposition for all k ≥ 1, l ≥ 1). Lecture 15: Branching processes 69

Example 98 (Simple birth-and-death processes) If individuals have lifetimes with parameter µ and give birth at rate β to single oﬀspring repeatedly during their lifetime, then we recover the case µ β λ = µ + β and ξ0 = , ξ2 = . µ + β µ + β

In fact, we have to reinterpret this model by saying each transition is a death, giving birth to either two or no oﬀspring. These parameters arise since, if only one individual is present, the time to the next transition is the minimum of the exponential birth time and the exponential death time.

The fact that all jump sizes are independent and identically distributed is reminiscent of compound Poisson processes, but for high population sizes j we have high parameters to the exponential times between two jumps – the process Y moves faster than a compound Poisson process at rate λ. Note however that for H ∼ Exp(jλ) we have jH ∼ Exp(λ). Let us use this observation to specify a time-change to slow down Y .

Proposition 99 Let (Yt)t 0 be a continuous-time Galton-Watson process with oﬀspring ≥ distribution ξ and lifetime distribution Exp(λ). Then for the piecewise linear functions

t Jt = Yudu, t ≥ 0, ϕs = inf{t ≥ 0 : Jt >s}, 0 ≤ s

≤ Xs = Yϕs , 0 s

Proof: Given Y0 = i, the ﬁrst jump time T1 = inf{t ≥ 0 : Yt =6 i} ∼ Exp(iλ), so

≤ ≤ JT1 = iT1 and ϕs = s/i, 0 s iT1, so we identify the ﬁrst jump of Xs = Ys/i, 0 ≤ s ≤ iT1 at time iT1 ∼ Exp(λ). Now the strong Markov property (or the lack of memory property of all other lifetimes) implies that given k oﬀspring are produced at time T1, the process (YT1+t)t 0 is ≥ a continuous-time Galton-Watson process starting from j = i + k − 1, independent of − ∼ ≥ (Yr)0 r T1 . We repeat the above argument to see that T2 T1 Exp(jλ), and for j 1, ≤ ≤ − ≤ ≤ − JT2 = iT1 + j(T2 T1) and ϕiT1+s = T1/i + s/j, 0 s j(T2 T1), ≤ ≤ − and the second jump of XiT1+s = YT1+s/j, 0 s j(T2 T1), happens at time iT1 + j(T2 − T1), where j(T2 − T1) ∼ Exp(λ) is independent of iT1. An induction as long as

YTn > 0 shows that X is a compound Poisson process run until the ﬁrst hitting time of 0. 2 70 Lecture Notes – MS3b L´evy Processes and Finance – Oxford HT 2008

Corollary 100 Let (Xs)s 0 be a compound Poisson process starting from l ≥ 1 with jump ≥ distribution (ξk+1)k 1 and jump rate λ> 0. Then for the piecewise linear functions ≥− s 1 ϕs = dv, 0 ≤ s < T 0 , and Jt = inf{s ≥ 0 : ϕs > t}, t ≥ 0, 0 X { } Z v the process ≥ Yt = XJt , t 0, is a continuous-time Galton-Watson process with oﬀspring distribution ξ and lifetime distribution Exp(λ).

15.3 Continuous-state branching processes

Population-size processes with state space N are natural, but for large populations, it is often convenient to use continuous approximations and use a state space [0, ∞). In view of Corollary 100 it is convenient to deﬁne as follows.

Definition 101 (Continuous-state branching process) Let (Xs)s 0 be a Lévy pro- ≥ cess with no negative jumps starting from x > 0, with E(exp{−γXs}) = exp{sφ(γ)}. Then for the functions s 1 ϕs = dv, 0 ≤ s < T 0 , and Jt = inf{s ≥ 0 : ϕs > t}, t ≥ 0, 0 X { } Z v the process ≥ Yt = XJt , t 0, is called a continuous-state branching process with branching mechanism φ. We interpret upward jumps as birth events and continuous downward movement as (infinitesimal) deaths. The behaviour is accelerated at high population sizes, so fluctuations will be larger. The behaviour is slowed down at small population sizes, so fluctuations will be smaller.

Example 102 (Pure death process) For Xs = x − cs we obtain s 1 1 x ct ϕs = dv = − log(1 − cs/x), and Jt = (1 − e− ), 0 x − cv c c Z ct and so Yt = xe− . Example 103 (Feller diffusion) For φ(γ)= γ2 we obtain Feller’s diffusion. There are lots of parallels with Brownian motion. There is a Donsker-type result which says that rescaled Galton-Watson processes converge to Feller’s diffusion. It is the most popular model in applications. A lot of quantities can be calculated explicitly. Proposition 104 Y is a Markov process. Let Y (1) and Y (2) be two independent continuous- state branching processes with branching mechanism φ starting from x > 0 and y > 0. Then Y (1) + Y (2) is a continuous-state branching process with branching mechanism φ starting from x + y. Lecture 16

The two-sided exit problem

Reading: Kyprianou Chapter 8, Bertoin Aarhus Notes, Durrett Sections 7.5-7.6

16.1 The two-sided exit problem for L´evy processes with no negative jumps

Let X be a L´evy process with no negative jumps. As we have studied processes with no positive jumps (such as −X) before, it will be convenient to use compatible notation and write

0 γXt tφ(γ) 1 2 2 γx E(e− )= e , φ(γ) = a X γ + σ γ + (e − 1 − γx1 x 1 )g X (x)dx − 2 {| |≤ } − Z−∞ 1 2 2 ∞ γx = −aX γ + σ γ − (1 − e− − γx1 x 1 )gX (x)dx, 2 0 {| |≤ } Z where a X = −aX and gX (x) = g X (−x), x > 0. Then we deduce from Section 11.4 − − that, if E(X1) < 0, E βX∞ β (X1) E(e− )= − , β ≥ 0. (1) φ(β) The two-sided exit problem is concerned with exit from an interval [−a, b], notably the time

c T = T[ a,b]c = inf{t ≥ 0 : Xt ∈ [−a, b] } − and the probability to exit at the bottom P(XT = −a). Note that an exit from [−a, b] at the bottom happens necessarily at −a, since there are no negative jumps, whereas an exit at the top may be due to a positive jump across the threshold b leading to XT > b.

Proposition 105 For any L´evy process X with no negative jumps, all a> 0, b> 0 and T = T[ a,b]c , we have −

W (b) ∞ βx 1 P(XT = −a)= , where W is such that e− W (x)dx = . W (a + b) 0 φ(β) Z 71 72 Lecture Notes – MS3b L´evy Processes and Finance – Oxford HT 2008

Proof: We only prove the case E(X1) < 0. By (1), we can identify (the right-continuous function) W , since

βE(X1) ∞ − E βX∞ βx = (e− )= e− fX∞ (x)dx φ(β) 0 Z ∞ βx = βe− P(X ) ≤ x)dx, 0 ∞ Z by partial integration, and so by the uniqueness of moment generating functions, we have W (x)= cP(X ≤ x), where c = −E(X1) > 0. ∞ Now deﬁne τa = inf{t ≥ 0 : Xt < −a} and apply the strong Markov property at τa to get a post-τa process X = (Xτa+s + a)s 0 independent of (Xr)r τa , in particular of Xτa , ≥ ≤ so that e P ≤ P ≤ ≤ cW (b) = (X b)= (Xτa b, X a + b) ∞ ∞ P ≤ P ≤ P ≤ = (Xτa b) (X a + b)= (Xτa b)cW (a + b), ∞ e and the result follows. 2

Example 106 (Stable processes) Let X be a stable process of index α ∈ (1, 2] with no negative jumps, then we have

∞ λx α α 1 e− W (x)dx = λ− ⇒ Γ(α)W (x)= x − . 0 Z We deduce that

α 1 b − P(X = −a)= . T a + b

16.2 The two-sided exit problem for Brownian motion

In the Brownian case, we can push the analysis further without too much eﬀort.

Proposition 107 For Brownian motion B, all a> 0, b> 0 and T = T[ a,b]c , we have −

qT Vq(b) qT Vq(a) E(e− |BT = −a)= and E(e− |BT = b)= , Vq(a + b) Vq(a + b) where √ sinh( 2qx) V (x)= . q x Lecture 16: The two-sided exit problem 73

Proof: We condition on BT and use the strong Markov property of B at T to obtain

a√2q qT{− } e− = E(e− a )

qT{−a} qT{−a} = P(BT = −a)E(e− |BT = −a)+ P(BT = b)E(e− |BT = b) e b qT a q(T +T{− − }) = E(e− |B = −a)+ E(e− a b |B = b) a + b T a + b T b qT a qT (a+b)√2q = E(e− |B = −a)+ E(e− |B = b)e− a + b T a + b T and, by symmetry,

b√2q a qT b qT (a+b)√2q e− = E(e− |B = b)+ E(e− |B = −a)e− . a + b T a + b T These can be written as

b + a 1 qT a√2q 1 qT b√2q = a− E(e− |B = −a)e + b− E(e− |B = b)e− ab T T b + a 1 qT a√2q 1 qT b√2q = a− E(e− |B = −a)e− + b− E(e− |B = b)e ab T T and suitable linear combinations give, as required,

b + a 1 qT 2 sinh(a 2q) = 2 sinh((a + b) 2q)b− E(e− |B = b) ab T p b + a p 1 qT 2 sinh(b 2q) = 2 sinh((a + b) 2q)a− E(e− |B = −a). ab T p p 2

Corollary 108 For Brownian motion B, all a> 0 and T = T[ a,a]c , we have −

qT 1 E(e− )= √ cosh(a 2q)

Proof: Just calculate from the previous proposition

√2qa − √2qa qT Vq(a) e e− 1 E(e− )= =2 = √ . √2qa 2 − √2qa 2 Vq(2a) (e ) (e− ) cosh(a 2q) 2

16.3 Appendix: Donsker’s Theorem revisited

We can now embed simple symmetric random walk (SSRW) into Brownian motion B by putting { ≥ | − | } ≥ T0 =0, Tk+1 = inf t Tk : Bt BTk =1 , Sk = BTk , k 0, √ √ (n) { ≥ (n) | − | } and for step sizes 1/ n modify Tk+1 = inf t Tk : Bt B (n) =1/ n . Tk 74 Lecture Notes – MS3b L´evy Processes and Finance – Oxford HT 2008

Theorem 109 (Donsker for SSRW) For a simple symmetric random walk (Sn)n 0 ≥ and Brownian motion B, we have

S[ ] √nt → B , locally uniformly in t ≥ 0, in distribution as n →∞. n t Proof: We use a coupling argument. We are not going to work directly with the original random walk (Sn)n 0, but start from Brownian motion (Bt)t 0 and deﬁne a family of ≥ ≥ embedded random walks (n) ≥ ≥ ⇒ (n) ∼ S√[nt] Sk := B (n) , k 0, n 1, S[nt] . Tk 0 t n t 0 ≥ ≥ To show convergence in distribution for the processes on the right-hand side, it suﬃces to establish convergence in distribution for the processes on the left-hand side, as n →∞. To show locally uniform convergence we take an arbitrary T ≥ 0 and show uniform convergence on [0, T ]. Since (Bt)0 t T is uniformly continuous (being continuous on a ≤ ≤ compact interval), we get in probability

(n) − − ≤ | − |→ sup S[nt] Bt = sup BT (n) Bt sup Bs Bt 0 0 t T 0 t T [nt] 0 : sup (n) ≤ ≤ ≤ ≤ s t T s t 0≤r≤T T[nr] r ≤ ≤ ≤ | − |≤ | − |

→∞ | (n) − |→ as n , if we can show (as we do in the lemma below) that sup0 t T T[nt] t 0. This establishes convergence in probability, which “implies” convergence≤ ≤ in distribution for the embedded random walks and for the original scaled random walk. 2 | (n) − |→ Lemma 110 In the setting of the proof of the theorem, sup0 t T T[nt] t 0 in probability. ≤ ≤ Proof: First for ﬁxed t, we have

(n) n [nt] qT qT ( ) 1 qt E(e− [nt] )= E(e− 1 ) = → e− , (cosh( 2q/n))[nt] ∼ p (n) → since cosh( 2q/n) 1+ q/n + O(1/n). Therefore, in probability T[nt] t. For unifor- mity, let ε > 0. Let δ > 0. We ﬁnd n0 ≥ 0 such that for all n ≥ n0 and all t = kε/2, p k 1 ≤ k ≤ 2T/ε we have ( ) P(|T n − t | > ε/2) < δε/2T, [ntk] k then 2T/ε P | (n) − | ≤ P | (n) − | ε ≤ P | (n) − | ε sup T[nt] t >ε sup T[nt ] tk > T[nt ] tk > < δ. 0 k k t T 1 k 2T/ε 2! =1 2 ≤ ≤ ≤ ≤ k X 2

We can now describe the recipe for the full proof of Donsker’s√ Theorem. In fact, we can embed every standardized random walk ((Sk − kE(S1))/ nVarS1)k 0 in Brownian − (n) (n) ≥ ∼ motion X, by ﬁrst exits from independent random intervals [ Ak , Bk ] so that X (n) Tk (n) (Sk − kE(S1))/ nVar(S1), and the embedding time change (T[ ])t 0 can still be shown nt ≥ to converge uniformly to the identity. p Appendix A

Assignments

Assignment sheets are issued on Tuesdays of weeks 1-7. They are made available on the website of the course at

http://www.stats.ox.ac.uk/∼winkel/ms3b.html.

Classes take place in weeks 2 to 8 at times and locations to be determined probably on Wednesdays 2.30pm, Thursdays 3.05pm and/or Fridays 8.55am. The class allocation can be accessed from the course website. Only undergraduates and MSc students in Mathematical and Computational Finance can sign up for classes. All others should talk to me after one of the first two lectures. Scripts are to be handed in probably by Tuesdays 11am or Wednesdays 4pm in the Department of Statistics, but one of the classes will probably be in OCIAM. Exercises on the problem sheets vary in style and difficulty. If you find an exercise difficult, please do not deduce that you cannot solve the following exercises, but aim at giving each exercise a serious try. Model solutions will be provided on the course website. There are lecture notes available. Please print these so that we can make best use of the lecture time. I gradually replace last year’s notes by an updated version. The beginning has not changed much, but some typos and unclear passages have been improved. Below are some comments on the recommended Reading and Further Reading literature.

Kyprianou: Introductory lectures on ﬂuctuations of L´evy processes with applications. Springer 2006

This is the treatment that is closest to the course. It is based on a Masters course and has been written for Masters students. The book assumes a background in measure- theoretic probability, but is written in a friendly way suitable for a wide range of diﬀerent backgrounds. The text contains some worked examples and exercises. I II Assignments – MS3b L´evy Processes and Finance – Oxford HT 2008

Kingman: Poisson processes. OUP 1993 This is a gentle introduction to (general, higher-dimensional) Poisson processes and contains a thorough discussion of the tools leading up to and including the study increasing Lévy processes (Section 8.4). Measure-theoretic arguments are isolated in a few proofs, and the reader can take the measure theory for granted. For consistency of terminology with other Oxford courses and most of the literature, in our course, we will reserve the term “Poisson process” to the one-dimensional process. What Kingman calls “Poisson process” is the associated counting measure, which we call “Poisson counting measure”. Schoutens: Lévy processes in finance. Wiley 2003 This is not a textbook, but a monograph advertising Lévy process for finance applications. All models for financial stock prices that we study in our course, are discussed in detail, both their properties and how they can be calibrated to fit financial market data. There are also sections on simulation and option pricing. Sato: Lévy processes and infinitely divisible distributions. CUP 1999 This is a graduate textbook on Lévy processes. The focus is on distributional properties and analytic methods. Bertoin: Lévy processes. CUP 1996 This is a research monograph on Lévy processes. The approach is sample path based. Grimmett and Stirzaker: Probability and Random Processes. OUP 2001 This is the standard probability reference book used in Oxford, overarching 1st, 2nd and 3rd year probability courses and more. It contains a section on spatial Poisson processes, a treatment of martingales. Williams: Probability with Martingales. CUP 1991 This is the standard reference on measure theory and martingales used in Oxford. Ross: Applied Probability Models. Academic Press 1989 We only use the simulation chapter. Durrett: Probability – Theory and Examples. Duxbury 2004 This is a good graduate textbook on Probability. We essentially refer to Durrett only for a proof of Donsker’s theorem, but there is also a section on infinite divisibility. Kallenberg: Foundations of Modern Probability. Springer 1997 This is a standard reference on Probability. The presentation is extremely concise and economical and the amount of material is encyclopaedic. We only refer to it for rigorous statements and proofs of convergence theorems of random walks to Lévy processes. Assignment 1 – MS3b Lévy Processes and Finance – Oxford HT 2008 III

A.1 Inﬁnite divisibility and limits of random walks

If in doubt, hand in scripts by Tuesday 22 January 2007, 11am, Department of Statistics 1. (a) Show that Y ∼ Gamma(α, β) with density

α α 1 β x − βx g(x)= e− , x ≥ 0, Γ(α) is an infinitely divisible distribution and that the independent and identically distributed “divisors” Yn,j in Yn,1 +. . .+Yn,n ∼ Y are also Gamma distributed. (b) Show that the G ∼ geom(p) distribution with probability mass function P(G = n)= pn(1 − p), n ≥ 0, is infinitely divisible and that “divisors” are not geometrically distributed. Hint: Study sums of geometric variables and guess the “divisor” distribution. (c) Show that the uniform distribution on [0, 1] is not infinitely divisible. 2. (a) Let X and Y be independent Lévy processes and a, b ∈ R. Show that aX +bY is also a Lévy process. (b) Let C and √D be two independent Gamma Lévy processes with C1 ∼ D1 ∼ Gamma(α, 2µ). Determine the moment generating function of Cs − Ds, s ≥ 0. We will see later that the process C − D has in fact the same distribution as ≥ Zs = BTs , s 0, for a Brownian motion B and a Gamma Lévy process T . It | is called Variance Gamma process, because Var(Bt) = t implies Var(BTs Ts) = Ts ∼ Gamma(αs,µ). It is a popular model for financial stock prices. 3. A large number N of policy holders in a given time period make claims independently of one another with small probability pN . Denote by SN the total number of policy holders who make a claim in the time period. Assume that claim amounts A1, A2,... are independent and identically distributed. (a) State the Poisson limit theorem and use probability generating functions to prove it.

(b) Explain why SN is approximately Poisson distributed and give its parameter.

(d) Show that the distribution of TN is well-approximated by a compound Poisson distribution, by precisely formulating and proving a limit theorem of the form

SN S∞

TN = An → T = An in distribution, as N →∞. ∞ n=1 n=1 X X 4. (a) Let A1, A2,... be independent and identically distributed random variables 2 with µ = E(A1) and σ = Var(A1) ∈ (0, ∞). Deﬁne − n Ak√ µ Yn,k = and Vn = Yn,k. σ n =1 Xk IV Assignments – MS3b L´evy Processes and Finance – Oxford HT 2008

(i) Formulate the Central Limit Theorem for A1, A2,... in terms of Vn. (ii) Let x > 0. Apply Tchebychev’s inequality to the random variable B1 = | 1 − | → A µ 1 A1 µ σx√n to show that there is a sequence γn(x) 0 as n →∞ with{| − |≥ } √ γn(x) P(|A1 − µ| > σx n) ≤ . n

(iii) Deﬁne Mn = max{|Yn,1|,..., |Yn,n|}. Show that P(Mn ≤ x) → 1 as n →∞, for all x> 0. Deduce that Mn → 0 in probability. (b) Consider an urn initially containing r red and s black balls, r, s ≥ 1. One ball is drawn with replacement (stage 1). After this, a black ball is added to the urn and two balls are drawn, each with replacement (stage 2). After this, another black ball is added and three balls drawn with replacement (stage 3). Continue so that n balls are drawn at stage n followed by the addition of a single black ball. Let Yn,k = 1 resp. 0 if the kth ball of stage n is red resp. black, 1 ≤ k ≤ n and Wn = Yn,1 + . . . + Yn,n.

(i) Show that Wn → Poi(r).

(ii) Show that P(Yn,k = 0) → 1 as n →∞. r (iii) Deﬁne Mn = max{|Yn,1|,..., |Yn,n|}. Show that P(Mn = 0) → e− as n →∞. Deduce that Mn 6→ 0 in probability. (c) (i) Formulate Donsker’s theorem and the process version of the Poisson limit theorem in the settings of (a) and (b). Hint: Consider only t ∈ [0, 1] and evaluate the discrete processes V and W at [nt], t ∈ [0, 1].

(ii) Show that in both cases Mn converges in distribution to the size of the biggest jump of the limit process during the time interval [0, 1].

Hint for 3.(c)-(d): If (Xn)n 0 or (Xt)t 0 is a stochastic process and N or T an indepen- ≥ ≥ dent random time, then for real-valued functions g for which the expectations exist, we have

∞ ∞ E(g(XN )) = E(g(Xn))P(N = n) and E(g(XT )) = E(g(Xt))fT (t)dt. 0 n=0 X Z To prove the ﬁrst formula for integer-valued (Xn)n 0 just note that ≥

∞ E(g(XN )) = g(k)P(XN = k) k= X−∞ ∞ ∞ = g(k)P(XN = k, N = n) k= n=0 X−∞ X ∞ ∞ = g(k)P(Xn = k)P(N = n) k= n=0 X−∞ X ∞ = P(N = n)E(g(Xn)). n=0 X Assignment 2 – MS3b L´evy Processes and Finance – Oxford HT 2008 V

A.2 Poisson counting measures

1. (a) Let (Xt)t 0 be a Poisson process with rate λ ∈ (0, ∞) and arrival times ≥ T1, T2,.... Show that N((c,d]) = #{n ∈ N : c < Tn ≤ d} is a Poisson counting measure on [0, ∞) with constant intensity λ. (b) Let N be a Poisson counting measure on [0, ∞) with time-varying intensity λ(t), t ≥ 0, continuous. Denote Xt = N([0, t]) and Tj = inf{t ≥ 0 : Xt = j}, j ≥ 1.

(i) Show that (Xt)t 0 has independent increments. ≥ (ii) Show that (Xt)t 0 has stationary increments if and only if the intensity ≥ function λ(t) is constant.

(iii) Show that (Xt)t 0 has right-continuous paths with left limits. ≥ (iv) Calculate the distribution of Xt − Xs. (v) Calculate the survival function P(T1 >s), s ≥ 0, of T1. (vi) Show that T2 −T1 is independent of T1 if and only if the intensity function λ is constant. Calculate the joint density of (T1, T2 − T1). 2. Let Π be a spatial Poisson process with constant intensity λ on the ball {(x, y, z) ∈ R3 : x2 + y2 + z2 ≤ 1}. Let P be the process given by the (x, y) coordinate of the points (think of the points as being projected onto the (x, y) plane passing through the centre of the ball). Show that P is a spatial Poisson process and ﬁnd its intensity function. Hint: For a rectangle A in the (x, y) plane, what points of the ball is P counting?

2 2 3. Let (Xt)t 0 be a L´evy process with E(X1 ) < ∞. Denote µ = E(X1), σ = Var(X1) ≥ ψ(λ) iλX1 γX1 Ψ(γ) γX1 and e− = E(e ). If E(e ) < ∞, denote e = E(e ). Show that the following processes are martingales.

γX1 (a) exp{γXt − tΨ(γ)}, if E(e ) < ∞. Hint: First show that E(exp{γXt}) = etΨ(γ) for all t = 1/m, then for all t ∈ Q ∩ [0, ∞), and ﬁnally, using right- continuity, for all t ∈ [0, ∞).

(b) exp{iλXt + tψ(λ)}.

∞ γx 1 βx γ (e − 1) e− dx = − log 1 − 0 x β Z e.g. by suitable (and well-justiﬁed) diﬀerentiation under the integral sign. 1 βx (b) Let (∆t)t 0 be a Poisson point process with intensity function αx− e− . Use ≥ the exponential formula for Poisson point processes to show that αt β αt 1 βx C = ∆ has a Gamma distribution, density x − e− , x ≥ 0. t s Γ(αt) s t X≤

(c) Show that (Ct)t 0 as defined in (b) is a Lévy process. ≥ VI Assignments – MS3b Lévy Processes and Finance – Oxford HT 2008

5. Let X and Y be two independent increasing compound Poisson processes. Denote the respective jump rates by λX and λY , assume that the jump size distributions are continuous with densities hX and hY . Denote D = X − Y . (a) Show that X and Y have no jump times in common.

(b) Show that D has jump times according to a Poisson process with rate λX +λY . (c) Calculate the distribution of the ﬁrst jump size of D.

(d) Show that (∆Xt)t 0 is a Poisson point process and specify its intensity func- ≥ tion.

(e) Show that (∆Dt)t 0 is a Poisson point process and specify its intensity func- ≥ tion. (f) Deduce from (d) and (e) that D is also a compound Poisson process. (g) Show that every real-valued compound Poisson process C can be written uniquely as the diﬀerence of two independent increasing compound Poisson processes. Note that the theory of Poisson point processes applied in (d)-(f) is neater than the conditioning in (c) that could also be developed and iterated to establish (f). Intensity functions just add, jump size distributions are mixtures/weighted averages. Remark: Interchanging limits and expectation/integration/summation is not always per- mitted, and while we do not develop in this course the reasons why we may interchange, we add “by monotone convergence”, whenever we have increasing or decreasing limits of ﬁnite quantities. General measure-theoretic statements have been established in B10a, special cases for Lebesgue integrals have been established in Part A Integration, which is also not a prerequisite for this course. As in BS3a, it is enough for our purposes to formulate special cases, whose statements do not require any of the formal technical setup. For convergence as n →∞ these are:

• Zn ↑ Z and E(|Zn|) < ∞ for all n ∈ N implies

E(Zn) ↑ E(Z) ∈ R ∪ {∞}. • ↑ | | ∞ ∈ N fn f and R fn(x) dx < for all n implies R fn(x)dx ↑ f(x)dx ∈ R ∪ {∞}. ZR ZR • (n) ↑ ∈ N | (n)| ∞ ∈ N am am for all m and m∞=0 am < for all n implies

∞ ∞ P (n) ↑ ∈ R ∪ {∞} am am . m=0 m=0 X X • (n) ↑ | (n)| ∞ ∈ N ∆s ∆s and s t ∆s < for all n implies ≤ (n) ↑ ∈ R ∪ {∞} P ∆s ∆s . s t s t X≤ X≤ The last statement is useful to show right-continuity and the existence of left limits of sums of Poisson point processes, such as in 4.(c) Assignment 3 – MS3b L´evy Processes and Finance – Oxford HT 2008 VII

A.3 Construction of L´evy processes

κ x 1. Let (∆t)t 0 be a Poisson point process with intensity function g(x) = x e− for a ≥ parameter κ ∈ R.

(a) Let κ ∈ (−1, ∞). Show that

Ct = ∆s s t X≤ is a compound Poisson process. Specify its jump rate λ and jump density h. ≤ − (n) ≥ (b) Let κ 1. Show that ∆t = ∆t1 ∆t>1/n , t 0, is a Poisson point process. { } Specify its intensity function. Show that

(n) (n) Ct = ∆s s t X≤ is a compound Poisson process. (n) (n) → ∞ (c) For Ct as deﬁned in (b), show that Ct converges to a limit Ct < as n →∞ if and only if κ> −2. Specify the moment generating function of Ct. (d) Show that for κ> −2, we have

| (n) − | | (n) − | sup Cs Cs = Ct Ct . s t ≤ (n) → Deduce that Ct Ct a.s. (or in probability) locally uniformly. (n) − E (n) − (e) Show that Ct (Ct ) converges for κ> 3. Show that the limit is a L´evy process and that it has unbounded variation.

1/α 2. Let (Xt)t 0 be a stable subordinator in the sense that (c Xt/c)t 0 ∼ X for all ≥ ≥ c> 0 (scaling relation) for some α ∈ R.

µXt (a) Show that for all µ ≥ 0 and t ≥ 0, we have E(e− ) ∈ (0, 1]. Denote µXt Φt(µ)= − ln(E(e− )) and Φ=Φ1.

(b) Show that Φt(µ)= tΦ(µ) for all t ≥ 0. Deduce from the scaling relation that Φ(µ) = Φ(1)µα for all µ ≥ 0. ∂ E µY ≤ ∂2 E µY ≥ (c) Show that ∂µ (e− ) 0 and ∂µ2 (e− ) 0 for any nonnegative random variable Y and for all µ> 0 with equality if and only if P(Y = 0) = 1. Deduce that α ∈ (0, 1] or P(X ≡ 0) = 1.

(d) By letting µ ↓ 0 in (c), show that E(Xt)= ∞ for all t> 0 and α ∈ (0, 1). (e) For α ∈ (0, 1), calculate g : (0, ∞) → (0, ∞) such that

∞ µx Φ(µ)= (1 − e− )g(x)dx. 0 Z (f) For every α ∈ (0, 1] and Φ(1) = b > 0, show that there exists a stable subordinator (Xt)t 0. ≥ VIII Assignments – MS3b L´evy Processes and Finance – Oxford HT 2008

3. (a) Let (Xt)t 0 and (Yt)t 0 be independent stable subordinators with common ≥ ≥ index α and intensities bX = ΦX (1) and bY = ΦY (1). Show that Z = X − Y 1/α is also a stable process with index α in the sense that (c Xt/c)t 0 ∼ X for ≥ all c> 0. (b) Let H be a real-valued random variable with symmetric distribution, i.e. H ∼ −H. Show that E(eiλH ) ∈ R for all λ ∈ R. Hint: eix = cos(x)+ i sin(x).

(c) In the setting of (a), for the special case bX = bY , show that E(exp{iλZt})= α exp{−bX |λ| }, λ ∈ R. Hint: Show that Zt is symmetric and that all symmetric stable processes have a characteristic function of this form for some b> 0. You may assume without proof that all characteristic functions are continuous in λ ∈ R and that those of inﬁnitely divisible distributions have no zeros. (d) Fix b> 0. Show that for α ∈ (0, 2), the function

∞ α 1 e ψ(λ)= (cos(λx) − 1)b|x|− − dx Z−∞ e1 e has the property ψ(λc /α)= cψ(λ) for all c> 0 and λ ∈ R. Deduce that

e e ψ(λ)= b|λ|α

for some b ∈ R. e (e) Using (d) or otherwise, show that a symmetric stable process R of index α ∈ (0, 2] has bounded variation if and only if α ∈ (0, 1), and deduce from the previous parts of the exercise that it can then be written as a diﬀerence of two stable subordinators. Show that E(Rt) exists if and only if α ∈ (1, 2], and that then Var(Rt) < ∞ if and only if α = 2.

Warning: Densities of stable processes are only known in closed form for some special cases α ∈{1/2, 1, 2}. It is known, however, that they all have smooth probability density functions. You may use results from the lectures and previous assignment sheets without proof if you state them clearly, except that the L´evy density g of stable processes is to be derived here, and its form should not be assumed. Question A.3.1 is the most relevant on this sheet for MSc MCF students – if pushed for time, please focus on this one. Assignment 4 – MS3b L´evy Processes and Finance – Oxford HT 2008 IX

A.4 Simulation

1. Let U ∼ Unif(0, 1) and F : R → [0, 1] right-continuous (weakly) increasing with 1 F (−∞) = 0 and F (∞) = 1. Deﬁne F − (u) = inf{x ∈ R : F (x) > u} ∈ [−∞, ∞] 1 for u ∈ [0, 1]. Show that F − (U) is a random variable with cumulative distribution function F .

2. (a) Let(Xt)t 0 be a Gamma process with Xt ∼ Gamma(t, 1) for all t > 0. Con- ≥ sider A = Xa and B = Xa+b − Xa for some a > 0 and b > 0. Show that R = A/(A + B) and S = A + B are independent and that R ∼ Beta(a, b), where Gamma and Beta densities are recalled as follows:

a+b 1 s s − e− Γ(a + b) a 1 b 1 f (s)= , s ∈ (0, ∞), f (r)= r − (1 − r) − , r ∈ (0, 1). S Γ(a + b) R Γ(a)Γ(b)

Deduce, vice versa, the distribution of (SR,S(1 − R)) for independent S ∼ Gamma(1,c) and R ∼ Beta(cp,c(1 − p)), for some c> 0 and p ∈ (0, 1). (b) Let U ∼ Unif(0, 1) and a> 0. Show that X = U 1/a ∼ Beta(a, 1). (c) Let U ∼ Unif(0, 1) and V ∼ Unif(0, 1) be independent and a ∈ (0, 1). Calcu- 1/a 1/(1 a) late for Y = U and Z = V −

Y P ≤ t, Y + Z ≤ 1 Y + Z and deduce that the conditional distribution of W = Y/(Y +Z) given Y +Z ≤ 1 is Beta(a, 1 − a). Hint: Write both inequalities as constraints on Z to ﬁnd the bounds when writing the probability as a double integral. (d) In the setting of (c), show that the conditional distribution of T W given Y + Z ≤ 1, for an independent T ∼ Exp(1) = Gamma(1, 1) random variable, is Gamma(a, 1). (e) Consider the following procedure due to Johnk. Let a ∈ (0, 1). 1. Generate two independent random numbers U ∼ Unif(0, 1) and V ∼ Unif(0, 1). 1/a 1/(1 a) 2. Set Y = U and Z = V − . 3. If Y + Z ≤ 1 go to 4., else go to 1. 4. Generate an independent C ∼ Unif(0, 1) and set T == ln(C). 5. Return the number TY/(Y + Z). What is this procedure doing? Explain its relevance for simulations.

3. (a) In the light of the previous exercise, explain how you can generate a Beta(a, b) random variable from a sequence of Unif(0, 1) random variables, for any a> 0 and b > 0. Hint: Consider a ∈ (0, 1) ﬁrst and use the additivity of Gamma variables to generate Gamma(a, 1) variables, from which the Beta variable can be constructed. X Assignments – MS3b L´evy Processes and Finance – Oxford HT 2008

(b) Consider the following method to generate a Gamma process on the time interval [0, 1]. Set X0 = 0 and generate X1 ∼ Gamma(1, 1). For n ≥ 0, if n you have generated Xk2−n , k =0,..., 2 , generate Bk,n ∼ Beta(an, bn) and set − − − − − − ≤ ≤ n X(2k 1)2 n 1 = X(k 1)2 n + Bk,n(X(k+1)2 n Xk2 n ), 1 k 2 . For what − − choices of an > 0 and bn > 0 does this procedure yield Gamma distributions for all Xk2−n , and for what choice do you get stationary increments? Hint: n 1 an = bn =2− − works, but are there other choices? (c) What are the advantages of this method when compared with the plain version of the time discretisation method (Method 1)?

4. Consider a variant of the Variance Gamma process of the form Vt = at + Gt − Ht where a ∈ R, G1 ∼ Gamma(α+, β+) and H1 ∼ Gamma(α , β ) − − (a) For what values of a, α , β is V a martingale? ± ± (b) Write out the steps needed to simulate Vt

• by Method 1 (using a random walk with increment distribution ∼ Vδ) • by Method 1 (applied to G and H separately) • by the refinement of Method 1 given in A.4.3 • by Method 2 (simulating the Poisson point process of jumps truncated at ε) (c) Carry out 9 simulations for a range of parameters α+ ∈{1, 10, 100} and α ∈ − {10, 100, 1000}, β = α2 /2 and a such that V is a martingale. This part of this question is optional.± ± t a 1 x Warning: The incomplete Gamma function Γt(a) = 0 x − e− dx cannot be simplified into closed form (nor expressed in terms of the Gamma function), except for some special R values of a such as a ∈ N. There are, however, numerical procedures to evaluate Γt(a), which we will not address in this course. If you have not used R, but would like to, you will find the “First steps with R” at http://www.stats.ox.ac.uk/∼myers/stats materials/R intro/WA5 R.pdf useful. Following are brief explanations of the commands used in the sample file http://www.stats.ox.ac.uk/∼winkel/gammavgamma.R This is a script file, which has to be run in the command window “R Console” to make the new commands available, e.g. select “Run all” in the drop-down menu “Edit”. • runif(n,a,b) generates an n-vector of uniform variables on [a, b]. 1 • qgamma(u,a,b) evaluates F − (u) for the Gamma(a, b) inverse distribution function 1 F − at u. If u is a vector, qgamma is applied to each component. • 1:n generates the vector (1, 2,...,n). Multiplication of vectors v by scalars a can be written as a*v, similarly for addition and subtraction of vectors. • plot(x,y,pch=".",sub=paste("text")) produces a scatter plot of pairs (xi,yi) for vectors x and y, with . marking the points, and text in the caption. • psum <- function(vector){...} defines a new command psum that takes a vector vector as an argument. When this line is executed, the command is just made available. To execute the command, type psum(v) for a vector v to get the partial sums of v displayed, or s=psum(v) to create a new vector s containing the partial sums of v. Assignment 5 – MS3b Lévy Processes and Finance – Oxford HT 2008 XI

A.5 Financial models

1. Consider a one-period model with three assets, a risk-free asset that increases from δ A0 =1to A1 = e and two risky assets B and C that can each move up to or down from B0 = C0 = 1, so that there are four scenarios ω1 = (up, up), ω2 = (up, down), up ω3 = (down, up) and ω4 = (down, down). Suppose that B1 = B1(ω1)= B1(ω2) > down up B1(ω3) = B1(ω4) = B1 and C1 = C1(ω1) = C1(ω3) > C1(ω2) = C1(ω4) = down up up C1 . Assume w.l.o.g. that B1

(a) For a portfolio (T0, U0,V0) of T0 units of A, U0 units of B and V0 units of C, specify the value W0 and W1(ωi) of the portfolio at times 0 and 1, i =1, 2, 3, 4.

(b) Show that this model is arbitrage-free if and only if B1(ω1) > A1 > B1(ω4) > C1(ω4).

Consider the arbitrage-free case now.

(d) Show that contingent claims of the form W1(ω1)= W1(ω2), W1(ω3)= W1(ω4) δ can be hedged and priced as e− E(W1), where you should specify

up down qB = P(B1 = B1 ) and 1 − qB = P(B1 = B1 ).

δt In particular, e− Bt, t =0, 1, is then a martingale. Is qB unique? (e) State the result analogous to (d) for contingent claims relating to C only rather than B only. (f) Show that there are inﬁnitely many possibilities to choose

up up p1 = P(ω1)= P(B1 = B1 ,C1 = C1 ), p2 = P(ω2), p3 = P(ω3) and p4 = P(ω4)

δt δt so that e− Bt and e− Ct, t =0, 1 are martingales.

(g) Consider the contingent claim W1(ω1) = 1, W1(ω2) = W1(ω3) = W1(ω4) = 0. Using the range of possibilities for p = (p1,p2,p3,p4), give the range of δ proposed (arbitrage-free) prices W0 = e− Ep(W1).

2. Consider Xt = Nt − µt for a Poisson process N of rate λ ∈ (0, ∞) and a drift coeﬃcient µ ∈ (0, ∞). Let

n (ε) (ε) Sn = Xi , i=1 X (ε) where (Xi )i 1 is a sequence of independent random variables with common prob- ≥ ability mass function given by

P (ε) − − λε P (ε) − − (Xi =1 µε)=1 e− =: pε and (Xi = µε)=1 pε. (ε) → ↓ (a) Show that S[t/ε] Xt in distribution as ε 0. Hint: You can either prove this (ε) (ε) directly or consider Tn = Sn + nµε ﬁrst. XII Assignments – MS3b L´evy Processes and Finance – Oxford HT 2008

(ε) δεn Sn (b) Show that the market model (e , e )n 0 is arbitrage-free if and only if ≥ −µ<δ< 1/ε − µ and then also complete.

Consider the arbitrage-free case now.

ε(δ+µ) (ε) e − 1 q = P(S1 =1 − µε)= . ε e − 1

(d) Show that under the martingalee probabilities

(ε) → − St/ε Xt = Nt µt,

where (Nt)t 0 is a Poisson processe withe ratee (δ + µ)/(e − 1). ≥ δt (e) Show that in the notation of part (d), the discounted process e− Rt associated e e Nt µt with Rt = e − is a martingale. By conditioning on Nt and Nt, respectively, e Nt µt explain brieﬂy why the distribution of e − can be seen as providing martin- Nt µt gale probabilities for e − . e (f) Using the following subsets of the set of right-continuous paths with left limits

Dµ = {f ∈ D([0, 1], (0, ∞)) : ∆log f(t)=1 or (log f)′(t)= −µ for all t ≤ 1},

e Nt µt show that (e − )0 t 1 is the only exponential L´evy process that has the ≤ ≤ Nt µt same set of possible paths as (e − )0 t 1 and whose discounted process is a ≤ ≤ martingale. Remark: In fact, it is the only process that has the same set of Nt µt possible paths as (e − )0 t 1 and whose discounted process is a martingale, so the market model is complete.≤ ≤ Assignment 6 – MS3b L´evy Processes and Finance – Oxford HT 2008 XIII

A.6 Time change and subordination

1. Consider Brownian motion (Bt)t 0 and a continuous increasing function f : [0, ∞) → ≥ [0, ∞) with f(0) = 0. Set Zy = Bf(y), y ≥ 0.

(a) Show that Z has quadratic variation

[2ny] 2 [Z]y := p-lim (Zj2−n − Z(j 1)2−n ) = f(y), n − →∞ j=1 X where p-lim denotes a limit in probability of random variables. (b) Assume that f is piecewise diﬀerentiable on [0, ∞) with piecewise constant 2 2 derivative σ (s) := f ′(s), say taking values σj on intervals [yj 1,yj) for some − 0= y0

s j − − Zy = σ(r)dWr := σi(Wyi Wyi−1 )+ σj+1(Wy Wyj ), 0 i=1 Z X e yj ≤ y

(c) Give an example of a L´evy process and a function f as in (b) for which Xf(y) y does not have the same distribution as 0 f ′(s)dXs. In fact, Brownian motion is the only process for which this holds for all such R p functions f.

2. Let (Xt)t 0 be a L´evy process with probability density function ft and (τy)y 0 a ≥ ≥ subordinator with characteristics (0,gτ ) (sum of jumps, no compensation!). Deﬁne

∞ g(z)= ft(z)gτ (t)dt, z ∈ R \{0}. 0 Z (a) In the case Var(X1) < ∞ and Var(τ1) < ∞, show that g satisfies the requirements of a Lévy density of a Lévy process. (b) In the case where either τ or X is compound Poisson, show that g also satisfies the requirements of a Lévy density of a Lévy process. More specifically, if X is λt a compound Poisson process with intensity λ, then we have P(Xt = 0) ≥ e− ; P λt P ∈ b assume that, in fact, (Xt = 0) = e− and that (Xt (a, b)) = a ft(x)dx for (a, b) 6∋ 0. R (c) If X is a stable process of index α > 1, show that g is the Lévy density of a bounded variation Lévy process if

∞ 1/α x gτ (x)dx < ∞. 0 Z XIV Assignments – MS3b L´evy Processes and Finance – Oxford HT 2008

3. Show that all L´evy processes X that can be obtained by subordination of Brownian motion with an independent subordinator are symmetric in the sense X ∼ −X, but that not all symmetric L´evy processes can be obtained in this way.

4. (a) Let C√ and D be two independent Gamma L´evy processes with parameters α and 2λ for C1 ∼ D1. Let T be a Gamma process with parameters α and λ, − ∼ and let B be an independent Brownian motion. Show that Xs Ys BTs . This result was mentioned in Question A.1.2. as an explanation for the name Variance Gamma process. (b) Let B be Brownian motion, S an independent stable subordinator with index ∈ ≥ α (0, 1). Show that Rt = BSt , t 0, is a stable process with index 2α. (c) Write down procedures to simulate the processes in (a) and (b) using Method 3 (Subordination).

2 5. Let X be a L´evy process with E(X1) = µ and Var(X1) = σ . Let S be an in- E 2 dependent subordinator with (S1) = m and Var(S1) = q . Denote Zt = XSt , t ≥ 0.

(a) Show that E(Zt)= mµt, t ≥ 0. 2 2 2 ≥ E 2 (b) Show that Var(Zt)=(σ m + µ q )t, t 0. Hint: Consider (Zt ) first. (c) Check this formula for the Variance Gamma process, using A.6.4.(a). For what E E 4 E 2 values of α and λ is (Z1) = 0 and Var(Z1) = 1? Show that (Zt )=3 (St ) 4 and deduce the range of E(Z1 ) for these values of α and λ. Standardized fourth moments (curtosis) give an indication of heavy tails. They reflect why Lévy processes such as the Variance Gamma process can better fit financial price processes. Assignment 7 – MS3b Lévy Processes and Finance – Oxford HT 2008 XV

A.7 Level passage events

≥ 1. Let (Xs)s 0 be a L´evy process and Xt = inf0 s t Xs, t 0. ≥ ≤ ≤

(a) For a ﬁxed time t> 0, show that the process (Xs)0 s t given by ≤ ≤ (t) − ≤ ≤ Xs = Xt Xt s , 0e s t, − − −

is a L´evy process withe the same distribution as (Xs)0 s t. ≤ ≤ (b) Show that this implies that for an independent random time τ with probability (τ) density function fτ (x), x ∈ (0, ∞), we have (Xs )0 s τ ∼ (Xs)0 s τ in the ≤ ≤ ≤ ≤ sense that for all 0 = s0

(c) Using results and/or arguments from the lectures show that Xτ is independent − ∼ of (Xτ Xτ ) for an independent τ Exp(q).

(d) Suppose now that X has no positive jumps. Calculate the distribution of Xτ .

2. Let (Xt)t 0 be an α-stable L´evy process with no positive jumps for some α ∈ (1, 2], ≥ α i.e. such that E(eγXt ) = etcγ . For α = 2 this is Brownian motion, for α ∈ (1, 2), α 1 we have a3 = 0 and g(x)= c|x|− − . For x ≥ 0 denote Tx = inf{t ≥ 0 : Xt > x}.

(a) Using the strong Markov property of (Xt)t 0 at Tx, show that (Tx)x 0 is a e ≥ ≥ stable subordinator with index 1/α. (b) Let Y have probability density function

√ b b2/(2z) fb(z)= e− , z> 0. 2πz3

Calculate the distribution of aY and deduce that (fb)b 0 is the family of den- ≥ sities of stable distributions on (0, ∞) of index 1/2. (c) Deduce that there is a constant c> 0 such that

∞ γx cb√γ e fb(x)dx = e− . 0 Z √ In fact, c = 2.

3. (a) Let A1, A2,... be identically distributed and Sn = A1 +. . .+An the associated random walk. Let (Nm)m 0 be an independent random walk. Denote the ≥ moment generating function of A1 by M(γ)= E(exp{γA1}) and assume that it is ﬁnite for γ ∈ (−ε,ε). Denote the probability generating function of N1 E N1 ≥ by G(s)= (s ). Show that Rm = SNm , m 0, is also a random walk (with independent and identically distributed increments). XVI Assignments – MS3b L´evy Processes and Finance – Oxford HT 2008

(b) Let (Xt)t 0 be a Lévy process and (Ts)s 0 an independent increasing Lévy ≥ ≥ ≥ process. Show that Ys = XTs , s 0, is also a Lévy process.

(c) (i) Let (Bt)t 0 be Brownian motion. For s ≥ 0 define Ts = inf{t ≥ 0 : ≥ Bt + bt >s}, where b ≥ 0 is fixed. Using the strong Markov property at Ts, show that (Ts)s 0 is an increasing Lévy process. { ≥− 1 2 } ∈ (ii) Show that exp γBt 2 γ t is a martingale for all γ R. Use the Optional Stopping Theorem to show that

2 E(exp{ρTs}) = exp{s(b − b − 2ρ)}. p This distribution is called the inverse Gaussian distribution (note that BTs = s means that s 7→ Ts is the right inverse of t 7→ Bt.) For an independent ≥ Brownian motion (Xt)t 0, the process Zs = XTs , s 0, obtained as in (b) has the so-called Normal Inverse≥ Gaussian (NIG) distribution. This is another popular process to model ﬁnancial price processes.

The last question is optional:

γXt 4. Let(Xt)t 0 be standard Brownian motion with moment generating function E(e )= tγ2/2 ≥ e . Denote Xt = sup0 s t Xs. ≤ ≤

(a) For y > 0 denote Ty = inf{t ≥ 0 : Xt = y}. Use the strong Markov property to deduce that the process

Xt t ≤ Ty X∗ = t 2y − X t ≥ T t y is a Brownian motion. (b) Show that P ≤ P − ∈ ∞ ∈ −∞ (Xt x, Xt >y)= (Xt∗ > 2y x), y (0, ), x ( ,y).

and deduce that 2(2y − x) (2y − x)2 f (x, y)= √ exp − , y ∈ (0, ∞), x ∈ (−∞,y). Xt,Xt 3 2t 2πt (c) Show that

√ x x2/(2z) fT (z)= e− , z> 0. x 2πz3

Feedback on the various topics and how you perceived them given your background (no BS3a, no B10a, MScMCF etc.) will be most gratefully received: [email protected]. Appendix B

Solutions

B.1 Inﬁnite divisibility and limits of random walks

1. (a) Recall that for independent A1 ∼ Gamma(α1, β) and A2 ∼ Gamma(α2, β) we have A1 + A2 ∼ Gamma(α1 + α2, β). A quick proof can be given using moment generating functions. The Gamma distribution has moment generating function α α 1 α E { } ∞ γx β x − βx β (exp γA )= e e− dx = α , γ<β. 0 Γ(α) (β − γ) Z We see that α1+α2 E { } E { } E { } β (exp γ(A1 + A2) )= (exp γA1 ) (exp γA2 )= + (β − γ)α1 α2

and recognise the moment generating function of the Gamma(α1 + α2, β) distribution. By the Uniqueness Theorem for moment generating functions, A1 + A2 ∼ Gamma(α1 + α2, β).

If we now choose Yn,1,...,Yn,n ∼ Gamma(α/n, β) independent, we obtain, by induction in n, that Yn,1 + . . . + Yn,n ∼ Gamma(α, β). Since this holds for all n ≥ 1, a random variable Y ∼ Gamma(α, β) has an inﬁnitely divisible distribution. (b) First calculate for B1, B2 ∼ geom(p) independent that n n k n k P(B1 + B2 = n) = P(B1 = k, B2 = n − k)= p (1 − p)p − (1 − p) =0 =0 Xk Xk = (n + 1)pn(1 − p)2,

and, e.g. by induction, for Am = B1 + . . . + Bm = Am 1 + Bm a negative − binomial distribution. Alternatively, consider independent Bernoulli trials until the mth success, then {Am = n} means there have been n failures and m successes, the m − 1 first successes chosen from the first n + m − 1 trials, and we get n + m − 1 (n + m − 1)! P(A = n) = pn(1 − p)m = pn(1 − p)m m m − 1 (m − 1)!n! Γ(n + m) = pn(1 − p)m. Γ(m)n! XVII XVIII Solutions – MS3b Lévy Processes and Finance – Oxford HT 2008

This formula makes sense for m ∈ (0, ∞), and we refer to this probability mass function as NB(m, p). Then we calculate the probability generating function for A ∼ NB(m, p) Γ(n + m) (1 − p)m E(sA)= (sp)n(1 − p)m = , s ∈ [0, 1], Γ(m)n! (1 − sp)m n 0 X≥ and if B ∼ NB(r, p) is independent, we obtain

m+r + (1 − p) E(sA B)= , (1 − sp)m+r the probability generating function of the NB(m + r, p) distribution, so we conclude by the Uniqueness Theorem for probability generating functions that A + B ∼ NB(m + r, p).

If we now choose Yn,1,...,Yn,n ∼ NB(1/n,p) independent, we obtain, by induction in n, that Yn,1 + . . . + Yn,n ∼ NB(1,p) = geom(p). Since this holds for all n ≥ 1, a random variable Y ∼ geom(p) has an inﬁnitely divisible distribution.

(c) Assume that a random variable U ∼ Unif(0, 1) can be written as U = Y1 + Y2 for some independent and identically distributed Y1 and Y2. Then for x ∈ [0, 1], √ 1 − x = P(U ≥ x) ≥ P(Y1 ≥ x/2,Y2 ≥ x/2) ⇒ P(Y1 ≥ x/2) ≤ 1 − x and √ 2 x = P(U ≤ x) ≥ P(Y1 ≤ x/2) ⇒ P(Y1 ≤ x/2) ≤ x.

For x = 1 and x = 0, respectively, we deduce P(Y1 ≥ 1/2)=0= P(Y1 ≤ 0). Now for x ∈ (0, 1/2) √ x = P(U ≤ x) ≤ P(Y1 ≤ x, Y2 ≤ x) ⇐⇒ P(Y1 ≤ x) ≥ x and the inequality on the left is an equality if and only if the inequality on the right is an equality. Similarly, √ 2 x = P(U ≥ 1 − x) ≤ P(Y1 ≥ 1/2 − x) ⇐⇒ P(Y1 ≥ 1/2 − x) ≥ x

For x = 1/4, we get P(Y1 ≤ 1/4) ≥ 1/2 and P(Y1 ≥ 1/4) ≥ 1/2. If both inequalities were equalities, we would deduce from the left-hand equalities that P(Y1 ∈ (1/8, 3/8)) = 0 and this is incompatible with P(U ∈ (1/4, 3/8)) > 0, so the assumption that U = Y1 + Y2 must have been wrong. 2. (a) (i) Independence of increments. By the independence of increments of X and Y and by the independence of X and Y we have for all 0 ≤ t0 < t1 < ...

(ii) Stationarity of increments. We have that Xt+s − Xt and Yt+s − Yt are independent, and also that Xs and Ys are independent. By the stationarity of increments we have that Xt+s − Xt ∼ Xs and Yt+s − Yt ∼ Ys and so the joint distributions of (Xt+s − Xt,Yt+s − Yt) is the same as the joint distribution of (Xs,Ys). If we apply the same linear function to the random vectors, these will also have the same distribution, i.e.

a(Xt+s − Xt)+ b(Yt+s − Yt) ∼ aXs + bYs. (iii) Right-continuity and left limits of paths. Linear combinations of such functions still have these properties. (b) We calculated the moment generating function of the Gamma(α, β) distribution in Exercise 1 as α α 1 α E { } ∞ γx β x − βx β (exp γA )= e e− dx = α , γ<β. 0 Γ(α) (β − γ) Z √ √ If C1 ∼ D1 ∼ Gamma(α, 2µ), then C ∼ D ∼ Gamma(αs, 2µ). Hence s√ s √ αs αs αs γ(Cs Ds γCs γDs 2µ 2µ µ E(e − )= E(e )E(e− )= √ √ = ( 2µ − γ)αs ( 2µ + γ)αs µ − 1 γ2 √ √ 2 for all − 2µ<γ< 2µ.

3. (a) Let Wn ∼ Binomial(n, pn) with npn → λ, then Wn → Poi(λ) in distribution as n →∞. To prove this, check n n n np (1 − s) (1 ) E Wn k k − n k − n → λ s (s )= s pn(1 pn) − = 1 e− − , =0 k n Xk and this is the probability generating function of Poi(λ). By the Uniqueness Theorem and by the Continuity Theorem for probability generating functions, Wn converges in distribution to a Poi(λ) distribution.

(b) Since pN is small, the Poisson limit theorem is appropriate, and since N is large, it will give a reasonably good approximation. As parameter of the Poisson distribution, NpN is appropriate, since NpN → λ in the limit theorem for a Poi(λ) limit.

(c) Denote by B1,...,BN the Bernoulli random variables so that Bj = 1 if policy holder j makes a claim. Then SN = B1 + . . . + BN ∼ Binomial(N,pN ). We calculate the moment generating function

E(exp{γTN }) = E exp γ Aj ( j=1 )! X N k N k N k = E exp γ A p (1 − p ) − j k N N =0 ( j=1 )! Xk X N k N E γA1 k − N k = e pN (1 pN ) − =0 k Xk γA1 N = (1 − pN + pN E(e )) , by the binomial theorem, for all γ ∈ R for which E(eγA1 ) < ∞. XX Solutions – MS3b L´evy Processes and Finance – Oxford HT 2008

(d) First, we can continue the argument in (b) to suggest that the actual random variable

An n=1 X has a distribution that is close to the distribution where SN is replaced by a Poisson random variable.

Then we formulate the limit statement. Suppose that SN ∼ Binomial(N,pN ) for all N ∈ N, that S ∼ Poi(λ) and that NpN → λ ∈ (0, ∞) as N → ∞ ∞. Suppose that A1, A2,... are nonnegative, independent and identically distributed, independent of SN , N ∈ N ∪ {∞}. Then

SN S∞

An → An in distribution, as N →∞. n=1 n=1 X X To prove this, consider the moment generating functions

N Np (1 − E(eγA1 )) E(exp{γT })= 1 − N → exp −λ(1 − E(eγA1 )) , N N and this is the moment generating function of the compound Poisson distribution, which we calculate as follows

S∞ k ∞ k λ λ E exp γ A = E exp γ A e− j j k! ( j=1 )! =0 ( j=1 )! X Xk X ∞ k k λ λ = (E (exp {γA1})) e− =0 k! Xk λ γA1 γA1 = e− exp λE(e ) = exp −λ(1 − E(e )) . 4. (a) (i) Note that

n − E n − n k=1 Ak n (A1) Ak√ µ = = Yn,k = Vn. nVar(A1) =1 σ n =1 P Xk Xk p Thus, the Central Limit Theorem in terms of Vn states Vn → Normal(0, 1) in distribution as n →∞. (ii) Markov’s inequality P(|X| >y) ≤ E(X2)/y2 yields √ √ P | 1 − | P | 1 − | ( A µ > σx n) = ( A µ 1 A1 µ σx√n > σx n) 2{| − |≥ } E | 1 − | ( A µ 1 A1 µ σx√n ) ≤ {| − |≥ } . σ2x2n Now note that, as n →∞,

2 2 2 E | 1 − | → E | 1 − | A µ 1 A1 µ <σx√n ( A µ )= σ , {| − | } Solutions 1 – MS3b L´evy Processes and Finance – Oxford HT 2008 XXI

(by monotone convergence) and so

1 2 E | 1 − | γn(x) := 2 2 ( A µ 1 A1 µ σx√n ) σ x {| − |≥ } 1 2 2 − E | 1 − | → = 2 2 σ ( A µ 1 A1 µ <σx√n ) 0. σ x {| − | } (iii) For all x> 0, calculate using (ii)

n P(Mn ≤ x) = P(|Yn,1| ≤ x, . . . , |Yn,n| ≤ x)=(P(|Yn,1| ≤ x)) γ (x) n ≥ 1 − n → e0 =1. n

This implies that P(|Mn| > ε)=1 − P(|Mn| ≤ ε) → 0 for all ε > 0, so Mn → 0 in probability. (b) (i) At stage n there are r red balls and s + n − 1 black balls in the urn. So

r Y ∼ Bernoulli ⇒ W ∼ Binomial (n, p ) , n,k r + s + n − 1 n n

where pn = r/(r + s + n − 1). Note that npn → r, so that the Poisson limit theorem yields Wn → Poi(r).

(ii) Clearly P(Yn,k =0)=1 − pn =1 − r/(r + s + n − 1) → 1, as n →∞. (iii) Now, as n →∞,

n n npn r P(M =0)= P(Y 1 =0,...,Y =0)=(1 − p ) = 1 − → e− , n n, n,n n n If Mn → 0, then P(|Mn| >ε)=1 − P(Mn = 0) → 0 for all 0 <ε< 1, and this is incompatible with the limit above. So, Mn 6→ 0 in probability. (n) ≥ ≥ (c) (i) Deﬁne Sk = Yn,1 + . . . + Yn,k, k 0, n 1. (n) (n) → Donsker’s theorem says in the setting of (a), where Vn = Sn , that S[nt] Bt locally uniformly in distribution for a Brownian motion (Bt)t 0. ≥ The process version of the Poisson limit theorem says in the setting of (b), (n) (n) → where Wn = Sn , that S[nt] Nt in the Skorohod sense in distribution for a Poisson process (Nt)0 t 1 with rate r. ≤ ≤ (ii) Clearly, the size of the biggest jump of Brownian motion is 0, and we have Mn → 0 in probability, hence also in distribution. The number of jumps of (Nt)0 t 1 is Poisson distributed with parameter ≤ ≤ r. The size J of the biggest jump of (Nt)0 t 1 is 1 if there is a jump, with r ≤ ≤ r probability P(J =1)=1−e− , and P(J =0)= e− is the probability that there is no jump. This is the limit distribution that we wish to establish. We have shown that

r P(Mn = 0) → e− = P(J = 0)

r and this implies P(Mn =1)=1 − P(Mn = 0) → 1 − e− = P(J = 1), as required. XXII Solutions – MS3b L´evy Processes and Finance – Oxford HT 2008

B.2 Poisson counting measures

1. (a) The distribution of Π is speciﬁed in terms of the associated counting measure

N((a, b]) = #Π ∩ (a, b]= {j ≥ 1 : a < Tj ≤ b} = Xb − Xa, 0 ≤ a < b.

Clearly, N satisﬁes property hom(b) of a Poisson counting measure: N((a, b]) = Xb − Xa ∼ Poi(λ(b − a)) by the stationarity (ii) and Poisson (iv) properties for increments of X, and we identify the constant intensity function λ(t)= λ, t ≥ 0.

N also satisﬁes (a), since for disjoint intervals (aj, bj], j = 1,...,n, we have − N((aj , bj]) = Xbj Xaj increments of X over disjoint time intervals. By property (i) of the Poisson process, these are independent, as required. ≤ − (b) (i) Let 0 t0 < t1 <...

(iii) Clearly t 7→ Xt is an increasing function, so all left and right limits exists. Denote by Π the associated spatial Poisson process, then Π= {t ≥ 0 : ∆Xt > 0} = {t ≥ 0 : ∆Xt = 1}. The set Π cannot have accumulation points since λ is locally integrable, so Π = {Tj, j ≥ 1} and Xt = N([0, t]) = j for t ∈ [Tj, Tj+1) is right-continuous at jump times, continuous elsewhere. − t inhom (iv) Xt Xs = N((s, t]) = Poi( s λ(x)dx), by property (b) of the Poisson counting measure. R {− s } ≥ (v) P(T1 >s)= P(N([0,s]) = 0) = exp 0 λ(x)dx for all s 0. T1 (vi) The density of is obtained by diﬀerentiatingR the survival function: s − fT1 (s)= λ(s) exp λ(x)dx . 0 Z To calculate the joint distribution of (T1, T2 − T1), ﬁrst calculate the joint distribution of (T1, T2), from

P(T1 > s,T2 > t)= P(N([0,s])=0, N((s, t]) ≤ 1) s t t = exp − λ(x)dx 1+ λ(x)dx exp − λ(x)dx 0 Z Zs Zs and diﬀerentiation, ﬁrst with respect to s then with respect to t

t − fT1,T2 (s, t)= λ(s)λ(t) exp λ(x)dx 0 Z Solutions 2 – MS3b L´evy Processes and Finance – Oxford HT 2008 XXIII

and the transformation formula for (T1, T2) 7→ (T1, T2 − T1) gives

s+r − fT1,T2 T1 (s,r)= λ(s)λ(s + r) exp λ(x)dx − 0 Z and then

s+r − fT2 T1 T1=s(r) = λ(s + r) exp λ(x)dx − | s Z s+r ⇒ P(T2 − T1 >r|T1 = s) = exp − λ(x)dx Zs is independent of s for all r ≥ 0 if and only x 7→ λ(x) is constant, by the argument given in (ii).

2. For simplicity think of Π as a spatial Poisson process in R3 with intensity function λ(x, y, z)= λ if x2 + y2 + z2 ≤ 1, λ(x, y, z) = 0 otherwise. We check the properties inhom (a) and (b) of a spatial Poisson process. Denote by N and NP the associated counting measures of Π and P . Then note that

b d NP ((a, b] × (c,d]) = N((a, b] × (c,d] × R) ∼ Poi λ(x, y, z)dzdydx Za Zc ZR and we see that the intensity function of P will have to be √ 1 x2 y2 − − 2 2 λP (x, y)= λ(x, y, z)dz = √ λdz =2λ 1 − x − y R 1 x2 y2 Z Z− − − p 2 2 2 for (x, y) ∈ R such that x + y ≤ 1, and λP (x, y) = 0 otherwise.

Property (a) also holds since for disjoint (aj, bj] × (cj,dj], j = 1,...,n, the sets (aj, bj] × (cj,dj] × R are also disjoint, and so independence of NP ((aj, bj] × (cj,dj]), j =1,...,n, follows from the corresponding property N.

Ψ( ) Ψ( ) 3. (a) First note that e γ = E(eγX1 ) implies that E(eγX1/m ) = e γ /m since sta- Ψ( ) tionarity and independence of increments implies E(eγX1/m )m = e γ , then Ψ( ) E(eγXq ) = eq γ , and then the right-continuity of sample paths implies that Xq → Xt almost surely and hence also in distribution, as q ↓ t. Therefore, Ψ( ) Ψ( ) characteristic functions converge and E(eγXq )= eq γ → et γ . Now we use the independence and stationarity of increments to see

E(exp{γXt}|Fs) = exp{γXs}E(exp{γ(Xt − Xs)})

= exp{γXs} exp{(t − s)Ψ(γ)}.

(b) The argument in (a) applies, with γ = iλ and ψ instead of Ψ as appropriate. Recall that moment generating functions do not exist for all random variables, but characteristic functions always exist (because x 7→ eiλx is bounded). XXIV Solutions – MS3b L´evy Processes and Finance – Oxford HT 2008

(c) The following argument can more easily be carried out for moment generating functions, but applies more generally if done for characteristic functions. tψ(λ) Diﬀerentiate E(exp{iλXt}) = e− with respect to λ at λ = 0 to get iE(Xt) = −tψ′(0) (see Grimmett-Stirzaker 5.7 for a statement and reference to the proof). The claim follows since µ = E(X1) must now be the slope of this linear function. Now, we use the independence and stationarity of increments to see

E(Xt − tµ|Fs)= E(Xs +(Xt −Xs) − tµ|Fs)= Xs +(t − s)µ − tµ = Xs − sµ.

tψ(λ) (d) Diﬀerentiate E(exp{iλXt}) = e− twice with respect to λ at λ = 0 to −E 2 − − 2 2 get (Xt ) = t(ψ′′(0) t(ψ′(0)) ), so Var(Xt) = tψ′′(0), where now σ = Var(X1)= ψ′′(0). Now we use the independence and stationarity of increments to see

2 2 E((Xt − tµ) |Fs) = E((Xs − sµ) + 2(Xs − sµ)(Xt − Xs − (t − s)µ) 2 +(Xt − Xs − (t − s)µ) |Fs) 2 = (Xs − sµ) + 2(Xs − sµ)E(Xt − Xs − (t − s)µ)

+Var(Xt − Xs) 2 2 = (Xs − sµ) +(t − s)σ .

4. (a) Fix β > 0. Note that the formula reduces to 0 = 0 for γ = 0. It is therefore suﬃcient to show that the γ-derivatives of both sides coincide. To diﬀerentiate the left hand side, note that

∂ γx 1 βx γx βx βx (e − 1) e− = e e− ≤ e− , ∂γ x

βx for γ ≤ 0, where x 7→ e− is integrable on [0, ∞). Therefore, we may interchange γ-diﬀerentiation and x-integration and have to show that for all γ < 0

∞ γx βx 1 1 e e− dx = , 0 1 − γ/β β Z which clearly is true.

(β γ0) The argument works for γ ≤ γ0 if we choose e− − as integrable upper bound. Clearly, for every ﬁxed γ < β, any γ0 ∈ (γ, β) will do. (b) We apply the exponential formula for Poisson point processes and (a) to obtain

αt ∞ γx 1 βx β E exp γ ∆s = exp t (e − 1)αx− e− dx = . 0 β − γ ( s t )! X≤ Z We recognise the last expression as the moment generating function of the Gamma distribution with the required density. By the Uniqueness Theorem

for moment generating functions, s t ∆s has this Gamma distribution. ≤ P Solutions 2 – MS3b L´evy Processes and Finance – Oxford HT 2008 XXV

and so are the sums tj−

~~lim ∆s = ∆s and lim ∆s = ∆s. t T t T ↑ s t s~~

~~(c) We condition on whether T1 < T1′ or T1′ < T1 and get for the ﬁrst jump size D J1 of D D X Y P(J1 ∈ A) = P(T1 < T1′)P(J1 ∈ A|T1 < T1′)+ P(T1 > T1′)P(−J1 ∈ A)~~

~~λX X λY Y = P(J1 ∈ A)+ P(−J1 ∈ A). λX + λY λX + λY This is a mixture of the jump size distributions of X and Y . We deduce that the density is~~

λX λX λY + hX (x) x> 0 − λX λY hD(x)= hX (x)+ hY ( x)= λY λ + λ λ + λ hY (−x) x< 0 X Y X Y ( λX +λY (d) This is bookwork, see Lecture 3, Example 18. The intensity function is λX hX (x), x> 0. (e) By the previous part, we have two Poisson point process ∆X and ∆Y in (0, ∞). Y It is easy to see that ∆− = −∆Y is a Poisson point process in (−∞, 0) with intensity function λY hY (−x), x < 0. It is easy to see that the associated Poisson counting measures on [0, ∞) × (0, ∞) and [0, ∞) × (−∞, 0) together form a Poisson counting measure on [0, ∞) × R \{0} via

~~N(A × B)= NX (A × (B ∩ (0, ∞))) + NY (A × (B ∩ (−∞, 0))).~~

~~The intensity function is λX hX (x), x> 0 and λY hY (−x), x< 0. XXVI Solutions – MS3b L´evy Processes and Finance – Oxford HT 2008~~

(f) Since (∆Dt)t 0 is a Poisson point process and Dt = ∆Ds, D is a com- ≥ s t pound Poisson process. ≤ P (g) For every real-valued compound Poisson process C we can deﬁne the processes X and Y of positive and negative jumps. Since the associated processes (∆Xt)t 0 and (∆Yt)t 0 inherit the properties of Poisson point processes (via ≥ ≥ their Poisson counting measures), this provides the required decomposition into two independent increasing compound Poisson processes. It is unique because any other decomposition must have more jumps, which must happen at the same time and cancel each other, but by (a), this is incompatible with independence. Solutions 3 – MS3b L´evy Processes and Finance – Oxford HT 2008 XXVII

~~B.3 Construction of L´evy processes~~

~~1. (a) If κ ∈ (−1, ∞), then~~

~~∞ ∞ κ x g(x)dx = x e− dx = Γ(κ + 1) < ∞. 0 0 Z Z~~

~~The Poisson point process is hence of the form of Example 18 and so (Ct)t 0 ≥ is a compound Poisson process with intensity Γ(κ + 1) and Gamma(κ +1, 1) jump distribution with density~~

~~1 κ x h(x)= x e− , x> 0. Γ(κ + 1)~~

~~(n) (b) The counting measures associated to (∆t)t 0 and (∆t )t 0 are ≥ ≥~~

~~N((a, b] × (c,d]) = #{t ∈ (a, b] : ∆t ∈ (c,d] d ∼ Poi (b − a) g(x)dx , 0 ≤ a < b, 0 1/n dx , 0 ≤ a < b, 0~~

Nn inherits the properties of a Poisson counting measure from N. We read oﬀ the intensity function gn(x) = g(x), x > 1/n, gn(x) = 0, x ≤ 1/n. The (n) argument of (a) shows that Ct is a compound Poisson process. (n) → ∞ (c) Ct increases as n . We can study the limit of moment generating functions, whether or not the limit is ﬁnite. We get, as n →∞,

~~(n) ∞ ∞ E(eγCt ) = exp (eγx − 1)g(x)dx ↓ exp (eγx − 1)g(x)dx 1 0 Z /n Z and because for γ < 0~~

∞ ∞ (eγx − 1)g(x)dx < ∞ ⇐⇒ (1 ∧ x)g(x)dx < ∞, 0 0 Z Z and by Lemma 21, we need to investigate the right hand condition. We check that 1 ∞ g(x)dx < ∞, and xg(x)dx < ∞ ⇐⇒ κ +1 > −1, 1 0 Z Z as required. (d) We can write

~~− (n) ≤ − (n) Cs Cs = ∆r1 ∆r 1/n ∆r1 ∆r 1/n = Ct Ct , { ≤ } { ≤ } r s r t X≤ X≤ XXVIII Solutions – MS3b L´evy Processes and Finance – Oxford HT 2008~~

and putting a supremum over s ≤ t on the left hand side, we get the required estimate (as an equality because we can take s = t on the left. Now we showed (n) → in (c) that Ct Ct a.s., and so we deduce here that | (n) − |→ →∞ sup Cs Cs 0 as n , s t ≤ i.e. that the convergence is locally uniform. (e) By Proposition 40(ii), we have for m ≤ n E | (n) − E (n) − (m) − E (m) |2 (n) − (m) ( Ct (Ct ) (Ct (Ct )) ) = Var(Ct Ct ) 1/m = x2g(x)dx, 1 Z /n ≥ → ∞ 1 2 ∞ and this decreases to zero as n m if and only if 0 x g(x)dx < , − (n) − E (n) i.e. κ > 3. In this case, (Ct (Ct ))n 1 is a Cauchy sequence that ≥ R converges by completeness of R (and the associated L2 space of R-valued random variables).

The limiting process includes all jumps (∆s)s t (intuitively, a more formal ≤ argument uses uniform convergence that preserves the jumps, see Theorem 42), and by (c), these are not summable for κ ∈ (−3, −2]. By Proposition 36, the limiting process has unbounded variation.

2. (a) Just note that for subordinators 0 ≤ Xt < ∞ a.s., and this implies that µXt µXt 1 ≥ e− > 0 a.s. and then also 1 ≥ E(e− ) > 0 as required. Therefore, Φt is well-defined. (b) This follows as in A.2.3, first for rational t ≥ 0 and then, by right-continuity of paths and since a.s. convergence implies convergence in distribution, hence of moment generating functions. The scaling relation for fixed t translates to

~~1/α 1/α µXt Φt/c(c µ)= − ln(E(exp{−µc Xt/c})) = − ln(E(e− ))=Φt(µ).~~

~~α and therefore, for t = 1, c = µ− , as required~~

~~α µ Φ(1) = Φ1/c(1) = Φ(µ).~~

µXt µXt (c) Clearly µ 7→ e− is a.s. decreasing and so is hence µ 7→ E(e− ), strictly α decreasing if Xt > 0 with positive probability. Now, Φ(µ) = Φ(1)µ is clearly diﬀerentiable for µ> 0, and so

~~α α ∂ µXt ∂ tΦ(1)µ α 1 tΦ(1)µ E(e− )= e− = −tΦ(1)αµ − e− ∂µ ∂µ~~

µXt and this is negative only for α > 0 (or α = 0 but then µ 7→ E(e− ) is µXt constant). To show that also α ≤ 1 note that µ 7→ e− is also a.s. convex, µXt and hence so is µ 7→ E(e− ). Now, Φ(µ) is also twice diﬀerentiable so that 2 α ∂ µXt α 2 tΦ(1)µ α E(e− )= tΦ(1)αµ − e− (tΦ(1)αµ − (α − 1)), ∂µ2 and this is nonnegative for all µ> 0 if and only if α ≤ 1. Solutions 3 – MS3b L´evy Processes and Finance – Oxford HT 2008 XXIX

~~(d) Note that, (by monotone convergence), as µ ↓ 0,~~

~~α α 1 tΦ(1)µ µXt tΦ(1)αµ − e− = E(Xte− ) ↑ E(Xt),~~

where the left-hand side increases to ∞ for α ∈ (0, 1). (e) Note that Φ(0) = 0 implies that the equation holds for µ = 0 no matter what g is. Now diﬀerentiate both sides with respect to µ to get

α 1 ∞ µx Φ(1)αµ − = e− xg(x)dx. 0 Z Remember that the density of the Gamma(1 − α,µ) distribution is f(x) = 1 1 α α µx (Γ(1 − α))− µ − x− e− . Therefore, we can (and have to, by the Uniqueness Theorem for moment generating functions) take

~~Φ(1)α α 1 g(x)= x− − , x> 0. Γ(1 − α)~~

(f) For α ∈ (0, 1), the Construction Theorem for subordinators (Theorem 8) shows that we can construct the stable subordinator from a Poisson point process with intensity function g as speciﬁed in (e). Note that g satisﬁes the integrability condition

~~∞ (1 ∧ x)g(x)dx < ∞ 0 Z α 1 α since x− − is integrable at x = ∞ and x− is integrable at x = 0.~~

~~For α = 1 note that Φt(µ) = Φ(1)tµ. The associated subordinator is the deterministic drift Xt = Φ(1)t.~~

~~3. (a) Just note that~~

~~1/α 1/α 1/α c Zt/c = c Xt/c − c Yt/c ∼ Xt − Yt = Zt~~

~~for ﬁxed t, and that, as processes in t ≥ 0, both the left-hand side and the right-hand side are L´evy processes. Therefore, the distributions as processes coincide. (b) H ∼ −H implies~~

~~iλH iλH E(cos(λH))+iE(sin(λH)) = E(e )= E(e− )= E(cos(λH))−iE(sin(λH))~~

~~and so the imaginary part E(sin(λH)) must vanish for all λ ∈ R.~~

(c) Clearly Zt = Xt − Yt ∼ Yt − Xt = −Zt, so Zt has a symmetric distribution. iλZt By (b), its characteristic function ϕt(λ)= E(e ) is real-valued. By the hint, we may assume that ϕt is continuous, and since Zt is inﬁnitely divisible, that ϕt(λ) =6 0, so it must stay positive everywhere (note that ϕ(0) = 1). Deﬁne

~~ψt(λ)= − ln(ϕt(λ)), ψ(λ)= ψ1(λ), λ ∈ R. XXX Solutions – MS3b L´evy Processes and Finance – Oxford HT 2008~~

~~By A.2.3.(b), we have ψt(λ)= tψ(λ). The scaling relation implies~~

~~1/α 1/α iλZt ψt/c(c λ)= − ln(E(exp{iλc Zt/c})) = − ln(E(e )) = ψt(λ),~~

~~and as in 2.(b), this implies ψ(λ)= ψ(1)λα for all λ ≥ 0. For λ< 0 note that~~

~~iλZt iλZt ψ(λ)= − ln(E(e )) = − ln(E(e− )) = ψ(−λ),~~

so we have ψ(λ)= ψ(1)|λ|α. (d) Before we start, note that the integral deﬁning ψ(λ) converges for α ∈ (0, 2) 1 α α 1 since the integrand behaves like x − at x = 0 and like x− − at |x| = ∞. We 1/α 1 1 then check, by change of variables y = c x (hencee x− dx = y− dy, that

1/α ∞ 1/α α 1 ψ(λc ) = (cos(λc x) − 1)b|x|− − dx Z−∞ ∞ α 1 e = (cos(λy) − 1)bc|ye|− − dy = cψ(λ). Z−∞ e e The argument of (c) shows that this implies λ = b|λ|α for some b ≥ 0 – the argument did not depend on α ∈ (0, 1). (e) Let R be a symmetric stable process R of indexe α. In (d) we expressed the α 1 characteristic exponent in terms of the Lévy density g(x)= |x|− − . Therefore, (Rt)t 0 can be constructed from a Poisson point process of jumps with this ≥ density. We see from the criterion ∞ (1 ∧|x|)g(x)dx < ∞ that jumps are absolutely summable if and only if α −∞∈ (0, 1). In that case, we expressed R as R the difference of two subordinators in (a), so R has indeed bounded variation. If α ∈ [1, 2), jumps are not absolutely summable, so variation is unbounded, by Proposition 36. If α = 2, then R is a multiple of Brownian motion, and it was shown in the lectures, that Brownian motion has also unbounded variation. We differentiate ψ(λ) in λ = 0 to see that E(Xt)= tψ′(0) < ∞ if and only if α ∈ (1, 2], and that Var(Xt)= tψ′′(0) < ∞ if and only if α = 2. Solutions 4 – MS3b Lévy Processes and Finance – Oxford HT 2008 XXXI

~~B.4 Simulation~~

~~1 1. Note that the deﬁnition F − (u) = inf{x ∈ R : F (x) >u} implies~~

~~1 F (t) >u ⇒ F − (u) ≤ t ⇒ ∀ε>0F (t + ε>u) ⇒ F (t) ≥ u~~

~~Therefore for all t ∈ R~~

~~1 F (t)= P(F (t) > U) ≤ P(F − (U) ≤ t) ≤ P(F (t) ≥ U)= F (t).~~

~~2. (a) Independent A ∼ Gamma(a, 1) and B ∼ Gamma(b, 1) have joint density~~

~~xa 1e x yb 1e y f (x, y)= − − − − A,B Γ(a) Γ(b)~~

~~The transformation (R,S) = T (A, B)=(A/(A + B), A + B) is bijective T : 2 1 (0, ∞) → (0, 1) × (0, ∞) with inverse transformation (A, B) = T − (R,S) = (SR,S(1 − R)) that has the Jacobian~~

~~s r J(r, s)= ⇒ | det(J(r, s))| = s −s 1 − r and so the transformation formula yields~~

a 1 sr b 1 s(1 r) 1 (sr) − e− (s(1 − r)) − e− − f (r, s) = | det(J(r, s))|f (T − (r, s)) = s R,S A,B Γ(a) Γ(b) a+b 1 s Γ(a + b) a 1 b 1 s − e− = r − (1 − r) − , Γ(a)Γ(b) Γ(a + b) as required. 1 Vice versa, for c = a + b and p = a/(a + b), we recognise T − (R,S)=(A, B), which has joint distribution

~~xa 1e x yb 1e y xcp 1e x yc(1 p) 1e y f (x, y)= − − − − = − − − − − . A,B Γ(a) Γ(b) Γ(cp) Γ(c(1 − p))~~

and so any random variable (R, S) with joint distribution as (R,S) will be 1 1 such that T − (R, S) ∼ T − (R,S)=(A, B). 1/a e e a a a 1 (b) P(X ≤ x)= P(U ≤ x)= P(U ≤ x )= x , x ∈ (0, 1) and so fX (x)= ax − , x ∈ (0, 1). We recognisee e X ∼ Beta(a, 1). (c) We calculate

Y Y (1 − t) P ≤ t, Y + Z ≤ 1 = P( ≤ Z ≤ 1 − Y ) Y + Z t t 1 y − a 1 a = ay − (1 − a)z− dzdy 0 y(1 t)/t Z Z − t a 1 1 a 1 a 1 a a 1 = ay − (1 − y) − − y − (1 − t) − t − dy. 0 Z XXXII Solutions – MS3b L´evy Processes and Finance – Oxford HT 2008

~~We diﬀerentiate with respect to t to get~~

~~a a 1 1 a a 2 at((1 − a)(1 − t)− t − − (a − 1)(1 − t) − t − ) fW Y +Z 1(t) = | ≤ P(X + Y ≤ 1) a(1 − a)ta 1(1 − t) a((1 − t)+ t) = − − P(X + Y ≤ 1)~~

and we recognise the density of Beta(a, 1−a), up to the normalisation constant, but we have calculated a conditional density which integrates to 1, so the normalisation constant must be the one of Beta(a, 1 − a). (d) Given Y + Z ≤ 1, W is Beta(a, 1 − a)-distributed. Since T is independent of (Y,Z,W ), its conditional distribution given Y +Z ≤ 1 is still Gamma(1, 1) and it is conditionally independent of W given Y + Z ≤ 1. Therefore P(T W ≤ h|Y + Z ≤ 1) = P(SR ≤ h), and we can apply (a) for c = 1 and p = a to deduce that, SR ∼ Gamma(a, 1), i.e. the conditional distribution of W T given Y + Z ≤ 1 is Gamma(a, 1). (e) This procedure generates a Gamma(a, 1) random variable. Speciﬁcally, the conditioning on Y + Z ≤ 1 is realised by repeated trials until Y + Z ≤ 1, see Lemma 57. The procedure is easily implemented and gives a more eﬃcient way of simulating Gamma random variables from uniform random variables than inverting the distribution function of the Gamma distribution numerically.

3. (a) From 2.(a) we take that we obain A/(A+B) ∼ Beta(a, b) for independent A ∼ Gamma(a, 1) and B ∼ Gamma(b, 1). Johnk’s procedure works for a ∈ (0, 1). To generate Gamma variables for higher parameters, we can write a = [a]+{a} for integer part and fractional part and then represent

~~[a]~~

~~A = Ek + A0 =1 Xk~~

~~where (Ek)1 k [a] is a sequence of independent Exp(1) random variables and ≤ ≤ A0 ∼ Gamma({a}, 1). To summarise, the following procedure generates a Beta(a, b) random variable:~~

~~1.-5. Run Johnk’s Gamma generator for parameter {a}. Set A0 = TY/(Y +Z). 6.-10. Independently of 1.-5., run Johnk’s Gamma generator for parameter {b}. Set B0 = TY/(Y + Z).~~

~~11. Generate independent U1,...,U[a]+[b] ∼ Unif(0, 1) and set~~

~~[a] [a]+[b]~~

A = A0 − ln Uk and B = B0 − ln Uk .  =1   =[ ]+1  kY k Ya     12. Return the number A/(A + B). (b) The procedure generates a stochastic process successively at reﬁning lattices of dyadic times. The key step (for n = 0 and then inductively for n ≥ 1) is n − − − ∼ n to take a 2− -increment Yk,n = Xk2 n X(k 1)2 n Gamma(2− , 1) and a − Solutions 4 – MS3b L´evy Processes and Finance – Oxford HT 2008 XXXIII

Bk,n ∼ Beta(an, bn) random variable to split Yk,n into two increments Bk,nYk,n n n and (1 − Bk,n)Yk,n. For 2.(a) to apply, we need an + bn =2− , i.e. an =2− p n n 1 and bn =2− (1 − p). In order to get identically Gamma(2− − , 1) increments, further p =1/2. This now yields stationary independent increments.

This is the only way to achieve Gamma distributions for all Xk2−n (at least in the framework of 2.(a), but in fact, in general) for Beta parameters not depending on k. If we were to use Bk,n ∼ Beta(ak,n, bk,n) for parameters that may depend on k, then we get more general processes with independent Gamma increments that are not stationary, as soon as we have the consistency condition that ak,n + bk,n is the parameter of Yk,n. (c) Johnk’s Gamma generator is more efficient than the inverse distribution function computation. The method is less liable to accumulating errors since time 1 is most accurate and errors only accumulate along the dyadic expansions, i.e. with local rather than global impact. Furthermore, we get an iterative procedure for which we do not have to fix the time lag δ in advance, but can continue to fill in extra points until a satisfactory result is obtained.

~~4. (a) Since Gt ∼ Gamma(α+t, β+) and Ht ∼ Gamma(α t, β ), we have − − α+t α t α α+ E(at + Gt − Ht)= at + − − =0 ⇐⇒ a = − − . β+ β β β+ − −~~

~~(b) • Denote Fδ(x)= P(Vδ ≤ x). 1. Set S0 = 0 and n = 1.~~

~~2. Generate Un ∼ Unif(0, 1). 1 3. Set Sn = Sn 1 + Fδ− (Un). If enough steps have been performed, go − to 4., otherwise increase n by 1 and go to 2.~~

~~4. Return (Sn)n 0 as simulation of (Vδn)n 0. ≥ ≥ • Denote F (x; α, β)= P(G ≤ x) for G ∼ Gamma(α, β). 1. Set S0 = 0 and n = 1.~~

2. Generate two independent random numbers U2n 1 ∼ Unif(0, 1) and − U2n ∼ Unif(0, 1). 1 1 3. Set Sn = Sn 1 + aδ + F − (U2n 1; α+δ, β+) − F − (U2n; α δ, β ). If − − − − enough steps have been performed, go to 4., otherwise increase n by 1 and go to 2.

4. Return (Sn)n 0 as simulation of (Vδn)n 0. ≥ ≥ • Fix t = 1, iterate for further time units if needed. Denote G(x; a, b) = P(B ≤ x) for B ∼ Beta(a, b). 1. Set V0 = 0 and n = 0. 2. Generate 2 independent random numbers U1 ∼ Unif(0, 1) and U2 ∼ Unif(0, 1). 1 1 3. Set P1 = F − (U1; α+, β+), N1 = F − (U2; α , β ) and V1 = a+P1 −N1. − − n ∼ 4. Generate 2 independent random numbers U2n+1+k Unif(0, 1), k = 1,..., 2n. 1 n 1 n 1 n 1 5. Set Bn,k = G− (U2n+1+k;2− − α+, 2− − α+), k = 1,..., 2 − and 1 n 1 n 1 n 1 Cn,k = G− (U2n+1+2n−1+k;2− − α , 2− − α ), k =1,..., 2 − − − XXXIV Solutions – MS3b L´evy Processes and Finance – Oxford HT 2008

6. Set P(2k 1)2−n = Bn,kP(2k 2)2−n + (1 − Bn,k)P(2k)2−n , N(2k 1)2−n = − − − − − − − − n Cn,kN(2k 2)2 n + (1 Cn,k)N(2k)2 n and V(2k 1)2 n = (2k 1)2− a + − n 1 − P(2k 1)2−n − N(2k 1)2−n for k = 1,..., 2 − . If the resolution is ﬁne enough,− go to 7.,− otherwise increase n by 1 and go to 4.

7. Return (Vk2−n )k=1,...,2n . 1 1 Instead of F − and G− , one can use Johnk’s Gamma generator of A.3.2. and the associated Beta generator of A.3.3. x 1 1 • βy ∞ βy Denote H(x; β)= ε y− e− dy/ ε y− e− dy. Also denote

~~R ∞ R ∞ 1 β+y 1 β−y λ = α+ y− e− dy + α y− e− dy ε − ε Z ∞ Z 1 1 β+y and p = λ− α+ y− e− dy. Zε 1. Set V0 = 0, T0 = 0 and n = 1.~~

2. Generate three independent random numbers U3n 2 ∼ Unif(0, 1) and − U3n 1 ∼ Unif(0, 1) and U3n ∼ Unif(0, 1). − 3. Set Zn = − ln(U3n)/λ. 1 1 4. If U3n 1 >p, let Jn = −H− (U3n; β ), otherwise let Jn = H− (U3n; β+). − − 5. Set Tn = Tn 1 + Zn and VTn = VTn−1 + aZn + Jn. If Tn is big enough, − go to 6., otherwise increase n by 1 and go to 2.

6. Return (VTn )n 0. ≥ (c) Below are 9 simulations for α+ ∈{1, 10, 100} (rows) and α ∈{10, 100, 1000} − (columns). Note the big positive jumps for α+ = 1, the cases α+ = α with − a = 0 and convergence to Brownian motion from top left to bottom right. The code is a similar to the symmetric case and is available on the course website. space space space −0.5 0.0 0.5 1.0 1.5 2.0 2.5 −1.5 −1.0 −0.5 0.0 0.5 1.0 −3 −2 −1 0 1

~~0 2 4 6 8 10 0 2 4 6 8 10 0 2 4 6 8 10~~

time time time Variance Gamma process with shape parameters 0.5 and 50 and scale parameters 1 and 10 Variance Gamma process with shape parameters 0.5 and 5000 and scale parameters 1 and 100 Variance Gamma process with shape parameters 0.5 and 5e+05 and scale parameters 1 and 1000 space space space −1 0 1 2 3 −3 −2 −1 0 1 2 −0.2 −0.1 0.0 0.1

~~0 2 4 6 8 10 0 2 4 6 8 10 0 2 4 6 8 10~~

time time time Variance Gamma process with shape parameters 50 and 50 and scale parameters 10 and 10 Variance Gamma process with shape parameters 50 and 5000 and scale parameters 10 and 100 Variance Gamma process with shape parameters 50 and 5e+05 and scale parameters 10 and 1000 space space space −5 −4 −3 −2 −1 0 −1.0 −0.5 0.0 0.5 1.0 1.5 2.0 −4 −3 −2 −1 0 1

~~0 2 4 6 8 10 0 2 4 6 8 10 0 2 4 6 8 10~~

time time time Variance Gamma process with shape parameters 5000 and 50 and scale parameters 100 and 10 Variance Gamma process with shape parameters 5000 and 5000 and scale parameters 100 and 100Variance Gamma process with shape parameters 5000 and 5e+05 and scale parameters 100 and 1000 Solutions 5 – MS3b L´evy Processes and Finance – Oxford HT 2008 XXXV

~~B.5 Financial models~~

δ up up δ up 1. (a) W0 = T0 + U0 + V0, W1(ω1)= T0e + U0B1 + V0C1 , W1(ω2)= T0e + U0B1 + down δ down up δ down V0C1 , W1(ω3)= T0e + U0B1 + V0C1 and W1(ω4)= T0e + U0B1 + down V0C1 . (b) By general reasoning, there is arbitrage if one asset is uniformly better than another asset. In particular:

• If B1(ω1) ≤ A1, then (1, −1, 0) is an arbitrage portfolio, since W0 = 0 and W1 ≥ 0 with W1(ω3)= W1(ω4) > 0. • If A1 ≤ B1(ω4), then (−1, 1, 0) is an arbitrage portfolio, since W0 = 0 and W1 ≥ 0 with W1(ω1)= W1(ω2) > 0. • Similarly (1, 0, −1) or (−1, 0, 1) are arbitrage portfolios if C1(ω1) ≤ A1 or A1 ≤ C1(ω4). These can also be deduced from the standard two-asset binary model (A, B) or up down up down (A, C). Now let B1 > A1 > B1 and C1 > A1 > C1 . Since the model (A, B) has no arbitrage, there is no arbitrage portfolio of the form (T0, U0, 0). Assume that (T0, U0, 1) is an arbitrage portfolio. Then 0 = W0 = T0 + U0 + 1, W1(ω1) > W1(ω2) ≥ 0 and W1(ω3) > W1(ω4) ≥ 0. down down • If U0 ≥ 0, then we have 0 ≤ W1(ω4) = T1A1 + U0B1 + C1 < (T1 + U0 + 1)A1 = 0, which is a contradiction. up down • If U0 ≤ 0, then we have 0 ≤ W1(ω2) = T1A1 + U0B1 + C1 < (T1 + U0 + 1)A1 = 0, which is a contradiction.

Similarly, now assume that (T0, U0, −1) is an arbitrage portfolio, then 0 = W0 = T0 + U0 − 1, W1(ω2) > W1(ω1) ≥ 0 and W1(ω4) > W1(ω3) ≥ 0. down up • If U0 ≥ 0, then we have 0 ≤ W1(ω3) = T1A1 + U0B1 − C1 < (T1 + U0 − 1)A1 = 0, which is a contradiction. up up • If U0 ≤ 0, then we have 0 ≤ W1(ω1)= T1A1 + U0B1 − C1 < (T1 + U0 + 1)A1 = 0, which is a contradiction. So there is no arbitrage portfolio.

~~(c) The contingent claim W1(ω1)=1, W1(ω2)= W1(ω3)= W1(ω4) = 0 cannot be hedged, since we would require~~

~~up down 0 = T0A1 + U0B1 + V0C1 down down down up = T0A1 + U0B1 + V0C1 = T0A1 + U0B1 + V0C1 ,~~

~~for ω2,ω3,ω4, and these imply T0 = U0 = V0 = 0, but then the fourth equation up up 1= T0A1 + U0B1 + V0C1 fails.~~

(d) Since the contingent claim does not change as C1 varies, we should consider portfolios of the form (T0, U0, 0). Since the model (A, B) with scenarios “up” and “down” is complete, the contingent claim W1(up) = W1(ω1), W1(up) = W1(ω3) can be hedged. Speciﬁcally, f f up down W1(up) = T0A1 + U0B1 and W1(down) = T0A1 + U0B1

~~f f XXXVI Solutions – MS3b L´evy Processes and Finance – Oxford HT 2008~~

has solution up down W1(down)B1 − W1(up)B1 W1(up) − W1(down) T0 = up down and U0 = up down , A1(B1 − B1 ) B1 − B1 and so wef read oﬀ from f f f down up A1 − B1 B1 − A1 W0 = T0 + U0 = up down W1(up) + up down W1(down) (1) A1(B1 − B1 ) A1(B1 − B1 thatf f f − down A1 B1 ∈ qB = up down (0, 1). B1 − B1

The martingale property is equation (1) for the contingent claim W1(down) = down up B1 and W1(up) = B1 . The martingale probability qB is unique and does not depend on W1. f (e) By symmetry,f contingent claims of the form W1(ω1) = W1(ω3), W1(ω2) = δ W1(ω4) can bef hedged and priced as e− E(W1), where − down up A1 C1 ∈ qC = P(C1 = C1 )= up down (0, 1). C1 − C1 δt The process e− Ct, t =0, 1, is a martingale under these probabilities. δt δt (f) In order for both e− Bt and e− Ct to be martingales, we need up up up down qB = P(B1 = B1 ,C1 = C1 )+ P(B1 = B1 ,C1 = C1 )= p1 + p2 and up up down down qC = P(B1 = B1 ,C1 = C1 )+ P(B1 = B1 , B1 = B1 )= p1 + p3.

Together with the normalisation condition p1 + p2 + p3 + p4, we have three equations (of rank three) for four unknowns, so there is a one-dimensional solution space. δ (g) The range of arbitrage-free prices W0 = e− p1 depends on qB and qC as follows.

~~• If qB + qC ≤ 1, then p1 can be arbitrarily close to zero, and then W0 will be arbitrarily close to zero.~~

• If qB +qC > 1, then qB +qC =2p1 +p2 +p3 qB +qC −1 δ and so W0 > e− (qB + qC − 1). δ • Clearly p1 < min{qB, qC } and so W0 < e− min{qB, qC }. δ So we get e W0 ∈ (max{0, qB + qC − 1}, min{qB, qC }). Note that this range is always non-empty. (ε) 2. (a) The direct proof is to calculate the moment generating function of Xi

γX(ε) γµε λε γ(1 µε) λε γµε λε γ E(e i )= e− e− + e − (1 − e− )= e− (1+(1 − e− )(e − 1)) and to see [t/ε] (ε) − λε γ − γS γµε[t/ε] [t/ε](1 e− )(e 1) γµt λ(eγ 1) E(e [t/ε] )= e− 1+ → e− e− − [t/ε] which we recognise as being the moment generating function of Xt = Nt − µt. Solutions 5 – MS3b L´evy Processes and Finance – Oxford HT 2008 XXXVII

~~(b) This is a special case of the n-step generalisation of the two-asset model (A, S) on two scenarios. Since A0 = S0 = 1, we have no arbitrage if and only if up S1(down) < A1 < S1 . Here, this is~~

~~µε δε 1 µε e− < e < e − ⇐⇒ −µ<δ< 1/ε − µ.~~

~~(c) We need~~

(ε) δε (ε) εδ µε 1 µε 1= S0 = e− Eq(S1 )= e− e− (1 − qε)+ e − qε and so e e eδε − e µε eµε δε − 1 q = − = − . ε µε − − e− (e 1) e 1 (d) This is in complete analogy to (a). We deduce this from the Poisson limit (ε) (ε) theorem considering Tn = Sn − nµε, a Bernoulli random walk with success probability qε. Noting that e e ( + ) 1 1 eε δ µ − 1 δ + µ q = → as ε ↓ 0, ε ε e − 1 ε e − 1 (ε) → → we obtain T[t/ε] Nt in distribution, as required. Now, clearly [t/ε]µε µt, and taking diﬀerences in the two limit results completes the argument. (e) Note frome the momente generating function of the Poisson distribution that

e δ+µ Nt t (e 1) δt+µt E(e )= e e−1 − = e e δt Nt µt and so Mt = e− e − is a martingale, because for s < t e e e δs Ns µs δ(t s) (Nt Ns) µ(t s) E(Mt|Fs) = E(e− e − e− − e − − − |Fs) e e e δs Ns µs (δ+µ)(t s) Nt Ns = e− e − e− − E(e − )= Ms. e Ns µs Ns µs Given Nt = k or Nt = k, the two processes (e − )0 s t and (e − )0 s t ≤ ≤ ≤ ≤ have the same conditional distribution, since the k jump times of N and N occur at independente uniform times on [0, t]. Since also P(Nt = k) > 0 if and only if P(Nt = k) > 0, the same paths are possible for the two processes.e e δt Nt µt Since the discounted process e− e − is a martingale, it provides martingale Nt µt probabilities fore the equivalent process e − .

(f) (Nt)t 0 only has jumps of size 1, all other jumps are impossible, and the only ≥ L´evy processes with this property are Poisson processes with drift. If (Yt)t 0 ≥ is a Poisson process with drift −νt, then we have

Ys P((e )0 s 1 ∈ Dν)=1. ≤ ≤ Yt Since Dν ∩ Dµ = ∅ for µ =6 ν, we must have µ = ν in order that e has the Nt µt same possible paths as e − . We can now check that of all intensities λ> 0 δt Yt of Y , only λ =(δ + µ)/(e − 1) is such that Mt = e− e is a martingale:

δs Ys δ(t s) Yt Ys E(Mt|Fs) = E(e− e e− − e − |Fs) δs Ys δ(t s) Yt Ys (δ+µ)(t s)+λ(e 1) = e− e e− − E(e − )= Mse− − − . XXXVIII Solutions – MS3b L´evy Processes and Finance – Oxford HT 2008

~~B.6 Time change and subordination~~

~~1. (a) First note that~~

[2ny] [2ny] 2 − − − − − − E (Zj2 n Z(j 1)2 n ) = Var (Bf(j2 n ) Bf((j 1)2 n)  −  − j=1 j=1 X X   [2ny] n n = (f(j2− ) − f((j − 1)2− )) j=1 X n n n n = f([2 y]2− ) − f(0) = f([2 y]2− ) → f(y), as n →∞. For L2-convergence we then calculate 2 [2ny] 2 − − − − E (Zj2 n Z(j 1)2 n ) f(y)  −   j=1 X  [2ny]   2 n n 2 ≤ Var (Zj2−n − Z(j 1)2−n ) +(f([2 y]2− ) − f(y))  −  j=1 X [2ny]  n n 2 2 n n 2 ≤ (f(j2− ) − f((j − 1)2− )) Var(B1 )+(f([2 y]2− ) − f(y))   j=1 X 2 →[f]yVar(B1 )=0,  provided that f is continuous (and increasing). Convergence in L2 implies convergence in probability. (b) Note ﬁrst that both Z and Z are continuous. For marginal distributions, note that Zy = Bf(y) ∼ Normal(0, f(y)) and, for yj ≤ y

~~j e − − Zy = σi(Wyi Wyi−1 )+ σj+1(Wy Wyj ) i=1 X e − ∼ 2 is the sum of independent σi(Wyi Wyi−1 ) Normal(0, τi ), where~~

~~yi 2 2 − − τi = σi (yi yi 1)= f ′(s)ds = f(yi) f(yi 1), − − Zyi−1 and these variances add up to f(y), as well. As for joint distributions, Z and~~

~~Z have independent increments: for 0 = u0~~

rk 1 − − − − − Z Z − = σ (W W − )+ σ (W W − )+ σ (W W − ) uk uk 1 lk ylk uk 1 i yi yi 1 rk ui yri 1 = +1 i Xlk e e ∼ Normal(0, f(uk) − f(uk 1)). − Solutions 6 – MS3b L´evy Processes and Finance – Oxford HT 2008 XXXIX

(c) Take a Poisson process X of rate λ. Then the process Z =(Xf(y))y 0 has jumps ≥ of size 1 only, for all continuous functions f : [0, ∞) → [0, ∞). However, if for a function as in (b), we have σj =6 1 for some j ≥ 1, then there is positive y probability that Zy = 0 f ′(s)dXs has jumps of size σj, speciﬁcally, there will be a Poi(λ(y +1 −y )) number of such jumps in the time interval (y ,y +1]. j jR p j j e 2. (a) If Var(X1) < ∞ (and hence Var(Xt) = tVar(X1) and E(Xt) = tE(X1)) and 2 ∞ Var(τ1)= 0∞ t gτ (t)dt < , we check the stronger integrability condition R ∞ 2 ∞ 2 ∞ z g(z)dz = z ft(z)gτ (t)dtdz 0 Z−∞ Z−∞ Z ∞ ∞ 2 = z ft(z)dzgτ (t)dt 0 Z Z−∞ ∞ 2 = (Var(Xt)+(E(Xt)) )gτ (t)dt 0 Z ∞ 2 2 = (tVar(X1)+ t (E(X1)) )gτ (t)dt < ∞. 0 Z ∞ (b) If τ is a compound Poisson process, i.e. 0∞ gτ (t)dt < , then R ∞ ∞ ∞ ∞ g(z)dz = ft(z)dzgτ (t)dt = gτ (t)dt < ∞. 0 0 Z−∞ Z Z−∞ Z

If X is a compound Poisson process with intensity λ and such that P(Xt ∈ b 6∋ P 6 − λt (a, b)) = a ft(x)dx for all (a, b) 0 and (Xt =0)=1 e− , then R ∞ ∞ λt g(z)dz = ft(z)dzgτ (t)dt = (1 − e− )gτ (t)dt < ∞. R 0 0 R 0 0 Z \{ } Z Z \{ } Z

Note that (a) and (b) deal, respectively, with the integrability condition for small z and large z. The general case, when neither the conditions of (a) nor of (b) are satisfied, we know that we still obtain a Lévy density, from the calculation of characteristic functions in the lectures, but the integrability condition for Lévy densities is difficult to check directly.

~~(c) A L´evy density is associated with a bounded variation L´evy process if and only if~~

~~∞ (1 ∧|x|)g(x)dx < ∞ Z−∞~~

~~1/α First note that the scaling relation Xt ∼ t X1 implies, by the transformation formula, that~~

~~1/α 1/α ft(y)= t− f1(t− y), y ∈ R. XL Solutions – MS3b L´evy Processes and Finance – Oxford HT 2008~~

~~Therefore, we can check that~~

∞ ∞ ∞ |y|g(y)dy = |y|ft(y)dygτ (t)dt 0 Z−∞ Z Z−∞ ∞ ∞ 1/α 1/α = |y|t− f1(t− y)dygτ (t)dt 0 Z Z−∞ ∞ ∞ 1/α = |x|t f1(x)dxgτ (t)dt 0 Z Z−∞ ∞ 1/α ∞ = t gτ (t)dt |x|f1(x)dx . 0 Z Z−∞ (1) (2) The last integral is E(|X1|). To see that it is ﬁnite, split X = X +X where (2) (1) X is the compound Poisson process of big jumps ∆Xt1 ∆Xt 1 , and X (1) {| |≥ } (2) is the martingale of small jumps. Now E(|X1|) ≤ E(|X1 |)+ E(|X1 |) < ∞, since the moment generating function of X(1) is an entire function, hence all moments are ﬁnite, and

~~(2) α 1 E(|X1 |) ≤ E |∆s| = |x||x|− − < ∞. 1 0 s 1 ! x X≤ ≤ Z| |≥ ∼ − 3. Since B B, we clearly have for Xy = Bτy~~

P ≤ ∞ P ≤ ∞ P ≥ − P ≥ − (Xy a)= (Bt a)fτy (t)dt = (Bt a)fτy (t)dt = (Xy a), 0 0 Z Z and this is enough, since both X and −X have stationary, independent increments and right-continuous paths with left limits. On the other hand, the symmetric process X = P −N for two independent Poisson processes of rate λ cannot be obtained as Brownian motion subordinated by an independent subordinator, since for every random time T with [or without] probability density function we have

~~∞ P(BT =1)= P(Bt = 1)fT (t)dt =0 or P(Bt = 1)P(T ∈ dt)=0. 0 [0, ) Z Z ∞ ~~

~~4. (a) Calculate~~

αs αs 1 E { } ∞ E { } λ t − λt (exp γBTs ) = (exp γBt ) e− dt 0 Γ(αs) Z αs αs 1 ∞ { 1 2} λ t − λt = E(exp 2 tγ ) e− dt 0 Γ(αs) Z λ αs = E(exp{ 1 γ2T })= . 2 s λ − 1 γ2 2

~~and identify this with the formula for Cs − Ds in Exercise A.1.2.(b). Solutions 6 – MS3b L´evy Processes and Finance – Oxford HT 2008 XLI~~

~~(b) We condition on St to get~~

~~∞ E iλRt E iλBSt E iλBs (e ) = (e )= fSt (s) (e )ds 0 Z ∞ 1 2 2 λ s = fSt (s)e− ds 0 Z 1 2 1 2 1 α 2α λ St tΦ( λ ) tΦ(1)( ) λ = E(e− 2 )= e− 2 = e− 2 | | .~~

~~We identify the symmetric 2α-stable distribution.~~

(c) Now recall Method 3: Let (τy)y 0 be an increasing process that we can simu- ≥ late, and let (Xt)t 0 be a L´evy process with cumulative distribution function ≥ Ft of Xt. Fix a time lag δ > 0. Then the process

~~n (3,δ) 1 Z = S[ ], where S = Y and Y = F − (U ) y y/δ n k k τkδ τ(k−1)δ k =1 − Xk~~

is the time discretisation of the subordinated process Zy = Xτy . and Example 84: We can use Method 3 to simulate the Variance Gamma process, since we can simulate the Gamma process τ and we can simulate the Yk. Actually, we can use the Box-Muller method to generate standard Normal random variables Nk and then use

~~Yk ∼ τkδ − τ(k 1)δNk, k ≥ 1, − ≥ p instead of Yk, k 1. e~~

E ∞ E ∞ 5. (a) (Zt)= 0 (Xs)fSt (s)ds = 0 µsfSt (ds)= µmt. 2 2 2 2 2 2 2 2 2 2 2 E ∞ E ∞ (b) (Zt )=R 0 (Xs) fSt (s)ds =R 0 (σ s+µ s )fSt (ds)= σ mt+µ q t+µ m t and then Var(Z )= σ2mt + µ2q2t + µ2m2t2 − µ2m2t2 =(σ2m + µ2q2)t. R t R (c) For the Variance Gamma process, we have αt αt Var(Ct − Dt) = Var(Ct) + Var(Dt)=2√ 2 = . 2λ λ On the other hand, for σ2 = 1, µ = 0, m = α/λ α Var(Z )= t. t λ

~~Clearly α = λ corresponds to Var(Z1) = 1. Diﬀerentiate the moment generat- 2 2 4 2 E γBt γ t/ E ing function (e )= e four times to get (Bt )=3t . Then~~

E 4 ∞ E 4 ∞ 2 E 2 (BSt )= (Bs )fSt (s)ds = 3s fSt (s)ds =3 (St ) 0 0 Z Z Since St ∼ Gamma(α,λ), we get α α2 1 E(B4 )=3 + =3 1+ ∈ (3, ∞). S1 λ2 λ2 λ XLII Solutions – MS3b L´evy Processes and Finance – Oxford HT 2008

~~B.7 Level passage events~~

(t) 1. (a) The process X has independent increments: for all 0 = s0 < s1 < ... < sn = t, we have e (t) − (t) − − ≤ ≤ Xs Xs − = Xt sj−1 Xt sj = Xtn−j+1 Xtn−j , 1 j n, j j 1 − − − − − − − P ≤ ≤ wheree tn j+1e = t sj, and furthermore (Xtk = Xtk , 1 k n) = 1; now − − we conclude by the independence of increments of X. The same argument yields the stationarity of increments and identity in distribution with X, since any increment of X(t) has been identiﬁed with an increment of X of the same length. Right-continuity and left limits are also deduced, using the following remarks. First, fore a right-continuous function f with left limits the function g(s)= f(s−) is left-continuous with right limits. Second, for a left-continuous function with right limits, the function h(s)= g(t−s) is right-continuous with left limits. (b) We condition on τ to get for B = [a, b]

(τ) (τ) P ∈ 1 ∈ ∈ +1 ∩ (Xs1 A ,..., Xsm Am, τ [sm,sm ) B) sm+1 b ∧ (τ) (τ) P ∈ 1 ∈ | =e (Xes1 A ,..., Xsm Am τ = t)fτ (t)dt sm a Z ∨ sm+1 b ∧ (t) (t) P e ∈ 1 e ∈ = (Xs1 A ,..., Xsm Am)fτ (t)dt sm a Z ∨ sm+1 b ∧ P e ∈ e ∈ = (Xs1 A1,...,Xsm Am)fτ (t)dt sm a ZP ∨ ∈ ∈ ∈ ∩ = (Xs1 A1,...,Xsm Am, τ [sm,sm+1) B),

where we applied (a) to pass from X(t) to X. (τ) (c) Since (Xs )0 s τ ∼ (Xs)0 s τ , we deduce from Proposition 86 that the fol- ≤ ≤ ≤ ≤ e lowing two random variables are independent: e (τ) X = sup (X − X )= X − inf X = X − X τ τ τ s τ 0 r τ τ 0 s τ − − − r τ − ≤ ≤ ≤ ≤ (τ) e − (τ) − − − Xτ Xτ = Xτ Xτ Xτ = Xτ , − P where we also usede that (eXτ = Xτ )=1. − (d) We calculate from Corollary 89

~~e (τ) e (τ) − βX β(X Xτ ) q(Φ(q) β) E(e τ )= E e− τ − = . Φ(q)(q − φ(β)) ~~

2. (a) Let us ﬁrst show that (Tx)x 0 is a subordinator. See the proof of Proposition ≥ 90 for the stationarity of increments and the independence of two consecutive − increments. For 0 = x0 < x1 < ...< xn we now show that Txk Txk−1 , Solutions 7 – MS3b L´evy Processes and Finance – Oxford HT 2008 XLIII

~~k = 1,...,n, are independent and proceed by induction on n. Suppose in-~~

~~dependence holds for n, then the strong Markov property at Txn shows that − F Txn+1 Txn is independent of Txn , where~~

n 1 { ≤ }∩{ ≤ ≤ } − { ≤ ∧ }∈F Txn t Tx1 s1,...,Txn sn = Txk sk t t =1 k\ { ≤ }∩{ ≤ 1 ≤ }∈F and so Txn t Tx1 s ,...,Txn sn Txn . We deduce that P ≤ ≤ ≤ − ≤ (Txk sk, 1 k n, Txn+1 Txn sn+1) P ≤ ≤ ≤ ≤ = (Tx sk, 1 k n, Txn+1 xn sn+1) k − P ≤ ≤ ≤ P ≤ = (Txk sk, 1 k n) (Txn+1 xn sn+1), e − − so that (Tx1 ,...,Txn ) (and by linear transformation (Tx1 ,...,Txn Txn−1 )) is − independent of Txn+1 Txn . By the induction hypothesis, we conclude that all − variables Tx1 ,...,Txn+1 Txn are independent.

For the right-continuity let xn = x + δn ↓ x and note that the deﬁnition Tx = inf{t ≥ 0 : Xt > x} implies that there is a sequence ε ↓ 0 such that ≤ ≤ → XTx+εn > x + δn, but then Tx Tx+δn Tx + εn implies that Tx+δn Tx. The existence of left limits is trivial for increasing x 7→ Tx. Then we apply Theorem 87 to see that E {− } {− } (exp qTx1 Tx< ) = exp xΦ(q) , { ∞} where we calculate Φ(q) by inverting q = φ(Φ(q)) = c(Φ(q))α to get Φ(q) = 1/α 1/α c− q , and we identify the distribution of a stable subordinator with index 1/α. (b) We apply the transformation formula to get √ b/a b2/(2z/a) √ ab (√ab)2/(2z) faY (z)= e− = e− = f√ab(z). 2πz3/a3 2πz3 p From Exercise 2.(c) we know that fb is the distribution of Tb in the case ∼ α = 1/2, c √= 1/2, so we have just shown that aTb T√ab, or indeed, for a = d2, x = ab that

~~2 Tx ∼ d Tx/d, x ≥ 0.~~

2 Since both processes (Tx)x 0 and (d Tx/d)x 0 are subordinators, we have iden- ≥ ≥ tity of joint distributions and hence another derivation, independent of (a) of the fact that (Tx)x 0 is a stable subordinator of index 1/2. ≥ (c) We now deduce from (b) and (a) respectively that for all γ ≥ 0

~~∞ γx γTb cb√γ e− fb(x)dx = E(e− )= e− 0 Z for some c ∈ (0, ∞). XLIV Solutions – MS3b L´evy Processes and Finance – Oxford HT 2008~~

3. (a) Let us denote the increments of (Nm)m 0 and of (Rm)m 0 by Bj = Nj − Nj 1 ≥ ≥ − and Cj = Rj − Rj 1, j ≥ 1. Then we calculate − E { } E { } P E { } (exp γC1 ) = (exp γSB1 )= (B1 = n) (exp γSn ) n N X∈ n = P(B1 = n)M(γ) = G(M(γ)). n N X∈ For m ≥ 1, the analogous calculation yields

~~E(exp{γ1C1 + . . . + γmCm})~~

~~= P(B1 = n1,...,Bm = nm) n ,...,n N 1 m E { 1 1 + + − + + } X ∈ (exp γ S + . . . + γm(Sn1 ... nm Sn1 ... nm−1 ) )~~

~~n1 nm = P(B1 = n1) . . . P(Bm = nm)M(γ1) ...M(γm)~~

~~n1,...,nm N X ∈ = G(M(γ1)) ...G(M(γm)),~~

~~and we can deduce that C1,...,Cm are independent and identically distributed, as required. (b) We apply the same argument as in (a) to ﬁrst calculate~~

E { } E { } ∞ E { } (exp iλYs ) = (exp γXTs )= fTs (t) (exp iλXt )dt 0 Z ∞ {− } − = fTs (t) exp tψ(λ) dt = MTs ( ψ(λ)), 0 Z ( ) E iλX1 ψ λ where we assumed that Ts has a density fTs and that (e )= e− . Now, for r, s ≥ 0,

E(exp{iλYs + iµ(Ys+r − Ys)}) ∞ ∞ E { − } = fTs,Ts+r Tr (t, u) (exp iλXt + iµ(Xt+u Xt) )dtdu 0 0 − Z Z ∞ ∞ tψ(λ) uψ(µ) − − = fTs (t)fTr (u)e− e− dtdu = MTs ( ψ(λ))MTr ( ψ(µ)) 0 0 Z Z

~~so that we deduce that Ys and Ys+r −Ys are independent, and that Ys+r −Ys ∼ Yr. For the right-continuity of paths, note that~~

~~lim Ys+ε = lim XTs+ε = XTs = Ys ε 0 ε 0 ↓ ↓ ↓ → since Ts + δ := Ts+ε Ts and therefore XTs+δ XTs . For left limits, the same argument applies.~~

(c) (i) First note that Xt = Bt +bt is also a Lévy process, as a linear combination of two Lévy processes. Because of the path continuity of Brownian motion, we have { ≥ ∈ ∞ } XTs = s, where Ts = inf t 0 : Xt (s, ) . Solutions 7 – MS3b Lévy Processes and Finance – Oxford HT 2008 XLV

The strong Markov property of X at the stopping time Ts (ﬁrst hitting ∞ (s) − time of (s, )) yields that X =(XTs+u s)u 0 is independent of Ts and ≥ has the same distribution as X. Note further that

~~(s) (s) { ≥ (s) } Ts+r = Ts + Tr , where Tr = inf t 0 : Xt >r ,~~

~~since we can wait for X to exceed level s + r by ﬁrst waiting for level s and then waiting to increase by a further r. Now, we get from the strong Markov property that~~

~~(s) ∼ (s) − Tr Tr and Tr = Ts+r Ts is independent of Ts.~~

This yields the stationarity and independence of increments (n = 2 but can now be generalised by induction). As an increasing process, we automatically get the existence of left and right limits. Assume now that at s the path s 7→ Ts is not right-continuous. Then there is ε > 0 so that for all δ > 0

− ⇒ ≤ ≤ ≤ Ts+δ Ts >ε XTs+u s + δ for all 0 u ε ≤ ≤ ≤ but then XTs+u s for all 0 u ε, and this contradicts the deﬁnition of Ts. So, the path s 7→ Ts must be right-continuous at s. { − 1 2 } (ii) We check for Et = exp γBt 2 γ t that |F { − − 1 2 }|F E(Et s) = E(exp γBs + γ(Bt Bs) 2 γ t s) { − 1 2 } { − } = exp γBs 2 γ t E(exp γ(Bt NB) ) { − 1 2 } {− 1 2 − } = exp γBs 2 γ t exp 2 γ (t s) = Es.

~~Now, applying the Optional Stopping Theorem at Ts (which can be shown − − to be a stopping time for (Et)t 0), we get from BTs = XTs bTs = s bTs ≥ that~~

~~E E { − 1 2 } γsE {− 1 2 } 1= (ETs )= (exp γBTs 2 γ Ts )= e (exp (bγ + 2 γ )Ts ) − 1 2 and this yields the claim by choosing ρ = ρ(γ) = (bγ + 2 γ ), i.e. γ = b2 − 2ρ − b~~

p − ≥ 4. (a) By the strong Markov property, the post-Ty process Xs = XTy+s y, s 0, is F a Brownian motion independent of Ty . Note that we can write e Xt t ≤ Ty Xt = ≥ y + Xt Ty t Ty − and that X ∼ −X (by the symmetrye of the centred multivariate Normal ∼ − − distribution), so it remains to replace y + Xt Ty y Xt Ty = 2y Xt. − − Formally, wee checke for 0 = t0 < t1 <...

n tk+1 P ∈ ∈ = (Xt1 A1,...,Xtk Ak, =0 tk Xk Z − ∈ − ∈ y Xt t Ak+1,...,y Xtn t An)fTy (t)dt k+1− − n tk+1 P ∈ ∈ = (Xt1 eA1,...,Xtk Ak) e =0 tk Xk Z P − ∈ − ∈ (y Xt t Ak+1,...,y Xtn t An)fTy (t)dt k+1− − n tk+1 P ∈ ∈ = (Xt1 eA1,...,Xtk Ak) e =0 tk Xk Z P ∈ ∈ (y + Xt t Ak+1,...,y + Xtn t An)fTy (t)dt k+1− − P ∈ ∈ = (Xt1 A1,...,Xtn An). e e ≤ − ≥ (b) Note that Xt >y implies Ty < t and so, if also Xt x, then Xt∗ =2y Xt − − − 2y x. Vice versa, if Xt∗ > 2y x, then xy and so − Ty < t and hence Xt >y, but also Xt∗ =2y Xt, so Xt < x. This means that { } { − } ⇒ P P − Xt < x, Xt >y = Xt∗ > 2y x (Xt < x, Xt >y)= (Xt∗ > 2y x)

~~We apply (a) to get upon diﬀerentiation, ﬁrst w.r.t. x then y that~~

~~∂ 1 (2y − x)2 2(2y − x) (2y − x)2 f (x, y)= √ exp − = √ exp − Xt,Xt ∂x πt 2t 3 2t 2 2πt for y ∈ (0, ∞), x ∈ (−∞,y).~~

~~(c) Since Ty < t if and only if Xt >y, we ﬁrst calculate~~

~~y 2(2y − x) (2y − x)2 2 y2 f (y)= √ exp − dx = √ exp − . Xt 2πt3 2t 2πt 2t Z−∞ ~~

and then diﬀerentiate the distribution function of Ty d d d f (t) = P(T ≤ t)= P(X >y)= 2P(X >y) Ty dt y dt t dt t √ 2 d 2 y /t 1 3/2 = 2P(X1 > y/ t)= √ exp − t− dt π 2 y 2 y y2 = √ exp − , 3 2t 2πt for all t> 0.