Math 288 - Probability Theory and Stochastic Process

Home , Adapted process, Bessel process, Stochastic process

Taught by Horng-Tzer Yau Notes by Dongryul Kim Spring 2017

This course was taught by Horng-Tzer Yau. The lectures were given at MWF 12-1 in Science Center 310. The textbook was Brownian Motion, Martingales, and Stochastic Calculus by Jean-Fran¸coisLe Gall. There were 11 undergrad- uates and 12 graduate students enrolled. There were 5 problem sets and the course assistant was Robert Martinez.

Contents

1 January 23, 2017 5 1.1 Gaussians ...... 5 1.2 Gaussian vector ...... 6

2 January 25, 2017 8 2.1 Gaussian process ...... 8 2.2 Conditional expectation for Gaussian vectors ...... 8

3 January 27, 2017 10 3.1 Gaussian white noise ...... 10 3.2 Pre-Brownian motion ...... 11

4 January 30, 2017 13 4.1 Kolmogorov’s theorem ...... 13

5 February 1, 2017 15 5.1 Construction of a Brownian motion ...... 15 5.2 Wiener measure ...... 16

6 February 3, 2017 17 6.1 Blumenthal’s zero-one law ...... 17 6.2 Strong Markov property ...... 18

1 Last Update: August 27, 2018 7 February 6, 2017 19 7.1 Reﬂection principle ...... 19 7.2 Filtrations ...... 20

8 February 8, 2017 21 8.1 Stopping time of ﬁltrations ...... 21 8.2 Martingales ...... 22

9 February 10, 2017 23 9.1 Discrete martingales - Doob’s inequality ...... 23

10 February 13, 2017 25 10.1 More discrete martingales - upcrossing inequality ...... 25

11 February 15, 2017 27 11.1 Levi’s theorem ...... 27 11.2 Optional stopping for continuous martingales ...... 28

12 February 17, 2017 29

13 February 22, 2017 30 13.1 Submartingale decomposition ...... 30

14 February 24, 2017 31 14.1 Backward martingales ...... 31 14.2 Finite variance processes ...... 32

15 February 27, 2017 33 15.1 Local martingales ...... 33

16 March 1, 2017 35 16.1 Quadratic variation ...... 35

17 March 3, 2017 37 17.1 Consequences of the quadratic variation ...... 37

18 March 6, 2017 39 18.1 Bracket of local martingales ...... 39 18.2 Stochastic integration ...... 40

19 March 8, 2017 41 19.1 Elementary process ...... 41

20 March 20, 2017 43 20.1 Review ...... 43

2 21 March 22, 2017 45 21.1 Stochastic integration with respect to a Brownian motion . . . . 45 21.2 Stochastic integration with respect to local martingales . . . . . 45

22 March 24, 2017 47 22.1 Dominated convergence for semi-martingales ...... 47 22.2 Itˆo’sformula ...... 48

23 March 27, 2017 50 23.1 Application of Itˆocalculus ...... 50

24 March 29, 2017 52 24.1 Burkholder–Davis–Gundy inequality ...... 52

25 March 31, 2017 54 25.1 Representation of martingales ...... 54 25.2 Girsanov’s theorem ...... 55

26 April 3, 2017 56 26.1 Cameron–Martin formula ...... 56

27 April 5, 2017 58 27.1 Application to the Dirichlet problem ...... 58 27.2 Stochastic diﬀerential equations ...... 59

28 April 7, 2017 61 28.1 Existence and uniqueness in the Lipschitz case ...... 61 28.2 Kolomogorov forward and backward equations ...... 62

29 April 10, 2017 64 29.1 Girsanov’s formula as an SDE ...... 65 29.2 Ornstein–Uhlenbeck process ...... 65 29.3 Geometric Brownian motion ...... 66

30 April 12, 2017 67 30.1 Bessel process ...... 67

31 April 14, 2017 69 31.1 Maximum principle ...... 69 31.2 2-dimensional Brownian motion ...... 70

32 April 17, 2017 71 32.1 More on 2-dimensional Brownian motion ...... 71

33 April 19, 2017 73 33.1 Feynman–Kac formula ...... 73 33.2 Girsanov formula again ...... 74

3 34 April 21, 2017 76 34.1 Kipnis–Varadhan cutoﬀ lemma ...... 76

35 April 24, 2017 78 35.1 Final review ...... 78

4 Math 288 Notes 5

1 January 23, 2017

The textbook is J. F. Le Gall, Brownian Motion, Martingales, and Stochastic Calculus. n A Brownian motion is a random function X(t): R≥0 → R . This is a model for the movement of one particle among many in “billiard” dymanics. For instance, if there are many particles in box and they bump into each other. Robert Brown gave a description in the 19th century. X(t) is supposed to describe the trajectory of one particle, and it is random. To simplify this, we assume X(t) is a random walk. For instance, let

N ( X 1 with probability 1/2 SN = Xi,Xi = −1 with probability 1/2. i=1 For non-integer values, you interpolate it linearly and let

bNtc X SN (t) = Xi + (Nt − bNtc)XbNtc+1. i=1

2 Then E(SN ) = N. As we let N → ∞, we have three properties: 2 SN (t) • lim E √ = t N→∞ N E(SN (t)SN (s)) • lim √ 2 = t ∧ s = min(t, s) N→∞ N √ • SN (t)/ N → N(0, t) This is the model we want.

1.1 Gaussians Deﬁnition 1.1. X is a standard Gaussian random variable if

1 Z 2 P (X ∈ A) = √ e−x /2dx 2π A for a Borel set A ⊆ R. For a complex variable z ∈ C, Z zX 1 zx −x2/2 z2/2 E(e ) = √ e e dx = e . 2π This is the moment generating function. When z = iξ, we have

∞ 2 k 2 X (−ξ ) 2 (iξ) = e−ξ /2 = 1 + iξ X + X2 + ··· . k! E 2 E k=0 Math 288 Notes 6

From this we can read oﬀ the moments of the Gaussian EXn. We have (2n)! X2n = , X2n+1 = 0. E 2nn! E Given X ∼ N (0, 1), we have Y = σX + m ∼ N (m, σ2). Its moment generating function is given by

zY imξ 2 2 Ee = e − σ ξ /2.

2 2 Also, if Y1 ∼ N (m1, σ1) and Y2 ∼ N (m2, σ2) and they are independent, then

z(Y +Y ) zY zY i(m +m )ξ−(σ2+σ2)ξ2/2 Ee 1 2 = Ee 1 Ee 2 = e 1 2 1 2 .

2 2 So Y1 + Y2 ∼ N (m1 + m2, σ1 + σ2). 2 2 Proposition 1.2. If Xn ∼ N (mn, σn) and Xn → X in L , then 2 2 (i) X is Gaussian with mean m = limn→∞ mn and variance σ = limn→∞ σn. p (ii) Xn → X in probability and Xn → X in L for p ≥ 1.

2 Proof. (i) By deﬁnition, E(Xn − X) → 0 and so |EXn − EX| ≤ (E|Xn − X|2)1/2 → 0. So EX = m. Similarly because L2 is complete, we have Var X = σ2. Now the characteristic function of X is

iξX iξX im ξ−σ2 ξ2/2 imξ−σ2ξ2/2 e = lim e n = lim e n n = e . E n→∞ n→∞

So X ∼ N (m, σ2). (ii) Because EX2p can be expressed in terms of the mean and variance, we 2p 2p p have supn E|Xn| < ∞ and so supn E|Xn − X| < ∞. Deﬁne Yn = |Xn − X| . 2 Then Yn is bounded in L and hence is uniformly integrable. Also it converges 1 to 0 in probability. It follows that Yn → 0 in L , which means that Xn → X in p L . It then follows that Xn → X in probability.

Deﬁnition 1.3. We say that Xn is uniformly integrable if for every > 0 there exists a K such that supn E[|Xn| · 1|Xn|>K ] < . 1 Proposition 1.4. A sequence of random variables Xn converge to X in L if and only if Xn are uniformly integrable and Xn → X in probability.

1.2 Gaussian vector

Deﬁnition 1.5. A vector X ∈ Rn is a Gaussian vector if for every n ∈ Rn, hu, Xi is a Gaussian random variable.

Example 1.6. Deﬁne X1 ∼ N (0, 1) and = ±1 with probability 1/2 each. Let X2 = X1. Then X = (X1,X2) is not a Gaussian vector. Math 288 Notes 7

Note that u 7→ E[u · X] is a linear map and so E[u · X] = u · mX for some n Pn mX ∈ R . We call mX the mean of X. If X = i=1 Xiei for an orthonormal n normal basis ei of R , then the mean of X is

N X mX = (EXi)ei. i=1 Likewise, the map u 7→ Var u · X is a quadratic form and so Var uX = hu, qX ui for some matrix qX . We all qX the covariant matrix of X. We write qX (n) = hu, qX ui. Then the characteristic function of X is

zi(u·X) 2 Ee = exp(izmX − z qX (u)/2).

Proposition 1.7. If Cov(Xj,Xk)1≤j,k≤n is diagonal, and X = (X1,...,Xn) is a Gaussian vector, then Xis are independent. Proof. First check that

n X qX (u) = uiuj Cov(Xi,Xj). i,j=1

Then

n 1 1 X exp(iu · X) = exp − q (u) = exp − u2 Var(X ) E 2 X 2 i i i=1 n Y 1 = exp − u2 Var(X ) . 2 i i i=1 This gives a full description of X.

Theorem 1.8. Let γ : Rn → Rn be a positive deﬁnite symmetric n × n matrix. Then there exists a Gaussian vector X with Cov X = γ.

Proof. Diagonalize γ and let λj ≥ 0 be the eigenvalues. Choose v1, . . . , vn an n orthonormal basis of eigenvectors of R and let wj = λjvj = γvj. Choose Yj to be Gaussians N (0, 1). Then the vector

n X X = λjvjYj j=1 is a Gaussian vector with the right covariance. Math 288 Notes 8

2 January 25, 2017

We start with a probability space (Ω, F,P ) and joint Gaussians (or a Gaussian vector) X1,...,Xd. We have

Cov(Xj,Xk) = E[XjXk] − E[Xj]E[Xk], γX = (Cov(Xj,Xk))jk = C. Then the probability distribution is given by

1 −hx,C−1xi/2 PX (x) = √ e (2π)d/2 det C where C > 0 and the characteristic function is

itx −ht,Cti/2 E[e ] = e .

2.1 Gaussian process A (centered) Gaussian space is a closed subspace of L2(Ω, F,P ) containing only Gaussian random variables.

Deﬁnition 2.1. A Gaussian process indexed by T (∈ R+) is one such that

{Xt1 ,...,Xtk } is always a Gaussian vector. Deﬁnition 2.2. For a collection S of random variables, we denote by σ(S) the smallest σ-algebra on Ω such that all variables in S are measurable.

Proposition 2.3. Suppose H1,H2 ⊆ H where H is the centered Gaussian space. Then H1 ⊥ H2 if and only if H1 and H2 are independent, i.e., the generated σ-algebras σ(H1) and σ(H2) are independent.

Example 2.4. If we take X1 = X to be a Gaussian and X2 = X where = ±1 2 with probability 1/2 each, then E[X1X2] = E[X ] = 0. That is, X1 and X2 are uncorrelated. But X1 and X2 are dependent. This is because {X1,X2} are not jointly Gaussian.

2.2 Conditional expectation for Gaussian vectors Lt X be a random variable in a Gaussian space H. If H is ﬁnite dimensional P with (jointly Gaussian) basis X1,...,Xd, then we may write X = j ajXj. Let K ⊆ H be a closed subspace. We want to look at the conditional expectation

E[X|σ(K)] = E[X|K], where σ(K) is the σ-algebra generated by elements in K. Deﬁnition 2.5. The conditional expectation is deﬁned as E[X|K] = ϕ(K), the measurable function with respect to K so that for any function g(K) we have

E[Xg(K)] = E[ϕ(K)g(K)], or alternatively, E[(X − ϕ(K))g(K)] = 0, i.e., ϕ(K) is the L2-orthogonal projection of X to the space σ(K). Math 288 Notes 9

We also write E[X|σ(K)] = pK (X). Let Y = X − pK (X). Note that Y is a Gaussian that is orthogonal to K. Then

2 2 Z e−y /2σ E[f(X)|σ(K)] = E[f(Y + pK (X))|σ(K)] = f(y + pK (X)) √ 2πσ 2 2 Z e−(y−pK (X)) /2σ = f(y) √ dy, 2πσ

2 where σ is the variance of Y = X − pK (X). We remark that if X is a Gaussian with E[X] = 0 and E[X2] = σ2, then

2 0 E[Xf(X)] = σ E[f (X)] by integration by parts. It follows that E[X2n] corresponds to the number of ways to pair elements in a set of size 2n. There is a more general interpretation. In general, we have

X E[Y f(X1,...,Xd)] = Cov(Y,Xj)E[∂jf]. j Math 288 Notes 10

3 January 27, 2017

For a Gaussian process Xs indexed by s ∈ R+, let us deﬁne

Γ(s, t) = Cov(Xs,Xt) ≥ 0, where being positive semidefinite means Z ZZ Var α(s)Xsds = α(s)Γ(s, t)α(t)dsdt ≥ 0 for any α. Theorem 3.1. Suppose Γ(s, t) is a positive semidefinite function. Then there exists a Gaussian process with Γ as the covariant matrix. Theorem 3.2 (Kolmogorov extension theorem). Given a family of consistent finite dimensional distributions, there exists a measure with this family of finite dimensional distributions as the finite dimensional marginals.

Consistency means that if we are given the distribution of (Xs1 ,Xs2 ,Xs3 ), then the distribution of (Xs1 ,Xs3 ) must be the integration. Kolmogorov’s theorem is giving you an abstract construction of the measure. So we are not going to be worry too much about the existence of these inﬁnite dimensional measure spaces. Example 3.3. Suppose µ is a probability measure on R. Then Z Γ(s, t) = eiξ(s−t)µ(dξ) is positive semideﬁnite. This is because ZZZ Z Z Z α(s)eiξ(s−t)α(t)dsdtµ(dξ) = α(s)eiξs α(t)e−iξt µ(dξ) ≥ 0.

3.1 Gaussian white noise Suppose µ is a σ-finite measure. Definition 3.4. We call a Gaussian space G a Gaussian white noise with intensity µ if Z E(G(f)G(g)) = fgdµ for every f, g ∈ L2(µ). Informally, this is the same as saying E[G(x)G(y)] = δ(x − y) with the definition G(x) = G(δx). Then Z Z E[G(f)G(g)] = E G(x)f(x)dx G(y)g(y)dy ZZ Z = f(x)E[G(x)G(y)]g(y)dx dy = f(x)g(x)dx. Math 288 Notes 11

Theorem 3.5. For every µ a σ-ﬁnite measure there exists a Gaussian white noise with intensity µ. Proof. Assume L2(µ) is separable. There then exists an orthonormal basis in 2 P∞ L (µ). For any f there exists a presentation f = i=1 αifi. Now deﬁne ∞ X G(f) = αiXi, i=1 where X1,X2,...are independent in N (0, 1). Then we obvious have the desired condition.

3.2 Pre-Brownian motion + Deﬁnition 3.6. Let G be a Gaussian white noise on (R , dx). Deﬁne Bt = G(1[0,t]). Then Bt is a pre-Brownian motion.

Proposition 3.7. The covariance matrix is E[BtBs] = s ∧ t. (Here, s ∧ t denotes min(s, t).) Proof. We have

E[BtBs] = E[G([0, t])G([0, s])] = s ∧ t.

Because Bt = G([0, t]), we may formally write dBt = G(dt). Then we can also write Z G(f) = f(s)dBs.

We can also think Z t G([0, t]) = dBs = Bt − B0 = Bt. 0

Roughly speaking, dBt is the white noise. This idea can be rigorously formu- lated as that if 0 = t0 < t1 < ··· < tn then Bt1 −Bt0 ,Bt2 −Bt1 ,...,Btn −Btn−1 are independent Gaussian. Proposition 3.8. The following statements are equivalent:

(i) (Xt) is a Brownian motion.

(ii) (Xt) is a centered Gaussian with covariance s ∧ t.

(iii) (Xt − Xs) is independent of σ(Xr, r ≤ s) and Xt − Xs ∼ N (0, t − s).

(iv) {Xti − Xti−1 } are independent Gaussian with variance ti − ti−1. Proof. (ii) ⇒ (iii) We have

E[(Bt − Bs)Br] = t ∧ r − s ∧ r = 0 for r ≤ s. The other implications are obvious. Math 288 Notes 12

Note that the probability density of Bt1 ,...,Btn is given by

1 2 Y −(xi−xi−1) /2(ti−ti−1) p(x1, . . . , xn) = e . n/2p (2π) t1(t2 − t1) ··· i It is important to understand this as a Wiener measure. Math 288 Notes 13

4 January 30, 2017

Let me remind you of some notions. A Gaussian space is a closed linear subspace 2 of L (Ω,P ). A random process is a sequence (Xt)t≥0 of random variables. A

Gaussian process is a process such that (Xt1 ,...,Xtk ) is Gaussian for any choice 2 2 t1, . . . , tk. A Gaussian white noise with intensity µ is a map G : L (µ) → L (P ) R such that G(f) is Gaussian and E[G(f)G(g)] = fgdµ. A pre-Brownian motion is a process with Bt = G(1[0,t]). Sometimes people write Bt(ω). In this case, this ω is an element of the probability space Ω. So then Bt(ω) is actually something concrete map.

4.1 Kolmogorov’s theorem

Theorem 4.1 (Kolmogorov). Let I = [0, 1] and (Xt)t∈I be a process. Suppose q 1+ for all t, s ∈ I, E[|Xt − Xs| ] ≤ C|t − s| . Then for all 0 < α < /q there exists a “modiﬁcation” X˜ of X such that

α |X˜t(ω) − X˜s(ω)| ≤ Cα(ω)|s − t| almost surely. Here we say that X˜ is a modiﬁcation of X if for every t ∈ T ,

P (X˜t = Xt) = 1.

Definition 4.2. X˜ and X are indistinguishable if there exists a set N of probability zero such that for all t and ω ∈ N, X˜t(ω) = Xt(ω). There is a slight difference, in the case of indistinguishability, N is some set that does not depend on t. That is, indistinguishable implies being a modification. If Xt and X˜t have continuous sample paths with probability 1, the converse holds. But nobody worries about these. For a pre-Brownian motion, we have Xt − Xs ∼ N (0, t − s) and so

q q/2 E[|Xt − Xs| ] = Cα|t − s| . Then letting = q/2 − 1 and α < /q = 1/2 − 1/q, we get α % 1/2 as q → ∞. That is, t → Xt(ω) is H¨oldercontinuous of order . Corollary 4.3. A Brownian motion sample path is H¨olderof order α < 1/2.

n n n n Proof of Theorem 4.1. Let Dn = {0/2 , 1/2 ,..., (2 − 1)/2 }. We have

2n [ −nα n q nαq P (|Xi/2n − X(i−1)/2n | ≥ 2 ) ≤ 2 E|Xi/2n − X(i−1)/2n | 2 i=1 ≤ C2nqα−(1+)n2n = C2nqα−n. Math 288 Notes 14

So − qα > 0 implies X [ −nα P (|Xi/2n − X(i−1)/2n | ≥ 2 ) < ∞. n i

Borel–Cantelli then implies that there exists a set N with P (N) = 0 such c that for every ω ∈ N there exists an n0(ω) such that

−nα Xi/2n (ω) − X(i−1)/2n (ω) < 2 for all n > n0(ω). Now

Xi/2n (ω) − X(i−1)/2n (ω) Kα(ω) = sup sup −nα < ∞ n≥1 i 2 almost surely.

We will ﬁnish the proof next time. Math 288 Notes 15

5 February 1, 2017

We start with Kolmogorov’s theorem.

5.1 Construction of a Brownian motion Proof of Theorem 4.1. We have proved that

|Xi/2n − X(i−1)/2n | Kα(ω) = sup sup −nα < ∞ n≥1 i 2 almost surely using Markov’s inequality and the Borel–Cantelli lemma. We look at the set n i o D = : 0 ≤ i ≤ 2n, n = 1,... . 2n

Suppose we have a functionf : D → R such that |f(i/2n) − f((i − 1)/2n)| ≤ K2−nα. This condition implies 2K |f(s) − f(t)| ≤ |t − s|α 1 − 2−α for t, s ∈ D. You can prove this using successive approximations. Now deﬁne X˜t(ω) = lims→t,s∈D Xs(ω) if Kα(ω) < ∞ and otherwise just assign an arbitrary value. You can check that

α |X˜t(ω) − X˜s(ω)| ≤ Cω|t − s| for some constant Cω depending only on ω, i.e., X˜t is H¨oldercontinuous of order α. Finally we see that

P (|Xt − X˜t| > ) = lim P (|Xt − Xs| > ) = 0 s→t,s∈D and so P (Xt = X˜t) = 1. This gives a construction of a Brownian motion, which is the modification of the pre-Brownian motion. We first constructed a Gaussian white noise in an abstract space, then constructed a pre-Brownian motion by Xt = G([0, t]). Then by Kolmogorov’s theorem defined a Brownian motion. Then a sample path is Höldercontinuous up to order 1/2. There is another construction of a Brownian motion, due to Lévy. This construction is Exercise 1.18. Math 288 Notes 16

5.2 Wiener measure

The space of continuous paths C(R+, R) can be given a structure of a σ-algebra by declaring that w 7→ w(t) is measurable for all t. Now there is a measurable map

Ω → C(R+, R), w 7→ (t 7→ Bt(ω)). The Wiener measure is the push-forward of this measure. Don’t think too hard about what this looks like.

Theorem 5.1 (Blumenthal’s zero-one law). Let F = σ(Bs, s ≤ t) and let T F0+ = t>0 Ft. Then F0+ is trivial, i.e., for any A ∈ F0+ , either P (A) = 0 or P (A) = 1. T Example 5.2. Let A = n{sup0≤s≤n−1 Bs > 0}. Then P (A) = 0 or P (A) = 1. The right answer is P (A) = 1. This is because 1 P (A) = lim P sup Bs > 0 ≥ lim P (Bn−1 > 0) = . n→∞ 0≤s≤n−1 n→∞ 2

Sketch of proof. Take a A ∈ F0+ and let g be a bounded measurable function. If I can show that

E[1Ag(Bt1 ,...,Btk )] = E[1A]E[g(Bt1 ,...,Btk )] then F0+ ⊥ σ(Bt, t > 0). Taking the limit, we see that F0+ is independent of itself. Then we get the result. Math 288 Notes 17

6 February 3, 2017

Using Kolmogorov’s theorem we defined a Brownian motion by modifying the pre-Brownian motion. A sample path of a Brownian motion is then Hölder continuous with order α for all α < 1/2. Using it we defined the Wiener measure on C(R+, R).

6.1 Blumenthal’s zero-one law Let us deﬁne + \ \ F0 = Ft = σ(Bs, s ≤ t). t>0 t>0

+ Theorem 6.1 (Blumenthal). The σ-algebra F0 is independent of itself. In + other words, P (A) = 0 or 1 for every A ∈ F0 . + Proof. Take an arbitrary bounded continuous function g. For an A ∈ F0 , we claim that

E[1Ag(Bt1 ,...,Btk )] = E[1A]E[g(Bt1 ,...,Btk )], for 0 < t1 < ··· < tk. Let us only work with the case k = 1 for simplicity. Note that Bt1 − B is independent of F. So we have

[1Ag(Bt )] = lim [1Ag(Bt − B)] = lim P (A) [g(Bt )] = [1A] [g(Bt )]. E 1 →0 E 1 →0 E 1 E E 1

+ + This implies that F0 is orthogonal to σ(Bt, t > 0). So F0 ⊥ Ft, but + + + F0 ⊆ Ft. So F0 ⊥ F0 .

Proposition 6.2. Deﬁne Ta = inf{t ≥ 0,Bt = a}. Then almost surely for all a ∈ R, Ta < ∞. That is, lim supt→∞ Bt = ∞ and lim inft∈∞ Bt = ∞. Proof. We have 1 = P sup Bs > 0 = lim P sup Bs > δ . 0≤s≤1 δ&0 0≤s≤1

−1 But by the scaling property of the Brownian motion, Bt ∼ λ Bλ2t, we can write P sup Bs > δ = P sup Bs > 1 . 0≤s≤1 0≤s≤δ−2

It follows that P (sup0≤s Bs > 1) = 1. Math 288 Notes 18

6.2 Strong Markov property Deﬁnition 6.3. A random variable T ∈ [0, ∞] is a stopping time if the set {T ≤ t} is in Ft. Intuitively, a stopping time cannot depend on the future.

Example 6.4. The random variable Ta = inf{t : Bt = a} is a stopping time. Definition 6.5. We define the σ-field

FT = {A : A ∩ {T ≤ t} ∈ Ft for all t ≥ 0} generated by T . Roughly speaking, this is all the information collected up to the stopping time.

(T ) Theorem 6.6. Deﬁne Bt = 1{T <∞} · (BT +t − BT ). Then under the measure (T ) P (· : T < ∞), the process Bt is a Brownian motion independent of FT . We will prove this next time. Math 288 Notes 19

7 February 6, 2017

(T ) Theorem 7.1. Suppose P (T < ∞) > 0. Deﬁne Bt = 1T <∞(BT +t − BT ). (T ) Then respect to P (· : T < ∞), the process Bt is a Brownian motion independent of FT .

Here, A ∈ FT is deﬁned as A ∩ {T ≤ t} ∈ Ft.

Proof. Let A ∈ FT . We want to show that for any continuous bounded function F ,

(T ) (T ) (T ) (T ) E[1AF (Bt1 ,...,Btp )] = E[1A]E[F (Bt1 ,...,Btp ].

2 2 (T ) ([T ]n) Deﬁne [t]n = min{k/n : k/n > t}. We have F (Bt ) = limn→∞ F (Bt ). Then

(T ) ([Tn]) [1AF (Bt )] = lim [1AF (Bt )] E n→∞ E ∞ X (k/2n) = lim [1A1{(k−1)2−n

Here we are using the fact that 1A1{(k−1)2−n

7.1 Reﬂection principle

Let St = sups≤t Bs.

Theorem 7.2 (Reflection principle). We have P (St ≥ a, Bt ≤ b) = P (Bt ≥ 2a − b). The idea is that you can reflect with respect to the first tie you meet a.

(T ) Proof. Let Ta = ﬁrst time St ≥ a. Then Bt is independent of FTa by the strong Markov property. This is why reﬂection works.

Proposition 7.3. The probability density of (St = a, Bt = b) is

2(2a − b) −(2a−b)2/2t √ e 1(a>0,b

The probability density of Ta is

a 2 f(t) = √ e−a /2t. 2πt3 Math 288 Notes 20

Proof. Exercise. We have

P (Ta ≤ t) = P (St ≥ a) = P (St ≥ a, Bt ≤ a) + P (St ≥ a, Bt ≥ a) 2 2 = 2P (Bt ≥ a) = P (tB1 ≥ a ).

Then compute the density. Proposition 7.4. Let 0 < tn < ··· < tn = t. Then 0 pn

pn X 2 (B n − B n ) → t ti ti−1 i=1

2 n n in L as n → ∞ and maxi|ti − ti−1| → 0.

R t 2 This is saying that 0 (dBt) → t. Proof. We have

n pn 2 p 2 X 2 X 2 n n (B n − B n ) − t = (B n − B n ) − (t − t ) E ti ti−1 E ti ti−1 i i−1 i=1 i=1 n p pn X 2 n n 2 X n n 2 = (B n − B n ) − (t − t ) ≤ C (t − t ) → 0, E ti ti−1 i i−1 i i−1 i=1 i=1

2 n n since the (B n − B n ) − (t − t ) are independent with expectation 0. ti ti−1 i i−1

7.2 Filtrations Now we are going to talk about martingales.

Definition 7.5. A filtration is an increasing sequence Ft. Define \ Ft+ = Fs ⊇ Ft. s>t

If Ft+ = Ft, then Ft is called right continuous. By deﬁnition, Ft+ is right continuous.

If you work with Ft+ , then you are going to be ﬁne. Math 288 Notes 21

8 February 8, 2017

The first homework is 1.18, 2.29, 2.31, 2.32. Last time we talked about the strong Markov property for Brownian motion. This implies the reflection principle, and so we were able to effectively compute the distributions of max0≤s≤t Bs.

8.1 Stopping time of filtrations ∞ + A filtration consists of an increasing sequence (Ft)t=0, and we define Ft = T s>t Fs = Gt.

Deﬁnition 8.1. Xt is adapted with respect to (Ft) if Xt is Ft measurable. Xt is progressively measurable if (ω, s) 7→ Xs(ω) is measurable with respect to Ft × B([0, t]).

Proposition 8.2. If the sample path is right continuous almost surely, then Xt is adapted if and only if Xt is progressively measurable. Proof. It is obvious that progressively measurable implies adapted. Assume that it is adapted. Fix a time t. Deﬁne

h(k − 1)t kt Xn = X for s ∈ , . s kt/n n n

n n Then Xs is now discrete, and limn→∞ Xs = Xs almost surely. Clearly (ω, s) → n Xs (ω) is measurable with respect to Ft × B([0, t]).

A stopping time is a random variable T such that {T ≤ t} ∈ Ft. We similarly deﬁne FT = {A : A ∩ {T ≤ t} ∈ Ft}.

+ Proposition 8.3. T is a stopping time with respect Gt = Ft if and only if {T < t} ∈ Ft for all t > 0. T S Proof. We have {T ≤ t} = s

We deﬁne Gt = {A : A ∩ {T ≤ t} ∈ FT } = FT + . Here are some facts: • If S and T are stopping times, then S ∨ T and S ∧ T are also stopping times.

• If Sn is a sequence of stopping times and increasing, then S = limn→∞ Sn is also a stopping time.

• If Sn is a decreasing sequence of stopping times converging to S, then S is a stopping time with respect to G. Math 288 Notes 22

8.2 Martingales

Deﬁnition 8.4. A process Xt is a martingale if E[|Xt|] < ∞ and E[Xt|Fs] = Xs for every t ≥ s. If E[Xt|Fs] ≥ Xs, then it is called a submartingale. 2 2 Example 8.5. Bt is a martingale. But Bt is not a martingale because E[Bt |Fs] = 2 2 (t − s) + Bs . But Bt − t is a martingale.

θB2−θ2/2t Example 8.6. The process e t is a martingale for all θ. This is because

2 2 2 2 2 2 θB −θ t/2 θ(Bt−Bs) +2θ(Bt−Bs)Bs θ B −θ t/2 E[e t |Fs] = E[e |Fs]e s . Example 8.7. For any f(s) ∈ L2[0, ∞), the process

Z t X f(s)dBs = lim f(si−1)(Bs − Bs ) n→∞ i i−1 0 i is a martingale. Note that this is well-deﬁned for simple functions f and you can extend for general f ∈ L2 because

Z t 2 Z t 2 E f(s)dBs = f(s) ds 0 0 if f is simple. Furthermore,

Z t 2 Z t 2 f(s)dBs − f(s) ds 0 0 is also a martingale. Math 288 Notes 23

9 February 10, 2017

Chapter VII of Shirayayev, Probability is a good reference for discrete martingales.

9.1 Discrete martingales - Doob’s inequality We are now going to look at martingales with discrete time.

Theorem 9.1 (Optional stopping theorem). Suppose Xn is a submartingale and σ ≤ τ < ∞ (almost surely) are two stopping times. If E[|Xn|1τ>n] → 0 and E[|Xτ | + |Xσ|] < ∞, then E[Xτ |Fσ] ≥ Xσ. If Xn is a martingale, then E[Xτ |Fσ] = Xσ.

Proof. To prove E[Xτ |Fσ] ≥ Xσ, it suﬃces to prove E[Xτ A] ≥ E[XσA] for all A ∈ Fσ. In this case, A will look like ∞ [ A = Ak, where Ak = A ∩ 1σ=k ∈ Fk. k=1

So it suffices to check for Am where m is fixed. Define τn = τ ∧ n and σn = σ ∧ n. We automatically have

[XσAm] = lim [Xσ Am], E n→∞ E n because the sequence on the right side is eventually constant. Also

E[Xτ Am] = E[Xτ 1τ>nAm] + E[Xτ 1τ≤nAm] = E[Xτ 1τ>nAm] + E[Xτn 1τ≤nAm]

= E[Xτ 1τ>nAm] − E[Xn1τ>nAm] + E[Xτn Am]. Here as n → ∞, the ﬁrst two terms goes to 0 because of the condition. It is also clear that E[Xσn Am] ≤ E[Xτn Am], because we can only care about ﬁnitely many outcomes of σ and τ. So

[XσAm] = lim [Xσ Am] ≤ lim [Xτ Am] = [Xτ Am]. E n→∞ E n n→∞ E n E

Theorem 9.2 (Doob inequality). Let Xj be a nonnegative submartingale, and let Mn = max0≤j≤n Xj.

(1) For a > 0, aP (Mn > a) ≤ E[1Mn≥aXn]. (2) for p > 1, p p kM kp ≤ [|X |p]. n p p − 1 E n

Proof. (1) Let τ = min{j ≤ n : Xj ≥ a}. This is a stopping time, and τ = n if Mn < a. Then

E[Xn] ≥ E[Xτ ] = E[1Mn≥aXτ ] + E[1Mn

≥ aP (Mn ≥ a) + E[1Mn

Lemma 9.3. If λP (Y ≥ λ) ≤ E[X1Y ≥λ] for X,Y ≥ 0, then p p [Y p] ≤ [Xp]. E p − 1 E Proof. We have Z ∞ Z ∞ p p−1 p−2 E[Y ] = pλ P (Y ≥ λ)dλ ≤ E Xpλ 1Y ≥λdλ 0 0 h p i p = X Y p−1 ≤ kXk kY kp−1. E p − 1 p − 1 p p

Theorem 9.4 (Doop upcrossing inequality). Suppose Xn is a submartingale. Let β(a, b) be the number of upcrossings, i.e., the number times the martingale goes from below a to above b. Then

[max(X − a, 0)] [β(a, b)] ≤ E n . E b − a Note that if X is a submartingale, then max(X − a, 0) is a submartingale. This is because

E[f(Xn)|Fn−1] ≥ f(E[Xn|Fn−1]) ≥ f(Xn−1). Math 288 Notes 25

10 February 13, 2017 10.1 More discrete martingales - upcrossing inequality In the optional stopping theorem, we needed the stopping time τ to satisfy the inequality E[Xn1τ>n] → 0 as n → ∞.

Proposition 10.1. If Xn is uniformly integrable, i.e., Z lim sup |Xn| = 0, M→∞ n |Xn|≥M then E[Xn1τ>n] → 0. Proof. We have Z Z lim |Xn|1τ>n ≤ lim M1τ>n + lim |Xn|1|X |≥M 1τ>n n→∞ n→∞ n→∞ n

≤ lim |Xn|1|X |≥M = 0. n→∞ n

Corollary 10.2. Let Xn be a submartingale. Suppose (1) E[|X0|] < ∞, (2) E[τ] < ∞, and (3) supn E[|Xn+1 − Xn||Fn] ≤ C. Then E[Xτ ] ≥ E[X0] and E[Xτ |F0] ≥ X0. Proof. We want to check the three conditions in the optional stopping inequality. We have ∞ ∞ X X E[|Xτ − Xm|] ≤ E[|Xn+1 − Xn|1τ≥n+1] = E[1τ≥n+1E[|Xn+1 − Xn|Fn|] n=m n=m ∞ X ≤ C E[1τ≥n+1] ≤ CE[τ] < ∞. n=m

This shows E[|Xτ |] < ∞. We also have

lim [|Xn|1τ≥n+1] ≤ lim [|Xτ − Xn|] + lim [|Xτ |1τ≥n+1] = 0. n→∞ E n→∞ E n→∞ E Theorem 10.3 (Upcrossing inequality). For a < b, let β(a, b) be the number n of upcrossings in (Xj)j=1. Then [max(X − a, 0)] [β(a, b)] ≤ E n . E b − a

Proof. Replace Xj by max(Xj − a, 0). This is still a submartingale, and so we can assume Xi ≥ 0 and a = 0. Deﬁne a random variable φi such that φi = 1 if there exists a k ≤ i such that Xk = 0 and Xj < b for all k ≤ j ≤ i. Then we get n−1 −1 X −1 X E[β(a, b)] ≤ b E φi(Xi+1 − Xi) = b E[φiE[Xi+1 − Xi|Fi]] i=1 i

−1 X −1 ≤ b E[Xi+1 − Xi] = b E[Xn], i Math 288 Notes 26

because E[Xi+1 − Xi|Fi] ≥ 0 and so we can replace φi by 1.

Theorem 10.4 (Doob’s convergence theorem). If Xn is a submartingale with supi E[|Xi|] < ∞, then Xn converges almost surely to some X. Proof. We have [ P (lim sup Xn > lim inf Xn) = P (lim sup Xn > b > a > lim inf Xn) = 0 a

1 Note that Xn → X in L may fail. For instance, take Yi be independent Bernoulli with {0, 2} and Xn = Y1 ··· Yn → 0. Then EXn = 1.

Proposition 10.5. (1) Suppose Xn → X almost surely and {Xn} is uniformly 1 integrable. Then Xn → X in L . (2) Suppose Xn ≥ 0, Xn → X almost surely, and EXn = EX. Then Xn → X in L1.

1 Theorem 10.6 (Levi). Suppose X ∈ L be a random variable and let Fn be a S 1 ﬁltration with F∞ = Fn. Then E[X|Fn] → E[X|F∞] almost surely and L . Conversely, if {Xn} is uniformly integrable and F∞ is the full σ-algebra F, then 1 there exists a X ∈ L such that Xn = E[X|Fn]. Math 288 Notes 27

11 February 15, 2017

So far we have looked that Doob’s inequality, the optional stopping time, and Doob’s upcrossing inequality. There is a supermartingale version of the upcrossing inequality: Theorem 11.1. For a discrete supermartingale X, the number of upcrossings n Mab has expectation value hmax{0, a − x )}i [M n (x)] ≤ n . E ab E b − a Proof. Let

τ1 = n ∧ min{j > 0 : xj ≤ a}, τ2 = n ∧ min{j > τ1 : xj ≥ b}, and so forth. Then all τk are stopping times. Let

D = (Xτ2 − Xτ−1) + (Xτ4 − Xτ3 ) + ··· . n Then (b − a)Mab + R ≤ D, where R = 0 if the tail is a down-crossing and

R = Xn−a otherwise. Because X is a supermartinagle, we have E[Xτ2 −Xτ1 ] ≤ 0 and so n (b − a)EMab ≤ −ER ≤ E[max(0, a − Xn)].

11.1 Levi’s theorem 1 Theorem 11.2 (Levi). Suppose X ∈ L and Fn be a ﬁltration, with F∞ = S 1 Fn. Then E[X|Fn] → E[X|F∞] in L and almost surely. Conversely, if a martingale Xn is uniformly integrable, then there is an X such that E[X|Fn] = Xn.

Proof. Let Xn = E[X|Fn]. We claim that Xn is uniformly integrable, and you can check this. Also Xn is a martingale. So the convergence theorem tells 1 us that Xn → X∞ almost surely and in L . Now the question is whether X∞ = E[X|F∞]. Because Xn is a martingale, for any A ∈ Fn and m ≥ n, Z Z Xndµ = Xmdµ. A A Taking m → ∞, we get Z Z Z Xdµ = Xndµ = X∞dµ A A A 1 because Xm → X∞ in L . So this is true for all A ∈ F∞. So X∞ = E[X|F∞]. Now let us do the second half. Suppose Xn is uniformly integrable. Then 1 Xn → X for some X almost surely and in L . So basically by the same argu- ment, Z Z Z Xndµ = lim Xmdµ = Xdµ. A m→∞ A A

This shows that Xn = E[X|Fn]. Math 288 Notes 28

So essentially all martingales come from taking a ﬁltration and looking at the conditional expectations.

11.2 Optional stopping for continuous martingales

Theorem 11.3. Suppose Xt is a submartingale with respect to a right continuous ﬁltration. If σ ≤ τ ≤ C, then E[Xτ |Fσ] ≥ Xσ.

Proof. Let τn = (bnτc + 1)/n. Then τn is a stopping time and Fτn ↓ Fτ as n → ∞, because F is right continuous. By the discrete version of optional stopping time, we see that Xτn → X implies that E[XC |Fτn ] ≥ Xτn . Then 1 Xτn is uniformly integrable and Xτn → Xτ almost surely and in L . Similarly

Xσn → Xσ. So

[Xτ+|Fσ] = lim [Xτ+|Fσ ] E m→∞ E m

= lim lim [Xτ +|Fσ ] ≥ lim Xσ = Xσ. m→∞ n→∞ E n m m→∞ m Math 288 Notes 29

12 February 17, 2017

We proved the upcrossing inequality and Doob’s inequality. We also proved optional stopping theorem. There is also the martingale convergence theorem that says that if (Xj) is a submartingale or a supermartingale and supj E[|Xj|] ≤ C, then limJ→∞ Xj = X∞ for some X∞ almost surely. If (Xj) is uniformly 1 integrable, then Xj → X∞ in L .

Theorem 12.1 (3.18). Suppose Ft is right continuous and complete. Let Xt be a super-martingale and assume that t 7→ E[t] is right continuous. Then X has a modiﬁcation such that the sample paths are right continuous with left limits, which is also a supermartingale. Lemma 12.2 (3.16). Let D be a dense set and f be deﬁned on D such that

(i) f is bounded on D ∩ [0,T ], f (ii) Mab(D ∩ [0,T ]) < ∞.

Then f+(t) = lims↓t f(s) and f−(t) = lims↑t f(s) exist for all t ∈ D ∩ [0,T ]. Deﬁne g(t) = f+(t). Then g(t) is right continuous with left limits.

Theorem 12.3 (3.21). Suppose a martingale (Xt) has right continuous sample paths. Then the following are equivalent:

1 (i) Xt is closed, i.e., there exists a Z ∈ L such that E[Z|Ft] = Xt.

(ii) Xt is uniformly integrable. 1 (iii) Xt → W almost surely and L as t → ∞ for some W .

There is an application. Let Ta = inf{t : Bt = a} for a > 0 and λ > 0.

2 Proposition 12.4. E[e−(λ /2)Ta ] = eλa. Math 288 Notes 30

13 February 22, 2017

Deﬁnition 13.1. Let (Xt) by a submartingale or a supermartingale with right continuous sample paths. Assume Xt → X∞ almost surely. Suppose T is a stopping time which can take ∞ with positive probability. We deﬁne

XT (ω) = 1(T (ω)<∞)XT (ω) + 1T (ω)=∞)X∞(ω).

Theorem 13.2 (Optional stopping theorem). Let (Xt) be a submartingale. Let S ≤ T almost surely and P (T < ∞) = 1. (1) If the submartingale (Xt) is uniformly integrable, then E[|Xn|1(T >n)] → 0 and E[|XT | + |XS|] < ∞. (2) If E[|Xn|1(T >n)] → 0 and E[|XT | + |XS|] < ∞ then E[XT |FS] ≥ XS. (3) If S ≤ T ≤ C almost surely, then E[XT |FS] ≥ XS. (4) Suppose a martingale (Xn) is uniformly integrable, and let S ≤ T be stopping times taking values in R ∪ {∞}. Then E[|Xn|1(T >n)] → 0 and E[|XT | + |XS|] < ∞.

Theorem 13.3 (Levi). Let (Xn) be a uniformly integrable martingale. Then 1 there exists a X ∈ L such that E[X|Fn] = Xn.

If Xn is a submartingale and T is a stopping time with T ∈ R ∪ {∞}, then XT ∧n is a submartingale. That is, for m < n, E[XT ∧n|Fm] ≥ XT ∧m.

Example 13.4. Let X1 = 1 and Xn be a symmetric random walk. Let T be the ﬁrst hitting time of 0 and let Yn = XT ∧n. Then E[Yn] = E[Y1] = 1 but Yn → 0 almost surely. This means that Yn does not converge to 0 in L1, and hence Yn is not uniformly integrable.

13.1 Submartingale decomposition

Theorem 13.5 (Doob submartingale decomposition). Suppose (Xn) is a submartingale. Then there exists a martingale Mn and a predictable increasing sequence An such that Xn = Mn + An. Here we say that An is predictable if and only if An is a measurable with respect to Fn−1. If Xn is a supermartingale, then An is decreasing. Furthermore, such a decomposition is unique.

Proof. Because we want X2 = M2 + A2, we get E[X|F1] = M1 + A2 = X1 + A2. So A2 = E[X2|F1] − X1. Similarly we can deﬁne

Ak = E[Xk|Fk−1] − Mk−1,Mk = Xk − Ak.

You can check that Ak increases. For instance,

A2 = E[X2|F1] − M1 = E[X2|F1] − [X1 − A1] = E[X2|F1] − X1 + A1. Math 288 Notes 31

14 February 24, 2017

There is another version of optional stopping time Theorem 14.1 (Le Gall 3.25). Suppose X is a nonnegative supermartingale with right continuous sample paths. Let S ≤ T be two stopping times. Then E[XT |FS] ≥ XS. Proof. Consider only the discrete case, since we can approximate the continuous case with the discrete case. A nonnegative supermartingale is in L1 in the sense that E[Xj] ≤ C < ∞. The upcrossing inequality tells us that Xj → X∞ = X almost surely (but maybe not in L1). For any M > 0, we know that S ∧ M ≤ T ∧ M are stopping times. You can check that XS∧M → XS almost surely. Then by Fatou’s lemma,

lim E[XS∧M ] ≥ E[XS]. M→∞

1 So XS ∈ L . For any A ∈ FX , we want to prove E[1AXS] ≥ E[1AXT ]. This would immediately imply E[XT |FS] ≥ XS. Deﬁne the stopping time ( S(ω) if ω ∈ A SA(ω) = ∞ if ω∈ / A. and similarly T A(ω). Then

E[XSA∧M ] ≥ E[XT A∧M ] because both are ﬁnite. Now

E[1A∩{S≤M}XS∧M ] + E[1A∩{S>M}XM ] + E[1Ac XM ] = E[XSA∧M ] ≥ E[XT A∧M ] = E[1A∩{S≤M}XT ∧M ] + E[1A∩{S>M}XM ] + E[1Ac XM ].

Hence we get E[1A∩{S≤M}XS] ≥ E[1A∩{S≤M}XT ∧M ]. Taking the limit M → ∞, we get

E[1A∩{S<∞}XS] = lim E[1A∩{S≤M}XS] ≥ lim E[1A∩{S≤M}XT ∧M ] M→∞ M→∞

≥ E[1A∩{S<∞}XT ] by dominated convergence and Fatou’s lemma. On the other hand, we have E[1A∩{S=∞}XS] = E[1A∩{S=∞}XT ] clearly. This implies that E[1AXS] ≥ E[1AXT ].

14.1 Backward martingales In this case, we have a ﬁltration

· · · ⊆ F−3 ⊆ F−2 ⊆ F−1 ⊆ F0. Math 288 Notes 32

Deﬁnition 14.2. A process (Xn) is a backward supermartingale if

E[Xn|Fm] ≤ Xm for any m ≤ n ≤ 0.

1 Theorem 14.3. if supn<0 E[|Xn|] ≤ C then there exists an X−∞ ∈ L such 1 that Xn → X−∞ almost surely and in L . Note that we also have L1 convergence, which we did not have in the forward case.

Proof. The almost surely convergence is almost identical to the forward case, as a consequence of the upcrossing inequality. To show convergence in L1, deﬁne

An = E[X−1 − X−2|F2] + ··· + E[X−(n−1) − X−n|F−(n−1)].

Note that this is negative. Then Mn = Xn + An is a martingale. Then An → 1 A−∞ almost surely and in L , which comes from the monotone convergence theorem. Now Mn being a martingale simply means

M−n = E[M−1|F−n]. This is just the setting of Levi’s theorem and so immediately we have L1 convergence of M −n.

14.2 Finite variance processes We assume that sample paths are continuous.

Definition 14.4. A function a : [0,T ] → R is of finite variance if there exists a signed measure µ with finite |µ| such that a(t) = µ([0, t]) for all t ≤ T .

Definition 14.5. A process (At) is called a finite variance process if all sample paths have finite variance.

Deﬁnition 14.6. A sequence Xt with X0 = 0 is called a local martingale if there exist increasing stopping times Tn ↑ ∞ such that Xt∧Tn is a uniform integrable martingale for all n. If X0 6= 0, then Xt is a local martingale if Xt − X0 is a local martingale. Math 288 Notes 33

15 February 27, 2017

For a finite variance process As(ω), we can write As(ω) = µ([0, s]) for some finite measure µ. So we can define Z Z Hs(ω)dAs(ω) = Hs(ω)µ(ds)(ω).

Here we require Z |Hs(ω)||dAs(ω)| < ∞.

This is Section 4.1 and you can read about it.

15.1 Local martingales

We are always going to assume that everything is adapted to Ft and sample paths are continuous.

Deﬁnition 15.1. A sequence (Mt) is a local martingale with M0 = 0 if there exists an increasing sequence of stopping times Tn → ∞ such that Mt∧Tn is an uniformly integrable martingale for each n. If M0 6= 0, then Mt is called a local martingale if Mt − M0 = Nt is a local martingale.

1 Proposition 15.2. (i) If A is a nonnegative local martingale with M0 ∈ L , then it is a supermartingale. 1 (ii) If there exists a Z ∈ L such that |Mt| ≤ Z for all t, then the local martingale Mt is a uniformly integrable martingale. (iii) If M0 = 0, then Tn = inf{t : Mt ≥ n} reduces the martingale, i.e., satisﬁes the condition in the deﬁnition.

Proof. (i) Let us write Ms = M0 + Ns. By deﬁnition, there exists a Tn ↑ ∞ such that

lim [M0 + Nt∧T |Fs] = lim M0 + Ns∧T = Ms. n→∞ E n n→∞ n

Then by Fatou’s lemma, we get E[Mt|Fs] ≤ Ms. (ii) This is clear because Mt is uniformly integrable. (iii) This basically truncating the local martingale.

Proposition 15.3. A ﬁnite variance local martingale with M0 = 0 is identical to 0. (Again we are assuming continuous sample paths.) Proof. Deﬁne a stopping time R t τn = inf{t : 0 |dMs| ≥ n}, Math 288 Notes 34

and let Nt = Mτn∧t. Then we have Z t

|Nt| ≤ dMs ≤ n. 0

For any t and 0 = t0 < ··· < tp = t,

 2 2 X X 2 X h i E[Nt ] = E δNi = E (δNi) = E sup|δNi| |δNj| ≤ nE sup|δNi| . i i i i j

Taking the mesh to go to 0, we get that the right hand side goes to 0.

Theorem 15.4. Suppose that Mt is a local martingale. Then there exists a 2 unique increasing process hM,Mit such that Mt −hM,Mit is a local martingale. This hM,Mit is called the quadratic variation of M.

2 Example 15.5. If Mt = Bt then hM,Mit = t. Then Bt − t is a martingale. 2 2 0 Proof. Let us first prove uniqueness. Suppose Mt − At and Mt − At are local 0 0 martingales. Then At − At is also a local martingale. By definition, At and At 0 are increasing and less than ∞ almost surely. So At − At is of finite variance. 0 This shows that At − At = 0 for all t. Proving existence is rather difficult. Suppose M0 = 0 and Mt is bounded, i.e., supt|Mt| < C almost surely. Fix and interval [0, k] and subdivide it to 0 = t0 < ··· < tn = K. Define

n X Xt = Mti−1 (Mti∧t − Mti−1∧t). i You can check that this is a local martingale, and you can also check

j X M 2 − 2Xn = (M − M )2. tj tj ti ti−1 i=1 Now deﬁne

X 2 hM,Mit = lim (Mt − Mt ) . n→∞ i i−1 i We are going to continue next time. Math 288 Notes 35

16 March 1, 2017 16.1 Quadratic variation As always, we assume continuous sample paths. This is the theorem we are trying to prove: Theorem 16.1. Let M be a local martingale. Then there exists an increasing 2 process hM,Mit such that Mt − hM,Mit is a local martingale. We proved uniqueness using the fact that if there are two such A, A0 then A − A0 is a finite variation process that is also a local martingale. For existence, fix K and subdivide [0,K] into 0 = t0 < ··· < tn = K. We define

n Xt = M0(Mt1∧t − M0∧t) + ··· + Mtn−1 (Mtn∧t − Mtn−1∧t).

Then we immediately have

j X M 2 − 2Xn = (M − M )2. tj tj ti ti−1 i=1

Lemma 16.2. Assuming that |Mt| ≤ C (so that M is actually a martingale),

n m 2 lim E[(Xk − Xk ) ] = 0. m,n→∞,mesh→0

Proof. Let us consider the simple case where n = 5 and m = 3 and the subsequence is given by t1, t3, t5. Then we compute

5 2 n m 2 X E[(Xk − Xk ) ] = E Mj−1(Mj − mj−1) − M3(M5 − M3) − M1(M3 − M1) j=1 2 = E[((M4 − M3)(M5 − M4) + (M2 − M1)(M3 − M2)) ] 2 2 2 2 = E[(M4 − M3) (M5 − M4) + (M2 − M1) (M3 − M2) ]

h 2 X 2i ≤ max|Mj − Mj−1| (Mj − Mj−1) E j j 2 1/2 4 1/2 hX 2 i ≤ E[max(Mj − Mj−1) ] E (Mj − Mj−1) . j Math 288 Notes 36

Here, we see that

 2 X 2 X 4 E (Mj − Mj−1) ≤ E[(Mj − Mj−1) ] j j 5 2 X 2 + 2E(M1 − M0) (Mj − Mj−1) + ··· j=2

2 X 2 ≤ 4C E[(Mj − Mj−1) ] j n X 2 X 2 + 2 E (Mj − Mj−1) E (Mk − Mk−1) Fj

j k=j+1

2 X 2 2 2 = 4C E[(Mj − Mj−1) ] + 2E[(Mj − Mj−1) E[(MK − Mj) |Fj]] j 2 X 2 2 2 4 ≤ 12C E (Mj − Mj−1) = 12C E[(MK − M0) ] ≤ 48C . j

4 So taking n, m → ∞ we get E[max(Mj − Mj−1) ] → 0. 2 n Now let us go back to the original theorem. We check that Mt − 2Xt is an n 2 increasing process. Take n → ∞. By the lemma, Xt∧K → Yt∧K in L . You can also check that Yt∧K has continuous sample paths. Deﬁne

2 hM,Mit = Mt∧K − 2Yt∧K .

If {ti} is a subdivision of [0, t], we have

n X 2 (M n − M n ) → hM,Mi ti ti−1 t i=1

2 in L . Also, Xt∧K is a local martingale. K This construction works for all ﬁxed K. For every K, we get a At such 2 K K+1 K that Mt∧K − At∧K is a martingale. Then by uniqueness, At = At if t ≤ K K almost surely. This means that we can just take the union of all the At . Math 288 Notes 37

17 March 3, 2017

Last class we almost ﬁnished proved the following theorem. Theorem 17.1. Let M be a local martingale. Then there exists a unique in- 2 creasing process hM,Mit such that Mt − hM,Mit is a local martingale. 2 n In the proof, we haven’t check that the L limit Xt → Xt has continuous sample paths. To show this, it suﬃces to check that

h m n 2i E sup|Xs − Xs | → 0 s≤t as n, m → ∞. By Doob’s inequality,

h m n 2i m n 2 E sup|Xs − Xs | ≤ 4E[|Xt − Xt | ]. s≤t

Then this is clear from the L2 convergence. Tn For the general case, deﬁne Tn = inf{t : |Mt| ≥}. Then M is a bounded n Tn 2 n martingale and so there exists an increasing process At such that (M ) − At is a martingale. By uniqueness, An+1 = An. Now we can take the limit t∧Tn t n At = limn→∞ At . Then we have

(M )2 − A = (M Tn )2 − An Tn∧t Tn∧t t t∧Tn 2 is a martingale. This means that Mt − At is a local martingale.

17.1 Consequences of the quadratic variation

Corollary 17.2. Let M be a local martingale with M0 = 0. Then M = 0 if and only if hM,Mit = 0. 2 Proof. Suppose hM,Mit = 0. Then M is a nonnegative local martingale. Then 2 Mt is a supermartingale. Since M0 = 0, we immediately get M = 0. 2 Corollary 17.3. Assume M is a local martingale and M0 ∈ L . Then EhM,Mi∞ < 2 ∞ if and only if M is a true martingale and EMt < C for all t. 2 Proof. Suppose M is a martingale with M0 = 0 and EMt < C. Then 2 2 E sup Mt ≤ 4EMT ≤ 4C. t≤T

n Let S = inf{t : hM,Mit ≥ n}. Then

2 MSn∧t − hM,MiSn∧t ≤ sup Mt . t≥0

2 1 But supt Mt is in L . Then

2 2 lim [MSn∧t − hM,MiSn∧t] = [M0 − hM,Mi0] = 0. n→∞ E E Math 288 Notes 38

2 By monotone convergence, we have EhM,Mit = EMt < C. Conversely, suppose EhM,Mi∞ < ∞. Let Tn = inf{t : |Mt| ≥ n}. We have M 2 − hM,Mi ≤ n2 ∈ L1. Tn∧t Tn∧t Then M 2 − hM,Mi is a uniformly integrable martingale and so Tn∧t Tn∧t

2 lim [MT ∧t − hM,MiTn∧t] = 0 n→∞ E n

2 and so E[Mt ] ≤ E[hM,Mit] < C. This also implies that M is a true martingale. Math 288 Notes 39

18 March 6, 2017

Let us try to ﬁnish this chapter. 2 2 Theorem 18.1. If supt E[Mt ] < ∞, then EhM,Mi∞ < ∞ and Mt − hM,Mit is a uniformly integrable martingale. If M is a local martingale and EhM,Mi∞ < 2 ∞ then supt E[Mt ] < ∞.

18.1 Bracket of local martingales Definition 18.2. If M and N are two local martingales, we define their bracket as 1 hM,Ni = [hM + N,M + Ni − hM,Mi − hN,Ni]. t 2 Proposition 18.3 (Le Gall 4.15). In probability, X lim (Mti − Mti−1 )(Nti − Nti−1 ) = hM,Nit. i 2 Furthermore, if M and N are two L martingales, then EhM,Ni∞ = E[M∞N∞]. Definition 18.4. We say that M is orthogonal to N if hM,Ni = 0. Theorem 18.5 (Kunita–Watanabe). If M and N are two local martingales, then Z ∞ Z ∞ 1/2Z ∞ 1/2 2 |Hs(ω)Ks(ω)||dhM,Nis| ≤ |Hs|dhM,Mis |Ks| dhN,Nis . 0 0 0 Proof. You can just subdivide the intervals finely and use the formula for hM,Nit.

Definition 18.6. A process Xt is called a semimartingale if Xt = Mt + At where Mt is a local martingale and At is a finite variance process. Note that such a decomposition is unique since a finite variance local martingale is zero. Proposition 18.7. For two semimartingales X = M + A and Y = N + B, we define hX,Y i = hM,Ni. Then X (Xti − Xti−1 )(Yti − Yti−1 ) → hX,Y i = hM,Ni. i Proof. We only have to check that X (Nti − Nti−1 )(Ati − Ati−1 ) → 0 i We use the Schwartz inequality. We have

X 2 X (Ati − Ati−1 ) ≤ sup|Ati − Ati−1 | |Ati − Ati−1 |. i i i The ﬁrst term goes to zero and the second term is bounded. Math 288 Notes 40

18.2 Stochastic integration We deﬁne the space 2 M a continuous martingale with H = 2 M0 = 0 and supt EMt < ∞.

2 We then have hM,Mi∞ = EM∞ and [M∞|Ft] = Mt. We deﬁne the inner product as

(M,N) = E[hM,Ni∞] = EM∞N∞. Also deﬁne the space of elementary processes as

 X E = Hs(ω) = Hi(ω)1(ti,ti+1](s),Hi ∈ Fti . i We can then deﬁne the integral as

Z X Hs(ω)dMs = Hi(ω)(Mti − Mti−1 ). i This is just like Riemann integration.

Theorem 18.8. H2 is a Hilbert space. n m 2 n Proof. Suppose there is a Cauchy sequence E[(M∞ − M∞) ] → 0. Then M∞ → 2 Z in L . Our goal is to ﬁnd a Mt with continuous sample paths such that n Mt → Mt in some sense. Math 288 Notes 41

19 March 8, 2017

2 2 We define H to be the space of martingales with supt E[Mt ] < ∞. The inner product is defined as (M,N) = E[M∞N∞] = EhM,Ni∞. Proposition 19.1. H2 is a Hilbert space. m n n Proof. Suppose limm,n→∞ E[(M∞,M∞)] = 0 so that M is a Cauchy sequence. n 2 Then there exists a Z such that M∞ → Z in L (P ). We also have, for an fixed m n 2 ∞ ∞ t, E[(Mt − Mt ) ] → 0 and so Mt . What we now need to show is that M has continuous sample paths. We have

n m 2 n m 2 E[sup(Mt − Mt ) ] ≤ 4E[(M∞ − M∞) ] → 0. t This shows that you can ﬁnd a subsequence such that ∞ ∞ X nk+1 nk X nk+1 nk 2 1/2 E sup|Mt − Mt | ≤ 4 E[(M∞ − M∞ ) ] < ∞. t k=1 k=1 In particular,

X nk+1 nk Yt = (Mt − Mt ) k

n is a martingale with continuous sample paths. Also Y∞ = Z and so M → Y in H2.

19.1 Elementary process Deﬁnition 19.2. We deﬁne the space E of elementary processes as

p−1 X E = Hi(ω)1(ti,ti+1] : Hi ∈ Fti , with i=0 X 2 Hi (hM,Miti+1 − hM,Miti ) < ∞ . i We also deﬁne Z 2 n 2 o L (M) = Hs such that Hs dhM,Mis < ∞ .

Proposition 19.3. If M ∈ H2, then the space E is dense in L2(M). Theorem 19.4. For any M ∈ H2, we can deﬁne p−1 Z t X 2 (H · M)t = Hi(ω)(Mti+1∧t − Mti∧t) = HdM ∈ H . i=0 0 Math 288 Notes 42

This map E → H2 given by H 7→ H · M is an isometry from E to H: Z 2 2 kH · MkH = Hs dhM,Mis.

2 R So there exists a unique extension from L (M) → H, denoted by HdM. Fur- thermore, (1) hH · M,Ni = HhM,Ni, T T (2) (1[0,T ]H) · M = (H · M) = H · M . Roughly speaking, stochastic integration is just Riemann integration on elementary functions, which has a unique extension. You can check that

X 2 hH · M,H · Mit = Hi (hM,Miti+1∧t − hM,Miti∧t). i

Proof. Since (H · M)2 − H2hM,Mi is a martingale, we have Z ∞ 2 2 E[(H · M)∞] = E Hs dhM,Mis. 0 This shows that stochastic integration is and L2 isometry. Then there exists a unique extension. Now we want to check that hH · M,Ni = HhM,Ni. Note that hM,Ni is a ﬁnite variance process hM,Mi and hN,Ni are increasing and hM + N,M + Ni is decreasing, and also E|hM,Ni| < ∞. We want to check that DZ t E Z t HdM ,Nt = HdhM,Ni. 0 0 For elementary processes, we have

DZ t Z t E Z t HsdMs, dNs = HsdhM,Nis. 0 0 0 Then it extends to all processes. Math 288 Notes 43

20 March 20, 2017 20.1 Review Today we will review, because all the properties are hard to remember. All processes have continuous sample paths and are adapted.

1. Finite variation processes: At is a ﬁnite variation process if its sample paths have ﬁnite variation almost surely. Classical theory of integration applies here. Then

Z t Hs(ω)dAs(ω) 0

is deﬁned by Lebesgue integration. (Note that dAs is not positive.)

2. Continuous local martingales: Usually we assume M0 = 0. Mt is a local martingale if there exists an increasing sequence Tn ↑ ∞ almost surely Tn such that (M )t = MTn∧t is an uniformly integrable martingale. In this case, we say that Tn reduces the local martingale.

Why do we look at these objects ? If Xn is a sub/supermartingale, and p supn E|Xn| < ∞ for some p ≥ 1, then Xn → X∞ almost surely. Moreover p if p > 1 then Xn → X∞ in L (by Doob’s inequality). Furthermore, if Xn 1 is uniformly integrable then Xn → X∞ also in L . We really need uniform integrability to run the theory, and so we have to stick in a stopping time to make things uniformly integrability. 3. Quadratic variation: If M is a local martingale, then there exists a unique 2 increasing process hM,Mit such that Mt − hM,Mit is a local martingale. This is constructed as

X 2 hM,Mit = lim (Mti+1 − Mti ) . i

For M = B a Brownian motion, we have hB,Bit = t. 4. Some properties of local martingales:

• A nonnegative local martingale M with M0 ∈ L1 is a submartingale. (This follows from Fatou’s lemma.) 1 • If there exists a Z ∈ L such that |Mt| < Z for all t, then M is a uniformly integrable martingale.

• If M with M0 = 0 is both a local martingale and a ﬁnite variation process, then M = 0.

• For M a local martingale with M0 = 0, hM,Mi = 0 if and only if M = 0. Math 288 Notes 44

5. Stochastic integration: We ﬁrst deﬁne the class H2 of square-integrable martingales with the inner product

2 (M,N)H = E[M∞N∞] = E[hM,Ni∞].

Then we deﬁned, for any M ∈ H2, the space L2(M) of processes H such that Z ∞ 2 E H dhM,Mi < ∞. 0 The space E is the space of elementary processes, processes of the form Pp−1 i=0 Hi1(ti,ti+1] for Hi measurable with respect to Fti and bounded almost surely. You can check that for every M ∈ H2, the space E is dense in L2(M). (Note that the inner product on E depends on M.) R We can now deﬁne stochastic integration HdM on the space E by

X (H · M)t = Hi(Mti+1∧t − Mti∧t). i

This gives an isometry E → H2. So you can extend it to an isometry L2(M) → H2. L2(M) H2

So far we have defined stochastic integration for M ∈ H2. Now we want to define stochastic integration for M a local martingale. Then we would like to define

Z Tn Z HdM = HdM Tn .

We are also going to deﬁne, for a semi-martingale X = M + V , Z Z Z HdX = HdM + HdV.

Next time we are going to quickly go over these deﬁnitions and start actually doing some stuﬀ. Math 288 Notes 45

21 March 22, 2017

For any ﬁxed M ∈ H2, we consider the space Z ∞ 2 n 2 o L (M) = adapted processes (Ht) with E Ht dhM,Mi < ∞ . 0 There is an isometry

2 X X (−) · M : E → H ; Hi1(ti,ti+1] 7→ Hi(Mti+1∧t − Mti∧t). i i

This can be extend to an isometry L2(M) → H2.

21.1 Stochastic integration with respect to a Brownian motion We now want to look at stochastic integration with respect to Brownian motions. 2 But (Bt) is not in H . More generally, we are going to stochastic integration this to local martingales. We are going to do in the setting (1) M is a local martingale, 2 R t 2 (2) H ∈ Lloc(M), i.e., 0 Hs dhM,Mis < ∞ almost surely for any ﬁxed t.

Recall that for the Brownian motion, hB,Bit = t. This is easy to prove because p−1 2 hX 2 n n i X n n 2 ((B n − B n ) − (t − t )) = C (t − t ) → 0. E ti+1 ti i+1 i i+1 i i i=0 Then we can deﬁne, for t ≤ T ,

Z t X HdB = Hi(Bti+1∧t − Bti∧t). 0 i We check that

Z T 2 2 Z T X X 2 E HdB = E Hi(Bti+1 − Bti ) = Hi (ti+1 − ti) = Hsds. 0 i i 0

R t 2 That is, 0 HdB is an isometry in L . Then you can again extend it to all R T 2 adapted processes Hs such that 0 Hs ds < ∞ almost surely

21.2 Stochastic integration with respect to local martingales Let M be a local martingale, and we deﬁne Z t 2 n 2 o Lloc(M) = (Hs): Hs dhM,Mis < ∞ almost surely for all t . 0 Math 288 Notes 46

Theorem 21.1. There exists a unique continuous local martingale H · M = R HsdMs such that •h H · M,Ni = H · hM,Ni, • H · (K · M) = (HK) · M, R R R t •h HdM, KdNit = 0 HsKsdhM,Nis, • R Hd(R KdM) = R (HK)dM. Proof. We look at the stopping time

Z t 2 Tn = inf{t : (1 + Hs )dhM,Mis ≥ n}. 0

You can check that they actually are stopping times and Tn ↑ ∞. Consider Tn Tn Tn Tn (M )t. You can also check hM ,M i = hM,Mi . Then

Tn 2 Tn E(M )t ≤ E(hM,Mit ) ≤ n.

That is, M Tn ∈ H2, and so H · M Tn is well-deﬁned. We now want a process H · M such that H · M Tn = (H · M)Tn . We will continue next time.

If X is a semi-martingale, with X = M + V where V is a ﬁnite variation process, then we can further deﬁne Z Z Z HsdX = HsdMs + HsdVs.

Theorem 21.2 (Ito’s formula). Let F ∈ C2 be a function and X1,...,Xp be semi-martingales. Then

p X Z t ∂F F (X1,...,Xp) = F (X1,...,Xp) + (X1,...,Xp)dXi t t 0 0 ∂Xi s s s i=1 0 p X Z t ∂2F + (X1,...,Xp)dhXi,Xji . ∂Xi∂Xj s s s i,j=1 0 Math 288 Notes 47

22 March 24, 2017

Theorem 22.1. Let M be a local martingale, and consider the space Z t 2 n 2 o Lloc(M) = (Hs): Hs (ω)dhM,Mis < ∞ almost surely for any ﬁxed t . 0 R Then there exists a unique integration HsdMs giving a continuous local martingale such that hH · M,Ni = H · hM,Ni.

Proof. Assume M0 = 0 as always, and consider the stopping time Z t n 2 o Tn = inf t : (1 + Hs )dhM,Mis ≥ n . 0

Tn 2 These are actually stopping times and Tn ↑ ∞. We see that M is a L R t martingale because 0 dhM,Mis ≤ n. Then

Tn 2 Tn Tn E[(M )t ] = E[hM ,M it] = E[hM,Mit∧Tn ] ≤ n.

So the integration H · M Tn = R HdM Tn is well-deﬁned. For m ≥ n, you can check that (H · M Tm )Tn = H · M Tn . That is, this is a consistent family and so there exists a local martingale H · M such that (H · M)Tn = H · M Tn .

22.1 Dominated convergence for semi-martingales Proposition 22.2. Let X be a continuous semi-martingale, and H be an adapted process with continuous sample paths. For every t > 0 ﬁxed, and n n 0 = t0 < ··· < tp = t,

p−1 X Z t H n (X n − X n ) → H dX ti ti+1 ti s s i=0 0 in probability. Lemma 22.3 (Dominated convergence in stochastic integration). Let X = n M + V be a semi-martingale. Assume that H satisﬁes Hs → Hs almost surely n for s ∈ [0, t], and there exists a Ks ≥ 0 such that |Hs | ≤ Ks for all n and s ∈ [0, t], with the conditions Z t Z t 2 Ks dhM,Mis < ∞, Ks|dVs| < ∞ 0 0 almost surely. Then Z Z n Hs dXs → HsdXs in probability. Math 288 Notes 48

Proof. The case M = 0 is exactly the classical dominated convergence. Let us now assume V = 0. Deﬁne the stopping time Z γ n 2 o Tp = γ ≤ t : Ks dhM,Mis ≥ p ∧ t. 0 We have

Z Tp Z Tp 2 Z Tp h n i n 2 E Hs dMs − HsdMs ≤ E (Hs − Hs) dhM,Mis 0 0 0 Z Tp 2 ≤ E 2Ks dhM,Mis < C(p). 0 By dominated convergence, the left hand side goes to 0 as n → ∞. Since P (Tp = t) → 1 as p → ∞, we get the desired result. Proof of Proposition 22.2. Deﬁne the processes

( n n H ti < s ≤ t Hn = ti i+1 s 0 s > t.

Let Ks = max0≤r≤s|Hr|. Then H has continuous sample paths and so Ks < ∞ n almost surely. We trivially check that |Hs | ≤ Ks for all n and s. Also Z 2 Ks dhM,Mis < ∞

n because it is locally bounded. Hs → Hs almost surely as n → ∞, because sample paths are continuous. So we can apply dominated convergence.

22.2 Itˆo’sformula Theorem 22.4 (Itˆo’sformula). Let F ∈ C2 be a function and X1,...,Xp be semi-martingales. Then

p X Z t ∂F F (X1,...,Xp) = F (X1,...,Xp) + (X1,...,Xp)dXi 0 0 ∂Xi s s s i=1 0 p 1 X Z t ∂2F + (X1,...,Xp)dhXi,Xji . 2 ∂Xi∂Xj s s s i,j=1 0

Proof. Let us consider only the case p = 1. We have

p−1 X n n F (Xt) = F (X0) + (F (X n ) − F (X )) ti+1 ti i=0

X 0 1 X 00 n 2 = F (Xt )(Xtn − Xtn ) + F (X + θ(Xtn − Xtn ))(Xtn − Xtn ) i i+1 i 2 ti i+1 i i+1 i i i Math 288 Notes 49 by simple Taylor expansion, with 0 ≤ θ ≤ 1. Now the ﬁrst term converges to R F (Xs)dXs by the Proposition 22.2. To show that the second term goes to 1 R 00 2 F (Xs)dhX,Xis, we note that

2 hX 00 n 2 i F (B )[(X n − X n ) − (hX,Xi n − hX,Xi n )] E ti ti+1 ti ti+1 ti i

X 2 ≤ C (hX,Xi n − hX,Xi n ) → 0 ti+1 ti since all the cross terms vanish. So we get the formula. Math 288 Notes 50

23 March 27, 2017

Last time we prove Itˆo’sformula. For F ∈ C2 and semi-martingales X1,...,Xd, we have

d Z t X ∂F F (t, X1,...,Xd) = F (0,X1,...,Xd) + (t, X1,...,Xd)dXi t t 0 0 ∂Xi s s s 0 i=1 d Z t∂F 1 X ∂2F + dt + (s, X1,...,Xd)dhX ,X i . ∂t 2 ∂Xi∂Xj s s i j s 0 i,j=1

In the special case of d = 1 and X = B, we get

Z t ∂F Z t∂F ∆ F (t, B0) = F (0,B0) + (s, Bs)dBs + + F (s, Bs)ds. 0 ∂X 0 ∂s 2 Example 23.1. Let λ2 λMt− hM,Mit E (λM)t = e 2 , where M is a local martingale. For any λ ∈ C, E (λM)t is a local martingale and Z t λM0 E (λM)t = e + λ (λM)sdMs. 0

λ2 λMt− r Proof. Let F (x, r) = e 2 . If we set G(x, t) = F (x, hM,Mit), then Itˆo’s formula tells us that Z t ∂G Z t∂G 1 ∂2G G(Mt, t) = G(M0, 0) + dMs + + 2 (s, x)dhM,Mis. 0 ∂M 0 ∂s 2 ∂M

Here, we can treat hM,Mit as an external time dependence, because it is a ﬁnite variation process. Computing the second derivative gives

∂G G00 h λ2 dr λ2 i ds + dhM,Mi = − ds + dhM,Mi G(s, M ) = 0. ∂s 2 s 2 ds 2 s s Then we are left with the ﬁrst derivative.

23.1 Application of Itˆocalculus Theorem 23.2 (Levi). Let X1,...,Xd be adapted local martingales with con- i j 1 d tinuous ample paths. If hX ,X it = δijt, then X ,...,X are independent Brownian motions. Proof. Let us only look at the case d = 1. By the previous theorem,

2 iξX + ξ t E (t) = e t 2 Math 288 Notes 51 is a local martingale and further a martingale because it is bounded. This shows that ξ2 iξ(Xt−Xs)+ (t−s) E[e 2 |Fs] = 1 for all ξ. Then Xt − Xs is Gaussian with variance (t − s) when conditioned on Fs.

Theorem 23.3 (Burkholder–Davis–Gundy). Let Mt be a local martingale and ∗ set Mt = sup0≤s≤t|Ms|. For every p > 0 there exist constants cp,Cp depending only on p such that for any time T ,

p/2 ∗ p p/2 cpE[hM,MiT ] ≤ E[(MT ) ] ≤ CpE[hM,MiT ]. Proof. Let us ﬁrst consider the case p ≥ 2. By Itˆo’sformula,

Z t Z t p p−1 1 p−2 |Mt| = p|Ms| sgn MsdMs + p(p − 1)|Ms| dhM,Mis. 0 2 0

Sn We may assume that M is bounded because we can apply to M with Sn = R t p−1 2 inf{t : |Mt| ≥ n} and take the limit. Then 0 p|Ms| sgn MsdMs is an L martingale. Hence Z t p h p−2 i ∗ p−2 E[|Mt| ] ≤ p(p − 1)E |Ms| dhM,Mis ≤ p(p − 1)E[|Mt | hM,Mit] 0 ∗ p (p−2)/p p/2 2/p ≤ p(p − 1)E[|Mt | ] E[hM,Mit ] .

∗ p p/2 So by Doob’s inequality, we get an upper bound on E[|Mt | ] by E[hM,Mit ]. Math 288 Notes 52

24 March 29, 2017

Last time we were proving the Burkholder–Davis–Gundy inequality.

24.1 Burkholder–Davis–Gundy inequality Theorem 24.1 (Burkholder–Davis–Gundy inequality). For 0 < p < ∞, there exists a constant Cp such that

∗ p p/2 E[(Mt ) ] ≤ CpE[hM,Mit ]. We ﬁrst assumed that M is bounded. This is because we can prove the Tn inequality for M where Tn = inf{t : |Mt| ≥ n} and then take the limit n → ∞. Next, we assumed that p ≥ 2 and applied Itˆo’sformula.

2 Proof (cont.) Now let us consider the case 0 < p < 2. Deﬁne Tx = inf{t : Mt > x}. Then 1 P ((M ∗)2 ≥ x) = P (T ≤ t) = P ((M )2 ≥ x) ≤ [(M )2] t x Tx∧t xE Tx∧t 1 1 = [hM,Mi ] ≤ [hM,Mi ] xE Tx∧t xE t because M 2 − hM,Mi is a martingale. Let us deﬁne Sx = {inf t : hM,Mit ≥ x}. Our goal is to prove 1 P ((M ∗)2 ≥ x) ≤ [hM,Mi 1(hM,Mi ≤ x)] + 2P (hM,Mi ≥ x). t xE t t t Once we prove this, we get for q = p/2, Z ∞ ∗ 2q ∗ 2 q−1 E[(Mt ) ] = P ((Mt ) ≥ x)qx dx 0 Z ∞ 1 q−1 ≤ c E[hM,Mit1(hM,Mit≤x)]x dx 0 x Z ∞ q−1 + c P (hM,Mit ≥ x)x dx 0 q q = cqE[hM,Mit ] + cqE[hM,Mit ]. Now set us prove the inequality. We have {(M ∗)2 ≥ x} ⊆ {(M ∗ )2 ≥ x} ∪ {S ≤ t}. t Sx∧t x Then P ((M ∗)2 ≥ x) ≤ P ((M ∗ )2 ≥ x) + P (S ≤ t) t Sx∧t x 1 ≤ [hM,Mi ] + P (S ≤ t) xE Sx∧t x 1 ≤ [hM,Mi ∧ x] + P (hM,Mi ≥ x) xE t t 1 ≤ [hM,Mi 1 ] + 2P (hM,Mi ≥ x). xE t (hM,Mit≤x) t Math 288 Notes 53

1/2 Corollary 24.2. Let M be a local martingale with M0 = 0. If E[hM,Mi∞ ] < ∞ then M is uniformly integrable.

∗ 1 Proof. By the theorem, E[(M∞) ] < ∞. Then Mt is bounded by a ﬁxed L variable.

2 2 Previously we have shown that E[hM,Mi∞] < ∞ implies that M is L bounded. This is a signiﬁcantly better criterion for checking that M is uniformly R t 2 1/2 integrable. As a consequence, if E[( 0 Hs(ω) dhM,Mis )] < ∞ then Z s Hσ(ω)dMσ 0 is uniformly integrable for s ≤ t. There are a few more things I want to go over. The ﬁrst one is representation of martingales as stochastic integrals. The second is Girsanov’s theorem.

2 Theorem 24.3. Let B be a Brownian motion. If Z ∈ L (Ω, F∞,P ), then there exists a unique adapted (progressive) process h ∈ L2(B) such that Z ∞ Z = E[Z] + hsdBs. 0 More generally,

Theorem 24.4. If M is a local martingale with ﬁltration Ft, then there exists 2 a hs ∈ Lloc(B) such that Z t Mt = C + hsdBs. 0 Math 288 Notes 54

25 March 31, 2017 25.1 Representation of martingales Theorem 25.1. 2 (1) If Z ∈ L (Ω, F∞,P ) where Ft is a σ-ﬁeld of the Brownian motion, then R t 2 there exists a process h with E 0 hsds < ∞ such that Z ∞ Z = E[Z] + hsdBs. 0

(2) If Mt is a local martingale with respect to Ft, then there exists a process R t 2 h with 0 hsds < ∞ such that Z t Mt = C + hsdBs. 0 Pp Lemma 25.2. The vector space generated by {exp(i j=1 λj(Btj − Btj−1 )} is dense in L2 (Ω, F ,P ). C ∞ Proof. By Fourier transformation, it suﬃces to show that the set

{F (Bt1 ,...,Btp ) for F smooth with compact support} 2 is dense in L (Ω, F∞,P ). P R t Proof of Theorem 25.1. (1) Define f(s) = lj1(t ,t ](s) so that f(s)dBs = P j j+1 0 j λj(Btj+1∧t − Btj ∧t). Define Z t Z t 1 2 E(t) = exp i f(s)dBs + f (s)ds . 0 2 0 Itô’sformula implies h 1 1 i dE(t) = E(t) if(t)dB + hif(t)dB , if(t)dB i + f 2(t)dt = E(t)if(t)dB . t 2 t t 2 t This shows that Z t E(t) = 1 + i E(s)f(s)dBs. 0 R t P Now let H be the closure of { 0 hsdBs}. We have shown that {exp(i j λj(Btj+1 − 2 Btj ))} ⊆ H and so by the lemma, H = L (P ). ∞ 2 R (2) We know that if Mt is an L -martingale, then M∞ = 0 hsdBs. We R t further hand E[M∞|Ft] = 0 hsdBs. Tn If Mt is a local martingale, then consider M with Tn = inft{t : |Mt| ≥ n}. n We then have a hs such that Z t Tn n Mt = hs dBs. 0 n m Check the consistence of hs and hs and put the together to form hs. Then R t Mt = 0 hsdBs. Math 288 Notes 55

25.2 Girsanov’s theorem

Theorem 25.3 (Girsanov). Let Pt and Qt be two probability measures with the same {Ft}. Assume that P Q (P is absolutely continuous with respect to Q) and Q P . Deﬁne dQt dQ∞ Dt = ,D∞ = . dPt dP∞

Then Dt is a uniformly integrable martingale with E[D∞|Ft] = Dt. Further D∞ > 0 almost surely and there exists a local martingale L such that

1 Lt− hL,Lit Dt = e 2 .

If M is a continuous local martingale with respect to P , then

M˜ = M − hM,Li is a continuous local martingale with respect to Q. Proof. From Itˆo’sformula, 1 1 1 d log Dt = dDt − 2 dhD,Dit. Dt 2 Dt So we have Z t dDs 1 dhD,Dis log Dt = − 2 . 0 Ds 2 Ds R t If we deﬁne Lt = 0 dDs/Ds then Z t Z t Z t D dDs dDs E dhD,Dis hL, Lit = , = 2 . 0 Ds 0 Ds 0 Ds

1 So log Dt = Lt − 2 hL, Lit. Now let us prove the main part. If MD is a local martingale with respect to P then M is a local martingale with respect to Q. This is almost a tautology, because for G a measurable function with respect to Fs,

P P E [(MD)tG] = E [(MD)sG]

Q Q and this translates to E [MtG] = E [MsG]. So it suﬃces to check that MD˜ is a martingale with respect to P . Then it suﬃces to check that d(MD˜ ) is a stochastic integral of something. This is going to be a half-page computation. Math 288 Notes 56

26 April 3, 2017

Theorem 26.1 (Girsanov’s formula). Let P Q and Q P , where P and dQ Q are probability measures with the same Ft. If Dt = dP |Ft then Dt is a uniformly integrable martingale. There exists a local martingale Lt such that 1 Dt = exp(Lt − 2 hL, Lit). Furthermore, if Mt is a local martingale with respect to P then M˜ = M − hM,Li is a local martingale with respect to Q.

Proof. We have checked that if MD˜ t is a (local) martingale with respect to P , then M˜ is a (local) martingale with respect to Q. Now by Itˆo’sformula,

d(M˜ tDt) = dM˜ tDt + M˜ tdDt + hdM,˜ dDit

= DtdMt − DtdhM,Lit + M˜ tdDt + hdM, dDit, because Dt is a martingale and so its variation with respect to a ﬁnite variation process is hhM,Li,Di = 0. We can also compute

L − 1 hL,Li 1 1 dD = d(e t 2 t ) = D dL − dhL, Li + dhL, Li = D(dL ). t t 2 t 2 t t Therefore

d(M˜ tDt) = DtdMt − DdhM,Lit + M˜ tDt + DthdM, dLit = DtdMt + M˜ tdDt.

Because Mt and Dt are martingales, M˜ tDt is a martingale with respect to P .

26.1 Cameron–Martin formula R t For example, take Lt 0 b(Bs, s)dBs by setting

R t 1 R t 2 b(Bs,s)dBs− b (Bs,s)ds Qt = Pte 0 2 0 .

If Mt = Bt, then D Z t E Z t M˜ = Bt − Bt, b(Bs, s)dBs = Bt − b(Bs, s)ds 0 0 is a Brownian motion with respect to Q. If b(Bs, s) = g(s), then

R ∞ 1 R ∞ 2 g(s)dBs− g (s)ds D∞ = e 0 2 0 . R t If we write h(t) = 0 g(s)ds, then for any function Φ on C(R+, R),

W Q Q W E [Φ(B + h)] = E [Φ((B − h) + h)] = E [Φ(B)] = E [D∞Φ(B)] Math 288 Notes 57 where W is the Wiener measure with respect to P . This is the Cameron– Martin formula. Intuitively, this is true because, in the case when Φ is given by the simple discrete form F (B(t1))G(B(t2)),

W E [F (B(t1) + h(t1))G(B(t2) + h(t2))] Z 2 (x−y)2 − x − = e 2t1 e 2(t2−t1) F (x + h(t1))G(y + h(t2))

Z 2 (x−y)2 h(t )x (h(t )−h(t ))(x−y) h(t )2 (h(t )−h(t ))2 − x − + 1 + 1 2 − 1 ( 1 + 2 1 )2 = F (x)G(y)e 2t1 2(t2−t1) t1 (t2−t1) 2 t1 t2−t1 .

Here the last three terms correspond to D∞. There are some applications of Itˆocalculus in solving diﬀerential equations involving processes.

∆ 1. Suppose ∂tu = 2 u with the initial condition u0(x) = φ(x). Then

u(x, t) = Ex[φ(xt)]

for some xt. 2. Let ∆u(x) = 0 for x ∈ D and u(x) = g(x) on x ∈ ∂D. Then

u(x) = Ex[φ(xT )],T = inf{t : xt ∈/ D}.

3. Feynman–Kac formula. Let us prove the ﬁrst statement. Let v(s, x) = u(t − s, x). Then

∆ ∂ v + v = 0. s 2 Now Itˆo’sformula tells us Z t∂v ∆v Ex[v(t, X(t))] = Ex[v(0, x0)]+E + ds+E[martingale] = v(0, x) = u(t, x). 0 ∂s 2

So Ex[φ(X(t))] = u(t, x). Math 288 Notes 58

27 April 5, 2017 27.1 Application to the Dirichlet problem Let us look at the Dirichlet problem, given by

∆u(x) = 0 for x ∈ D, u(x) = g(x) for x ∈ ∂D.

Consider the Brownian motion Bt starting from x. Then

Z t ∆ u(Bt) = u(B0) + u(Bs)ds + (martingale term) = u(B0) + (martingale). 0 2

If we take T = inf{t : Bt ∈/ D} then stopping here and taking expectation gives, by the optional stopping theorem,

Ex[u(Bt∧T )] = u(x). Let us assume that P (T < ∞) = 1. This is true if, in particular, D is bounded. Then u(x) = lim x[Bt∧T ] = x[BT ] = x[g(BT )]. t→∞ E E E Let T(x, dy) be the probability measure of the Brownian motion starting from x to exit at y. Then Z u(x) = g(y)T(x, dy). (∗)

This arguments shows that if u solves the equation, it has to be given by this integral. But when does (∗) actually solve the problem? You need some condition on D. There is something called the exterior cone condition that implies the existence of the Dirichlet problem. The same condition implies that (∗) solves the equation. Consider the Dirichlet problem given by ( 0 if |x| = R, g(x) = 1 if |x| = .

The solution is going to be, in d = 2,

log R − log|x| uR(x) = , log R − log and for d = 3, 1 − 1 R R |x| u (x) = 1 1 . R − This has the interpretation that u(x) is the probability of the Brownian motion starting from x exits ≤ |x| ≤ R from S. Math 288 Notes 59

Taking R → ∞, we get ( 1 if d = 2 uR(x) → < 1 if d ≥ 3.

That is, for any > 0 the Brownian motion starting from x will visit B before going to inﬁnity, in the case d = 2. For d ≥ 3 this is not true.

27.2 Stochastic diﬀerential equations We want to look at equations of the form

dXt = σ(t, Xt)dBt + b(t, Xt)dt, with respect to a ﬁltration Ft. Something being a solution to the equation E(σ, b) means

(i)(Ω , Ft,P ) is a probability space,

(ii) Bt is a Brownian motion with respect to Ft,

(iii) there exists Xt a continuous progressively measurable with respect to Ft such that Z t Z t Xt = σ(s, Xs)dBs + b(s, Xs)ds + X0. 0 0

Definition 27.1. A solution X is called a strong solution if Ft is adapted with respect to the canonical filtration of the Brownian motion. Definition 27.2. A equation is pathwise unique if any two solutions X and 0 0 X for the same Bt has X = X almost surely. Theorem 27.3. If σ and b are uniformly Lipschitz then the SDE has a strong solution and is pathwise unique.

0 Proof. Iterate this process, i.e., let Xt = X and Z t Z t n n−1 n−1 Xt = σ(s, Xs )dBs + b(s, Xs )ds + X. 0 0 We want to check that this iteration converges. To show this, deﬁne

h n n−1 2i gn(t) = E sup |Xs − Xs | . 0≤s≤t

We claim that tn−1 g (t) ≤ C0 (C )n−1 . n T T (n − 1)! Math 288 Notes 60

We have Z t 2 h n+1 n 2i n n−1 E sup |Xs − Xs | ≤ 2E sup (σ(s, Xs ) − σ(s, Xs ))dBs 0≤s≤t 0≤s≤t 0 Z t 2 n n−1 + 2E sup (b(s, Xs ) − b(s, Xs ))ds . 0≤s≤t 0

Using the Lipschitz condition, we are going to bound this, next time. Math 288 Notes 61

28 April 7, 2017

We want to prove the existence and uniqueness of stochastic PDEs.

28.1 Existence and uniqueness in the Lipschitz case Theorem 28.1. Consider the stochastic diﬀerential equation

dX = σ(t, Xt)dBt + b(t, Xt)dt, (∗) where σ(t, x) and b(t, x) are uniformly Lipschitz with respect to x. Then there exists a pathwise unique solution to (∗), i.e., there exists a (Ω, F,P ) with continuous sample paths such that

Z t Z t Xt = X + σ(s, Xs)dBs + b(s, Xs)ds 0 0 almost surely.

0 Proof of existence. Let us define, Xt = X and Z t Z t n n−1 n−1 Xt = X + σ(s, Xs )dBs + b(s, Xs )ds. 0 0 Let us define h n n−1 2i gn(T ) = E sup |Xs − Xs | . 0≤s≤t For fixed T and t ∈ [0,T ], we have

Z s 2 h n+1 n 2i h n n−1 i gn+1(t) = E sup |Xs − Xs | ≤ 2E sup (σ(Xu ) − σ(Xu ))dBu 0≤s≤t 0≤s≤t 0 Z s 2 h n n−1 i + 2E sup (b(Xu ) − b(Xu ))du 0≤s≤t 0 Z t Z t h n 2 n−1 2 n n−1 2 i ≤ CE (σ(Xu ) − σ(Xu )) du + t (b(Xu ) − b(Xu )) du 0 0 Z t Z t h n n−1 2 i ≤ CT E sup |Xs − Xs | du ≤ CT gn(u)du, 0 0≤s≤u 0 by the Burkholder–Davis–Gundy inequality and the Lipschitz condition. From this inequality, you will inductively get tn g (t) ≤ (C )n . n+1 T n!

P∞ 1/2 That is, for any T ﬁxed, n=1 gn(T ) < ∞. Hence the process converges and it is a solution. Math 288 Notes 62

Proof of uniqueness. This actually just follows from the same arguments. If we deﬁne h 2i g(t) = E sup |Xs − Ys| , 0≤s≤t then Z t g(t) ≤ C g(s)ds. 0 Then by Gronwall’s inequality, g = 0.

Let us apply Itˆo’sformula to a solution Xt. We get Z t Z t 1 00 f(Xt) = f(X0) + ∇fsdXs + f (Xs)dhX,Xis 0 2 0 Z t Z t 2 = f(X0) + (∇fs)σ(s, Xs)dBs + (∇fs) b(s, Xs)ds 0 0 Z t 1 00 2 + f (Xs)σ (s, Xs)ds 2 0 Z t 1 2 = f(X0) + σ ∆ + b∇ f(Xs)ds + (martingale). 0 2 We call the operator 1 L = σ2∆ + b∇ 2 the generator of the process. 1 The Brownian motion is a solution to ∂tu = 2 ∆u. This is illustrated by Itˆo’sformula Z t 1 f(Bt) = f(B0) + ∆f(Bs)ds + M. 0 2 1 2 Likewise, to solve the equation ∂tu = Lu for L = 2 σ ∆ + b∇, you can solve the equation ds = σdB + ddt.

28.2 Kolomogorov forward and backward equations Theorem 28.2. Let P (s, x; t, y) be a transition probability density such that Z Ex,s[h(x(t))] = p(s, x; t, y)h(y)dy.

Then

∂sp(s, x; t, y) + Lxp(s, x; t, y) = 0, ∗ ∂tp(s, x; t, y) = Lyp(s, x; t, y). Math 288 Notes 63

Proof. Let f(t, x) be the probability density of x(t). Denote f0(x) = f(0, x). Then formally Z f(x, t) = p(0, z; t, x)f0(z)dz.

Then we get Z Z f(t, x)h(x)dx = f0(x)Ex[h(x(t))]dx.

The derivative with respect to t gives Z Z ∂t f(t, x)h(x)dx = ∂t f0(x)Ex[h(x(t))]dx Z h Z t i = ∂t f0(x)Ex h(x(0)) + (Lh)(x(s))ds 0 Z Z = f0(x)Ex(Lh)(x(t)) = (Lh)(x)f(t, x)dx Z = h(x)(L∗f)(t, x)dx.

This gives the forward equation, and its dual is the backward equation. Math 288 Notes 64

29 April 10, 2017

We have looked at the Kolmogorov’s equation. Given a diﬀerential equation

dXt = σ(Xt)dBt + b(Xt)dt, let f(t, x) be the probability density of the solution Xt. Then σ(x)2 ∂ f(t, x) = L∗f(t, x), where L = ∆ + b(x)∇. t 2 The Kolmogorov equations say that, if the transition probability is p(s, x; t, y), Z Ex;s[h(Xt)] = p(s, x; t, y)h(y)dy, then ∗ (∂s + Lx)p(s, x; t, y) = 0, (∂t − Ly)p(s, x; t, y) = 0. Proof. Recall that by Itˆo’sformula, Z t h(Xt) = h(Xs) + (Lh)(Xs)ds + (martingale). 0 So Z Z Z ∂t f(t, x)h(x) = ∂t f0(x)Ex(h(Xt)) = f0(x)Ex[(Lh)(Xt)] Z Z = f(t, x)(Lh)(x)dx = (L∗f)(t, x)h(x)dx.

More generally, if f(t, x) is the density relative to µ, then Z Z Z ∗ ∂t f(t, x)h(x)µ(dx) = f(t, x)(Lh)(x)µ(dx) = (Lµ)f(t, x)h(x)µ(dx) and so ∗ ∂tf = Lµf. (∗) ∗ Deﬁnition 29.1. µ is called an invariant measure of the process if Lµ1 = 0, i.e., 1 is a solution to (∗). In other words, µ is invariant if and only if Z (Lf)dµ = 0 for any f with compact support. ∗ Deﬁnition 29.2. µ is called a reversible measure of the process if Lµ = L, i.e., Z Z f(Lh)dµ = (Lf)hdµ.

In this case, Z − h(Lh)dµ = Dµ(h) is called the Dirichlet form of the process. Math 288 Notes 65

29.1 Girsanov’s formula as an SDE Girsanov’s formula says that if

dQ R t b(X )dB − 1 R t b2(X )ds = e 0 s s 2 0 s dP [0,t] and M is a local martingale with respect to P , then

D Z t E M˜ = M − M, b(Xs)dBs 0 is a local martingale with respect to Q. Take M = B with respect to P . Then this implies that

D Z t E Z t βt = M˜ = Bt − Bt, b(Bs)dBs = Bt − b(Bs)ds 0 0 is a martingale with respect to Q. We note that hM,˜ M˜ it = t. Therefore M˜ is in fact a Brownian motion with respect to Q. But Bt is no longer a Brownian motion in Q. So we change notation and write Xt instead of Bt. Then

dXt = dβt + b(Xt)dt, (†) i.e., Xt is the solution to the SDE (†), i.e., the probability measure Q is the solution to (†). This means that when you have a SDE that looks like dXt = σdBt + bdt, the really important term is σdBt, since the bdt comes from Girsanov’s formula by changing the probability measure. We will look at three examples of SDEs: (1) Ornsetin–Uhlenbeck process (2) Geometric Brownian motion - Black–Scholes formula (3) Bessel process

29.2 Ornstein–Uhlenbeck process Let us ﬁrst look at the Ornstein–Uhlenbeck process:

dXt = dBt − λXtdt, where the negative coeﬃcient means that we are pushing back the process into −λt the origin. If we deﬁne Xt = e Yt, then

λt λt λt dYt = d(e Xt) = λYtdt + e dBt − e λXtdt

λt and so dYt = e dBt. That is, Z t −λt −λt −λ(t−s) Xt = e Yt = e X0 + e dBs, 0 Math 288 Notes 66 and this is the exact solution. The generator is 1 ∂2 ∂ L = − λx , 2 ∂x2 ∂x −c x2 and you can check that the Gaussian measure µ = dλe λ is an invariant measure.

29.3 Geometric Brownian motion Let us now look at the geometric Brownian motion:

dXt = σXtdBt + rXtdt.

rt If we again make a substitution Xt = e Yt, then

−rt −rt dYt = −rYtdt + e rXtdt + e σXtdBt = σYtdBt.

If we simply guess Yt = f(Bt, t), then

f 00(B , t) dY = f 0(B , t)dB + t dt + (∂ f)(B , t)dt. t t t 2 t t

00 0 So we want ∂tf + f /2 = 0 and f (Bt, t) = σf. You can solve this, and you get

σ2 Y = f(B , t) = eσBt − t. t t 2 Therefore the exact solution is given by

σ2 rt σBt− t Xt = e e 2 + X0. Math 288 Notes 67

30 April 12, 2017

Last time we looked at the geometric Brownian motion, given by

dXt = σXtdBt + γXtdt.

There is something called a Black–Scholes model for stock prices. Let St be the stock price at t and let βt be the bond price at t. The value of investment is given by Vt = atSt + btβt. The stock dynamics is

dSt = µStdt + σStdBt, dβt = rβtdt, dVt = atdSt + btdβt.

The constants µ, σ, r are given, and we want to ﬁnd at, bt such that VT = h(ST ) is maximal. We have

dVt = atdSt + bdβt = at(µStdt + σStdBt) + btrβtdt

= (atµSt + brβt)dt + atσStdBt.

Suppose Vt = f(t, St) is a function of t and St. Then by Itˆo’sformula, 1 dV = ∂ f + f 00dhS, Si + f 0(t, S )dS t t 2 t t t h 1 i = ∂ f + σ2S2f 00 + µS f 0 dt + f 0σS dB . t 2 t t t Comparing this with the other equation gives

1 00 2 0 ft + 2 f σ St at = f (t, St), bt = . rβt

Using the formula f(t, St) = atSt + btβt gives 1 f + σ2x2f (t, x) + f (t, x)x − rf(t, x) = 0. t 2 xx x This is the Black–Scholes equation.

30.1 Bessel process Let us now look at the Bessel process given by p dXt = 2 XtdBt + mdt, m > 0.

1 d If we look at the d-dimensional Brownian motion βt = (β , . . . ,√ β ) and deﬁne 2 1 2 d 2 Xt = βt = (β ) + ··· + (β ) , then we can check that dXt = 2 XtdBt + ddt. 2 So m = d is exactly the process of |βt| . Let us deﬁne Tr = inf{t ≥ 0 : Xt = r}, where we assume x ≤ r. We now claim that P (Tr = ∞) = 0. Math 288 Notes 68

Suppose that < x < r and deﬁne

( 1−m/2 Xt m 6= 2 Mt = log Xt m = 2.

Then Itˆo’scalculus gives m 1 m m dM = 1 − X−m/2dX + 1 − − X−m/2−1dhX,Xi t∧T 2 t t∧T 2 2 2 t∧T h m 1 m m i = 1 − mX−m/2 + 1 − − X−m/2 dt 2 t∧T 2 2 2 t∧T m −m/2+1/2 m −m/2+1/2 + 2 1 − X dBt∧T = 2 1 − X dBt∧T . 2 t∧T 2 t∧T Thus Z t∧T m −m/2+1/2 Mt∧T = 2 1 − Xs dBs. 2 0 This is a martingale for ﬁxed, and just a local martingale if → 0. By optional 1−m/2 stopping, we have E[Mt∧T∧TA ] = x and if you solve the equation, we get

A1−m/2 − x1−m/2  m 6= 2  A1−m/2 − 1−m/2 P (T < TA) = log A − log x  m = 2. log A − log This shows that for m ≥ 2, as → 0, the probability that the Bessel process reach 0 is zero. Math 288 Notes 69

31 April 14, 2017

Last time we talked about the Bessel process. This is given by p dXt = 2 XtdBt + mdt.

d 2 1−m/2 An example, when m = d is an integer, is Xt = |Bt | . If we deﬁne Mt = Xt then m −m/2 p dM = 1 − X 2 X dB . t 2 t t t Then Mt is a local martingale. −1 −1 In the case of m = 3, we would have Mt = |Bt| , and so d|Bt| = −2 C|Bt| dBt. Then Z t −1 −2 |Bt| = |Bs| dBs 0 is a local martingale. R t −2 2 −1 Recall that E( 0 |Bs| dBs) < ∞ then |Bt| would be a martingale. How- ever, you can check that Z t h −4 i E |Bs| ds = ∞. 0 −1 Indeed, |Bt| is a local martingale but not a martingale. Likewise, in d = 2, Mt = log|Bt| is a local martingale but not a martingale.

31.1 Maximum principle Theorem 31.1. Suppose you want to solve the equation 1 ∂ u + ∆u + g = 0, u(T, x) = f(x) t 2 with boundary condition u(t, x) = k(t, x) for x ∈ ∂Ω and T > 0. Or more generally, consider the equation ∂tu + Lu + g = 0 (in this case, we use the stochastic process generated by L instead of the Brownian motion). Then

h Z τ∧T i u(0, x) = Ex f(x, T ))1(τ≥T ) + k(τ, x(τ))1(τ

Z t ∆ u(t, X(t)) = ∂tu + u ds + (martingale) + u(0, x). 0 2 If we deﬁneτ ˜ = τ ∧ T , then Z τ∧T Ex[u(τ ∧ T,X(τ ∧ T ))] + Ex g(s, Xs)ds = u(0, x). 0 This proves the theorem. Math 288 Notes 70

Corollary 31.2. If f2 ≥ f1, g2 ≥ g1, and k2 ≥ k1, then u2(s, x) ≥ u1(s, x), i.e., we get monotonicity. Moreover, if g = 0, the maximum of u(t, x) in [0,T ] × Ω can be achieved on the boundary. If the maximum is achieved in the interior, then u is a constant.

Proof. If sup f ∨ sup k ≤ M, then Ex ≤ M.

31.2 2-dimensional Brownian motion

Consider a 2-dimensional Brownian motion Bt = Xt + iYt. Theorem 31.3 (Conformal invariance, Theorem 7.8). Suppose Φ is analytic. Then Φ(B) is a Brownian motion up to a time change. Proof. Let us compute dΦ(B). We have

∂g ∂g ∂k ∂k dΦ(B) = d(g(X ,Y ) + ik(X ,Y )) = dX + dY + i dX + dY . t t t t ∂x t ∂y t ∂x t ∂y t The quadratic variation of one component is

h ∂g 2 ∂g 2i hdg, dgi = + dt. ∂x ∂y

So g(Xt,Yt) is a Brownian motion with time change. The similar thing holds for k. Moreover, D ∂g ∂g ∂k ∂k E ∂g ∂k ∂g ∂k hdg, dki = dX + dY, dX + dY = + dt = 0 ∂x ∂y ∂x ∂t ∂x ∂x ∂y ∂y from the Cauchy–Riemann equations. Math 288 Notes 71

32 April 17, 2017

Today we are going to look more at the 2-dimensional Brownian motion.

32.1 More on 2-dimensional Brownian motion

Theorem 32.1 (7.19). Let Bt be a Brownian motion. Then

Bt = exp(βHt + iγHt )

R t 2 where Ht = 0 ds/|Bs| and β and γ are independent Brownian motions.

Proof. We want βHt + iγHt = log Bt. Diﬀerentiate both sides, and we get

2 dXt + idYt 1 1 2 dXt + idYt d log Bt = − (dXt + idYt) = . Xt + iYt 2 Xt + iYt Xt + iYt Then we get XdX + Y dY XdY − Y dX dβ = , dγ = . t (X2 + Y 2) t X2 + Y 2 Now we have (XdX + Y dY )2 (X2 + Y 2)dt dt hdβ , dβ i = = = , t t (X2 + Y 2)2 (X2 + Y 2)2 (X2 + Y 2) dt hdγ , dγ i = ··· = , t t (X2 + Y 2) (XdX + yDY )(XdY − Y dX) hdβ , dγ i = = 0. t t (X2 + Y 2)2

What we want is actually βHt instead of βt. If β is a Brownian motion, then 1 hdβHt , dβHt i = dHt = 2 2 dt Xt + Yt and so we see that βHt for β a Brownian motion works. From this we get, we get 1 1 dHt = 2 dt = dt |Bt| exp(2βHt ) and so

exp(2βHt )dHt = dt.

(λ) (λ) (λ) −1 Lemma 32.2. Let T1 = inf{t ≥ 0; βt = 1}, where Bt = λ βλ2t is the rescaled Brownian motion. Then 4 H − T (log t/2) → 0 (log t)2 t 1 in probability as t → ∞. Math 288 Notes 72

Proof. From exp(2βHt )dHt = dt, we have nZ s o 2βu inf e du = t = Ht. s 0

Let λt = log t/2. Then the lemma tells us that

2 λt P (Ht > λt T1+) → 0. Then

2 (λt) Z λt T1+ 2 2βu P (Ht > λt T1+λt) = P e du < t 0 (λt) Z T1+ log λt 1 2λ β(λt) = P + log e t v dv < 1 λt 2λt 0 1 Z T1+ 2λtβv ≤ P log e dv < 1 = P max βv < 1 = 0 2λt 0 0≤v≤T1+

2 after the rescaling u = λt v.

Theorem 32.3. If Bs is a 2-dimensional Brownian motion starting from any z 6= 0, −a/2 1 lim P inf |Bs| ≤ t = . t→∞ 0≤s≤t 1 + a Proof. Due to scaling, we as may as well assume that z = 1. We have log inf |Bs| = inf βHs = inf βu 0≤s≤t 0≤s≤t 0≤u≤Ht by monotonicity. Now we have 2 1 (λt) log inf |Bs| = inf βu = inf βs . 0≤s≤t 0≤u≤Ht −2 log t λt 0≤s≤λt Ht

−2 (λt) By the lemma, λt Ht converges to T1 and also

(λt) inf βs = inf βs. (λt) 0≤s≤T1 0≤s≤T1 We will continue next time. Math 288 Notes 73

33 April 19, 2017

Last time we showed that a 2-dimensional Brownian motion can be expressed as Z t 1 βHt +iγHt Bt = e ,Ht = 2 ds, 0 |Bs| where β and γ are independent Brownian motions. Lemma 33.1 (7.21). In probability, 4 H − T (log t/2) → 0 (log t)2 t 1

(λ) (λ) as t → ∞, where T1 = inf{s ≥ 0, βt = 1}.

 −a/2 1 Proposition 33.2 (7.22). lim P inf |Bs| ≤ t = . t→∞ 0≤s≤t 1 + a

Proof. Because Ht is monotone, we have log inf |Bs| = inf βHs = inf βu. 0≤s≤t 0≤s≤t 0≤u≤Ht

Then 2 1 (λt) (λt) log inf |Bs| = inf βu = inf βv = inf βv 2 (λ ) log t 0≤s≤t λt 0≤u≤Ht 0≤v≤Ht/λ t t 0≤v≤T1

2 after the rescaling u = λt v. Then the probability is 1 (λt) P log inf |Bs| ≤ −a = P inf βv ≤ −a λt 0≤s≤t (λt) 0≤v≤T1 1 = P inf βv ≤ −a = P (T−a < T1) = 0≤v≤T1 1 + a because of the property of the 1-dimensional Brownian motion. (You can see this from solving the Laplace equation with u(−a) = 1 and u(1) = 0.)

33.1 Feynman–Kac formula Theorem 33.3. Consider an equation 1 ∂ u + ∆u + V (u) = 0 t 2 with the boundary condition u(T, x) = f(x). Then

 Z t u(0, x) = Ex exp V (x(s))ds f(x(T )) . 0 Math 288 Notes 74

Proof. Take a u solving the equation. Then

Z t 1 Z t d u(t, x(t)) exp V (x(s))ds = ∂tu + ∆u exp V (x(s))ds dt 0 2 0 Z t + ∇xu(t, x(t))dx(t) + u exp V (x(s))ds V (x(t))dt 0 Z t + du d exp V (x(s))ds 0 1 Z t = (∂tu + ∆u + V u) exp V (x(s))ds dt + (martingale) 2 0 is a martingale. So we get

 Z T u(0, x) = Ex exp V (x(s))ds f(x(T )) . 0

1 Here is an intuitive proof. The equation can be written as ∂tu = ( 2 ∆+V )(u) and so we can write t( 1 ∆+V ) u = e 2 u0. There is also the Trotter product formula

eA+B = lim (eA/neB/n)n. n→∞ Then our solution can be thought of as

(x −x )2 (x −x )2 ∆ t t V ∆ t t V 1 − 0 1 t V (x ) 0 1 2 t V (x ) e 2 n e n e 2 n e n ··· = √ e 2(t/n) e n 1 e 2(t/n) e n 2 ··· . 2π ···

Then path x0 → x1 → x2 → · · · becomes a discrete approximation of the Brownian motion, and the factor we pick up is

t (V (x )+···+V (x )) R t V (x(s))ds e n 1 n → e 0 under the limit n → ∞.

33.2 Girsanov formula again There is a nice proof of Girsanov’s formula. Let b(t, X(t)) be a function and suppose Q solves the equation

dX = bdt + dβ.

Let P be the law of β. Then we want to prove that

dQ Z t 1 Z t = exp b(s, X(s))dx(s) − b2ds . dP 0 2 0 Math 288 Notes 75

Let us deﬁne Z t Z(t) = b(s, X(s))dX(s) 0 and consider the object

1 Z t d exp λX(t) + Z(t) − (λ2 + 2λb(s, X(s)) + b2)ds . 2 0 This is going to be a martingale plus a dt term. Our claim is that the dt term vanish. We check that Itˆocalculus gives 1 1 1 λdX − (λ2 + 2λb + b2)dt + dZ + λdXdZ + λ2dXdX + dZdZ. 2 2 2

1 2 λ2 1 b2 Here λdXdZ = bdt and 2 λ dXdX = 2 dt and 2 dZdZ = 2 dt. So the dt term indeed vanishes. Everything follows from this. Math 288 Notes 76

34 April 21, 2017

The Girsanov formula says that

dQ Z t 1 Z t = exp b(s, X(s))dX(s) − b2(s, X(s))ds , dP 0 2 0 where Q solves dX = b(t, X(t))dt = dβ. The idea is that the Brownian motion measures something like

2 2 − (x2−x1) − (x3−x2) −··· (const)e 2t/n 2t/n .

Here the discrete diﬀerence Xt − Xt+1 is like dβ(t) = dX − b(t, X(t))dt and so this quantity is like 1 h t i2 (x − x )2 exp − x −x −b(t , x ) −· · · = exp − 1 2 +b(t , x )(x −x )−· · · . 2t/n 2 1 1 1 n 2t/n 1 1 2 1 P R t Here j b(tj, xj)(xj+1 − xj) is like 0 b(s, x(s))ds. So this is why we get Gir- sanov’s formula.

34.1 Kipnis–Varadhan cutoﬀ lemma Theorem 34.1. Suppose µ is a reversible measure with respect to a process with generator L. Let D(f) = − R fLfdµ be the Dirichlet form. Then

µ 1q 2 P sup |f(X(s))| ≥ ` ≤ kfkL2µ + D(f). 0≤s≤t `

Proof. Suppose ∂tu + Lu + g = 0. Then by Itˆocalculus,

du(t, x) = (∂tu + Lu)dt + (martingale). Now consider the stopping time τ = inf{s ≥ 0, f(X(s)) = `}. Then Z t∧τ u(t ∧ τ, x) = −g(s, X(s))ds + u(0, x) + (martingale). 0

Suppose a function φλ solves the equation ( (λ − L)φ = 0 x ∈ G = {y : f(y) < `},

φλ = 1 x ∈ ∂G.

−λt If we take u(t, x) = φλ(x)e then

−λt −λt ∂tu + Lu = −λφλ(x)e + λφλ(x)e = 0.

−λτ Then you can show that φλ = Exe . Then we get

λt µ λtp µ 2 P (τ < t) ≤ e E φλ(x) ≤ e E [φλ(x) ] Math 288 Notes 77 by Chebyshev’s inequality. But recall that φλ is an eigenfunction of L. But note that for any h with h = 1 on ∂G, we have hh, (λ − L)hiµ ≥ hφλ, (λ − L)φλiµ and so

2 2 1 kφ k 2 ≤ khk + D(h). λ L 2 λ Therefore r 1 P (τ < t) ≤ eλt khk2 + D(h). 2 λ Take λ = 1/t and h = `−1(|f| ∧ `) gives the theorem. Math 288 Notes 78

35 April 24, 2017

Let µ be a reversible (R gLfdµ = R (Lg)fdµ) measure of a process with Dirichlet form D = − R fLfdµ. Then we want to show

 cq 2 P sup |f(X(s))| ≥ ` ≤ kfkL2 + tD(f). 0≤s≤t ` µ

Proof. Let G = {x : |f(x)| ≥ `} and τ is the stopping time to exit the open set {y : f(y) < `}. Suppose φλ is the ground state such that

(λ − L)φλ(x) = 0 for x ∈ G, φλ(x) = 1 for x ∈ ∂G.

This being a ground state means that λ is minimal. So φλ is nonnegative. (The first eigenfunction of an elliptic problem is nonnegative.) −λt If we define u(t, x) = φλ(x)e then ∂tu + Lu = 0. Then Itô’sformula implies that Z t u(0, x) = −Ex (∂s + L)u(s, x)ds + Ex[u(t, x)] = Ex[u(t, x)]. 0 So

φλ(x) = Ex[u(t ∧ τ, x)] = Ex[u(τ, x)1(τ≤t)] + Ex[u(t, x)1(τ>t)] −λτ −λt = Ex[e 1(τ≤t) + e φλ(x)1(τ>t)]. Now we have, with respect to µ,

−λt µ −λτ µ µ 2 1/2 e Pµ(τ ≤ t) ≤ E [e 1(τ≤t) ≤ E φλ(x) ≤ (E (φλ(x) )) .

Since φλ is the minimizer, we have

hφλ, (λ − L)φλi ≤ hh, (λ − L)hi

2 for any h such that h(x) = 1 on ∂G. This implies that λkφλk + D(φλ) ≤ 2 1 λkhk + D(h). If we let h = ` min(|f(x)|, `), we have 1 1 1 kφ k2 ≤ khk2 + D(h) = kfk2 + D(f). λ 2 2 λ `2 2 `2λ Plugging this into the previous formula gives

eλt r 1 eq P (τ ≤ t) ≤ kfk2 + D(f) ≤ kfk2 + tD(f). µ ` 2 λ ` 2

35.1 Final review Here are the things we have done so far: • construction of Brownian motions (Gaussian processes, L´evy’sconstruc- tion) Math 288 Notes 79

• martingales (discrete) • local martingales (continuous time); roughly speaking, M is a local martingale if it can be appproximated by a martingale • a nonnegative local martingale is a supermartingale

• construction of quadratic variation hM,Mit

• if E[hM,Mit] < ∞ for each t then M is a martingale (Theorem 4.13) • stochastic integration R HdM, when R HsdhM,Mi < ∞ almost surely R s • if E[ H dhM,Mi∞] < ∞ then it is a marginale • Itˆo’sformula (in general, we only know that R ··· dM is a local martingale) • Burkholder–Davis–Gundy inequality

∗ p p/2 E[(MT ) ] ≤ CpE[hM,MiT ].

• Levi’s theorem for characterizing Brownian motions, hM,Mit = t R t n R t n • dominated convergence, 0 Hs dXs → 0 HsdXs in probability if Hs → n R t 2 Hs and |Hs | ≤ Ks and 0 Ks dhX,Xis < ∞ (Proposition 5.18) • existence and uniqueness of dX = bdt + σdβ for σ and β Lipschitz 1 2 • applications of Itô’sformula; dF = ∇F dM + 2 ∇ F dhM,Mi, check that ∇F dM is a martingale, and then take expectation • Feynman–Kac formaula, Girnasov, Dirichlet problem, Kipnis–Varadhan, ... You will not remember the details, and I will not remember the details. But it is important to have the big picture. Index adapted variable, 21 Kipnis–Varadhan cutoff lemma, 76 Kolmogorov’s theorem, 13 backward martingale, 32 Bessel process, 67 Levi’s theorem, 27 Black–Scholes model, 67 Levi’s theorem on local Blumenthal’s zero-one law, 17 martingales, 50 bracket, 39 local martingale, 33 Brownian motion, 11, 15 Burkholder–Davis–Gundy martingale, 22 inequality, 51 mean, 7 moment generating function, 5 Cameron–Martin formula, 57 conditional expectation, 8 optional stopping theorem, 23 covariant matrix, 7 Ornstein–Uhlenbeck process, 65 Dirichlet form, 64 pathwise uniqueness, 59 Doob inequality, 23 predictable, 30 Doob upcrossing inequality, 24, 25 progressive measurability, 21 elementary process, 41 quadratic variation, 34 filtration, 20 finite variance process, 32 reflection principle, 19 reversible measure, 64 Gaussian process, 8 right continuous, 20 Gaussian random variable, 5 Gaussian space, 8 semimartingale, 39 Gaussian vector, 6 stopping time, 18, 21 Gaussian white noise, 10 strong Markov property, 18 generator of a process, 62 strong solution, 59 geometric Brownian motion, 66 submartingale, 22 Girsanov’s theorem, 55 submartingale decomposition, 30 indistinguishable, 13 uniform integrability, 6 invariant measure, 64 Itô’sformula, 48 Wiener measure, 16