Stochastic Integration

Prakash Balachandran Department of Mathematics Duke University

June 11, 2008

These notes are based on Durrett’s Stochastic Calculus, Revuz and Yor’s Continuous Martingales and Brownian Motion, and Kuo’s Introduction to Stochastic Integration.

1 Preliminaries

Deﬁnition: A continuous-time process Xt is said to be a continuous local martingale w.r.t. {Ft, t ≥ 0} if there are stopping times Tn ↑ ∞ such that

 X on {T > 0}  Tn∧t n Tn Xt = 0 on {Tn = 0}

is a martingale w.r.t. {Ft∧Tn : t ≥ 0}. The stopping times {Tn} are said to reduce X.

Remarks:

T 1. I brooded over why we set Xt = 0 in the deﬁnition, and this is the only explanation I could ﬁnd:

If we deﬁned T Xt = XT ∧t on {T ≥ 0},

T T then X0 = X0, so according to the above deﬁnition of a local martingale, Xt a martingale implies T E[X0 ] = E[X0] < ∞.

T So, with the above deﬁnition of Xt , X0 has to be integrable.

1 Since we want to consider more general processes in which X0 need not be integrable, we set  XT ∧t on {T > 0} T  Xt = 0 on {T = 0}

so that

 X0 on {T > 0} T  X0 = 0 on {T = 0}

T and according to the deﬁnition of a local martingale, Xt a martingale implies that

T E[X0 ] = E[X0; T > 0] < ∞

which does not necessarily imply that X0 is integrable since E[X0; T > 0] ≤ E[X0].

Thus, our deﬁnition of a continuous local martingale frees us from integrability of X0.

2. We say that a process Y is locally A if there is a sequence of stopping times Tn ↑ ∞ so that the T stopped processes Yt has property A.

Now, you might ask why the hell we should care about continuous local martingales. Again, I thought about this a lot, and these are the only reasons I could salvage:

(1) (n) n (j)on Example 1: Let Bt = Bt ,...,Bt be n-dimensional Brownian motion, where Bt are j=1 independent Brownian motions on R. r (1)2 (n)2 Suppose we’re interested in the process ||Bt|| = Bt + ··· + Bt ; it can be shown that:

n Z t (j) Z t X Bs n − 1 1 ||B || = dB(j) + ds t ||B || s 2 ||B || j=1 0 s 0 s and that n Z t (j) X Bs W = dB(j) t ||B || s j=1 0 s is a Brownian motion, and hence, a martingale.

Now, what about the second integral above? It’s not immediately obvious how this integral behaves, but it’s certainly not a martingale. In fact, it can be shown that 1 for t ≥ 0 is a continuous local ||Bt|| martingale, and so R t ds is a continuous local martingale. 0 ||Bs||

2 Since a continuous martingale is certainly a continuous local martingale, it follows that ||Bt|| is a continuous local martingale.

Now, suppose that a particle is exhibiting Brownian motion. If w(t, ω) is any suitable process which represents a quantity that varies with the distance from the origin to the particle, the process

Z t W (ω) = w(s, ω)d||Bs||(ω) 0 represents the total accumulation of this quantity, along a path of Brownian motion. So, we need to know how to integrate processes w.r.t. continuous local martingales to evaluate this quantity (and ensure that it does, in fact, exist).

Example 2: Let Xt be a continuous martingale, and let φ be a convex function (imagine that Xt is the

−tXt interest rate at time t, so that φ(Xt) = e is the present value of a dollar made in t years).

Theorem 1 If E[|φ(Xt)|] < ∞, then φ(Xt) is a submartingale.

Proof: Jensen’s inequality for conditional expectation states that if φ is convex, and E[|Xt|],E[|[φ(Xt)|] < ∞ for each t, then for s < t:

φ(Xs) = φ (E[Xt|Fs]) ≤ E[φ(Xt)|Fs].

On the other hand, we have:

Theorem 2 φ(Xt) is always a local submartingale.

For the proof of this, see the corollary after Theorem 4.

R T −tXt R T So, if φ(t) is a cash ﬂow from now to time T , 0 φ(t)e dt = 0 φ(t)dYt is the net present value of −tXt this cash ﬂow, where we’ve set dYt = e dt. Again, we need to know how to integrate processes w.r.t. continuous local martingales to evaluate this quantity (and ensure that it does, in fact, exist).

Example 3:

2 Deﬁnition: Deﬁne Lad(Ω,L [a, b]) denote the space of stochastic processes f(t, ω) satisfying:

1. f(t, ω) is adapted to the ﬁltration {Ft} of Brownian motion.

R b 2 2. a |f(t, ω)| dt < ∞ almost surely.

2 R b 2 Deﬁnition: Deﬁne L ([a, b]×Ω) to be the space of {Ft} adapted processes f(t, ω) such that a E(|f(t, ω)| )dt < ∞.

3 2 hR b 2 i R b 2 Now, by Fubini’s Theorem if f ∈ L ([a, b] × Ω), E a |f(t, ω)| dt = a E[|f(t, ω)| ]dt < ∞. Thus, R b 2 2 a |f(t, ω)| dt < ∞ almost surely, so that f ∈ Lad(Ω,L [a, b]). Since f was arbitrary, we must have

2 2 L ([a, b] × Ω) ⊆ Lad(Ω,L [a, b]).

R t 2 Now, in Stochastic Calculus, one constructs the integral 0 f(s, ω)dBs for f ∈ L ([a, b] × Ω). In this R t case, we have that 0 f(s, ω)dBs is a martingale.

2 R t However, when f ∈ Lad(Ω,L [a, b]), 0 f(s, ω)dBs need not be a martingale. Now, in order to proceed, we need a couple of theorems about continuous local martingales, and theorems concerning variance and covariance processes. They may seem irrelevant, now, but they’ll come in handy later.

4 2 Continuous Local Martingales

The ﬁrst section was supposed to convince you why you should care about continuous local martingales. Now, we prove some theorems.

Theorem 3 (The Optional Stopping Theorem) Let X be a continuous local martingale. If S ≤ T are stopping times, and XT ∧t is a uniformly integrable martingale, then E[XT |FS] = XS.

Proof: The classic Optional Stopping Theorem states that:

If L ≤ M are stopping times and YM∧n is a uniformly integrable martingale w.r.t. Gn, then

E[YM |GL] = YL.

[2nS]+1 To extend the result from discrete to continuous time, let Sn = 2n . Applying the discrete time result n to the uniformly integrable martingale Ym = XT ∧m2−n with L = 2 Sn and M = ∞, we have

E[XT |FSn ] = XT ∧Sn .

Now, the dominated convergence theorem for conditional expectation states:

If Zn → Z a.s., |Zn| ≤ W for all n where E[W ] < ∞, and Fn ↑ F∞,

E[Zn|Fn] → E[Z|F∞] a.s.

Taking Zn = XT and Fn = FSn , and noticing that XT ∧t a uniformly integrable martingale implies that

E[|XT |] < ∞, we have

E[XT |FSn ] → E[XT |FS] a.s.. Since E[XT |FSn ] = XT ∧Sn → XS a.s., we have that XS = E[XT |FS].

5 Theorem 4 If X is a continuous local martingale, we can always take the sequence which reduces X to be

Tn = inf{t : |Xt| > n}

0 0 or any other sequence Tn ≤ Tn that has Tn ↑ ∞ as n ↑ ∞.

Proof: Let Sn be a sequence that reduces X. If s < t, then applying the optional stopping theorem to

Sn 0 0 Xr at times r = s ∧ Tm and t ∧ Tm gives:

0 0 0 E[Xt∧Tm∧Sn 1Sn>0|Fs∧Tm∧Sn ] = Xs∧Tm∧Sn 1Sn>0.

0 0 Multiplying by 1Tm>0 ∈ F0 ⊆ Fs∧Tm∧Sn :

0 0 0 0 0 E[Xt∧Tm∧Sn 1Tm>0,Sn>0|Fs∧Tm∧Sn ] = Xs∧Tm∧Sn 1Tm>0,Sn>0.

0 0 0 0 0 0 As n ↑ ∞, Fs∧Tm∧Sn ↑ Fs∧Tm , and Xr∧Tm∧Sn 1Sn>0,Tm>0 → Xr∧Tm 1Tm>0 for all r ≥ 0 and

0 0 |Xr∧Tm∧Sn 1Sn>0,Tm>0| ≤ m it follows from the dominated convergence theorem for conditional expectation that:

0 0 0 0 0 E[Xt∧Tm 1Tm>0|Fs∧Tm ] = Xs∧Tm 1Tm>0.

Corollary 1 If X is a continuous martingale, and φ is a convex function, then φ(Xt) is a continuous local submartingale.

Proof: By Theorem 4, we can let Tn = inf{t : |Xt| > n} to be a sequence of stopping times that reduce Tn Tn Tn Xt. By deﬁnition of Xt , we therefore have |Xt | ≤ n ⇒ E[|Xt |] ≤ n, t ≥ 0.

Tn Tn Now, since φ is convex, and |Xt | ≤ n, φ(Xt ) is contained in φ([−n, n]). Since φ is convex, it is Tn continuous, so that φ([−n, n]) is bounded. Hence, |φ(Xt )| ≤ M for some 0 < M < ∞, and so Tn E[|φ(Xt )|] ≤ M.

So, by Jensen’s inequality:

Tn Tn Tn E[φ(Xt )|FTn∧s] ≥ φ(E[Xt |FTn∧s]) = φ(Xs ).

Tn Thus, φ(Xt ) is a submartingale, so that φ(Xt) is a local submartingale.

6 Tn In the proof of Theorem 4, we used the fact that Xt is a martingale w.r.t. {Ft∧Tn , t ≥ 0}, as per the deﬁnition of a continuous local martingale. In general, we have

S Theorem 5 Let S be a stopping time. Then, Xt is a martingale w.r.t. Ft∧S, t ≥ 0 if and only if it is a martingale w.r.t. Ft, t ≥ 0.

Proof: (⇐) We begin by proving the following

Claim 1 If S ≤ T are stopping times, then FS ⊆ FT .

Proof of Claim: Recall that if S is a stopping time,

FS = {A : A ∩ {S ≤ t} ∈ Ft for all t ≥ 0}.

Now, let A ∈ FS. Then, A ∩ {S ≤ t} ∈ Ft. Since {T ≤ t} ⊆ {S ≤ t} since S ≤ T ,

{T ≤ t} = {T ≤ t} ∩ {S ≤ t}.

Finally, since T is a stopping time, {T ≤ t} ∈ Ft, so that we have:

A ∩ {T ≤ t} = (A ∩ {S ≤ t}) ∩ {T ≤ t} ∈ Ft.

Thus, A ∈ FT since t was arbitrary, and so FS ⊆ FT . 4

S Now, suppose that Xt is a martingale w.r.t. Ft, t ≥ 0. Then, for s ≤ t: Z Z S S S S E[Xt |Fs] = Xs ⇔ Xs = Xt A A for A ∈ Fs. By the claim, Fs∧S ⊆ Fs, so that for any A ∈ Fs∧S , the same holds, so that

S S E[Xt |Fs∧S] = Xs .

S Since s, t were arbitrary, Xt is a martingale w.r.t. Ft∧S, t ≥ 0.

S (⇒): Suppose that Xt us a martingale w.r.t. Ft∧S, t ≥ 0. Let A ∈ Fs.

Claim 2 A ∩ {S > s} ∈ FS∧s.

Proof of Claim: If r < s and ω ∈ {S > s}, then

(S ∧ s)(ω) = s > r.

Thus, ω ∈ {S ∧ s ≤ r}c ⇒ {S > s} ⊆ {S ∧ s ≤ r}c, so that

{S > s} ∩ {S ∧ s ≤ r} = ∅ ⇒ A ∩ {S > s} ∩ {S ∧ s ≤ r} = ∅ ∈ FS∧s for r < s.

7 c If r ≥ s, Fs ⊆ Fr. Thus, A ∈ Fs ⊆ Fr and {S > s} = {S ≤ s} ∈ Fs ⊆ Fr. Thus, since S ∧ s is a stopping time:

(A ∩ {S > s}) ∩ {S ∧ s ≤ r} ∈ Fr.

So, for all values of r, (A ∩ {S > s}) ∩ {S ∧ s ≤ r} ∈ Fr, so that A ∩ {S > s} ∈ FS∧s. 4

S By the above claim, A ∩ {S > s} ∈ FS∧s, so that since Xt is a martingale w.r.t. Ft∧S, we have for s ≤ t: Z Z S S S S E[Xt |Fs∧S] = Xs ⇔ Xt = Xs . A∩{S>s} A∩{S>s}

On the other hand, Z Z Z S S Xt = XS · 1S>0 = Xs . A∩{S≤s} A∩{S≤s} A∩{S≤s} So, for s ≤ t: Z Z Z Z Z Z S S S S S S Xt = Xt + Xt = Xs + Xs = Xs A A∩{S>s} A∩{S≤s} A∩{S>s} A∩{S≤s} A

⇒ E[Xt|Fs] = Xs by uniqueness of conditional expectation.

Tn Thus, in the deﬁnition of a continuous local martingale, Xt can be allowed to be a martingale w.r.t. Ft. We use this in Theorem 6.

When is a continuous local martingale a martingale?

Deﬁnition: A real valued process X is said to be of class DL if for every t > 0, the family of random variables XT where T ranges through all stopping times less than t is uniformly integrable.

Theorem 6 A local martingale is a martingale if and only if it is of class DL.

Proof: (⇒): Suppose that Xt is a local martingale that is a martingale, and let t > 0 be ﬁxed, and T be a stopping time s.t. T ≤ t. Since Xt is a (true) martingale, we have that:

Xs = E[Xt|Fs] for s ≤ t.

Now, Theorem 5.1 of Durrett states:

Given a probability space (Ω, F,P ) and an X ∈ L1, {E[X|G]: G is a σ-ﬁeld ⊆ F} is uniformly integrable.

8 So, {Xs∧t} = {E[Xt|Fs]}, which is a uniformly integrable martingale by Durrett’s theorem, since 1 Xt ∈ L (Ω, F,P ).

Thus, the Optional Stopping Theorem implies: XT = E[Xt|FT ], so that {XT }T ≤t = {E[Xt|FT ]}T ≤t, the latter of which is again uniformly integrable by Durrett’s Theorem.

Since t > 0 was arbitrary, the result follows.

(⇐): Suppose that Xt is a local martingale that is of class DL. Let {Tn} be a sequence of stopping times that reduce X.

Tn Tn For ﬁxed t ≥ 0, then, we have that Xt → Xt a.s. as n → ∞. By hypothesis, {Xt }n≥0 are uniformly Tn 1 1 integrable, so that we must have Xt → Xt in L . Thus, Xt ∈ L (Ω, F,P ), so that since t was arbitrary, this is true for all t > 0.

Let s ≤ t.

Tn Tn 1 Now, Xt → Xt a.s. and {Xt } is uniformly integrable implies that this happens in L .

Tn 1 Tn 1 Claim 3 Xt → Xt in L implies E[Xt |Fs] → E[Xt|Fs] in L .

Proof of Claim:

Tn Tn Tn ||E[Xt|Fs] − E[Xt |Fs]||1 = ||E[Xt − Xt |Fs]||1 ≤ ||Xt − Xt ||1 <

Tn 1 for n > N since Xt → Xt in L . 4.

Tn 1 By the claim E[Xt |Fs] → E[Xt|Fs] in L .

Tn Tn On the other hand, since Xt is a local martingale (recall theorem 5): E[Xt |Fs] = Xs → Xs a.s. as Tn Tn 1 n → ∞, so that since {Xs } are uniformly integrable, E[Xt |Fs] → Xs in L .

Thus, Xs = E[Xt|Fs].

9 Corollary 2 If Xt is a local martingale, and E sup |Xs| < ∞ 0≤s≤t for each t, then Xt is a martingale.

Proof: Let t > 0 be ﬁxed. For 0 ≤ k ≤ t, |Xk| ≤ sup0≤s≤t |Xs|. Thus, if T be a stopping time less than t,

|XT | ≤ sup |Xs|. 0≤s≤t

Since sup0≤s≤t |Xs| is an integrable function, it follows immediately that the family {XT } where T ranges through all stopping times less than t is uniformly integrable. Since t was arbitrary, the result follows from Theorem 6.

Corollary 3 A bounded local martingale is a martingale.

Proof: Obvious.

Theorem 7 Let Xt be a local martingale on [0, τ). If E sup |Xs| < ∞ 0≤s≤τ then Xτ = limt↑τ Xt exists and E[X0] = E[Xτ ]. Proof: |Xk| ≤ sup0≤s≤τ |Xs| for 0 ≤ k < τ, so that E[Xk] ≤ E sup0≤s≤τ |Xs| < ∞ for 0 ≤ k < τ.

Thus, {Xt}0≤t<τ is uniformly integrable.

The method of Theorem 6 (⇐) carries over, so that Xt is a martingale on [0, τ). By the Martingale Convergence Theorem (which we can apply since E[Xk] ≤ E sup0≤s≤τ |Xs| < ∞ for 0 ≤ k < τ implies sup0≤s<τ E[Xs] ≤ E sup0≤s≤τ |Xs| < ∞) Xτ (ω) := limt→τ Xt(ω) exists for a.e. ω ∈ Ω.

1 Since {Xt}0≤t<τ is uniformly integrable, this convergence occurs in L . But then,

E[Xτ ] = lim E[Xt] = E[X0] t→τ since Xt is a martingale.

10 3 The Doob-Meyer Decomposition: Motivation

In this section, we motivate the construction of variance and covariance processes for continuous local martingales, which is crucial in the construction of stochastic integrals w.r.t. continuous local martingales as we shall see.

In this section, unless otherwise specified, we fix a Brownian motion Bt and a filtration {Ft} such that:

1. For each t, Bt is Ft-measurable.

2. For and s ≤ t, the random variable Bt − Bs is independent of the σ-ﬁeld Fs.

Recall that for any Brownian motion, hBit = t where hBit is the quadratic variation of Bt. This immediately implies (2) in the following

2 Deﬁnition: Deﬁne Lad([a, b] × Ω) to be the space of all stochastic processes f(t, ω), a ≤ t ≤ b, ω ∈ Ω such that:

1. f(t, ω) is adapted to the ﬁltration {Ft}.

R b 2 R b 2 2. a E[|f(t)| ]dt = a E[|f(t)| ]d hBit < ∞. Also recall when constructing a theory of integration w.r.t. a Brownian motion, we begin with constructing the stochastic integral Z b f(t)dBt a 2 for f ∈ Lad([a, b] × Ω).

Now, we want a more general formalism of integrating a class of processes w.r.t. a generalized martingale that in the case Brownian motion will reduce to the above.

Definition: Let Gt be a right-continuous filtration. We define L denote the collection of all jointly measurable stochastic processes X(t, ω) such that:

1. Xt is adapted w.r.t. Gt.

2. Almost all sample paths of Xt are left continuous.

Furthermore, we deﬁne P to be the smallest σ-ﬁeld of subsets of [a, b] × Ω with respect to which all the stochastic processes in L are measurable. A stochastic processes Y (t, ω) that is P measurable is said to be predictible.

11 The motivation for the deﬁnition of a predictable process comes from the following argument: If Yt is a predictable process, then almost all its values at time t can be determined [with certainty] with the information available strictly before time t, since left continuity of the process Yt implies that for almost every ω ∈ Ω and any sequence tn ↑ t as n → ∞:

lim Yt (ω) = Yt(ω). n→∞ n

Now, we have the following theorem [which we shall prove a version of in the next section for continuous local martinagles]:

Theorem 8 (Doob-Meyer) Let Mt, a ≤ t ≤ b be a right continuous, square integrable martingale with left hand limits. Then, there exists a unique decomposition

2 (Mt) = Lt + At, a ≤ t ≤ b where Lt is a right-continuous martingale with left-hand limits, and At is a predictable, right continuous, increasing process such that Aa ≡ 0 and E[At] < ∞ for all a ≤ t ≤ b.

The above theorem certainly applies to the square integrable process Bt.

Claim 4 In the case Mt = Bt in Doob-Meyer, At = hBit = t.

2 Proof of Claim 4: WLOG, we may take a = 0 and b = t0. Deﬁne Pt = (Bt) − t. Then, for

0 ≤ s ≤ t ≤ t0:

2 2 2 2 E[(Bt) |Fs] = E[(Bt − Bs + Bs) |Fs] = E[(Bt − Bs) + 2Bs(Bt − Bs) + (Bs) |Fs]

2 2 2 = E[(Bt − Bs) ] + 2BsE[Bt − Bs] + (Bs) = t − s + (Bs)

2 2 ⇒ E[Pt|Fs] = E[(Bt) − t|Fs] = (Bs) − s = Ps.

2 2 Thus, Pt = (Bt) − t is a martingale, so that (Bt) = Pt + t. Clearly, t satisﬁes all the conditions that

At must satisfy in Doob-Meyer, so that by uniqueness of At, At = t = hBit. 4

So, another way of viewing the integral w.r.t. the martingale Mt w.r.t. the ﬁltration Gt is the following:

First, we look for the unique process (guaranteed by Doob-Meyer) hMit such that

2 Lt = (Mt) − hMit is a martingale. Then, we make the

12 2 Deﬁnition: Deﬁne Lpred([a, b]hMi × Ω) to be the space of all stochastic processes f(t, ω), a ≤ t ≤ b, ω ∈ Ω such that:

1. f(t, ω) is predictable w.r.t. {Gt}.

R b 2 2. a E[|f(t)| ]d hMit < ∞. Then, we proceed to construct the integral

Z b f(t)dMt a

2 for f ∈ Lpred([a, b]hMi × Ω).

It’s clear that in the case Mt = Bt and Gt = Ft that the above formulation coincides with the original construction of the stochastic integral w.r.t. Bt reviewed at the beginning of this section.

For right continuous, square integrable martingales Mt with left hand limits, at least, this process works.

In the case where Mt is a continuous local martingale, we do the same thing. However, it’s not immediately clear:

1. that we have a version of Doob-Meyer for continuous local martingales.

2. how the construction of the integral is affected by the stopping times Tn that reduce Mt, if at all.

In the next section, we deal with the ﬁrst problem. Then, we proceed to remedy the second.

13 4 Variance and Covariance Processes

We take L and P as deﬁned in section 1.

Theorem 9 If Xt is a continuous local martingale, then we deﬁne the variance process hXit to be the 2 unique continuous predictable increasing processes At that has A0 ≡ 0 and makes Xt − At a local martingale.

Deﬁnition: If X and Y are two continuous local martingales, we let 1 hX,Y i = (hX + Y i − hX − Y i ). t 4 t t

We call hX,Y it the covariance of X and Y .

Based on the discussion in the ﬁrst section, it’s clear why we’re interested in variance processes. It is convenient to deﬁne covariance processes since they are very useful and have quite nice properties, such as:

Theorem 10 h·, ·it is a symmetric bilinear form on the class of continuous local martinagles.

We might prove it this time around. If not, hopefully next time. Two questions I’m still pondering is

1. Can you turn this into an inner product?

2. If so, how you can characterize the class of processes that is the completion of this space?

The proof of theorem 9 is long, but it is instructive to go through it, since it develops techniques that will be useful later. In order to proceed, recall that any predictable discrete time martingale is constant [why?]. There is a result analogous to this in continuous time, and we use it to prove the uniqueness statement in theorem 9:

Theorem 11 Any continuous local martingale Xt that is predictable and locally of bounded variation is constant (in time).

Proof of theorem 11: By subtracting X0, WMA that X0 ≡ 0. Thus, we wish to show that Xt ≡ 0 for all t > 0 almost surely.

t Let Vt(ω) = supπ∈Πt Tπ(ω) be the variation of Xs(ω) on [0, t], where Π denotes the set of all (ﬁnite) partitions of [0, t], π = {0 = t0 < t1 < ··· < tN = t}, and where for a given partition of this sort,

N X Tπ(ω) = |Xtm (ω) − Xtm−1 (ω)|. m=1

Lemma 1 For almost all ω ∈ Ω, t 7→ Vt(ω) is continuous

Proof of Lemma 1: First, notice that for any ω ∈ Ω, t 7→ Vt(ω) is increasing:

14 For s < t, [0, s] ⊂ [0, t], so that any finite partition π = {0 = t0 < t1 < ··· < tN = s} of [0, s] gives a 0 finite partition π = {0 = t1 < ··· < tN = s < tN+1 = t} of [0, t]. Thus, for any finite partition π of 0 [0, s], Tπ(ω) ≤ Tπ0 (ω), where π is a finite partition of [0, t], so that

Tπ(ω) ≤ Tπ0 (ω) ≤ sup Tπ(ω) = Vt(ω) π∈Πt

⇒ Vs(ω) = sup Tπ(ω) ≤ sup Tπ(ω) = Vt(ω). π∈Πs π∈Πt Since ω was arbitrary, this is true for all ω ∈ Ω.

Thus, to show that t 7→ Vt is continuous a.s., it sufﬁces to show that for almost all ω ∈ Ω, t 7→ Vt(ω) has no discontinuities (of the ﬁrst kind).

u u Claim 5 For any ω ∈ Ω, Vu(ω) = Vs(ω) + Vs (ω) where Vs (ω) is the variation of Xt(ω) on [s, u].

Proof of Claim 5: Take any two partitions {s = t0 < t1 < ··· < tN = u}, {0 = t−N 0 < t−N 0−1 <

··· < t0 = s}. Then:

0 N X X u |Xtm (ω) − Xtm−1 (ω)| + |Xtm (ω) − Xtm−1 (ω)| ≤ Vs(ω) + Vs (ω). m=−N 0+1 m=1

Now, the LHS is Tπ(ω) for π = {0 = t−N 0 < ··· < t0 = s < ··· < tN = u}. Given an arbitrary partition π0 of [0, u], it’s clear that there exists a partition π00 such that π0 ⊂ π00 and π00 is of the form π. 0 u Since Tπ0 (ω) ≤ Tπ00 (ω), and π was arbitrary, we have that Vu(ω) ≤ Vs(ω) + Vs (ω).

For the other inequality, note that {0 = t−N 0 < ··· < t0 = s < ··· < tN = u} is a partition of [0, u]. Thus:

Now, ﬁxing one of the partitions on the RHS, we may take the supremum of the remaining, and then u proceed to take the supremum of the ﬁnal term. Thus: Vu(ω) ≥ Vs(ω) + Vs (ω), so that Vu(ω) = u Vs(ω) + Vs (ω). 4

Now, by hypothesis, Xs is of locally bounded variation. So, there exists a sequence of stopping times

Tn Tn ↑ ∞ a.s. such that Xs (ω) is of bounded variation in time. Let

A = {ω ∈ Ω: Tn(ω) ↑ ∞}

B = {ω ∈ Ω: Xt(ω) is continuous}.

By deﬁnition, P [A] = P [B] = 1 ⇒ P [A ∩ B] = 1. Now, let ω ∈ A ∩ B be ﬁxed, and suppose that s 7→ Vs(ω) has a discontinuity at t. Choosing n large enough so that Tn(ω) > t, there exists s0 ≤ t < u0 such that Xs(ω) is of bounded variation on [s0, u0].

15 Since s 7→ Vs(ω) has a discontinuity at t, there exists > 0 such that for every δ > 0, u − s < δ implies u Vu(ω) − Vs(ω) > 3 where s < t < u. By Claim 5 then, for every δ > 0, u − s < δ implies Vs (ω) > 3 where s < t < u.

0 0 Pick δ > 0 so that if |r − s| < δ then |Xs − Xr| < (using uniform continuity of Xs(ω) on [s0, u0]).

Assuming sn and un have been deﬁned, pick a partition of [sn, un] not containing t with mesh less than 0 u δ and variation greater than 2 (this is possible since for every δ > 0, u − s < δ implies Vs (ω) > 3 where s < t < u).

Let sn+1 be the largest point in the partition less than t, and un+1 be the smallest point in the partition 0 larger than t. Then un+1 − sn+1 < δ ⇒ |Xsn+1 (ω) − Xun+1 (ω)| < . Thus:

By omitting the points sn+1 and un+1 from the partition, we obtain a partition for [sn, un]−[sn+1, un+1]. Thus, after taking supremums:

un un+1 V[sn,un]−[sn+1,un+1](ω) = Vsn (ω) − Vsn+1 (ω) > .

u0 Thus, Vs0 (ω) > M for arbitrarily large integer values M. Thus, it must be inﬁnity, contradicting that Xs(ω) has bounded variation on [s0, u0].

Thus, t 7→ Vt(ω) must be continuous for every ω ∈ Ω, since ω ∈ A was arbitrary. 4

Now, we needed Lemma 1 in order to guarantee that the functions

Sn(ω) = inf{s : Vs(ω) ≥ n} are stopping times (why?).

Lemma 2 {Sn} reduce Xt.

Proof of Lemma 2: Recall theorem 4:

If X is a continuous local martingale, we can always take the sequence which reduces X to be

Tn = inf{t : |Xt| > n}

0 0 or any other sequence Tn ≤ Tn that has Tn ↑ ∞ as n ↑ ∞.

Now, suppose that t satisﬁes n < |Xt| = |Xt − X0| ≤ Vt. Then, Vt > n, so that

{t : |Xt| > n} ⊆ {t : Vt ≥ n} ⇒ Sn = inf{t : Vt ≥ n} ≤ inf{t : |Xt| ≥ n} = Tn.

16 Certainly, Vs ≥ n + 1 ≥ n ⇒ Vs ≥ n so that

{t : Vs ≥ n + 1} ⊆ {t : Vs ≥ n} ⇒ Sn = inf{{t : Vs ≥ n} ≤ {t : Vs ≥ n + 1} = Sn+1.

Finally, since t 7→ Vt is continuous a.s., it’s clear that limn→∞ Sn(ω) = ∞ almost surely. Thus, Sn reduce Xt. 4

Now, ﬁx some n > 0. Then, t ≤ Sn implies |Xt| ≤ n. By Lemma 2, Mt = Xt∧Sn is a bounded martingale.

Now, if s < t:

(we refer to this relationship as orthogonality of martingale increments). If 0 = t0 < t1 < ··· < tN = t is a partition of [0, t], we have:

" N # " N # 2 X 2 2 X 2 E[Mt ] = E Mtm − Mtm−1 = E (Mtm − Mtm−1 ) ≤ E Vt∧Sn sup |Mtm − Mtm−1 | m m=1 m=1

≤ nE sup |Mtm − Mtm−1 | m

n n n Taking a sequence of partitions ∆n = {0 = t0 < t1 < ··· < tk(n) = t} in which the mesh |∆n| =

n n n n supm |tm − tm−1| → 0 continuity of sample paths imply supm |Mtm − Mm−1| → 0 a.s.

Since supm |Mtm − Mtm−1 | ≤ 2n, the bounded convergence theorem implies n n E sup |Mtm − Mtm−1 | → 0. m

2 Thus, E[Mt ] = 0 so that Mt = 0 a.s.

Let At = {ω ∈ Ω: Mt(ω) 6= 0}. Then, since t above was arbitrary, P [At] = 0 for any t, so that   [ P  At = 0. t∈Q, t≥0

Thus, with probability 1, Mt = 0 for all rational t. By continuity of sample paths, we have that Mt = 0 with probability 1 for all t.

0 Uniqueness in theorem 9: Suppose that At and At are two continuous, predictable, increasing processes 0 2 2 0 2 0 that have A0 = A0 ≡ 0, and make Xt − At, Xt − At local martingales. If Tn reduce Xt − At and Tn 2 0 0 2 2 0 0 0 reduce Xt − At it’s clear that Tn ∧ Tn reduce X − At − (X − At) = At − At, so that At − At is a continuous local martingale.

0 0 It’s clear that At − At is predictable, since each At and At are predictable.

17 0 0 Finally, At − At is locally of bounded variation. To see this, take the stopping times Sn = Tn ∧ Tn. 0 0 Clearly, Tn ∧ T ↑ ∞, and the stopped processes A 0 − At∧T ∧T 0 are of bounded variation for n t∧Tn∧Tn n n 0 each ω, being the difference of two increasing processes on the random interval [0,Tn(ω) ∧ Tn(ω)].

0 0 0 Thus, by theorem 11, At − At must be constant, so that since A0 = A0 = 0, At − At = 0 for all t. Thus, 0 At = At for all t.

The existence proof is a little more difﬁcult, but uses some great analysis.

Existence in theorem 9:

We proceed in steps:

Step 1: Proof of existence in theorem 9 when Xt is a bounded martingale: (note that uniqueness follows from the previous argument)

Given a partition ∆ = {0 = t0 < t1 < ···} with limn→∞ tn = ∞, let k(t) = sup{k : tk < t} be the index of the last point before t; note that k(t) is not a random variable, but a number.

Deﬁne k(t) 2 ∆ X 2 Qt (X) = (Xtk − Xtk−1 ) + Xt − Xtk(t) . k=1 2 ∆ Lemma 3 If Xt is a bounded continuous martingale, then Xt − Qt (X) is a martingale.

Proof of Lemma 3: First, notice that

k(t) k(s) 2 2 ∆ ∆ X 2 X 2 Qt − Qs = Xtk − Xtk−1 + Xt − Xtk(t) − Xtk − Xtk−1 − Xs − Xtk(s) k=1 k=1

k(t) 2 2 X 2 2 = Xtk(s)+1 − Xtk(s) − Xs − Xtk(s) + Xtk − Xtk−1 + Xt − Xtk(t) . k=k(s)+2

∆ ∆ ∆ ∆ Deﬁne ui = ti for k(s) ≤ i ≤ k(t) and uk(t)+1 = t. Then, writing Qt = Qs + (Qt − Qs ):

2 ∆ E[Xt − Qt (X)|Fs]

 k(t)+1  2 2 2 ∆ X 2 = E[Xt |Fs]−Qs (X)−E  Xui − Xui−1 Fs−E Xtk(s)+1 − Xtk(s) |Fs +E Xs − Xtk(s) |Fs i=k(s)+2

 k(t)+1  X h i = E[X2|F ] − Q∆(X) − E X2 − X2 F − E X2 − 2X X + X2 |F t s s  ui ui−1 s tk(s)+1 tk(s)+1 tk(s) tk(s) s i=k(s)+2 h i +E X2 − 2X X + X2 |F s s tk(s) tk(s) s

18 h i h i = −Q∆(X) + E X2 |F − E X2 |F + 2X X − X2 + X2 − 2X X + X2 s tk(s)+1 s tk(s)+1 s s tk(s) tk(s) s s tk(s) tk(s) 2 ∆ = Xs − Qs (X)

∆ where in the ﬁrst equality, we have used the fact that Qs (X) is Fs measurable, and in the second equality, we have used the orthogonality of martingale increments. 4

Lemma 4 Let Xt be a bounded continuous martingale. Fix r > 0 and let ∆n be a sequence of partitions

n n 0 = t0 < ··· < tkn = r

n n ∆n 2 of [0, r] with mesh |∆n| = supk |tk − tk−1| → 0. Then, Qr (X) converges to a limit in L (Ω, F,P ).

Proof of Lemma 4: First, we begin with some notation. If ∆ and ∆0 are two partitions of [0, r], we let ∆∆0 denote the partition obtained by taking all the points in ∆ and ∆0.

Now, by lemma 3, for ﬁxed partitions ∆ and ∆0 of [0, r], we have that for a bounded continuous martingale Xt: 2 ∆0 2 ∆ ∆ ∆0 Yt = (Xt − Qt ) − (Xt − Qt ) = Qt − Qt

∆ ∆0 is again a bounded martingale (Since |Xt| ≤ M for all t ≥ 0 implies |Qt − Qt | ≤ KM since the 0 2 ∆∆0 partitons ∆ and ∆ are ﬁxed). Thus, again by lemma 3: Zt = (Yt) − Qt (Y ) is a martingale with

Z0 = 0, so that 2 ∆ ∆0 2 h ∆∆0 i E[Zr] = 0 ⇒ E Qr − Qr = E (Yr) = E Qr (Y ) .

Now, 2a2 + 2b2 − (a + b)2 = (a − b)2 ≥ 0 for any real numbers a and b, so that

(a + b)2 ≤ 2(a2 + b2) for any real numbers a, b. Thus:

k(r) k(r) 0 X 2 2 X h 0 0 i2 Q∆∆ (Y ) = Y − Y + Y − Y = Q∆ − Q∆ − Q∆ − Q∆ r tk tk−1 r tk(r) tk tk tk−1 tk−1 k=1 k=1

h 0 0 i2 + Q∆ − Q∆ − Q∆ − Q∆ r r tk(r) tk(r)

k(r) X h 0 0 i2 h 0 0 i2 = Q∆ − Q∆ − Q∆ − Q∆ + Q∆ − Q∆ − Q∆ − Q∆ tk tk−1 tk tk−1 r tk(r) r tk(r) k=1 k(r) X 2 0 0 2 2 0 0 2 h 0 0 0 i ≤ 2 Q∆ − Q∆ + Q∆ − Q∆ + Q∆ − Q∆ + Q∆ − Q∆ = 2 Q∆∆ Q∆ + Q∆∆ Q∆ . tk tk−1 tk tk−1 r tk(r) r tk(r) r r k=1

19 Putting it all together then, we have:

2 ∆ ∆0 h ∆∆0 i h ∆∆0 ∆ ∆∆0 ∆0 i E Qr − Qr = E Qr (Y ) ≤ 2E Qr Q + Qr Q .

∆n 2 Thus, to show that {Qr (X)} is Cauchy in L (Ω, F,P ), and hence converges in this space since it is complete, it is sufﬁcient to show that

0 h ∆∆0 ∆i |∆| + |∆ | → 0 ⇒ E Qr Q → 0.

n 0 0 To do this, let {sk}k=1 = ∆∆ and {tj} = ∆. Let sk ∈ ∆∆ and tj ∈ ∆ such that tj ≤ sk < sk+1 ≤ tj+1. Then:

Q∆ − Q∆ = (X − X )2 − (X − X )2 = (X − X )2 + 2(X − X )(X − X ) sk+1 sk sk+1 tj sk tj sk+1 sk sk+1 sk sk tj

= (Xsk+1 − Xsk )(Xsk+1 + Xsk − 2Xtj ) ∆∆0 ∆ ∆∆0 2 ⇒ Qr (Q ) ≤ Qr (X) sup(Xsk+1 + Xsk − 2Xtj(k) ) k where j(k) = sup{j : tj ≤ sk}. By the Cauchy-Schwartz inequality:

1 1 4 2 h ∆∆0 ∆i h ∆∆0 2i 2 E Qr Q ≤ E Qr (X) E sup Xsk+1 + Xsk − 2Xtj(k) . k

4 Since the sample paths of Xt are continuous almost surely, supk Xsk+1 + Xsk − 2Xtj(k) → 0 almost 4 0 4 surely as |∆| + |∆ | → 0. Since supk Xsk+1 + Xsk − 2Xtj(k) ≤ (4M) , the bounded convergence theorem implies that 1 4 2 E sup Xsk+1 + Xsk − 2Xtj(k) → 0 k as |∆| + |∆0| → 0.

h ∆∆0 2i Thus, it remains to show that E Qr (X) is bounded. To do this, note that:

n !2 n n−1 n ∆∆0 2 X 2 X 4 X 2 X 2 Qr (X) = (Xsm − Xsm−1 ) = (Xsm −Xsm−1 ) +2 (Xsm −Xsm−1 ) (Xsj −Xsj−1 ) m=1 m=1 m=1 j=m+1

n n−1 X 4 X 2 ∆∆0 ∆∆0 = (Xsm − Xsm−1 ) + 2 (Xsm − Xsm−1 ) Qr (X) − Qsm (X) m=1 m=1 " n # " n−1 # h ∆∆0 2i X 4 X 2 ∆∆0 ∆∆0 ⇒ E Qr (X) = E (Xsm − Xsm−1 ) +2E (Xsm − Xsm−1 ) Qr (X) − Qsm (X) m=1 m=1

To bound the ﬁrst term on the RHS, note that |Xt| ≤ M for all t implies:

" n # " n # " n # X 4 2 X 2 2 X 2 2 2 2 4 E (Xsm − Xsm−1 ) ≤ (2M) E (Xsm − Xsm−1 ) = 4M E Xsm − Xsm−1 ≤ 4M E Xr ≤ 4M m=1 m=1 m=1

20 where in the ﬁrst equality, we have used that orthogonality of martingale increments:

2 2 2 E[(Xsm − Xsm−1 ) |Fsm−1 ] = E[Xsm − Xsm−1 |Fsm−1 ] implies: 2 2 2 E[(Xsm − Xsm−1 ) ] = E[Xsm − Xsm−1 ].

2 For the second term on the RHS, note that (Xsm − Xsm−1 ) ∈ Fsm . By lemma 3, and orthogonality of martingale increments:

∆ ∆ 2 2 h 2 i E Qt (X) − Qs (X)|Fs = E Xt − Xs |Fs = E (Xt − Xs) |Fs .

So:

h 2 ∆∆0 ∆∆0 i 2 h ∆∆0 ∆∆0 i E (Xsm − Xsm−1 ) Qr (X) − Qsm (X) |Fsm = (Xsm −Xsm−1 ) E Qr (X) − Qsm (X) |Fsm

2 2 2 2 = (Xsm − Xsm−1 ) E (Xr − Xsm ) |Fsm ≤ (Xsm − Xsm−1 ) (2M) " n−1 # " n−1 # X h 2 ∆∆0 ∆∆0 i 2 X 2 ⇒ E E (Xsm − Xsm−1 ) Qr (X) − Qsm (X) |Fsm ≤ 4M E (Xsm − Xsm−1 ) m=1 m=1 " n−1 # " n−1 # X 2 ∆∆0 ∆∆0 2 X 2 2 2 2 4 ⇒ E (Xsm − Xsm−1 ) Qr (X) − Qsm (X) ≤ 4M E Xsm − Xsm−1 ≤ 4M E Xr ≤ 4M . m=1 m=1

Thus, h ∆∆0 2i 4 4 4 E Qr (X) ≤ 4M + 2 · 4M = 12M .

n ∆nk o Lemma 5 Let {∆n} be as in Lemma 4. Then, there exists a subsequence {∆nk } such that Qt converges uniformly a.s. on [0, r].

∆n 2 Proof of Lemma 5: Since Qr converges in L (Ω, F,P ), it is Cauchy in this space. So, choose a n ∆nk o subsequence Qr such that for m ≥ nk,

∆ 2 ∆m nk −k E Qr − Qr < 2 .

Let ∆n ∆ 1 k+1 nk Ak = ω ∈ Ω : sup Qt (ω) − Qt (ω) > 2 . t≤r k By Chebyshev’s inequality:

2 4 ∆n ∆ 1 ∆n ∆ k k+1 nk 4 k+1 nk P [Ak] = P sup Qt − Qt ≥ 2 ≤ k E Qr − Qr < k . t≤r k 2

21 Since the RHS is summable, Borel-Cantelli implies that P [lim supk Ak] = 0. So, for almost all ω ∈ Ω, there exists Nω such that k > Nω implies

∆n ∆ 1 k+1 nk sup Qt (ω) − Qt (ω) < 2 . t≤r k

0 So, for m > m > Nω:

m−1 m−1 ∆n ∆n ∆ 1 ∆nm m0 X k+1 nk X sup Qt (ω) − Qt (ω) ≤ sup Qt (ω) − Qt (ω) < 2 . t≤r t≤r k k=m0 k=m0 P∞ 1 0 Since the series k=1 k2 converges, we have that given > 0, there exists N such that m, m > N implies m−1 X 1 < . k2 k=m0 0 thus, for m, m > max{N,Nω}:

∆ ∆nm nm0 sup Qt (ω) − Qt (ω) < . t≤r

n ∆nk o Thus, Qt converges uniformly almost surely on [0, r]. 4

r In what follows, call the limiting function in Lemma 5 At , and deﬁne it to be zero outside [0, r].

∆ ∆0 [0, r + 1] |∆0 | → 0 Now, for each nk in lemma 5, we can extend it to a partition nk of , such that nk . ∆0 nk 2 Then, for this sequence of partitions, lemma 4 implies that Qr+1 converges to a limit in L (Ω, F,P ). ∆0 nkj Repeating the procedure in lemma 5 then, we can select a subsequence Qt such that it converges r+1 uniformly almost surely on [0, r + 1]. Call the limiting function At , and similarly deﬁne it to be zero outside [0, r + 1].

∆0 ∆ nkj nkj r r+1 It’s clear that for t ≤ r, Qt = Qt , so that At = At for t ≤ r.

r+j ∞ r+j Repeating the procedure above, we obtain a sequence of functions {At }j=0 such that At is contin- r+j r+k uous on [0, r + j] a.s., and At = At for t ≤ min{r + j, r + k}. So, we can unambiguously deﬁne r+j At(ω) = limj→∞ At (ω). r+j S∞ If Dj = {ω ∈ Ω: At (ω) is not continuous on [0, r + j]}, then clearly D = j=0 Dj has measure c zero. Thus, for ω ∈ D , it’s clear that At(ω) will be continuous, so that At is continuous a.s.

It’s clear from the construction that At is predictable.

r+j To show that At is increasing, it is sufﬁcient to show that each At is increasing. To this end, let ∆n be the partition of [0, r + j] with points at k2−n(r + j) for 0 ≤ k ≤ 2n; clearly, taking this sequence of

∆n r+j partitions doesn’t alter the above arguments, so that Qt → At uniformly a.e. on [0, r + j].

22 S∞ Clearly, ∆n+1 is a reﬁnement of ∆n and n=1 ∆n is dense in [0, r + j]. Thus, for any pair s, t, s < t, in S∞ ∆n ∆n n=1 ∆n there exists n0 such that s and t belong to ∆n for n ≥ n0. Thus, Qs ≤ Qt for n ≥ n0, so r+j r+j S∞ that As ≤ At . Since this is true for any s < t, s, t ∈ n=1 ∆n, by continuity of the process, it must hold everywhere on [0, r + j].

2 Thus, At is continuous, predictable, and increasing. All we need to verify now is that (Xt) − At is a martingale.

r+j ∆nk Now, for each j, At is the limit of processes of the form Qt which converge uniformly almost r+j surely to At . Thus, we have convergence in probability.

∆nk 2 Similarly, since Qt was obtained as a subsequence of a sequence converging in L (Ω, F,P ) to say Qt, ∆nk 2 the subsequence Qt converges to Qt in L (Ω, F,P ), so that we also have convergence in probability.

R+j ∆nk r+j 2 Thus, Qt = At with probability 1, so that Qt converges to At in L (Ω, F,P ).

n n Claim 6 Suppose that for each n, Zt is a martingale w.r.t. Ft, and that for each t, Zt → Zt in p L (Ω, F,P ) where p ≥ 1. Then, Zt is a martingale.

Proof of Claim 6: Recall that since we’re working over the ﬁnite measure space (Ω, F,P ), convergence in Lp(Ω, F,P ) implies convergence in L1(Ω, F,P ) (why?).

n n Now, the martingale property implies that for s < t, E[Zt |Fs] = Zs , so that for any A ∈ Fs Z Z n n E[Zt |Fs] = Zs . A A n p 1 Since Zs → Zs in L (and hence in L ) we have that Z Z Z n n lim E[Zt |Fs] = lim Zs = Zs. n→∞ A n→∞ A A Now,

n p n p n p n p E[|E[Zt |Fs] − E[Zt|Fs]| ] = E[|E[Zt − Zt|Fs]| ] ≤ E[E[|Zt − Zt| |Fs]] = E[|Zt − Zt| ] where we have used the conditional Jensen inequality, and linearity of conditional expectation.

E[Zt|Fs] = Zs. 4

2 ∆nk ∆nk r+j 2 2 r+j Thus, by lemma 3, since (Xt) −Qt is a martingale, and Qt → At in L (Ω, F,P ), (Xt) −At 2 is a martingale. Thus, (Xt) − At is obviously a martingale.

23 Step 2: Proof of existence in theorem 2 when Xt is a local martingale

Lemma 6 Let X be a bounded martingale, and T be a stopping time. Then, XT = hXiT .

2 T Proof of Lemma 6: By the construction in step 1, Mt = (Xt) − hXit is a martingale. Then, Mt = T 2 T T T T (Xt ) − hXi is a martingale, so that by uniqueness of the process X , X = hXi . 4

Now, let Xt be a continuous local martingale, with a sequence of stopping times {Tn} that reduce it.

WLOG, we may take the stopping times to be the canonical times: Tn = inf{t : |Xt| > n}.

n Tn Then, Y = X · 1Tn>0 is a bounded martingale.

n By the results in step 1, there is a unique, continuous predictable, increasing process At such that n 2 n n n+1 (Yt ) − At is a martingale. By lemma 6, for t ≤ Tn, At = At , so that we may unambiguously n deﬁne hXit = At for t ≤ Tn.

Clearly, hXit is continuous, predictable, and increasing. By deﬁnition:

X2 · 1 − hXi Tn∧t Tn>0 Tn∧t

2 is a martingale, so that (Xt) − hXit is a local martingale.

We proceed now to prove the analogue above for the covariance process. In particular, it is very useful in computing hX,Y it.

Theorem 12 Suppose that Xt and Yt are continuous local martingales. hX,Y it is the unique continuous predictable process At that is locally of bounded variation, has A0 = 0, and makes XtYt − At a local martingale.

Proof of Theorem 5: By deﬁnition: 1 X Y − hX,Y i = (X + Y )2 − hX + Y i − (X − Y )2 − hX − Y i t t t 4 t t t t t t is [obviously] a continuous local martingale.

0 To prove uniqueness, notice that if At and At are two processes with the desired properties, then 0 0 At − At = (XtYt − At) − (XtYt − At) is a continuous local martingale that is of locally bounded 0 variation. Hence, by theorem 4, this must be identically zero, so that At = At for all t.

24 5 Integration

5.1 Integration w.r.t. Bounded Continuous Martingales

It’s time to put all our hard work in section 4 to use. In this section, we establish how to integrate predictable processes w.r.t. bounded martingales. Once we establish the Kunita-Watanabe inequality, we will be able to integrate w.r.t. continuous local martingales, and then ultimately, continuous semimartingales.

We proceed in three steps: 1) define the integral for basic integrands, 2) extend the definition to simple integrands, 3) take limits of simple integrands and define the integral for square integrable integrands.

Throughout the integration section, we will continually verify the following three (desirable) properties for each class of integrands and integrators:

1. If H and K are predictable, then ((H + K) · X)t = (H · X)t + (K · X)t.

2. If H is predictable and X,Y are continuous bounded martingales, then (H · (X + Y ))t = (H ·

X)t + (H · Y )t. R t 3. For H,K predictable and X,Y bounded continuous martingales, hH · X,K · Y it = 0 HsKsd hX,Y is .

5.1.1 Basic Integrands

Deﬁnition: We say that H(s, ω) is a basic predictable process if H(s, ω) = 1(a,b]C(ω) where C ∈ Fa.

We set Π0 to be the set of basic predictable processes, and bΠ0 to be the set of bounded basic predictable processes.

Deﬁnition: When Xt is a continuous martingale, and H ∈ Π0, we deﬁne Z ∞ HsdXs = C(ω)(Xb(ω) − Xa(ω)) 0 Z t Z (H · X)t = HsdXs = Hs1[0,t](s)dXs. 0

Theorem 13 If Xt is a continuous martingale, and H ∈ bΠ0, (H · X)t is a continuous martingale.

Proof of Theorem 13:

 0 0 ≤ t ≤ a   (H · X)t = C(Xt − Xa) a ≤ t ≤ b   C(Xb − Xa) b ≤ t < ∞.

25 so it’s clear that (H ·X)t is continuous, (H ·X)t ∈ Ft, and E[|(H ·X)t|] < ∞. To verify the martingale property, let s < t. There are three cases we need to check:

1. t < a : Since s < t < a, this implies that (H · X)t = 0 so that E[(H · X)t|Fs] = 0.

2. a ≤ t ≤ b : In the ﬁrst case where s < a ≤ t ≤ b,

E[(H · X)t|Fs] = E[E[(H · X)t|Fa]|Fs] = E[E[C(Xt − Xa)|Fa]|Fs]

= E[CE[Xt − Xa|Fa]|Fs] = 0 = (H · X)s

where we have used the martingale property E[Xt|Fa] = Xa.

In the second case where a ≤ s < t ≤ b,

E[(H · X)t|Fs] = E[C(Xt − Xa)|Fs] = CE[Xt − Xa|Fs] = C(Xs − Xa) = (H · X)s

since Xa,C ∈ Fa ⊆ Fs and E[Xt|Fs] = Xs.

3. b < t < ∞ : In the ﬁrst case s < a,

E[(H·X)t|Fs] = E[C(Xb−Xa)|Fs] = E[E[C(Xb−Xa)|Fa]|Fs] = E[CE[Xb−Xa|Fa]|Fs] = 0

since E[Xb|Fa] = Xa by the martingale property.

In the second case a ≤ s ≤ b,

E[(H · X)t|Fs] = E[C(Xb − Xa)|Fs] = CE[Xb − Xa|Fs] = C(Xs − Xa) = (H · X)s

since Xa,C ∈ Fa ⊆ Fs and E[Xb|Fs] = Xs.

In the third case b < s < ∞,

E[(H · X)t|Fs] = E[C(Xb − Xa)|Fs] = C(Xb − Xa).

Corollary 4 If Yt is constant for t∈ / [a, b] and E[Yt|Fs] = Ys when a ≤ s ≤ t ≤ b then Yt is a martingale.

26 5.1.2 Simple Integrands

Deﬁnition: We say that H(s, ω) is a simple predictable process if H can be written as the sum of a

ﬁnite number of basic predictable processes. We set Π1 to be the set of simple predictable processes, and bΠ1 to be the set of simple predictable processes.

It’s clear that if H ∈ Π1, then H can be written:

m X H(s, ω) = 1(tj−1,tj ](s)Cj(ω) j=1

where t0 < t1 < ··· < tm and Cj ∈ Ftj−1 . In this case, we make the

Deﬁnition: m Z ∞ X HsdXs = Cj(Xtj − Xtj−1 ). 0 j=1

Remark: while the representation of H above is not unique, it’s clear that this deﬁnition of the integral does not depend on the choice of representation of H.

Theorem 14 Suppose X and Y are continuous martinagles. If H,K ∈ Π1 then

((H + K) · X)t = (H · X)t + (K · X)t

(H · (X + Y ))t = (H · X)t + (H · Y )t.

Proof of Theorem 14: Let H1 = H, H2 = K. By subdividing the intervals if necessary, we may j Pm j assume Hs = k=1 1(tk−1,tk](s)Ck for j = 1, 2. Thus,

m X j j ((H + K) · X)t = (C1 + C2)(Xtj − Xtj−1 ) = (H · X)t + (K · X)t. j=1

Pm In the second case, writing Hs = i=1 1(ti−1,ti](s)Ci, we have:

m X (H · (X + Y ))t = Ci(Xti + Yti − Xti−1 − Yti−1 ) = (H · X)t + (H · Y )t. i=1

Using theorem 13, since the sum of a ﬁnite number of continuous martingales is again a continuous martingale, the ﬁrst part of theorem 15 implies:

Corollary 5 If X is a continuous martingale and H ∈ bΠ1, then (H · X)t is a continuous martingale.

Theorem 15 If X and Y are bounded continuous martingales and H,K ∈ bΠ1, then Z t hH · X,K · Y it = HsKsd hX,Y is . 0

27 hR t i Consequently, E[(H · X)t(K · Y )t] = E 0 HsKsd hX,Y is , and

Z t 2 2 E[(H · X)t ] = E Hs d hXis . 0

Remark: Here and in what follows, integrals w.r.t. hX,Y is are Lebesgue-Stieltjes integrals. That is, since hX,Y is is locally of bounded variation, for almost all ω ∈ Ω, hX,Y is (ω) is a function of bounded variation on the random time intervals [0,Tn(ω)], where Tn(ω) ↑ ∞, so that on each of these intervals, n n n n we may write hX,Y it (ω) = At (ω) − Bt (ω) where At and Bt are increasing functions, and hence R t make sense of the Lebesgue-Stieltjes integral 0 f(s, ω)d hX,Y is (ω) for almost all ω and any ﬁnite t. Proof of Theorem 15: To prove these results, we note that it is sufﬁcient to prove that

Z t Zt = (H · X)t(K · Y )t − HsKsd hX,Y is 0 is a martingale, since the ﬁrst result will follow from theorem 12, the second from taking expectation of both sides, and the third by taking H = K and X = Y .

To prove that Zt is a martingale, we note that

1 2 1 2 ((H + H ) · X)t(K · Y )t = (H · X)t(K · Y )t + (H · X)t(K · Y )t

Z t Z t Z t 1 2 1 2 (Hs + Hs )Ksd hX,Y is = Hs Ksd hX,Y is + Hs Ksd hX,Y is . 0 0 0 So, if the result holds for the pairs (H1,K) and (H2,K), then it holds for (H1 + H2,K). Similarly, if the result holds for (H,K1) and (H,K2), then it holds for (H,K1 + K2).

Thus, it is sufﬁcient to prove that Zt is a martingale, with H = 1(a,b]C, K = 1(c,d]D. WLOG, we may assume that 1) b ≤ c or 2) a = c, b = d (since otherwise we could decompose the interval into disjoint half open intervals, and apply the previous argument)

R t Case 1: In this case, 0 HsKsd hX,Y is = 0 for all t. Thus, we need to show that (H · X)t(K · Y )t is a martingale. To prove this, observe that if J = C(Xb − Xa)D1(c,d] then (H · X)t(K · Y )t = (J · Y )t is a martingale by theorem 13.

Case 2: In this case

 0 s ≤ a   Zs = CD[(Xs − Xa)(Ys − Ya) − (hX,Y is − hX,Y ia)] a ≤ s ≤ b   CD[(Xb − Xa)(Yb − Ya) − (hX,Y ib − hX,Y ia)] s ≥ b

28 so it sufﬁces to check the martingale property for a ≤ s ≤ t ≤ b (recall corollary 4). To do this, note that

Zt − Zs = CD[XtYt − XsYs − Xa(Yt − Ys) − Ya(Xt − Xs) − (hX,Y it − hX,Y is)].

Taking expected values and noting Xa ∈ Fa, E[Yt − Ys|Fs] = 0 we have

E[Zt − Zs|Fs] = CDE[XtYt − hX,Y it − (XsYs − hX,Y is)|Fs] = 0.

5.1.3 Square integrable integrands

Deﬁnition: For any martingale X, let Π2(X) denote the set of all predictable processes H such that

1 Z 2 2 ||H||X = E Hs d hXis < ∞.

Since integrating w.r.t. hXis for each ω in the Lebesgue-Stieltjes sense is the same as integrating w.r.t. a

σ-ﬁnite Borel measure generated by hXis (ω) in the Lebesgue sense, it’s clear that

Claim 7 || · ||X is a norm on Π2(X).

Remark: In this deﬁntion, we have used the Doob-Meyer decomposition for the bounded continuous martingale Xt developed in section 4.

2 Deﬁnition: We deﬁne M to be the set of all martingales adapted to {Ft}t≥0 such that

1 2 2 ||X||2 = sup E Xt < ∞. t≥0

Theorem 16 If X is a bounded continuous martingale and H ∈ bΠ1, then ||H · X||2 = ||H||X .

Proof of Theorem 16: From theorem 15: Z Z t 2 2 2 2 2 ||H||X = E Hs d hXis = sup E Hs d hXis = sup E (H · X)t = ||H · X||2. t≥0 0 t≥0

Now, we show that we can deﬁne the integral of H ∈ Π2(X) w.r.t. X by taking limits in bΠ1:

i 2 Tk i Lemma 7 If for 1 ≤ i ≤ k we have X ∈ M and H ∈ i=1 Π2(X ) then there is a sequence n n H ∈ bΠ1 with ||H − H||Xi → 0 for 1 ≤ i ≤ k.

Proof of Lemma 7: Fix i, and let X = Xi. Since X ∈ M2, X is a martingale, and hence is a continuous 2 local martingale, so that we have the local martingale Zt = (Xt) − hXit. Now, let Tn be the canonical stopping times that reduce Xt; it was shown that in the construction of the process hXit that Tn reduces

2 Zt1Tn>0 = (Xt) 1Tn>0 − hXit 1Tn>0

29 so that

Tn Tn 2 Tn Zt 1Tn>0 = (Xt ) 1Tn>0 − hXit 1Tn>0 is a martingale.

Thus:

Tn E[Z0] = E[Zt 1Tn>0] ⇒ E[X21 ] = E[X2 1 ] − E[hXi 1 ]. 0 Tn>0 Tn∧t Tn>0 Tn∧t Tn>0 Now, by the L2 maximal inequality:

" 2# 2 E sup |Xt| ≤ 4 sup E[Xt ] < ∞. t≥0 t≥0

2 Thus, sup |X | ∈ L2 so that |X |2 ≤ sup |X | . Since X2 → X2 almost surely, the t≥0 t Tn∧t t≥0 t Tn∧t t dominated convergence theorem, and monotone convergence theorems imply:

2 2 E[X0 ] = E[Xt ] − E[hXit]

2 2 so that E[X0 ],E[Xt ] < ∞ for all t imply that E[hXit] < ∞.

So, if H ∈ bΠ1, we may write |Ht| ≤ M for all t ≥ 0 so that, Z 2 2 i 2 i ||H||Xi = E Hs d X s ≤ M E X s < ∞.

Tk i i Thus, bΠ1 ⊂ i=1 Π2(X ) since X in the above argument was arbitrary. We need this in order to guarantee that ||H||Xi makes sense for H ∈ bΠ1.

Now, let

( k ) \ i Ht = G ∈ Π2(X ): G vanishes on (t, ∞) and the conclusion holds i=1 and recall the

Theorem 17 (Monotone Class Theorem) Let A be a collection of subsets of a measure space X that contains X and is closed under intersection. Let H be a vector space of real valued functions on X satisfying

1. If A ∈ A, 1A ∈ H.

2. If 0 ≤ fn ∈ H and fn ↑ f, where f is a bounded function, then f ∈ H.

Then, H contains all the bounded functions on Ω that are measurable w.r.t. σ(A).

30 We want to apply the monotone class theorem to Ht, the space [0, t] × Ω and the collection of sets A of the form (r, s] × A r < s ≤ t A ∈ Fr. Clearly, this class of sets is closed under intersections and contains [0, t] × Ω.

Also, any set S ∈ A has 1S ∈ Ht since 1(r,s]1A ∈ Ht. This also shows that Ht is nonempty.

By definition, it’s clear that Ht is a vector space. In what follows, we use the definitions of L and P as defined in section 3.

Tk i Now, suppose that 0 ≤ Gn ∈ Ht and that Gn ↑ G where G is bounded and in i=1 Π2(X ). Since each

Gn is predictable, each Gn is P-measurable, so that G is P-measurable, and hence predictable, since the convergence is a.e. Also, since Gn ∈ Ht implies that Gn vanishes outside (t, ∞) for every n, the monotone convergence implies that G vanishes outside (t, ∞).

Now, the dominated convergence theorem implies that Z 2 n 2 i ||G − Gn||Xi = E (Gs − Gs ) d X s → 0

i 1 for each 1 ≤ i ≤ k, so we can pick n large enough such that ||G − G i || i < j−1 . Letting nj = j nj X 2 i 1 max1≤i≤k nj we have that ||G − Gnj ||Xi < 2j+1 for all 1 ≤ i ≤ k.

nj ,m nj ,m nj Now, since Gnj ∈ Ht, we can ﬁnd a sequence of functions H ∈ bΠ1 such that ||H −G ||Xi → j−1 0 for all 1 ≤ i ≤ k. Assuming {mi}i=1 have been deﬁned, we can pick mj > max1≤i≤j−1 mi and large nj ,mj nj 1 enough such that ||H − G ||Xi < 2j+1 . By construction then, we have a sequence of functions n ,m H j j ∈ bΠ1 such that 1 nj ,mj nj ,mj nj ,mj ||H − G|| i ≤ ||H − G || i + ||H − G || i < X nj X nj X 2j for all 1 ≤ i ≤ k. Thus, G ∈ Ht.

So, the monotone class theorem implies that Ht contains all bounded predictable processes that vanish on (t, ∞).

Tk i n If K ∈ i=1 Π2(X ), and we deﬁne K = K1|K|≤n1[0,n] then the dominated convergence theorem implies n ||K − K||Xi → 0

i 1 for each 1 ≤ i ≤ k. So we can pick n large enough such that ||K − K i || i < j−1 . Letting j nj X 2 i 1 nj = max1≤i≤k nj we have that ||K − Knj ||Xi < 2j+1 for all 1 ≤ i ≤ k.

nj nj Since K is bounded, predictable, and vanishes on (n, ∞) for each j, K ∈ Hnj , so that there exists a sequence of functions Knj ,m ∈ bΠ such that

nj ,m nj ||K − K ||Xi → 0

31 j−1 for all 1 ≤ i ≤ k. Assuming {mi}i=1 have been deﬁned, we can pick mj > max1≤i≤j−1 mi and large nj ,mj nj 1 nj ,mj enough such that ||K − K ||Xi < 2j+1 . Thus, we obtain a sequence of functions K such that 1 nj ,mj nj ,mj nj ,mj ||K − K|| i ≤ ||K − K || i + ||K − K || i < X nj X nj X 2j for all 1 ≤ i ≤ k.

Theorem 18 M2 is complete.

Proof of Theorem 18:

In order to proceed, we prove the following

2 Lemma 8 Let Mt be a continuous martingale, for which Mt ∈ L (Ω, F,P ) for all t. Then,

∞ 2 X 2 sup E[Mt ] < ∞ ⇔ E[(Mtk − Mtk−1 ) ] < ∞ t≥0 k=1

∞ for any sequence {tk}k=0 such that tk ↑ ∞ strictly.

2 In particular, when this occurs, we have that Mt → M∞ almost surely, and in L (Ω, F,P ).

∞ Proof of Lemma 8: Let {tk}k=0 be a sequence of real numbers s.t. tk ↑ ∞ strictly.

Now, for ti < tj < tm < tn:

E[Mtn |Ftm ] = Mtm

2 2 implies that Mtn − Mtm is orthogonal to L (Ftm ) (recall that for functions f ∈ L (F), E[f|Ft] is the 2 projection of f onto the closed subspace L (Ft)).

So, since Mtj − Mti ∈ Ftm , Mtm − Mtn ,Mtj − Mti = 0. Pj Thus, since we may write Mtj = Mt0 + k=1(Mtk − Mtk−1 ), this implies

j 2 2 X 2 E[Mtj ] = E[Mt0 ] + E(Mtk − Mtk−1 ) . (1) k=1

(⇒): By (1):

j j 2 X 2 2 2 X 2 2 E[Mt0 ]+ E(Mtk −Mtk−1 ) = E[Mtj ] ≤ sup E[Mt ] ⇒ E(Mtk −Mtk−1 ) ≤ sup E[Mt ] < ∞. t≥0 t≥0 k=1 k=1

nPj 2o Since this is true for all j and the sequence k=1 E(Mtk − Mtk−1 ) is monotonically increasing, we must have that ∞ X 2 2 E[(Mtk − Mtk−1 ) ] ≤ sup E[Mt ] < ∞. t≥0 k=1

32 Since the sequence {tk} was arbitrary, the result follows.

(⇐): By (1):

j ∞ 2 2 X 2 2 X 2 E[Mtj ] = E[Mt0 ] + E(Mtk − Mtk−1 ) ≤ E[Mt0 ] + E(Mtk − Mtk−1 ) = K < ∞. k=1 k=1

Since the sequence I = {tk} was arbitrary, we may it to be a strictly ascending countable sequence of 2 numbers dense in R. In that case: E[Mt ] ≤ K < ∞ for all t ∈ I. By continuity of sample paths, the dominated convergence theorem immediately implies that

2 E[Mt ] ≤ K

2 for all t ≥ 0. Thus, supt≥0 E[Mt ] ≤ K < ∞.

2 To prove the ﬁnal claim, note that by the above, either condition implies that Mt is bounded in L (Ω, F,P ), and hence in L1(Ω, F,P ) since L2(Ω, F,P ) ⊆ L1(Ω, F,P ). So, by the Martingale Convergence The- orem, M∞ = limt→∞ Mt exists for almost all ω ∈ Ω.

∞ 2 Pn+r h 2i For any {tk}k=0 tk ↑ ∞ strictly, E[(Mtn+r −Mtn ) ] = k=n+1 E Mtk − Mtk−1 by the discussion above. Letting r → ∞ and using Fatou’s lemma:

h 2i X h 2i E (M∞ − Mtn ) ≤ E Mtk − Mtk−1 . k≥n+1

P∞ 2 The RHS is simply the statement of convergence of the series k=1 E[(Mtk − Mtk−1 ) ] so that

h 2i lim E (M∞ − Mt ) = 0. n→∞ n

Since the sequence {tn} was arbitrary, the result follows immediately. 4

Now, given a martingale M ∈ M2, deﬁne the function

2 2 Φ:(M , || · ||2) → L (Ω, F∞,P ) by Φ(M) = M∞, where F∞ = σ (F : t ≥ 0), which we can do by lemma 8. Again by lemma 8, we 2 have that Mt → M∞ in L (Ω, F,P ). So that:

2 2 2 E[M∞] = lim E[Mt ] = sup E[Mt ] t→∞ t≥0

(the ﬁrst equality is the statement of the convergence of the L2 norms; the latter equality comes from 2 the fact that Mt is a continuous submartingale: use the conditional Jensen inequality and the fact that 2 Mt ∈ L for all t ≥ 0). Thus, Φ preserves norms.

To prove that Φ is an isomorphism, it sufﬁces to show 1) injectivity, 2) surjectivity since it’s obvious that Φ is linear.

33 1. Suppose X∞ = Φ(X) = Φ(Y ) = Y∞. Then, since the martingales Xt and Yt can be recovered

from Yt = E[Y∞|Ft] = E[X∞|Ft] = Xt, it’s clear that X = Y .

2 1 1 2. Suppose Y ∈ L (F∞). Then, Y ∈ L (F∞) ⊆ L (F) so that Yt = E[Y |Ft] is a martingale (it’s certainly L1 by construction, adapted by deﬁnition, and satisﬁed the martingale property since for s ≤ t:

E[Yt|Fs] = E[E[Y |Ft]|Fs] = E[Y |Fs] = Ys).

By the conditional Jensen inequality:

2 2 2 2 E[Yt ] = E[E[Y |Ft] ] ≤ E[E[Y |Ft]] = E[Y ] < ∞

2 2 2 so that ||Y ||2 = supt≥0 E[Xt ] < ∞ and so Y ∈ M .

Finally, since Yt → Y∞ a.e. we have that Y∞ = limn→∞ E[Y |Ft] = E[Y |F∞] = Y . Thus,

Φ([Yt]) = Y .

2 2 Thus, Φ is an isometry, so that since L (F∞) is complete, M must be complete.

We may now deﬁne the integral for integrands in Π2(X). To deﬁne H · X when X is a continuous n n bounded martingale and H ∈ Π2(X), let H ∈ bΠ1 so that ||H − H||X → 0 the existence of which is n n guaranteed by lemma 7. Since H ∈ bΠ1, H · X is a continuous martingale for each n and Z t h n 2i n 2 E (H · X)t = E (H )sd hXis 0 by corollary 5 and theorem 15. So: Z n m 2 n m 2 n m 2 E ((H − H ) · X)t ≤ E (Hs − Hs ) d hXis = ||H − H ||X

n m 2 n m 2 n m 2 ⇒ ||H · X − H · X||2 = sup E ((H − H ) · X)t ≤ ||H − H ||X t≥0 n n 2 Thus, since the sequence {H } is Cauchy in Π2(X), the sequence (H · X) is Cauchy in M , and so must converge to a limit in this space by theorem 18, which we call H · X. It’s clear by using a shufﬂe sequence that this deﬁnition does not depend on the sequence Hn.

2 Theorem 19 If X is a bounded continuous martingale and H ∈ Π2(X) then H · X ∈ M and is continuous.

Proof of Theorem 19: By deﬁnition, H · X ∈ M2.

n n n To show that H · X is continuous, note that if H ∈ bΠ1 have ||H − H||X → 0, then H · X are n 2 continuous by corollary 5. Since ||H · X − H · X||2 → 0, the Chebyshev and L maximal inequality

34 imply:

h n 2i E supt≥0 |(H · X)t − (H · X)t| n 2 n 4||(H · X)t − (H · X)||2 P sup |(H · X)t − (H · X)t| > ≤ 2 ≤ 2 → 0. t≥0

n Thus, supt≥0 |(H · X)t − (H · X)t| converge to 0 in probability, so that there exists a subsequence nj n nj (H · X)t of (H · X)t such that supt≥0 |(H · X)t − (H · X)t| → 0 almost surely.

Thus, for almost all ω ∈ Ω, we have that

nj nj |(H · X)t(ω) − (H · X)t(ω)| ≤ sup |(H · X)t(ω) − (H · X)t(ω)| → 0 t≥0

n for all t ≥ 0. Thus, (H j · X)t converge uniformly almost surely to (H · X)t, so that (H · X)t is continuous almost surely.

Theorem 20 If X is a bounded continuous martingale, and H,K ∈ Π2(X) then H + K ∈ Π2(X) and

((H + K) · X)t = (H · X)t + (K · X)t.

Proof of Theorem 20: Since Π2(X) is a normed linear space

||H + K||X ≤ ||H||X + ||K||X < ∞ since by hypothesis, H,K ∈ Π2(X). Thus, H + K ∈ Π2(X).

n n n n Now, let H ,K ∈ bΠ1 such that ||H − H||X → 0 and ||K − K||X → 0. By a trivial application of the triangle inequality:

n n n n ||(H + K ) − (H + K)||X ≤ ||H − H||X + ||K − K||X → 0.

Let ((H + K) · X) denote the martingale in M2 such that ((Hn + Kn) · X) converges to.

By theorem 14, n n n n ((H + K ) · X)t = (H · X)t + (K · X)t

⇒ ||((H + K) · X)t − (H · X)t − (K · X)t||2

n n n n ≤ ||((H + K) · X)t − ((H + K ) · X)t||2 + ||((H + K ) · X)t − (H · X)t − (K · X)t||2

n n n n = ||((H + K) · X)t − ((H + K ) · X)t||2 + ||(H · X)t + (K · X)t − (H · X)t − (K · X)t||2

n n n n ≤ ||((H + K) · X)t − ((H + K ) · X)t||2 + ||(H · X)t − (H · X)t||2 + ||(K · X)t − (K · X)t||2 → 0

Thus,

((H + K) · X)t = (H · X)t + (K · X)t almost surely.

35 5.2 The Kunita -Watanabe Inequality

The Kunita-Watanabe inequality states:

Theorem 21 For any two continuous local martingales M and N and measurable processes H and K, the inequality 1 1 Z t Z t 2 Z t 2 2 2 |HsKs|d |hM,Ni|s ≤ Hs d hMis Ks d hNis 0 0 0 holds a.s. for t ≤ ∞.

Proof of Theorem 21: First, we show that

1 1 Z t Z t 2 Z t 2 2 2 HsKsd hM,Nis ≤ Hs d hMis Ks d hNis (2) 0 0 0 for simple integrands K and H.

Pm Towards this end, let K = j=1 1(tj−1,tj ](s)Kj where Kj ∈ Ftj−1 . By subdividing intervals if neces- Pm sary, we may assume that H = j=1 1(tj−1,tj ](s)Hj where Hj ∈ Ftj−1 .

t Now, we deﬁne hM,Nis = hM,Nit − hM,Nis. Since

t t t 2 t 0 ≤ hM + rN, M + rNis = hM,Mis + 2r hM,Nis + r hN,Nis for every r ∈ R, we have that after minimizing this function w.r.t. r that

1 1 t t 2 t 2 | hM,Nis | ≤ hMis hNis a.s.

So:

Z t m m 1 1 X tj X tj 2 tj 2 HsKsd hM,Ni ≤ |HjKj| hM,Ni ≤ |Hj||Kj| hMi hNi s tj−1 tj−1 tj−1 0 j=1 j=1

1 1  m  2  m  2 1 1 Z t 2 Z t 2 X t X t ≤ H2 hMi j K2 hNi j = H2d hMi K2d hNi .  j tj−1   j tj−1  s s s s j=1 j=1 0 0

Now, if H and K are bounded measurable processes, then there exists Hn → H and Kn → K everywhere, where Hn,Kn are simple processes.

Since the sets over which we integrated in the preceding argument are of the form [0, t] which are compact, hMis and hNis are ﬁnite measures a.e. Thus, we may apply the bounded convergence theorem so that (2) holds for bounded measurable processes H and K.

Now, we have the following

36 Proposition 1 Let µ be a complex measure on a σ-algebra M on X. Then there is a measurable function h such that |h(x)| = 1 for all x ∈ X and such that

dµ = hd|µ|.

For the proof, see Rudin’s Real and Complex Analysis, pg. 124.

In our case, we want to apply the result to the real signed measure d hM,Nis and its total variation measure d |hM,Ni|s. The theorem then implies that for almost all ω ∈ Ω there exists a Radon-Nikodym derivative Js(ω) taking values in {−1, 1} such that d hM,Nis (ω) = Js(ω)d |hM,Ni|s.

So, when H and K are bounded measurable functions, by replacing H by HJsgn(HK) on the LHS of (2), which is again a bounded measurable process, we obtain

1 1 Z t Z t 2 Z t 2 2 2 |HsKs|d |hM,Ni|s ≤ Hs d hMis Ks d hNis (3) 0 0 0

Finally, for arbitrary measurable functions H and K, we can ﬁnd an increasing sequence of bounded functions that converge to |H| and |K| respectively. The monotone convergence theorem then estab- lishes (3) for arbitrary measurable functions, and so the proof is complete.

Theorem 22 If H ∈ Π2(X) ∩ Π2(Y ) then H ∈ Π2(X + Y ) and

(H · (X + Y ))t = (H · X)t + (H · Y )t.

Proof of Theorem 22: Recall that hX,Y is is the unique continuous predictably process At that is locally of bounded variation, has A0 = 0, and makes XtYt − At a local martingale.

So, in the case when Yt = Xt, it’s clear that hX,Xis = hXis. Furthermore, bilinearity of h·, ·it is immediate from its uniqueness and deﬁnition, so that:

hX + Y is = hX + Y,X + Y is = hXis + hY is + 2 hX,Y is .

Now, the Kunita-Watanabe inequality implies:

1 hXit + hY it | hX,Y i | ≤ (hXi hY i ) 2 ≤ t t t 2 2 2 2 the latter inequality which comes from the fact that 0 ≤ (a − b) ⇒ 2ab ≤ a + b for all a, b ∈ R.

Thus: | hX + Y it | ≤ 2(hXit + hY it). Thus, for H ∈ Π2(X) ∩ Π2(Y ), we have that H ∈ Π2(X + Y ).

n n Now, by lemma 7, we can ﬁnd H ∈ bΠ1 such that ||H −H||Z → 0 for Z = X,Y,X +Y . By theorem 14, n n n (H · (X + Y ))t = (H · X)t + (H · Y )t.

37 So:

n n ||(H·(X+Y ))t−(H·X)t−(H·Y )t||2 ≤ ||(H·(X+Y ))t−(H ·(X+Y ))t||2+||(H ·(X+Y ))t−(H·X)t−(H·Y )t||2

n n n ≤ ||(H · (X + Y ))t − (H · (X + Y ))t||2 + ||(H · X)t − (H · X)t||2 + ||(H · Y )t − (H · Y )t||2

n n n ≤ ||H − H ||X+Y + ||H − H ||X + ||Y − Y ||Y → 0.

Thus,

(H · (X + Y ))t = (H · X)t + (H · Y )t a.e.

Theorem 23 If X,Y are bounded martingales, H ∈ Π2(X), K ∈ Π2(Y ) then

Z t hH · X,K · Y it = HsKsd hX,Y is . 0

Proof of Theorem 23: By the discussion in the proof of theorem 15, it is sufﬁcient to show that

Z t Zt = (H · X)t(K · Y )t − HsKsd hX,Y is (4) 0

n n is a martingale. Let H and K be sequences of elements of bΠ1 that converge to H and K in Π2(X) n n n and Π2(Y ) respectively, and let Zt be the quantity that results when H and K replace H and K in (4).

n n t 1 By theorem 15, Zt is a martingale. By claim 6, it is sufﬁcient to show that Zt → Z in L to show that

Zt is a martingale.

To take care of the ﬁrst term, note that: n n n n E sup |((H − H) · X)t(K · Y )t| ≤ E sup |((H − H) · X)t| sup |(K · Y )t| t≥0 t≥0 t≥0

1 1 " 2# 2 " 2# 2 n n ≤ E sup |((H − H) · X)t| E sup |(K · Y )t| t≥0 t≥0 n n n n ≤ 4||(H − H) · X||2||K · Y ||2 = 4||H − H||X ||K ||Y → 0

38 n n as n → ∞ since ||H − H||X → 0 ||K ||Y → ||K||Y < ∞ as n → ∞ and where we have used the Cauchy-Schwartz inequality and the L2 maximal inequality.

Similarly, to take care of the second term, note that: n n E sup |(H · X)t((K − K) · Y )t| ≤ E sup |(H · X)t| sup |((K − K) · Y )t| t≥0 t≥0 t≥0

1 1 " 2# 2 " 2# 2 n ≤ E sup |(H · X)t| E sup |((K − K) · Y )t| t≥0 t≥0 n n ≤ 4||(H · X)t||2||((K − K) · Y )t||2 = 4||H||X ||K − K ||Y → 0

n as n → ∞ since ||K − K ||Y → 0 as n → ∞ and ||H||X < ∞.

n n 1 Thus, we have that (H · X)s(K · Y )s → (H · X)s(K · Y )s in L .

Now, Z t Z t Z t n n n n Hs Ks d hX,Y is − HsKsd hX,Y is ≤ |Hs Ks − HsKs|d |hX,Y i|s 0 0 0 Z t Z t n n n ≤ |Hs − Hs||Ks |d |hX,Y i|s + |Hs||Ks − Ks|d |hX,Y i|s . 0 0 By the Kunita-Watanabe inequality:

1 1 Z t Z t 2 Z t 2 n n n 2 n 2 |Hs − Hs||Ks |d |hX,Y i|s ≤ |Hs − Hs| d hXis |Ks | d hY is 0 0 0

( 1 1 ) Z t Z t 2 Z t 2 n n n 2 n 2 ⇒ E |Hs − Hs||Ks |d |hX,Y i|s ≤ E |Hs − Hs| d hXis |Ks | d hY is 0 0 0

1 1 Z t 2 Z t 2 n 2 n 2 ≤ E |Hs − Hs| d hXis E |Ks | d hY is → 0 0 0 n n as n → ∞ since ||H − H||X → 0 and ||K ||Y → ||K||Y < ∞.

Similarly:

1 1 Z t Z t 2 Z t 2 n 2 n 2 |Hs||Ks − Ks|d |hX,Y i|s ≤ |Hs| d hXis |Ks − Ks| d hY is 0 0 0

( 1 1 ) Z t Z t 2 Z t 2 n 2 n 2 ⇒ E |Hs||Ks − Ks|d |hX,Y i|s ≤ E |Hs| d hXis |Ks − Ks| d hY is 0 0 0

1 1 Z t 2 Z t 2 2 n 2 ≤ E |Hs| d hXis E |Ks − Ks| d hY is → 0 0 0 n as n → ∞ since ||H||X < ∞ and ||K − K||Y → 0 as n → ∞.

39 Thus Z t Z t n n Hs Ks d hX,Y is → HsKsd hX,Y is 0 0 in L1, so that Z t Z t n n n n n E[|Zt −Zt|] = E (H · X)t(K · Y )t − Hs Ks d hX,Y is − (H · X)t(K · Y )t − HsKsd hX,Y is 0 0 Z t Z t n n n n ≤ E [|(H · X)s(K · Y )s − (H · X)s(K · Y )s|]+E Hs Ks d hX,Y is − HsKsd hX,Y is → 0 0 0 n 1 as n → ∞, so that Zt → Zt in L .

40 5.3 Integration w.r.t. Local Martingales

In this section we will extend the integral so that the integrators are allowed to be continuous local martingales, and integrands are in Z t 2 Π3(X) = H predictable : Hs d hXis < ∞ a.s. for all t ≥ 0 . 0

It turns out that this is the largest possible class of integrands.

In order to extend the integral from bounded martingales to local martingales, we prove the:

Theorem 24 Suppose X is a bounded continuous martingale, H,K ∈ Π2(X) and Hs = Ks for s ≤ T where T is a stopping time. Then, (H · X)s = (K · X)s for s ≤ T .

1 2 1 2 Proof of Theorem 24: We may write H2 = Hs + Hs and Ks = Ks + Ks where

1 1 2 2 Hs = Ks = Hs1s≤T = K21s≤T ,Hs = Hs1s>T ,Ks = Ks1s>T .

1 2 1 2 Since H,K ∈ Π2(X), it’s clear that H ,H ,K ,K ∈ Π2(X), and that

1 1 (H · X)t = (K · X)t

2 2 for all t ≥ 0. Since Hs = Ks = 0 for s ≤ T theorem 23 implies:

2 2 H · X s = K · X s = 0, s ≤ T.

In order to proceed further, we need the following two claims:

Claim 8 Let T be a stopping time and X,Y continuous local martingales. Then

XT ,Y T = hX,Y iT .

Proof of Claim 8: First, notice that if X is a continuous local martingale, then XT is a continuous local martingale; so see this, let Tn be a sequence of stopping times that reduce the continuous local martingale X. Then,

T Tn T ∧Tn Tn T (Xt ) = Xt = (Xt ) .

Tn Since Xt is by deﬁnition a martingale for each n, the RHS is again a continuous martingale for each n, T so that the sequence of stopping times Tn that reduce X also reduce X .

2 Let Zt = (Xt) − hXit, so that since Xt is a continuous local martingale, so is Zt.

0 0 Tn 0 Let Tn reduce Z and Tn reduce X such that X is bounded. Let Sn = Tn ∧ Tn. Since Tn ↑ ∞ and 0 0 Tn ↑ ∞ it’s clear that Sn ↑ ∞. Since Sn ≤ Tn and Sn ≤ Tn, theorem 4, Sn reduces Z and X.

41 So: T ∧Sn 2 T ∧Sn 2 T ∧Sn T ∧Sn T ∧Sn Zt = (Xt) − hXit = Xt − hXit

2 T ∧Sn T ∧Sn = Xt − hXit .

T ∧Sn T ∧Sn Since Zt is a martingale, and Xt is a bounded continuous martingale, we must have that T ∧Sn T ∧Sn T ∧Sn X t = hXit by uniqueness of the process X t. Letting n → ∞ above, we have:

2 ZT = XT − lim XT ∧Sn t t n→∞ t

T so that by uniqueness of the process X t,

XT = lim XT ∧Sn = lim hXiT ∧Sn = hXiT t n→∞ t n→∞ t t

Thus, by deﬁnition of hX,Y i: 1 1 h i XT ,Y T = XT + Y T − XT − Y T = hX + Y iT − hX − Y iT = hX,Y iT .4 t 4 t t 4 t t t

Claim 9 Suppose that the continuous local martingale X has hXit = 0 for t ≤ T where T is a stopping time. Then, Xt = X0 for t ≤ T a.s.

Proof of Claim 9: WLOG, WMA X0 = 0. First, let’s assume that X is a continuous bounded martingale 2 that satisﬁes the hypotheses. Then, we know that Zt = Xt − hXit is a martingale.

Using the L2 maximal inequality, we obtain:

" 2# 2 2 E sup Xt∧T ≤ E sup Xt∧T ≤ 4E XT ∧n . t≤n t≤n

Now, for all t ≤ n, the optional stopping theorem applied to the martingale Zt and the stopping times t ∧ T ≤ n ∧ T yields:

2 2 2 E Xt∧T − hXit∧T = E Xn∧T − hXin∧T ⇒ E Xn∧T = E [hXin∧T ] where we let t = 0. Thus: 2 2 E sup Xt∧T ≤ 4E [hXin∧T ] = 0 ⇒ E sup Xt∧T = 0. t≤n t≤n

Since the martingale Xt is bounded, the bounded convergence theorem implies 2 E sup Xt∧T = 0. t≥0

42 Thus, Xt∧T = 0 almost surely, so that Xt = 0 for t ≤ T almost surely.

Now let X be a continuous local martingale that satisﬁes the hypotheses, reduced by the canonical stop-

Tn Tn Tn ping times Tn. Then by Claim 8, X t = hXit = 0 for t ≤ T ∧ Tn. Since Xt is a continuous Tn bounded martingale, for each n, the previous argument applies, so that we have Xt = X0 for t ≤ T ∧Tn almost surely. Since this is true for all n, letting n → ∞, we have that Xt = X0 for t ≤ T almost surely. 4

2 2 2 2 Applying claim 9 to the martingales H · X and K · X, we obtain that (H · X)t = (K · X)t = 0 for t ≤ T .

Thus: 1 2 1 2 (H · X)t = (H · X)t + (H · X)t, (K · X)t = (K · X)t + (K · X)t

⇒ (H · X)t = (K · X)t for t ≤ T . To extend the deﬁnition of the integral, let X be a continuous local martingale reduced by the canonical times Sn, and let H ∈ Π3(X). Deﬁne

Z t 2 Rn(ω) = inf 0 ≤ t < ∞ : Hs (ω)d hXis (ω) ≥ n . 0

Since H ∈ Π3(X), it’s clear that Rn ↑ ∞ almost surely as n ↑ ∞. Now, let Tn be stopping times such that Tn ↑ ∞ almost surely, and

Tn(ω) ≤ Rn(ω) ∧ Sn(ω), and deﬁne (n) Ht (ω) = Ht(ω)1t≤Tn(ω).

It’s clear by deﬁnition that Tn reduce X. Now,

Z 2 Z Z Tn(ω) (n) 2 2 Hs (ω) d hXis (ω) = Hs(ω)1s≤Tn(ω) d hXis (ω) = Hs (ω)d hXis (ω) ≤ n 0

Z 2 (n) ⇒ E Hs d hXis ≤ n < ∞.

(n) (n) Thus, H ∈ Π2 (X) for every n, and so H · X t is a quantity that we’ve previously constructed for all n.

(n) (m) Now, for n ≤ m, we have that Ht (ω) = Ht (ω) for t ≤ Tn(ω). So, by theorem 24:

(n) (m) (H · X)s = (H · X)s

43 for s ≤ Tn. Thus, we may unambiguously deﬁne

(n) (H · X)s(ω) = H · X (ω) s for s ≤ Tn(ω). Since Tn ↑ ∞ almost surely, this deﬁnition extends to all non-negative values of s.

0 To show that this is independent of the choice of stopping times Tn, suppose that Tn is another set of 0 0 stopping times such that Tn ↑ ∞ a.s. and Tn ≤ Rn ∧ Sn.

0 Deﬁne Qn = Tn ∧ Tn. Then: 0 HTn · X = HTn · X t t for t ≤ Qn. Since Qn ↑ ∞, it follows immediately that (H · X)t is independent of the sequence of stopping times.

Theorem 25 If X is a continuous local martingale and H ∈ Π3(X) then

HT · X = (H · X)T = H · XT = HT · XT .

Theorem 26 If X is a continuous local martingale and H ∈ Π3(X), then (H · X)t is a continuous local martingale.

Proof of Theorem 26: By stopping at Rn ∧ Sn as deﬁned above, it sufﬁces to show that if X is a bounded martingale and H ∈ Π2(X) then (H · X)t is a continuous martingale b theorem 25. But this is immediate from theorem 19.

Theorem 27 Let X and Y be continuous local martingales. If H,K ∈ Π3(X) then H + K ∈ Π3(X) and

((H + K) · X)t = (H · X)t + (K · X)t .

If H ∈ Π3(X) ∩ Π3(Y ) then H ∈ Π3(X + Y ) and

(H · (X + Y ))t = (H · X)t + (H · Y )t .

Theorem 28 If X and Y are continuous local martingales, H ∈ Π3(X) and K ∈ Π3(Y ) then Z t hH · X,K · Y it = HsKsd hX,Y is . 0 2 Theorem 29 If X is a continuous local martingale and H ∈ Π2(X) then H ·X ∈ M and ||H ·X||2 =

||H||X .

5.4 Integration w.r.t. Semimartingales

Deﬁntion: X is said to be a semimartingale if Xt = Mt +At where Mt is a continuous local martingale and At is a continuous adapted process that is locally of bounded variation.

44 Theorem 30 Let Xt be a continuous semimartingale. If the continuous process Mt and At are chosen so that A0 = 0 then the decomposition Xt = Mt + At is unique.

0 0 0 0 0 Proof of Theorem 30: If Mt +At is another decomposition with A0 = 0, then Mt −At = Xt = Mt −At 0 0 so that At − At = Mt − Mt. Since this is a continuous local martingale of locally bounded variation, 0 0 0 0 theorem 11 implies that At − At is constant, so that At − At = 0 for all t. Thus, Mt = Mt and At = At for all t. Why should we care about semimartingales? First, we recall Ito’s formula for continuous local martingales:

Theorem 31 Suppose X is a continuous local martingale, and f is a function with two continuous derivatives. Then, with probability 1, for all t ≥ 0:

Z t Z t 0 1 00 f(Xt) − f(X0) = f (Xs)dXs + f (Xs)d hXis . 0 2 0

2 1. If X is a continuous local martingale and f is C then Ito’s formula shows that f(Xt) is always a semimartingale but is not a local martingale unless f 00(x) = 0 for all x. It can be shown that if X 2 is a continuous semimartingale and f is C then f(Xt) is again a semimartingale.

2. Deﬁne an easy integrand to be a process of the form

n X H = Hj1(Tj ,Tj+1] j=0

where 0 = T0 ≤ T1 ≤ · · · ≤ Tn+1 are stopping times and Hj ∈ FTj satisfy |Hj| < ∞ a.s.

Let bΠe,t be the collection of bounded easy predictable processes that vanish on (t, ∞) equipped with the uniform norm

||H||u = sup |Hs(ω)|. s,ω

Finally, let L0 be the collection of all random variables topologized by convergence in probability, which is induced by the metric |X| ||X|| = E . 0 1 + |X|

A result proved by Bichteler and Dellacherie states:

0 Theorem 32 If H 7→ (H · X) is continuous from bΠe,t → L for all t then X is a semimartingale.

The class of integrands we want to integrate are as follows:

Deﬁnition: We say H ∈ lbΠ =the set of locally bounded predictable processes if there is a sequence of stopping times Tn ↑ ∞ such that |H(s, ω)| ≤ n for s ≤ Tn.

45 To define the integral w.r.t. the semimartingale Xt = Mt + At, first note that we can define Z t (H · A)t(ω) = Hs(ω)dAs(ω) 0 as a Lebesgue-Stieltjes integral (which exists a.e. ω).

To integrate w.r.t. the continuous local martingale Mt, note that

Z Tn H2d hMi ≤ n2 hMi < ∞. s s Tn 0

Thus, since Tn ↑ ∞, we have that H ∈ Π3(M), so that since H was arbitrary, lbΠ ⊂ Π3(M). Thus, we can deﬁne (H · M)t, and let

(H · X)t = (H · M)t + (H · A)t .

From this deﬁnition, it’s immediate that

Theorem 33 If X is a continuous semimartingale and H ∈ lbΠ then (H · X)t is a continuous semimartingale.

Deﬁnition: If X = M +A and X0 = M 0 +A0 are continuous semimartingales, we deﬁne the covariance 0 0 hX ,Xit = hM,M it.

6 Concluding Remarks

1. Note that almost everything we’ve done above works when instead of continuous (local) martingales, we deal with square integrable cadlag (local) martingales (of course, care must be taken to deal with jumps). In particular, probably the most important square integrable cadlag process is the Poisson process:

Deﬁnition: A process Nt is a Poisson process with parameter λ if

(a) P [ω ∈ Ω: N0(ω) = 0] = 1. (b) If t1 < ··· tn, the increments Ntj − Ntj−1 1 ≤ j ≤ n are independent where we set

t0 = 0.

e−λ(t−s)(λ(t−s))k (c) For any 0 ≤ s < t, P [Nt − Ns = k] = k! , k = 0, 1, 2,....

(d) The sample paths of Nt are cadlag a.s.

So, appropriately replacing Xt with Nt when necessary, the theory developed above also lets us integrate the appropriate integrands w.r.t. the Possion process.

46 2. I’ve looked around, and it’s quite common to use semimartingales in Finance (see Shiryaev, A. N. (1999). Essentials of Stochastic Finance: Facts, Models, Theory. World Scientic.). The price

Ht process of a ﬁnancial asset St = e where Ht is a semimartingale.

3. We’ve already encountered an example of a continuous local martingale, namely the Bessel process

r n Z t (j) Z t (1)2 (n)2 X Bs n − 1 1 B + ··· + B = ||B || = dB(j) + ds. t t t ||B || s 2 ||B || j=1 0 s 0 s

k Another example of a continuous local martinagle used in applications is f(t) = eB(t) . For k ≥ 3, we have 2 ∞ −x h k i Z k e 2t E |f(t)|2 = E e2B(t) = e2x √ dx = ∞. −∞ 2πt

Thus, f(t) ∈/ Π2(B). On the other hand, since almost all sample paths of f(t) are continuous, Z t Z t 2 2 |f(s)| d hBis = |f(s)| ds < ∞ 0 0

almost surely for all t ≥ 0. Thus, f ∈ Π3(B).

4. What’s useful about what we’ve done is that in general it requires hard computation to check

whether a process H belongs to Π2(B), and hence hard to verify that H · B even exists. On the

other hand, it’s much easier to check that a process H belongs to Π3(B), so that it’s easier to verify the existence of H · B.

5. Consider the SDE

dXt = σ(Xt)dZt (5)

n n where Zt is a R valued continuous semimartingale, and σ : R → Mn,n, the space of n × n matrices.

2 n If this equation has a solution, and f ∈ C (R ) then Z t i α 1 t i j D α βE f(Xt) = f(X0) + fxi (Xs)σαdZs + ∈0 fxi,xj (Xs)σα(Xs)σβ(Xs)d Z ,Z (6) 0 2 (this is the generalized Ito formula).

Of course in order to use this, we need to prove the existence of the solution Xt.

Theorem 34 Suppose that σ is globally Lipschitz and X0 is square integrable. Then, (5) has a unique solution.

Notice how general (and constraining) this is! If Zt is obtained through some Ito formula calculu- ation, instead of worrying about the martingale and bounded variation processes separately [which

could be messy] we can just lump it all into one quantity Zt (which could be neater) and invoke theorem 34.

47 6. Let’s go back to to (5) and (6). Recall that if H ∈ Π2(X) and X is a martingale then H · X is

a martingale. Thus, E [(H · X)t] = 0. However, for H ∈ lbΠ where X is a semimartingale, we

know that H · X is a semimartingale, so it’s deﬁnitely not necessarily true that E [(H · X)t] =

0. Even if H ∈ Π3(X) and X is a continuous local martingale, we still can’t conclude that B(t)k E [(H · X)t] = 0 (to be more concrete, think of when H is a Bessel process, of the process e for k ≥ 3).

α Thus, taking expectations in (6), the integral w.r.t. Zs will contribute nontrivially. Thus, SDEs driven by semimartingales (continuous local martingales) yield more interesting dynamics.

7. Theorem 34 implies that the solution Xt runs for all time since the growth of σ is at most linear (recall the Lipschitz condition). When σ is only locally Lipschitz, we have to allow for the possibility of explosion. For example, the solution to dX t = X2,X = 1 dt t 0

1 is Xt = 1−t , which explodes at t = 1. In general, let M be a locally compact metric space, and let Mc = M ∪ {∂M } denote its one point compactiﬁcation.

Deﬁnition: An M valued path x with explosion time e = e(x) > 0 is a continuous map x :

[0, ∞) → Mc such that xt ∈ M for 0 ≤ t < e and xt = ∂M for all t ≥ e if e < ∞.

It can be shown that the exploding time e(Xt) of a continuous process Xt is a stopping time. We have the theorem

Theorem 35 Let Z be a semimartingale deﬁned up to a stopping time τ. Then, there is a unique solution X to (5) up to the stopping time e(X) ∧ τ. If T is a another solution up to a stopping time

η ≤ τ then η ≤ e(X) ∧ τ and Xt = Yt for 0 ≤ t < e(X) ∧ η.

8. Finally, let U(r) be some radially symmetric potential, and let Bt be the motion of a body under- R t going Brownian motion. Then, the integral 0 U(||Bt||)d||Bt|| represents the work done by that particle. Of course, we would need sufﬁcient hypotheses on U to guarantee that this integral exists, but that’s exactly what we’ve been doing all along.

A future topic: now that we know all about integration, how would you go about simulating the paths of Brownian motion, or any continuous local martingale?