Variance and Covariance Processes

Prakash Balachandran Department of Mathematics Duke University

May 26, 2008

These notes are based on Durrett’s Stochastic Calculus, Revuz and Yor’s Continuous Martingales and Brownian Motion, Karatzas and Shreve’s Brownian Motion and Stochastic Calculus, and Kuo’s Intro- duction to Stochastic Calculus

1 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 < ∞.

1 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.

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 1 (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.

2 Claim 1 In the case Mt = Bt in Doob-Meyer, At = hBit = t.

2 Proof of Claim 1: 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

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.

3 2 Variance and Covariance Processes

We take L and P as deﬁned in section 1.

Theorem 2 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 3 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 2 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 2:

Theorem 4 Any continuous local martingale Xt that is predictable and locally of bounded variation is constant (in time).

Proof: 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

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

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

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 2 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: 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

u Now, the LHS is Tπ(ω) for π = {0 = t−N 0 < ··· < tN = u}. Thus, 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) ↑ ∞}.

By deﬁnition, P [A] = 1. Now, let ω ∈ A 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].

5 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 2 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: We proved last time that

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.

6 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 2: 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.

7 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 4, 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 2:

We proceed in steps:

Step 1: Proof of existence in theorem 2 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

8 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

9 Putting it all together then, we have:

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

10 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 − Xt) |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 ≤ (2M) (Xsm − Xsm−1 ) " 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

11 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 meshes, 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 ∆n kj ∆nk 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].

12 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 3 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 3: 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?). 2 Now, the martingale property implies that n n 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.

13 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 5 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 At − 0 0 At = (XtYt − At) − (XtYt − At) is a continuous local martingale that is of locally bounded variation. 0 Hence, by theorem 4, this must be identically zero, so that At = At for all t.