Math 288 - Probability Theory and Stochastic Process€¦ · Math 288 - Probability Theory and...

Math 288 - Probability Theory and Stochastic

Process

Taught by Horng-Tzer YauNotes by Dongryul Kim

Spring 2017

This course was taught by Horng-Tzer Yau. The lectures were given at MWF12-1 in Science Center 310. The textbook was Brownian Motion, Martingales,and Stochastic Calculus by Jean-Francois Le Gall. There were 11 undergrad-uates and 12 graduate students enrolled. There were 5 problem sets and thecourse assistant was Robert Martinez.

Contents

1 January 23, 2017 51.1 Gaussians . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2 Gaussian vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2 January 25, 2017 82.1 Gaussian process . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.2 Conditional expectation for Gaussian vectors . . . . . . . . . . . 8

3 January 27, 2017 103.1 Gaussian white noise . . . . . . . . . . . . . . . . . . . . . . . . . 103.2 Pre-Brownian motion . . . . . . . . . . . . . . . . . . . . . . . . 11

4 January 30, 2017 134.1 Kolmogorov’s theorem . . . . . . . . . . . . . . . . . . . . . . . . 13

5 February 1, 2017 155.1 Construction of a Brownian motion . . . . . . . . . . . . . . . . . 155.2 Wiener measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

6 February 3, 2017 176.1 Blumenthal’s zero-one law . . . . . . . . . . . . . . . . . . . . . . 176.2 Strong Markov property . . . . . . . . . . . . . . . . . . . . . . . 18

1 Last Update: August 27, 2018

7 February 6, 2017 197.1 Reflection principle . . . . . . . . . . . . . . . . . . . . . . . . . . 197.2 Filtrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

8 February 8, 2017 218.1 Stopping time of filtrations . . . . . . . . . . . . . . . . . . . . . 218.2 Martingales . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

9 February 10, 2017 239.1 Discrete martingales - Doob’s inequality . . . . . . . . . . . . . . 23

10 February 13, 2017 2510.1 More discrete martingales - upcrossing inequality . . . . . . . . . 25

11 February 15, 2017 2711.1 Levi’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2711.2 Optional stopping for continuous martingales . . . . . . . . . . . 28

12 February 17, 2017 29

13 February 22, 2017 3013.1 Submartingale decomposition . . . . . . . . . . . . . . . . . . . . 30

14 February 24, 2017 3114.1 Backward martingales . . . . . . . . . . . . . . . . . . . . . . . . 3114.2 Finite variance processes . . . . . . . . . . . . . . . . . . . . . . . 32

15 February 27, 2017 3315.1 Local martingales . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

16 March 1, 2017 3516.1 Quadratic variation . . . . . . . . . . . . . . . . . . . . . . . . . . 35

17 March 3, 2017 3717.1 Consequences of the quadratic variation . . . . . . . . . . . . . . 37

18 March 6, 2017 3918.1 Bracket of local martingales . . . . . . . . . . . . . . . . . . . . . 3918.2 Stochastic integration . . . . . . . . . . . . . . . . . . . . . . . . 40

19 March 8, 2017 4119.1 Elementary process . . . . . . . . . . . . . . . . . . . . . . . . . . 41

20 March 20, 2017 4320.1 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

2

21 March 22, 2017 4521.1 Stochastic integration with respect to a Brownian motion . . . . 4521.2 Stochastic integration with respect to local martingales . . . . . 45

22 March 24, 2017 4722.1 Dominated convergence for semi-martingales . . . . . . . . . . . 4722.2 Ito’s formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

23 March 27, 2017 5023.1 Application of Ito calculus . . . . . . . . . . . . . . . . . . . . . . 50

24 March 29, 2017 5224.1 Burkholder–Davis–Gundy inequality . . . . . . . . . . . . . . . . 52

25 March 31, 2017 5425.1 Representation of martingales . . . . . . . . . . . . . . . . . . . . 5425.2 Girsanov’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 55

26 April 3, 2017 5626.1 Cameron–Martin formula . . . . . . . . . . . . . . . . . . . . . . 56

27 April 5, 2017 5827.1 Application to the Dirichlet problem . . . . . . . . . . . . . . . . 5827.2 Stochastic differential equations . . . . . . . . . . . . . . . . . . . 59

28 April 7, 2017 6128.1 Existence and uniqueness in the Lipschitz case . . . . . . . . . . 6128.2 Kolomogorov forward and backward equations . . . . . . . . . . 62

29 April 10, 2017 6429.1 Girsanov’s formula as an SDE . . . . . . . . . . . . . . . . . . . . 6529.2 Ornstein–Uhlenbeck process . . . . . . . . . . . . . . . . . . . . . 6529.3 Geometric Brownian motion . . . . . . . . . . . . . . . . . . . . . 66

30 April 12, 2017 6730.1 Bessel process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

31 April 14, 2017 6931.1 Maximum principle . . . . . . . . . . . . . . . . . . . . . . . . . . 6931.2 2-dimensional Brownian motion . . . . . . . . . . . . . . . . . . . 70

32 April 17, 2017 7132.1 More on 2-dimensional Brownian motion . . . . . . . . . . . . . . 71

33 April 19, 2017 7333.1 Feynman–Kac formula . . . . . . . . . . . . . . . . . . . . . . . . 7333.2 Girsanov formula again . . . . . . . . . . . . . . . . . . . . . . . 74

3

34 April 21, 2017 7634.1 Kipnis–Varadhan cutoff lemma . . . . . . . . . . . . . . . . . . . 76

35 April 24, 2017 7835.1 Final review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

4

Math 288 Notes 5

1 January 23, 2017

The textbook is J. F. Le Gall, Brownian Motion, Martingales, and StochasticCalculus.

A Brownian motion is a random function X(t) : R≥0 → Rn. This is amodel for the movement of one particle among many in “billiard” dymanics.For instance, if there are many particles in box and they bump into each other.Robert Brown gave a description in the 19th century. X(t) is supposed todescribe the trajectory of one particle, and it is random.

To simplify this, we assume X(t) is a random walk. For instance, let

SN =

N∑i=1

Xi, Xi =

1 with probability 1/2

−1 with probability 1/2.

For non-integer values, you interpolate it linearly and let

SN (t) =

bNtc∑i=1

Xi + (Nt− bNtc)XbNtc+1.

Then E(S2N ) = N . As we let N →∞, we have three properties:

• limN→∞

E(SN (t)√

N

)2

= t

• limN→∞

E(SN (t)SN (s))√N

2 = t ∧ s = min(t, s)

• SN (t)/√N → N(0, t)

This is the model we want.

1.1 Gaussians

Definition 1.1. X is a standard Gaussian random variable if

P (X ∈ A) =1√2π

∫A

e−x2/2dx

for a Borel set A ⊆ R.

For a complex variable z ∈ C,

E(ezX) =1√2π

∫ezxe−x

2/2dx = ez2/2.

This is the moment generating function. When z = iξ, we have

∞∑k=0

(−ξ2)k

k!= e−ξ

2/2 = 1 + iξEX +(iξ)2

2EX2 + · · · .

Math 288 Notes 6

From this we can read off the moments of the Gaussian EXn. We have

EX2n =(2n)!

2nn!, EX2n+1 = 0.

Given X ∼ N (0, 1), we have Y = σX + m ∼ N (m,σ2). Its moment gener-ating function is given by

EezY = eimξ − σ2ξ2/2.

Also, if Y1 ∼ N (m1, σ21) and Y2 ∼ N (m2, σ

22) and they are independent, then

Eez(Y1+Y2) = EezY1 EezY2 = ei(m1+m2)ξ−(σ21+σ2

2)ξ2/2.

So Y1 + Y2 ∼ N (m1 +m2, σ21 + σ2

2).

Proposition 1.2. If Xn ∼ N (mn, σ2n) and Xn → X in L2, then

(i) X is Gaussian with mean m = limn→∞mn and variance σ2 = limn→∞ σ2n.

(ii) Xn → X in probability and Xn → X in Lp for p ≥ 1.

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

EeiξX = limn→∞

eiξXn = limn→∞

eimnξ−σ2nξ

2/2 = eimξ−σ2ξ2/2.

So X ∼ N (m,σ2).(ii) Because EX2p can be expressed in terms of the mean and variance, we

have supn E|Xn|2p <∞ and so supn E|Xn−X|2p <∞. Define Yn = |Xn−X|p.Then Yn is bounded in L2 and hence is uniformly integrable. Also it convergesto 0 in probability. It follows that Yn → 0 in L1, which means that Xn → X inLp. It then follows that Xn → X in probability.

Definition 1.3. We say that Xn is uniformly integrable if for every ε > 0there exists a K such that supn E[|Xn| · 1|Xn|>K ] < ε.

Proposition 1.4. A sequence of random variables Xn converge to X in L1 ifand only if Xn are uniformly integrable and Xn → X in probability.

1.2 Gaussian vector

Definition 1.5. A vector X ∈ Rn is a Gaussian vector if for every n ∈ Rn,〈u,X〉 is a Gaussian random variable.

Example 1.6. Define X1 ∼ N (0, 1) and ε = ±1 with probability 1/2 each. LetX2 = εX1. Then X = (X1, X2) is not a Gaussian vector.

Math 288 Notes 7

Note that u 7→ E[u ·X] is a linear map and so E[u ·X] = u ·mX for somemX ∈ Rn. We call mX the mean of X. If X =

∑ni=1Xiei for an orthonormal

normal basis ei of Rn, then the mean of X is

mX =

N∑i=1

(EXi)ei.

Likewise, the map u 7→ Varu · X is a quadratic form and so VaruX =〈u, qXu〉 for some matrix qX . We all qX the covariant matrix of X. We writeqX(n) = 〈u, qXu〉. Then the characteristic function of X is

Eezi(u·X) = exp(izmX − z2qX(u)/2).

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

Proof. First check that

qX(u) =

n∑i,j=1

uiuj Cov(Xi, Xj).

Then

E exp(iu ·X) = exp(−1

2qX(u)

)= exp

(−1

2

n∑i=1

u2i Var(Xi)

)

=

n∏i=1

exp(−1

2u2i Var(Xi)

).

This gives a full description of X.

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

Proof. Diagonalize γ and let λj ≥ 0 be the eigenvalues. Choose v1, . . . , vn anorthonormal basis of eigenvectors of Rn and let wj = λjvj = γvj . Choose Yj tobe Gaussians N (0, 1). Then the vector

X =

n∑j=1

λjvjYj

is a Gaussian vector with the right covariance.

Math 288 Notes 8

2 January 25, 2017

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

Cov(Xj , Xk) = E[XjXk]− E[Xj ]E[Xk], γX = (Cov(Xj , Xk))jk = C.

Then the probability distribution is given by

PX(x) =1

(2π)d/2√

detCe−〈x,C

−1x〉/2

where C > 0 and the characteristic function is

E[eitx] = e−〈t,Ct〉/2.

2.1 Gaussian process

A (centered) Gaussian space is a closed subspace of L2(Ω,F , P ) containingonly Gaussian random variables.

Definition 2.1. A Gaussian process indexed by T (∈ R+) is one such thatXt1 , . . . , Xtk is always a Gaussian vector.

Definition 2.2. For a collection S of random variables, we denote by σ(S) thesmallest σ-algebra on Ω such that all variables in S are measurable.

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

Example 2.4. If we take X1 = X to be a Gaussian and X2 = εX where ε = ±1with probability 1/2 each, then E[X1X2] = E[εX2] = 0. That is, X1 and X2

are uncorrelated. But X1 and X2 are dependent. This is because X1, X2 arenot jointly Gaussian.

2.2 Conditional expectation for Gaussian vectors

Lt X be a random variable in a Gaussian space H. If H is finite dimensionalwith (jointly Gaussian) basis X1, . . . , Xd, then we may write X =

∑j ajXj . Let

K ⊆ H be a closed subspace. We want to look at the conditional expectation

E[X|σ(K)] = E[X|K],

where σ(K) is the σ-algebra generated by elements in K.

Definition 2.5. The conditional expectation is defined as E[X|K] = ϕ(K),the measurable function with respect to K so that for any function g(K) wehave

E[Xg(K)] = E[ϕ(K)g(K)], or alternatively, E[(X − ϕ(K))g(K)] = 0,

i.e., ϕ(K) is the L2-orthogonal projection of X to the space σ(K).

Math 288 Notes 9

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

E[f(X)|σ(K)] = E[f(Y + pK(X))|σ(K)] =

∫f(y + pK(X))

e−y2/2σ2

√2πσ

=

∫f(y)

e−(y−pK(X))2/2σ2

√2πσ

dy,

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

E[Xf(X)] = σ2E[f ′(X)]

by integration by parts. It follows that E[X2n] corresponds to the number ofways to pair elements in a set of size 2n. There is a more general interpretation.In general, we have

E[Y f(X1, . . . , Xd)] =∑j

Cov(Y,Xj)E[∂jf ].

Math 288 Notes 10

3 January 27, 2017

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

Γ(s, t) = Cov(Xs, Xt) ≥ 0,

where being positive semidefinite means

Var

(∫α(s)Xsds

)=

∫∫α(s)Γ(s, t)α(t)dsdt ≥ 0

for any α.

Theorem 3.1. Suppose Γ(s, t) is a positive semidefinite function. Then thereexists a Gaussian process with Γ as the covariant matrix.

Theorem 3.2 (Kolmogorov extension theorem). Given a family of consistentfinite dimensional distributions, there exists a measure with this family of finitedimensional distributions as the finite dimensional marginals.

Consistency means that if we are given the distribution of (Xs1 , Xs2 , Xs3),then the distribution of (Xs1 , Xs3) must be the integration. Kolmogorov’s the-orem is giving you an abstract construction of the measure.

So we are not going to be worry too much about the existence of these infinitedimensional measure spaces.

Example 3.3. Suppose µ is a probability measure on R. Then

Γ(s, t) =

∫eiξ(s−t)µ(dξ)

is positive semidefinite. This is because∫∫∫α(s)eiξ(s−t)α(t)dsdtµ(dξ) =

∫ (∫α(s)eiξs

)(∫α(t)e−iξt

)µ(dξ) ≥ 0.

3.1 Gaussian white noise

Suppose µ is a σ-finite measure.

Definition 3.4. We call a Gaussian space G a Gaussian white noise withintensity µ if

E(G(f)G(g)) =

∫fgdµ

for every f, g ∈ L2(µ).

Informally, this is the same as saying E[G(x)G(y)] = δ(x − y) with thedefinition G(x) = G(δx). Then

E[G(f)G(g)] = E[∫

G(x)f(x)dx

∫G(y)g(y)dy

]=

∫∫f(x)E[G(x)G(y)]g(y)dx dy =

∫f(x)g(x)dx.

Math 288 Notes 11

Theorem 3.5. For every µ a σ-finite measure there exists a Gaussian whitenoise with intensity µ.

Proof. Assume L2(µ) is separable. There then exists an orthonormal basis inL2(µ). For any f there exists a presentation f =

∑∞i=1 αifi. Now define

G(f) =

∞∑i=1

αiXi,

where X1, X2, . . .are independent in N (0, 1). Then we obvious have the desiredcondition.

3.2 Pre-Brownian motion

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

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

Proof. We have

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

Because Bt = G([0, t]), we may formally write dBt = G(dt). Then we canalso write

G(f) =

∫f(s)dBs.

We can also think

G([0, t]) =

∫ t

0

dBs = Bt −B0 = Bt.

Roughly speaking, dBt is the white noise. This idea can be rigorously formu-lated as that if 0 = t0 < t1 < · · · < tn then Bt1−Bt0 , Bt2−Bt1 , . . . , Btn−Btn−1

are independent Gaussian.

Proposition 3.8. The following statements are equivalent:

(i) (Xt) is a Brownian motion.

(ii) (Xt) is a centered Gaussian with covariance s ∧ t.(iii) (Xt −Xs) is independent of σ(Xr, r ≤ s) and Xt −Xs ∼ N (0, t− s).

(iv) Xti −Xti−1 are independent Gaussian with variance ti − ti−1.

Proof. (ii) ⇒ (iii) We have

E[(Bt −Bs)Br] = t ∧ r − s ∧ r = 0

for r ≤ s. The other implications are obvious.

Math 288 Notes 12

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

p(x1, . . . , xn) =1

(2π)n/2√t1(t2 − t1) · · ·

∏i

e−(xi−xi−1)2/2(ti−ti−1).

It is important to understand this as a Wiener measure.

Math 288 Notes 13

4 January 30, 2017

Let me remind you of some notions. A Gaussian space is a closed linear subspaceof L2(Ω, P ). A random process is a sequence (Xt)t≥0 of random variables. AGaussian process is a process such that (Xt1 , . . . , Xtk) is Gaussian for any choicet1, . . . , tk. A Gaussian white noise with intensity µ is a map G : L2(µ)→ L2(P )such that G(f) is Gaussian and E[G(f)G(g)] =

∫fgdµ. A pre-Brownian motion

is a process with Bt = G(1[0,t]).Sometimes people write Bt(ω). In this case, this ω is an element of the

probability space Ω. So then Bt(ω) is actually something concrete map.

4.1 Kolmogorov’s theorem

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

|Xt(ω)− Xs(ω)| ≤ Cα(ω)|s− t|α

almost surely.

Here we say that X is a modification of X if for every t ∈ T ,

P (Xt = Xt) = 1.

Definition 4.2. X and X are indistinguishable if there exists a set N ofprobability zero such that for all t and ω ∈ N , Xt(ω) = Xt(ω).

There is a slight difference, in the case of indistinguishability, N is some setthat does not depend on t. That is, indistinguishable implies being a modi-fication. If Xt and Xt have continuous sample paths with probability 1, theconverse holds. But nobody worries about these.

For a pre-Brownian motion, we have Xt −Xs ∼ N (0, t− s) and so

E[|Xt −Xs|q] = Cα|t− s|q/2.

Then letting ε = q/2− 1 and α < ε/q = 1/2− 1/q, we get α 1/2 as q →∞.That is, t→ Xt(ω) is Holder continuous of order ε.

Corollary 4.3. A Brownian motion sample path is Holder of order α < 1/2.

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

P

( 2n⋃i=1

(|Xi/2n −X(i−1)/2n | ≥ 2−nα)

)≤ 2n

(E|Xi/2n −X(i−1)/2n |q

)2nαq

≤ C2nqα−(1+ε)n2n = C2nqα−εn.

Math 288 Notes 14

So ε− qα > 0 implies∑n

P

(⋃i

(|Xi/2n −X(i−1)/2n | ≥ 2−nα)

)<∞.

Borel–Cantelli then implies that there exists a set N with P (N) = 0 suchthat for every ω ∈ N c there exists an n0(ω) such that∣∣Xi/2n(ω)−X(i−1)/2n(ω)

∣∣ < 2−nα

for all n > n0(ω). Now

Kα(ω) = supn≥1

supi

∣∣∣∣Xi/2n(ω)−X(i−1)/2n(ω)

2−nα

∣∣∣∣ <∞almost surely.

We will finish the proof next time.

Math 288 Notes 15

5 February 1, 2017

We start with Kolmogorov’s theorem.

5.1 Construction of a Brownian motion

Proof of Theorem 4.1. We have proved that

Kα(ω) = supn≥1

supi

|Xi/2n −X(i−1)/2n |2−nα

<∞

almost surely using Markov’s inequality and the Borel–Cantelli lemma.We look at the set

D = i

2n: 0 ≤ i ≤ 2n, n = 1, . . .

.

Suppose we have a functionf : D → R such that |f(i/2n) − f((i − 1)/2n)| ≤K2−nα. This condition implies

|f(s)− f(t)| ≤ 2K

1− 2−α|t− s|α

for t, s ∈ D. You can prove this using successive approximations.Now define Xt(ω) = lims→t,s∈DXs(ω) if Kα(ω) < ∞ and otherwise just

assign an arbitrary value. You can check that

|Xt(ω)− Xs(ω)| ≤ Cω|t− s|α

for some constant Cω depending only on ω, i.e., Xt is Holder continuous of orderα.

Finally we see that

P (|Xt − Xt| > ε) = lims→t,s∈D

P (|Xt −Xs| > ε) = 0

and so P (Xt = Xt) = 1.

This gives a construction of a Brownian motion, which is the modificationof the pre-Brownian motion. We first constructed a Gaussian white noise inan abstract space, then constructed a pre-Brownian motion by Xt = G([0, t]).Then by Kolmogorov’s theorem defined a Brownian motion. Then a samplepath is Holder continuous up to order 1/2.

There is another construction of a Brownian motion, due to Levy. Thisconstruction is Exercise 1.18.

Math 288 Notes 16

5.2 Wiener measure

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

Ω→ C(R+,R), w 7→ (t 7→ Bt(ω)).

The Wiener measure is the push-forward of this measure. Don’t think toohard about what this looks like.

Theorem 5.1 (Blumenthal’s zero-one law). Let F = σ(Bs, s ≤ t) and letF0+ =

⋂t>0 Ft. Then F0+ is trivial, i.e., for any A ∈ F0+ , either P (A) = 0 or

P (A) = 1.

Example 5.2. Let A =⋂nsup0≤s≤n−1 Bs > 0. Then P (A) = 0 or P (A) = 1.

The right answer is P (A) = 1. This is because

P (A) = limn→∞

P(

sup0≤s≤n−1

Bs > 0)≥ limn→∞

P (Bn−1 > 0) =1

2.

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

E[1Ag(Bt1 , . . . , Btk)] = E[1A]E[g(Bt1 , . . . , Btk)]

then F0+ ⊥ σ(Bt, t > 0). Taking the limit, we see that F0+ is independent ofitself. Then we get the result.

Math 288 Notes 17

6 February 3, 2017

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

6.1 Blumenthal’s zero-one law

Let us defineF+

0 =⋂t>0

Ft =⋂t>0

σ(Bs, s ≤ t).

Theorem 6.1 (Blumenthal). The σ-algebra F+0 is independent of itself. In

other words, P (A) = 0 or 1 for every A ∈ F+0 .

Proof. Take an arbitrary bounded continuous function g. For an A ∈ F+0 , we

claim that

E[1Ag(Bt1 , . . . , Btk)] = E[1A]E[g(Bt1 , . . . , Btk)],

for 0 < t1 < · · · < tk. Let us only work with the case k = 1 for simplicity. Notethat Bt1 −Bε is independent of Fε. So we have

E[1Ag(Bt1)] = limε→0

E[1Ag(Bt1 −Bε)] = limε→0

P (A)E[g(Bt1)] = E[1A]E[g(Bt1)].

This implies that F+0 is orthogonal to σ(Bt, t > 0). So F+

0 ⊥ Ft, butF+

0 ⊆ Ft. So F+0 ⊥ F

+0 .

Proposition 6.2. Define Ta = inft ≥ 0, Bt = a. Then almost surely for alla ∈ R, Ta <∞. That is, lim supt→∞Bt =∞ and lim inft∈∞Bt =∞.

Proof. We have

1 = P(

sup0≤s≤1

Bs > 0)

= limδ0

P(

sup0≤s≤1

Bs > δ).

But by the scaling property of the Brownian motion, Bt ∼ λ−1Bλ2t, we canwrite

P(

sup0≤s≤1

Bs > δ)

= P(

sup0≤s≤δ−2

Bs > 1).

It follows that P (sup0≤sBs > 1) = 1.

Math 288 Notes 18

6.2 Strong Markov property

Definition 6.3. A random variable T ∈ [0,∞] is a stopping time if the setT ≤ t is in Ft. Intuitively, a stopping time cannot depend on the future.

Example 6.4. The random variable Ta = inft : Bt = a is a stopping time.

Definition 6.5. We define the σ-field

FT = A : A ∩ T ≤ t ∈ Ft for all t ≥ 0

generated by T . Roughly speaking, this is all the information collected up tothe stopping time.

Theorem 6.6. Define B(T )t = 1T<∞ · (BT+t−BT ). Then under the measure

P (· : T <∞), the process B(T )t is a Brownian motion independent of FT .

We will prove this next time.

Math 288 Notes 19

7 February 6, 2017

Theorem 7.1. Suppose P (T < ∞) > 0. Define B(T )t = 1T<∞(BT+t − BT ).

Then respect to P (· : T <∞), the process B(T )t is a Brownian motion indepen-

dent of FT .

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

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

E[1AF (B(T )t1 , . . . , B

(T )tp )] = E[1A]E[F (B

(T )t1 , . . . , B

(T )tp ].

Define [t]n = mink/n2 : k/n2 > t. We have F (B(T )t ) = limn→∞ F (B

([T ]n)t ).

Then

E[1AF (B(T )t )] = lim

n→∞E[1AF (B

([Tn])t )]

= limn→∞

∞∑k=0

E[1A1(k−1)2−n<T≤k2−nF (B(k/2n)t)]

= limn→∞

∞∑k=0

E[1A1(k−1)2−n<T≤k2−n] · E[F (B(k/2n)t )]

= · · · = E[1A] · E[F (B(T )t )].

Here we are using the fact that 1A1(k−1)2−n<T≤k2−n is in Fk2−n and so is

independent to B(k2−n)t = Bk2−n+t −Bk2−n .

7.1 Reflection principle

Let St = sups≤tBs.

Theorem 7.2 (Reflection principle). We have P (St ≥ a,Bt ≤ b) = P (Bt ≥2a− b).

The idea is that you can reflect with respect to the first tie you meet a.

Proof. Let Ta = first time St ≥ a. Then B(T )t is independent of FTa by the

strong Markov property. This is why reflection works.

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

2(2a− b)√2πt3

e−(2a−b)2/2t1(a>0,b<a).

The probability density of Ta is

f(t) =a√2πt3

e−a2/2t.

Math 288 Notes 20

Proof. Exercise. We have

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

= 2P (Bt ≥ a) = P (tB21 ≥ a2).

Then compute the density.

Proposition 7.4. Let 0 < tn0 < · · · < tnpn = t. Then

pn∑i=1

(Btni −Btni−1)2 → t

in L2 as n→∞ and maxi|tni − tni−1| → 0.

This is saying that∫ t

0(dBt)

2 → t.

Proof. We have

E[ pn∑i=1

(Btni −Btni−1)2 − t

]2

= E[ pn∑i=1

((Btni −Btni−1

)2 − (tni − tni−1))]2

=

pn∑i=1

E[(Btni −Btni−1

)2 − (tni − tni−1)]2 ≤ C pn∑

i=1

(tni − tni−1)2 → 0,

since the (Btni −Btni−1)2 − (tni − tni−1) are independent with expectation 0.

7.2 Filtrations

Now we are going to talk about martingales.

Definition 7.5. A filtration is an increasing sequence Ft. Define

Ft+ =⋂s>t

Fs ⊇ Ft.

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

If you work with Ft+ , then you are going to be fine.

Math 288 Notes 21

8 February 8, 2017

The first homework is 1.18, 2.29, 2.31, 2.32.Last time we talked about the strong Markov property for Brownian motion.

This implies the reflection principle, and so we were able to effectively computethe distributions of max0≤s≤tBs.

8.1 Stopping time of filtrations

A filtration consists of an increasing sequence (Ft)∞t=0, and we define F+t =⋂

s>t Fs = Gt.

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

Proposition 8.2. If the sample path is right continuous almost surely, then Xt

is adapted if and only if Xt is progressively measurable.

Proof. It is obvious that progressively measurable implies adapted. Assumethat it is adapted. Fix a time t. Define

Xns = Xkt/n for s ∈

[ (k − 1)t

n,kt

n

).

Then Xns is now discrete, and limn→∞Xn

s = Xs almost surely. Clearly (ω, s)→Xns (ω) is measurable with respect to Ft ×B([0, t]).

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

Proposition 8.3. T is a stopping time with respect Gt = F+t if and only if

T < t ∈ Ft for all t > 0.

Proof. We have T ≤ t =⋂s<tT < s. Likewise T < t =

⋃s<tT ≤

s.

We define Gt = A : A ∩ T ≤ t ∈ FT = FT+ .Here are some facts:

• If S and T are stopping times, then S ∨ T and S ∧ T are also stoppingtimes.

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

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

Math 288 Notes 22

8.2 Martingales

Definition 8.4. A process Xt is a martingale if E[|Xt|] <∞ and E[Xt|Fs] =Xs for every t ≥ s. If E[Xt|Fs] ≥ Xs, then it is called a submartingale.

Example 8.5. Bt is a martingale. ButB2t is not a martingale because E[B2

t |Fs] =(t− s) +B2

s . But B2t − t is a martingale.

Example 8.6. The process eθB2t−θ

2/2t is a martingale for all θ. This is because

E[eθB2t−θ

2t/2|Fs] = E[eθ(Bt−Bs)2+2θ(Bt−Bs)Bs |Fs]eθ

2B2s−θ

2t/2.

Example 8.7. For any f(s) ∈ L2[0,∞), the process∫ t

0

f(s)dBs = limn→∞

∑i

f(si−1)(Bsi −Bsi−1)

is a martingale. Note that this is well-defined for simple functions f and youcan extend for general f ∈ L2 because

E(∫ t

0

f(s)dBs

)2

=

∫ t

0

f(s)2ds

if f is simple. Furthermore,(∫ t

0

f(s)dBs

)2

−∫ t

0

f(s)2ds

is also a martingale.

Math 288 Notes 23

9 February 10, 2017

Chapter VII of Shirayayev, Probability is a good reference for discrete martin-gales.

9.1 Discrete martingales - Doob’s inequality

We are now going to look at martingales with discrete time.

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

Proof. To prove E[Xτ |Fσ] ≥ Xσ, it suffices to prove E[XτA] ≥ E[XσA] for allA ∈ Fσ. In this case, A will look like

A =

∞⋃k=1

Ak, where Ak = A ∩ 1σ=k ∈ Fk.

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

E[XσAm] = limn→∞

E[XσnAm],

because the sequence on the right side is eventually constant. Also

E[XτAm] = E[Xτ1τ>nAm] + E[Xτ1τ≤nAm] = E[Xτ1τ>nAm] + E[Xτn1τ≤nAm]

= E[Xτ1τ>nAm]− E[Xn1τ>nAm] + E[XτnAm].

Here as n → ∞, the first two terms goes to 0 because of the condition. It isalso clear that E[XσnAm] ≤ E[XτnAm], because we can only care about finitelymany outcomes of σ and τ . So

E[XσAm] = limn→∞

E[XσnAm] ≤ limn→∞

E[XτnAm] = E[XτAm].

Theorem 9.2 (Doob inequality). Let Xj be a nonnegative submartingale, andlet Mn = max0≤j≤nXj.(1) For a > 0, aP (Mn > a) ≤ E[1Mn≥aXn].(2) for p > 1,

‖Mn‖pp ≤( p

p− 1

)pE[|Xn|p].

Proof. (1) Let τ = minj ≤ n : Xj ≥ a. This is a stopping time, and τ = n ifMn < a. Then

E[Xn] ≥ E[Xτ ] = E[1Mn≥aXτ ] + E[1Mn<aXτ ]

≥ aP (Mn ≥ a) + E[1Mn<aXn].

(2) It follows from the following lemma.

Math 288 Notes 24

Lemma 9.3. If λP (Y ≥ λ) ≤ E[X1Y≥λ] for X,Y ≥ 0, then

E[Y p] ≤( p

p− 1

)pE[Xp].

Proof. We have

E[Y p] =

∫ ∞0

pλp−1P (Y ≥ λ)dλ ≤ E∫ ∞

0

Xpλp−21Y≥λdλ

= E[X

p

p− 1Y p−1

]≤ p

p− 1‖X‖p‖Y ‖p−1

p .

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

E[β(a, b)] ≤ E[max(Xn − a, 0)]

b− a.

Note that if X is a submartingale, then max(X − a, 0) is a submartingale.This is because

E[f(Xn)|Fn−1] ≥ f(E[Xn|Fn−1]) ≥ f(Xn−1).

Math 288 Notes 25

10 February 13, 2017

10.1 More discrete martingales - upcrossing inequality

In the optional stopping theorem, we needed the stopping time τ to satisfy theinequality E[Xn1τ>n]→ 0 as n→∞.

Proposition 10.1. If Xn is uniformly integrable, i.e.,

limM→∞

supn

∫|Xn|≥M

|Xn| = 0,

then E[Xn1τ>n]→ 0.

Proof. We have

limn→∞

∫|Xn|1τ>n ≤ lim

n→∞M1τ>n + lim

n→∞

∫|Xn|1|Xn|≥M1τ>n

≤ limn→∞

|Xn|1|Xn|≥M = 0.

Corollary 10.2. Let Xn be a submartingale. Suppose (1) E[|X0|] < ∞, (2)E[τ ] < ∞, and (3) supn E[|Xn+1 − Xn||Fn] ≤ C. Then E[Xτ ] ≥ E[X0] andE[Xτ |F0] ≥ X0.

Proof. We want to check the three conditions in the optional stopping inequality.We have

E[|Xτ −Xm|] ≤∞∑n=m

E[|Xn+1 −Xn|1τ≥n+1] =

∞∑n=m

E[1τ≥n+1E[|Xn+1 −Xn|Fn|]

≤ C∞∑n=m

E[1τ≥n+1] ≤ CE[τ ] <∞.

This shows E[|Xτ |] <∞. We also have

limn→∞

E[|Xn|1τ≥n+1] ≤ limn→∞

E[|Xτ −Xn|] + limn→∞

E[|Xτ |1τ≥n+1] = 0.

Theorem 10.3 (Upcrossing inequality). For a < b, let β(a, b) be the numberof upcrossings in (Xj)

nj=1. Then

E[β(a, b)] ≤ E[max(Xn − a, 0)]

b− a.

Proof. Replace Xj by max(Xj − a, 0). This is still a submartingale, and so wecan assume Xi ≥ 0 and a = 0.

Define a random variable φi such that φi = 1 if there exists a k ≤ i suchthat Xk = 0 and Xj < b for all k ≤ j ≤ i. Then we get

E[β(a, b)] ≤ b−1En−1∑i=1

φi(Xi+1 −Xi) = b−1∑i

E[φiE[Xi+1 −Xi|Fi]]

≤ b−1∑i

E[Xi+1 −Xi] = b−1E[Xn],

Math 288 Notes 26

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

Theorem 10.4 (Doob’s convergence theorem). If Xn is a submartingale withsupi E[|Xi|] <∞, then Xn converges almost surely to some X.

Proof. We have

P (lim supXn > lim inf Xn) = P⋃a<ba,b∈Q

(lim supXn > b > a > lim inf Xn) = 0

by the upcrossing inequality.

Note that Xn → X in L1 may fail. For instance, take Yi be independentBernoulli with 0, 2 and Xn = Y1 · · ·Yn → 0. Then EXn = 1.

Proposition 10.5. (1) Suppose Xn → X almost surely and Xn is uniformlyintegrable. Then Xn → X in L1.(2) Suppose Xn ≥ 0, Xn → X almost surely, and EXn = EX. Then Xn → Xin L1.

Theorem 10.6 (Levi). Suppose X ∈ L1 be a random variable and let Fn be afiltration with F∞ =

⋃Fn. Then E[X|Fn] → E[X|F∞] almost surely and L1.

Conversely, if Xn is uniformly integrable and F∞ is the full σ-algebra F , thenthere exists a X ∈ L1 such that Xn = E[X|Fn].

Math 288 Notes 27


So far we have looked that Doob’s inequality, the optional stopping time, andDoob’s upcrossing inequality. There is a supermartingale version of the upcross-ing inequality:

Theorem 11.1. For a discrete supermartingale X, the number of upcrossingsMnab has expectation value

E[Mnab(x)] ≤ E

[max0, a− xn)b− a

].

Proof. Let

τ1 = n ∧minj > 0 : xj ≤ a, τ2 = n ∧minj > τ1 : xj ≥ b,

and so forth. Then all τk are stopping times. Let

D = (Xτ2 −Xτ−1) + (Xτ4 −Xτ3) + · · · .

Then (b − a)Mnab + R ≤ D, where R = 0 if the tail is a down-crossing and

R = Xn−a otherwise. BecauseX is a supermartinagle, we have E[Xτ2−Xτ1 ] ≤ 0and so

(b− a)EMnab ≤ −ER ≤ E[max(0, a−Xn)].

11.1 Levi’s theorem

Theorem 11.2 (Levi). Suppose X ∈ L1 and Fn be a filtration, with F∞ =⋃Fn. Then E[X|Fn] → E[X|F∞] in L1 and almost surely. Conversely, if a

martingale Xn is uniformly integrable, then there is an X such that E[X|Fn] =Xn.

Proof. Let Xn = E[X|Fn]. We claim that Xn is uniformly integrable, and youcan check this. Also Xn is a martingale. So the convergence theorem tellsus that Xn → X∞ almost surely and in L1. Now the question is whetherX∞ = E[X|F∞]. Because Xn is a martingale, for any A ∈ Fn and m ≥ n,∫

A

Xndµ =

∫A

Xmdµ.

Taking m→∞, we get∫A

Xdµ =

∫A

Xndµ =

∫A

X∞dµ

because Xm → X∞ in L1. So this is true for all A ∈ F∞. So X∞ = E[X|F∞].Now let us do the second half. Suppose Xn is uniformly integrable. Then

Xn → X for some X almost surely and in L1. So basically by the same argu-ment, ∫

A

Xndµ = limm→∞

∫A

Xmdµ =

∫A

Xdµ.

This shows that Xn = E[X|Fn].

Math 288 Notes 28

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

11.2 Optional stopping for continuous martingales

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

Proof. Let τn = (bnτc + 1)/n. Then τn is a stopping time and Fτn ↓ Fτ asn → ∞, because F is right continuous. By the discrete version of optionalstopping time, we see that Xτn → X implies that E[XC |Fτn ] ≥ Xτn . ThenXτn is uniformly integrable and Xτn → Xτ almost surely and in L1. SimilarlyXσn → Xσ. So

E[Xτ+ε|Fσ] = limm→∞

E[Xτ+ε|Fσm ]

= limm→∞

limn→∞

E[Xτn+ε|Fσm ] ≥ limm→∞

Xσm = Xσ.

Math 288 Notes 29


We proved the upcrossing inequality and Doob’s inequality. We also provedoptional stopping theorem. There is also the martingale convergence theoremthat says that if (Xj) is a submartingale or a supermartingale and supj E[|Xj |] ≤C, then limJ→∞Xj = X∞ for some X∞ almost surely. If (Xj) is uniformlyintegrable, then Xj → X∞ in L1.

Theorem 12.1 (3.18). Suppose Ft is right continuous and complete. Let Xt bea super-martingale and assume that t 7→ E[t] is right continuous. Then X hasa modification such that the sample paths are right continuous with left limits,which is also a supermartingale.

Lemma 12.2 (3.16). Let D be a dense set and f be defined on D such that

(i) f is bounded on D ∩ [0, T ],

(ii) Mfab(D ∩ [0, T ]) <∞.

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

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

(i) Xt is closed, i.e., there exists a Z ∈ L1 such that E[Z|Ft] = Xt.

(ii) Xt is uniformly integrable.

(iii) Xt →W almost surely and L1 as t→∞ for some W .

There is an application. Let Ta = inft : Bt = a for a > 0 and λ > 0.

Proposition 12.4. E[e−(λ2/2)Ta ] = eλa.

Math 288 Notes 30


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

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

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

Theorem 13.3 (Levi). Let (Xn) be a uniformly integrable martingale. Thenthere exists a X ∈ L1 such that E[X|Fn] = Xn.

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

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

13.1 Submartingale decomposition

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

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

Ak = E[Xk|Fk−1]−Mk−1, Mk = Xk −Ak.

You can check that Ak increases. For instance,

A2 = E[X2|F1]−M1 = E[X2|F1]− [X1 −A1] = E[X2|F1]−X1 +A1.

Math 288 Notes 31


There is another version of optional stopping time

Theorem 14.1 (Le Gall 3.25). Suppose X is a nonnegative supermartingalewith right continuous sample paths. Let S ≤ T be two stopping times. ThenE[XT |FS ] ≥ XS.

Proof. Consider only the discrete case, since we can approximate the continuouscase with the discrete case. A nonnegative supermartingale is in L1 in the sensethat E[Xj ] ≤ C < ∞. The upcrossing inequality tells us that Xj → X∞ = Xalmost surely (but maybe not in L1).

For any M > 0, we know that S ∧M ≤ T ∧M are stopping times. You cancheck that XS∧M → XS almost surely. Then by Fatou’s lemma,

limM→∞

E[XS∧M ] ≥ E[XS ].

So XS ∈ L1.For any A ∈ FX , we want to prove E[1AXS ] ≥ E[1AXT ]. This would

immediately imply E[XT |FS ] ≥ XS . Define the stopping time

SA(ω) =

S(ω) if ω ∈ A∞ if ω /∈ A.

and similarly TA(ω). Then

E[XSA∧M ] ≥ E[XTA∧M ]

because both are finite. Now

E[1A∩S≤MXS∧M ] + E[1A∩S>MXM ] + E[1AcXM ] = E[XSA∧M ]

≥ E[XTA∧M ] = E[1A∩S≤MXT∧M ] + E[1A∩S>MXM ] + E[1AcXM ].

Hence we get E[1A∩S≤MXS ] ≥ E[1A∩S≤MXT∧M ].Taking the limit M →∞, we get

E[1A∩S<∞XS ] = limM→∞

E[1A∩S≤MXS ] ≥ limM→∞

E[1A∩S≤MXT∧M ]

≥ E[1A∩S<∞XT ]

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

14.1 Backward martingales

In this case, we have a filtration

· · · ⊆ F−3 ⊆ F−2 ⊆ F−1 ⊆ F0.

Math 288 Notes 32

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

E[Xn|Fm] ≤ Xm

for any m ≤ n ≤ 0.

Theorem 14.3. if supn<0 E[|Xn|] ≤ C then there exists an X−∞ ∈ L1 suchthat Xn → X−∞ almost surely and in L1.

Note that we also have L1 convergence, which we did not have in the forwardcase.

Proof. The almost surely convergence is almost identical to the forward case,as a consequence of the upcrossing inequality.

To show convergence in L1, define

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

Note that this is negative. Then Mn = Xn + An is a martingale. Then An →A−∞ almost surely and in L1, which comes from the monotone convergencetheorem.

Now Mn being a martingale simply means

M−n = E[M−1|F−n].

This is just the setting of Levi’s theorem and so immediately we have L1 con-vergence of M−n.

14.2 Finite variance processes

We assume that sample paths are continuous.

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

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

Definition 14.6. A sequence Xt with X0 = 0 is called a local martingaleif there exist increasing stopping times Tn ↑ ∞ such that Xt∧Tn is a uniformintegrable martingale for all n. If X0 6= 0, then Xt is a local martingale ifXt −X0 is a local martingale.

Math 288 Notes 33


For a finite variance process As(ω), we can write As(ω) = µ([0, s]) for somefinite measure µ. So we can define∫

Hs(ω)dAs(ω) =

∫Hs(ω)µ(ds)(ω).

Here we require ∫|Hs(ω)||dAs(ω)| <∞.

This is Section 4.1 and you can read about it.

15.1 Local martingales

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

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

Proposition 15.2. (i) If A is a nonnegative local martingale with M0 ∈ L1,then it is a supermartingale.(ii) If there exists a Z ∈ L1 such that |Mt| ≤ Z for all t, then the local martingaleMt is a uniformly integrable martingale.(iii) If M0 = 0, then Tn = inft : Mt ≥ n reduces the martingale, i.e., satisfiesthe condition in the definition.

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

limn→∞

E[M0 +Nt∧Tn |Fs] = limn→∞

M0 +Ns∧Tn = Ms.

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

Proposition 15.3. A finite variance local martingale with M0 = 0 is identicalto 0. (Again we are assuming continuous sample paths.)

Proof. Define a stopping time

τn = inft :∫ t

0|dMs| ≥ n,

Math 288 Notes 34

and let Nt = Mτn∧t. Then we have

|Nt| ≤∣∣∣∣∫ t

0

dMs

∣∣∣∣ ≤ n.For any t and 0 = t0 < · · · < tp = t,

E[N2t ] = E

(∑i

δNi

)2

= E∑i

(δNi)2 = E

[supi|δNi|

∑j

|δNj |]≤ nE

[supi|δNi|

].

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

Theorem 15.4. Suppose that Mt is a local martingale. Then there exists aunique increasing process 〈M,M〉t such that M2

t −〈M,M〉t is a local martingale.This 〈M,M〉t is called the quadratic variation of M .

Example 15.5. If Mt = Bt then 〈M,M〉t = t. Then B2t − t is a martingale.

Proof. Let us first prove uniqueness. Suppose M2t − At and M2

t − A′t are localmartingales. Then At −A′t is also a local martingale. By definition, At and A′tare increasing and less than ∞ almost surely. So At − A′t is of finite variance.This shows that At −A′t = 0 for all t.

Proving existence is rather difficult. Suppose M0 = 0 and Mt is bounded,i.e., supt|Mt| < C almost surely. Fix and interval [0, k] and subdivide it to0 = t0 < · · · < tn = K. Define

Xnt =

∑i

Mti−1(Mti∧t −Mti−1∧t).

You can check that this is a local martingale, and you can also check

M2tj − 2Xn

tj =

j∑i=1

(Mti −Mti−1)2.

Now define

〈M,M〉t = limn→∞

∑i

(Mti −Mti−1)2.

We are going to continue next time.

Math 288 Notes 35

16 March 1, 2017

16.1 Quadratic variation

As always, we assume continuous sample paths. This is the theorem we aretrying to prove:

Theorem 16.1. Let M be a local martingale. Then there exists an increasingprocess 〈M,M〉t such that M2

t − 〈M,M〉t is a local martingale.

We proved uniqueness using the fact that if there are two such A,A′ thenA−A′ is a finite variation process that is also a local martingale.

For existence, fix K and subdivide [0,K] into 0 = t0 < · · · < tn = K. Wedefine

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

Then we immediately have

M2tj − 2Xn

tj =

j∑i=1

(Mti −Mti−1)2.

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

limm,n→∞,mesh→0

E[(Xnk −Xm

k )2] = 0.

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

E[(Xnk −Xm

k )2] = E[ 5∑

j=1

Mj−1(Mj −mj−1)−M3(M5 −M3)−M1(M3 −M1)

2]= E[((M4 −M3)(M5 −M4) + (M2 −M1)(M3 −M2))2]

= E[(M4 −M3)2(M5 −M4)2 + (M2 −M1)2(M3 −M2)2]

≤ E[maxj|Mj −Mj−1|2

∑j

(Mj −Mj−1)2]

≤ E[max(Mj −Mj−1)4]1/2E[(∑

j

(Mj −Mj−1)2)2]1/2

.

Math 288 Notes 36

Here, we see that

E[(∑

j

(Mj −Mj−1)2

)2]≤∑j

E[(Mj −Mj−1)4]

+ 2E(M1 −M0)25∑j=2

(Mj −Mj−1)2 + · · ·

≤ 4C2∑j

E[(Mj −Mj−1)2]

+ 2∑j

E[(Mj −Mj−1)2E

[ n∑k=j+1

(Mk −Mk−1)2

∣∣∣∣Fj]]= 4C2

∑j

E[(Mj −Mj−1)2] + 2E[(Mj −Mj−1)2E[(MK −Mj)2|Fj ]]

≤ 12C2E[∑

j

(Mj −Mj−1)2

]= 12C2E[(MK −M0)2] ≤ 48C4.

So taking n,m→∞ we get E[max(Mj −Mj−1)4]→ 0.

Now let us go back to the original theorem. We check that M2t − 2Xn

t is anincreasing process. Take n→∞. By the lemma, Xn

t∧K → Yt∧K in L2. You canalso check that Yt∧K has continuous sample paths.

Define

〈M,M〉t = M2t∧K − 2Yt∧K .

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

n∑i=1

(Mtni−Mtni−1

)2 → 〈M,M〉t

in L2. Also, Xt∧K is a local martingale.This construction works for all fixed K. For every K, we get a AKt such

that M2t∧K −AKt∧K is a martingale. Then by uniqueness, AK+1

t = AKt if t ≤ Kalmost surely. This means that we can just take the union of all the AKt .

Math 288 Notes 37

17 March 3, 2017

Last class we almost finished proved the following theorem.

Theorem 17.1. Let M be a local martingale. Then there exists a unique in-creasing process 〈M,M〉t such that M2

t − 〈M,M〉t is a local martingale.

In the proof, we haven’t check that the L2 limit Xnt → Xt has continuous

sample paths. To show this, it suffices to check that

E[sups≤t|Xm

s −Xns |2]→ 0

as n,m→∞. By Doob’s inequality,

E[sups≤t|Xm

s −Xns |2]≤ 4E[|Xm

t −Xnt |2].

Then this is clear from the L2 convergence.For the general case, define Tn = inft : |Mt| ≥. Then MTn is a bounded

martingale and so there exists an increasing process Ant such that (MTn)2−Antis a martingale. By uniqueness, An+1

t∧Tn = Ant . Now we can take the limitAt = limn→∞Ant . Then we have

(MTn∧t)2 −ATn∧t = (MTn)2

t −Ant∧Tn

is a martingale. This means that M2t −At is a local martingale.

17.1 Consequences of the quadratic variation

Corollary 17.2. Let M be a local martingale with M0 = 0. Then M = 0 ifand only if 〈M,M〉t = 0.

Proof. Suppose 〈M,M〉t = 0. Then M2 is a nonnegative local martingale. ThenM2t is a supermartingale. Since M0 = 0, we immediately get M = 0.

Corollary 17.3. Assume M is a local martingale and M0 ∈ L2. Then E〈M,M〉∞ <∞ if and only if M is a true martingale and EM2

t < C for all t.

Proof. Suppose M is a martingale with M0 = 0 and EM2t < C. Then

E supt≤T

M2t ≤ 4EM2

T ≤ 4C.

Let Sn = inft : 〈M,M〉t ≥ n. Then

MSn∧t − 〈M,M〉Sn∧t ≤ supt≥0

M2t .

But suptM2t is in L1. Then

limn→∞

E[M2Sn∧t − 〈M,M〉Sn∧t] = E[M2

0 − 〈M,M〉0] = 0.

Math 288 Notes 38

By monotone convergence, we have E〈M,M〉t = EM2t < C.

Conversely, suppose E〈M,M〉∞ <∞. Let Tn = inft : |Mt| ≥ n. We have

M2Tn∧t − 〈M,M〉Tn∧t ≤ n2 ∈ L1.

Then M2Tn∧t − 〈M,M〉Tn∧t is a uniformly integrable martingale and so

limn→∞

E[M2Tn∧t − 〈M,M〉Tn∧t] = 0

and so E[M2t ] ≤ E[〈M,M〉t] < C. This also implies that M is a true martingale.

Math 288 Notes 39

18 March 6, 2017

Let us try to finish this chapter.

Theorem 18.1. If supt E[M2t ] <∞, then E〈M,M〉∞ <∞ and M2

t − 〈M,M〉tis a uniformly integrable martingale. If M is a local martingale and E〈M,M〉∞ <∞ then supt E[M2

t ] <∞.

18.1 Bracket of local martingales

Definition 18.2. IfM andN are two local martingales, we define their bracketas

〈M,N〉t =1

2[〈M +N,M +N〉 − 〈M,M〉 − 〈N,N〉].

Proposition 18.3 (Le Gall 4.15). In probability,

lim∑i

(Mti −Mti−1)(Nti −Nti−1

) = 〈M,N〉t.

Furthermore, if M and N are two L2 martingales, then E〈M,N〉∞ = E[M∞N∞].

Definition 18.4. We say that M is orthogonal to N if 〈M,N〉 = 0.

Theorem 18.5 (Kunita–Watanabe). If M and N are two local martingales,then∫ ∞

0

|Hs(ω)Ks(ω)||d〈M,N〉s| ≤(∫ ∞

0

|Hs|d〈M,M〉s)1/2(∫ ∞

0

|Ks|2d〈N,N〉s)1/2

.

Proof. You can just subdivide the intervals finely and use the formula for〈M,N〉t.

Definition 18.6. A process Xt is called a semimartingale if Xt = Mt + Atwhere Mt is a local martingale and At is a finite variance process.

Note that such a decomposition is unique since a finite variance local mar-tingale is zero.

Proposition 18.7. For two semimartingales X = M +A and Y = N +B, wedefine 〈X,Y 〉 = 〈M,N〉. Then∑

i

(Xti −Xti−1)(Yti − Yti−1

)→ 〈X,Y 〉 = 〈M,N〉.

Proof. We only have to check that∑i

(Nti −Nti−1)(Ati −Ati−1)→ 0

We use the Schwartz inequality. We have∑i

(Ati −Ati−1)2 ≤ sup

i|Ati −Ati−1

|∑i

|Ati −Ati−1|.

The first term goes to zero and the second term is bounded.

Math 288 Notes 40

18.2 Stochastic integration

We define the space

H2 =

M a continuous martingale withM0 = 0 and supt EM2

t <∞.

We then have 〈M,M〉∞ = EM2

∞ and [M∞|Ft] = Mt. We define the innerproduct as

(M,N) = E[〈M,N〉∞] = EM∞N∞.

Also define the space of elementary processes as

E =

Hs(ω) =

∑i

Hi(ω)1(ti,ti+1](s), Hi ∈ Fti.

We can then define the integral as∫Hs(ω)dMs =

∑i

Hi(ω)(Mti −Mti−1).

This is just like Riemann integration.

Theorem 18.8. H2 is a Hilbert space.

Proof. Suppose there is a Cauchy sequence E[(Mn∞−Mm

∞)2]→ 0. Then Mn∞ →

Z in L2. Our goal is to find a Mt with continuous sample paths such thatMnt →Mt in some sense.

Math 288 Notes 41

19 March 8, 2017

We define H2 to be the space of martingales with supt E[M2t ] < ∞. The inner

product is defined as (M,N) = E[M∞N∞] = E〈M,N〉∞.

Proposition 19.1. H2 is a Hilbert space.

Proof. Suppose limm,n→∞ E[(Mm∞,M

n∞)] = 0 so that Mn is a Cauchy sequence.

Then there exists a Z such that Mn∞ → Z in L2(P ). We also have, for an fixed

t, E[(Mmt −Mn

t )2] → 0 and so M∞t . What we now need to show is that M∞

has continuous sample paths.We have

E[supt

(Mnt −Mm

t )2] ≤ 4E[(Mn∞ −Mm

∞)2]→ 0.

This shows that you can find a subsequence such that

E[ ∞∑k=1

supt|Mnk+1

t −Mnkt |]≤ 4

∞∑k=1

E[(Mnk+1∞ −Mnk

∞ )2]1/2 <∞.

In particular,

Yt =∑k

(Mnk+1

t −Mnkt )

is a martingale with continuous sample paths. Also Y∞ = Z and so Mn → Yin H2.

19.1 Elementary process

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

E =

p−1∑i=0

Hi(ω)1(ti,ti+1] : Hi ∈ Fti , with

∑i

H2i (〈M,M〉ti+1

− 〈M,M〉ti) <∞.

We also define

L2(M) =Hs such that

∫H2sd〈M,M〉s <∞

.

Proposition 19.3. If M ∈ H2, then the space E is dense in L2(M).

Theorem 19.4. For any M ∈ H2, we can define

(H ·M)t =

p−1∑i=0

Hi(ω)(Mti+1∧t −Mti∧t) =

∫ t

0

HdM ∈ H2.

Math 288 Notes 42

This map E → H2 given by H 7→ H ·M is an isometry from E to H:

‖H ·M‖H2 =

∫H2sd〈M,M〉s.

So there exists a unique extension from L2(M)→ H, denoted by∫HdM . Fur-

thermore,

(1) 〈H ·M,N〉 = H〈M,N〉,(2) (1[0,T ]H) ·M = (H ·M)T = H ·MT .

Roughly speaking, stochastic integration is just Riemann integration on el-ementary functions, which has a unique extension.

You can check that

〈H ·M,H ·M〉t =∑i

H2i (〈M,M〉ti+1∧t − 〈M,M〉ti∧t).

Proof. Since (H ·M)2 −H2〈M,M〉 is a martingale, we have

E[(H ·M)2∞] = E

∫ ∞0

H2sd〈M,M〉s.

This shows that stochastic integration is and L2 isometry. Then there exists aunique extension.

Now we want to check that 〈H ·M,N〉 = H〈M,N〉. Note that 〈M,N〉 is afinite variance process 〈M,M〉 and 〈N,N〉 are increasing and 〈M +N,M +N〉is decreasing, and also E|〈M,N〉| <∞. We want to check that⟨∫ t

0

HdM , Nt

⟩=

∫ t

0

Hd〈M,N〉.

For elementary processes, we have⟨∫ t

0

HsdMs,

∫ t

0

dNs

⟩=

∫ t

0

Hsd〈M,N〉s.

Then it extends to all processes.

Math 288 Notes 43

20 March 20, 2017

20.1 Review

Today we will review, because all the properties are hard to remember. Allprocesses have continuous sample paths and are adapted.

1. Finite variation processes: At is a finite variation process if its samplepaths have finite variation almost surely. Classical theory of integrationapplies here. Then ∫ t

0

Hs(ω)dAs(ω)

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

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

Why do we look at these objects ? If Xn is a sub/supermartingale, andsupn E|Xn|p <∞ for some p ≥ 1, then Xn → X∞ almost surely. Moreoverif p > 1 then Xn → X∞ in Lp (by Doob’s inequality). Furthermore, if Xn

is uniformly integrable then Xn → X∞ also in L1. We really need uniformintegrability to run the theory, and so we have to stick in a stopping timeto make things uniformly integrability.

3. Quadratic variation: If M is a local martingale, then there exists a uniqueincreasing process 〈M,M〉t such that M2

t −〈M,M〉t is a local martingale.This is constructed as

〈M,M〉t = lim∑i

(Mti+1 −Mti)2.

For M = B a Brownian motion, we have 〈B,B〉t = t.

4. Some properties of local martingales:

• A nonnegative local martingale M with M0 ∈ L1 is a submartingale.(This follows from Fatou’s lemma.)

• If there exists a Z ∈ L1 such that |Mt| < Z for all t, then M is auniformly integrable martingale.

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

• For M a local martingale with M0 = 0, 〈M,M〉 = 0 if and only ifM = 0.

Math 288 Notes 44

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

(M,N)H2 = E[M∞N∞] = E[〈M,N〉∞].

Then we defined, for any M ∈ H2, the space L2(M) of processes H suchthat

E∫ ∞

0

H2d〈M,M〉 <∞.

The space E is the space of elementary processes, processes of the form∑p−1i=0 Hi1(ti,ti+1] for Hi measurable with respect to Fti and bounded al-

most surely. You can check that for every M ∈ H2, the space E is densein L2(M). (Note that the inner product on E depends on M .)

We can now define stochastic integration∫HdM on the space E by

(H ·M)t =∑i

Hi(Mti+1∧t −Mti∧t).

This gives an isometry E → H2. So you can extend it to an isometryL2(M)→ H2.

L2(M) H2

E

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

HdM)Tn

=

∫HdMTn .

We are also going to define, for a semi-martingale X = M + V ,∫HdX =

∫HdM +

∫HdV.

Next time we are going to quickly go over these definitions and start actuallydoing some stuff.

Math 288 Notes 45

21 March 22, 2017

For any fixed M ∈ H2, we consider the space

L2(M) =

adapted processes (Ht) with E∫ ∞

0

H2t d〈M,M〉 <∞

.

There is an isometry

(−) ·M : E → H2;∑i

Hi1(ti,ti+1] 7→∑i

Hi(Mti+1∧t −Mti∧t).

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

21.1 Stochastic integration with respect to a Brownianmotion

We now want to look at stochastic integration with respect to Brownian motions.But (Bt) is not in H2. More generally, we are going to stochastic integrationthis to local martingales. We are going to do in the setting

(1) M is a local martingale,

(2) H ∈ L2loc(M), i.e.,

∫ t0H2sd〈M,M〉s <∞ almost surely for any fixed t.

Recall that for the Brownian motion, 〈B,B〉t = t. This is easy to provebecause

E[∑i

((Btni+1−Btni )2 − (tni+1 − tni ))

]2= C

p−1∑i=0

(tni+1 − tni )2 → 0.

Then we can define, for t ≤ T ,∫ t

0

HdB =∑i

Hi(Bti+1∧t −Bti∧t).

We check that

E[∫ T

0

HdB

]2

= E(∑

i

Hi(Bti+1−Bti)

)2

=∑i

H2i (ti+1 − ti) =

∫ T

0

Hsds.

That is,∫ t

0HdB is an isometry in L2. Then you can again extend it to all

adapted processes Hs such that∫ T

0H2sds <∞ almost surely

21.2 Stochastic integration with respect to local martin-gales

Let M be a local martingale, and we define

L2loc(M) =

(Hs) :

∫ t

0

H2sd〈M,M〉s <∞ almost surely for all t

.

Math 288 Notes 46

Theorem 21.1. There exists a unique continuous local martingale H ·M =∫HsdMs such that

• 〈H ·M,N〉 = H · 〈M,N〉,• H · (K ·M) = (HK) ·M ,

• 〈∫HdM,

∫KdN〉t =

∫ t0HsKsd〈M,N〉s,

•∫Hd(

∫KdM) =

∫(HK)dM .

Proof. We look at the stopping time

Tn = inft :

∫ t

0

(1 +H2s )d〈M,M〉s ≥ n.

You can check that they actually are stopping times and Tn ↑ ∞. Consider(MTn)t. You can also check 〈MTn ,MTn〉 = 〈M,M〉Tn . Then

E(MTn)2t ≤ E(〈M,M〉Tnt ) ≤ n.

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

continue next time.

If X is a semi-martingale, with X = M + V where V is a finite variationprocess, then we can further define∫

HsdX =

∫HsdMs +

∫HsdVs.

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

F (X1t , . . . , X

pt ) = F (X1

0 , . . . , Xp0 ) +

p∑i=1

∫ t

0

∂F

∂Xi(X1

s , . . . , Xps )dXi

s

+

p∑i,j=1

∫ t

0

∂2F

∂Xi∂Xj(X1

s , . . . , Xps )d〈Xi, Xj〉s.

Math 288 Notes 47

22 March 24, 2017

Theorem 22.1. Let M be a local martingale, and consider the space

L2loc(M) =

(Hs) :

∫ t

0

H2s (ω)d〈M,M〉s <∞ almost surely for any fixed t

.

Then there exists a unique integration∫HsdMs giving a continuous local mar-

tingale such that〈H ·M,N〉 = H · 〈M,N〉.

Proof. Assume M0 = 0 as always, and consider the stopping time

Tn = inft :

∫ t

0

(1 +H2s )d〈M,M〉s ≥ n

.

These are actually stopping times and Tn ↑ ∞. We see that MTn is a L2

martingale because∫ t

0d〈M,M〉s ≤ n. Then

E[(MTn)2t ] = E[〈MTn ,MTn〉t] = E[〈M,M〉t∧Tn ] ≤ n.

So the integration H ·MTn =∫HdMTn is well-defined.

For m ≥ n, you can check that (H · MTm)Tn = H · MTn . That is, thisis a consistent family and so there exists a local martingale H ·M such that(H ·M)Tn = H ·MTn .

22.1 Dominated convergence for semi-martingales

Proposition 22.2. Let X be a continuous semi-martingale, and H be anadapted process with continuous sample paths. For every t > 0 fixed, and0 = tn0 < · · · < tnp = t,

p−1∑i=0

Htni(Xtni+1

−Xtni)→

∫ t

0

HsdXs

in probability.

Lemma 22.3 (Dominated convergence in stochastic integration). Let X =M + V be a semi-martingale. Assume that H satisfies Hn

s → Hs almost surelyfor s ∈ [0, t], and there exists a Ks ≥ 0 such that |Hn

s | ≤ Ks for all n ands ∈ [0, t], with the conditions∫ t

0

K2sd〈M,M〉s <∞,

∫ t

0

Ks|dVs| <∞

almost surely. Then ∫Hns dXs →

∫HsdXs

in probability.

Math 288 Notes 48

Proof. The case M = 0 is exactly the classical dominated convergence. Let usnow assume V = 0. Define the stopping time

Tp =γ ≤ t :

∫ γ

0

K2sd〈M,M〉s ≥ p

∧ t.

We have

E[(∫ Tp

0

Hns dMs −

∫ Tp

0

HsdMs

)2]≤ E

∫ Tp

0

(Hns −Hs)

2d〈M,M〉s

≤ E∫ Tp

0

2K2sd〈M,M〉s < C(p).

By dominated convergence, the left hand side goes to 0 as n → ∞. SinceP (Tp = t)→ 1 as p→∞, we get the desired result.

Proof of Proposition 22.2. Define the processes

Hns =

Hnti ti < s ≤ tni+1

0 s > t.

Let Ks = max0≤r≤s|Hr|. Then H has continuous sample paths and so Ks <∞almost surely. We trivially check that |Hn

s | ≤ Ks for all n and s. Also∫K2sd〈M,M〉s <∞

because it is locally bounded. Hns → Hs almost surely as n → ∞, because

sample paths are continuous. So we can apply dominated convergence.

22.2 Ito’s formula

Theorem 22.4 (Ito’s formula). Let F ∈ C2 be a function and X1, . . . , Xp besemi-martingales. Then

F (X1, . . . , Xp) = F (X10 , . . . , X

p0 ) +

p∑i=1

∫ t

0

∂F

∂Xi(X1

s , . . . , Xps )dXi

s

+1

2

p∑i,j=1

∫ t

0

∂2F

∂Xi∂Xj(X1

s , . . . , Xps )d〈Xi, Xj〉s.

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

F (Xt) = F (X0) +

p−1∑i=0

(F (Xntni+1

)− F (Xnti))

=∑i

F ′(Xti)(Xtni+1−Xtni

) +1

2

∑i

F ′′(Xnti + θ(Xtni+1

−Xtni))(Xtni+1

−Xtni)2

Math 288 Notes 49

by simple Taylor expansion, with 0 ≤ θ ≤ 1. Now the first term converges to∫F (Xs)dXs by the Proposition 22.2. To show that the second term goes to

12

∫F ′′(Xs)d〈X,X〉s, we note that

E[∑i

F ′′(Bnti)[(Xtni+1−Xtni

)2 − (〈X,X〉tni+1− 〈X,X〉tni )]

]2≤ C

∑(〈X,X〉tni+1

− 〈X,X〉tni )2 → 0

since all the cross terms vanish. So we get the formula.

Math 288 Notes 50

23 March 27, 2017

Last time we prove Ito’s formula. For F ∈ C2 and semi-martingales X1, . . . , Xd,we have

F (t,X1t , . . . , X

dt ) = F (0, X1

0 , . . . , Xd0 ) +

∫ t

0

d∑i=1

∂F

∂Xi(t,X1

s , . . . , Xds )dXi

s

+

∫ t

0

(∂F

∂tdt+

1

2

d∑i,j=1

∂2F

∂Xi∂Xj(s,X1

s , . . . , Xds )d〈Xi, Xj〉s

).

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

F (t, B0) = F (0, B0) +

∫ t

0

∂F

∂X(s,Bs)dBs +

∫ t

0

(∂F∂s

+∆

2F)

(s,Bs)ds.

Example 23.1. Let

E (λM)t = eλMt−λ2

2 〈M,M〉t ,

where M is a local martingale. For any λ ∈ C, E (λM)t is a local martingaleand

E (λM)t = eλM0 + λ

∫ t

0

ε(λM)sdMs.

Proof. Let F (x, r) = eλMt−λ2

2 r. If we set G(x, t) = F (x, 〈M,M〉t), then Ito’sformula tells us that

G(Mt, t) = G(M0, 0) +

∫ t

0

∂G

∂MdMs +

∫ t

0

(∂G∂s

+1

2

∂2G

∂M2

)(s, x)d〈M,M〉s.

Here, we can treat 〈M,M〉t as an external time dependence, because it is afinite variation process. Computing the second derivative gives

∂G

∂sds+

G′′

2d〈M,M〉s =

[−λ

2

2

dr

dsds+

λ2

2d〈M,M〉s

]G(s,Ms) = 0.

Then we are left with the first derivative.

23.1 Application of Ito calculus

Theorem 23.2 (Levi). Let X1, . . . , Xd be adapted local martingales with con-tinuous ample paths. If 〈Xi, Xj〉t = δijt, then X1, . . . , Xd are independentBrownian motions.

Proof. Let us only look at the case d = 1. By the previous theorem,

E (t) = eiξXt+ξ2

2 t

Math 288 Notes 51

is a local martingale and further a martingale because it is bounded. This showsthat

E[eiξ(Xt−Xs)+ξ2

2 (t−s)|Fs] = 1

for all ξ. Then Xt −Xs is Gaussian with variance (t− s) when conditioned onFs.

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

cpE[〈M,M〉p/2T ] ≤ E[(M∗T )p] ≤ CpE[〈M,M〉p/2T ].

Proof. Let us first consider the case p ≥ 2. By Ito’s formula,

|Mt|p =

∫ t

0

p|Ms|p−1 sgnMsdMs +1

2

∫ t

0

p(p− 1)|Ms|p−2d〈M,M〉s.

We may assume that M is bounded because we can apply to MSn with Sn =inft : |Mt| ≥ n and take the limit. Then

∫ t0p|Ms|p−1 sgnMsdMs is an L2

martingale.Hence

E[|Mt|p] ≤ p(p− 1)E[∫ t

0

|Ms|p−2d〈M,M〉s]≤ p(p− 1)E[|M∗t |p−2〈M,M〉t]

≤ p(p− 1)E[|M∗t |p](p−2)/pE[〈M,M〉p/2t ]2/p.

So by Doob’s inequality, we get an upper bound on E[|M∗t |p] by E[〈M,M〉p/2t ].

Math 288 Notes 52

24 March 29, 2017

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

24.1 Burkholder–Davis–Gundy inequality

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

E[(M∗t )p] ≤ CpE[〈M,M〉p/2t ].

We first assumed that M is bounded. This is because we can prove theinequality for MTn where Tn = inft : |Mt| ≥ n and then take the limitn→∞. Next, we assumed that p ≥ 2 and applied Ito’s formula.

Proof (cont.) Now let us consider the case 0 < p < 2. Define Tx = inft : M2t >

x. Then

P ((M∗t )2 ≥ x) = P (Tx ≤ t) = P ((MTx∧t)2 ≥ x) ≤ 1

xE[(MTx∧t)

2]

=1

xE[〈M,M〉Tx∧t] ≤

1

xE[〈M,M〉t]

because M2 − 〈M,M〉 is a martingale.Let us define Sx = inf t : 〈M,M〉t ≥ x. Our goal is to prove

P ((M∗t )2 ≥ x) ≤ 1

xE[〈M,M〉t1(〈M,M〉t ≤ x)] + 2P (〈M,M〉t ≥ x).

Once we prove this, we get for q = p/2,

E[(M∗t )2q] =

∫ ∞0

P ((M∗t )2 ≥ x)qxq−1dx

≤ c∫ ∞

0

1

xE[〈M,M〉t1(〈M,M〉t≤x)]x

q−1dx

+ c

∫ ∞0

P (〈M,M〉t ≥ x)xq−1dx

= cqE[〈M,M〉qt ] + cqE[〈M,M〉qt ].

Now set us prove the inequality. We have

(M∗t )2 ≥ x ⊆ (M∗Sx∧t)2 ≥ x ∪ Sx ≤ t.

Then

P ((M∗t )2 ≥ x) ≤ P ((M∗Sx∧t)2 ≥ x) + P (Sx ≤ t)

≤ 1

xE[〈M,M〉Sx∧t] + P (Sx ≤ t)

≤ 1

xE[〈M,M〉t ∧ x] + P (〈M,M〉t ≥ x)

≤ 1

xE[〈M,M〉t1(〈M,M〉t≤x)] + 2P (〈M,M〉t ≥ x).

Math 288 Notes 53

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

Proof. By the theorem, E[(M∞)∗] < ∞. Then Mt is bounded by a fixed L1

variable.

Previously we have shown that E[〈M,M〉2∞] < ∞ implies that M is L2

bounded. This is a significantly better criterion for checking that M is uniformly

integrable. As a consequence, if E[(∫ t

0Hs(ω)2d〈M,M〉1/2s )] <∞ then∫ s

0

Hσ(ω)dMσ

is uniformly integrable for s ≤ t.There are a few more things I want to go over. The first one is representation

of martingales as stochastic integrals. The second is Girsanov’s theorem.

Theorem 24.3. Let B be a Brownian motion. If Z ∈ L2(Ω,F∞, P ), then thereexists a unique adapted (progressive) process h ∈ L2(B) such that

Z = E[Z] +

∫ ∞0

hsdBs.

More generally,

Theorem 24.4. If M is a local martingale with filtration Ft, then there existsa hs ∈ L2

loc(B) such that

Mt = C +

∫ t

0

hsdBs.

Math 288 Notes 54

25 March 31, 2017

25.1 Representation of martingales

Theorem 25.1.

(1) If Z ∈ L2(Ω,F∞, P ) where Ft is a σ-field of the Brownian motion, then

there exists a process h with E∫ t

0h2sds <∞ such that

Z = E[Z] +

∫ ∞0

hsdBs.

(2) If Mt is a local martingale with respect to Ft, then there exists a process

h with∫ t

0h2sds <∞ such that

Mt = C +

∫ t

0

hsdBs.

Lemma 25.2. The vector space generated by exp(i∑pj=1 λj(Btj − Btj−1

) is

dense in L2C(Ω,F∞, P ).

Proof. By Fourier transformation, it suffices to show that the set

F (Bt1 , . . . , Btp) for F smooth with compact support

is dense in L2(Ω,F∞, P ).

Proof of Theorem 25.1. (1) Define f(s) =∑lj1(tj ,tj+1](s) so that

∫ t0f(s)dBs =∑

j λj(Btj+1∧t −Btj∧t). Define

E(t) = exp

(i

∫ t

0

f(s)dBs +1

2

∫ t

0

f2(s)ds

).

Ito’s formula implies

dE(t) = E(t)[if(t)dBt +

1

2〈if(t)dBt, if(t)dBt〉+

1

2f2(t)dt

]= E(t)if(t)dBt.

This shows that

E(t) = 1 + i

∫ t

0

E(s)f(s)dBs.

Now letH be the closure of ∫ t

0hsdBs. We have shown that exp(i

∑j λj(Btj+1−

Btj )) ⊆ H and so by the lemma, H = L2(P ).

(2) We know that if Mt is an L2-martingale, then M∞ =∫∞

0hsdBs. We

further hand E[M∞|Ft] =∫ t

0hsdBs.

If Mt is a local martingale, then consider MTn with Tn = inftt : |Mt| ≥ n.We then have a hns such that

MTnt =

∫ t

0

hns dBs.

Check the consistence of hns and hms and put the together to form hs. Then

Mt =∫ t

0hsdBs.

Math 288 Notes 55

25.2 Girsanov’s theorem

Theorem 25.3 (Girsanov). Let Pt and Qt be two probability measures with thesame Ft. Assume that P Q (P is absolutely continuous with respect to Q)and Q P . Define

Dt =dQtdPt

, D∞ =dQ∞dP∞

.

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

Dt = eLt−12 〈L,L〉t .

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

M = M − 〈M,L〉

is a continuous local martingale with respect to Q.

Proof. From Ito’s formula,

d logDt =1

DtdDt −

1

2

1

D2t

d〈D,D〉t.

So we have

logDt =

∫ t

0

dDs

Ds− 1

2

d〈D,D〉sD2s

.

If we define Lt =∫ t

0dDs/Ds then

〈L,L〉t =⟨∫ t

0

dDs

Ds,

∫ t

0

dDs

Ds

⟩=

∫ t

0

d〈D,D〉sD2s

.

So logDt = Lt − 12 〈L,L〉t.

Now let us prove the main part. If MD is a local martingale with respect toP then M is a local martingale with respect to Q. This is almost a tautology,because for G a measurable function with respect to Fs,

EP [(MD)tG] = EP [(MD)sG]

and this translates to EQ[MtG] = EQ[MsG].So it suffices to check that MD is a martingale with respect to P . Then it

suffices to check that d(MD) is a stochastic integral of something. This is goingto be a half-page computation.

Math 288 Notes 56

26 April 3, 2017

Theorem 26.1 (Girsanov’s formula). Let P Q and Q P , where P andQ are probability measures with the same Ft. If Dt = dQ

dP |Ft then Dt is auniformly integrable martingale. There exists a local martingale Lt such thatDt = exp(Lt − 1

2 〈L,L〉t). Furthermore, if Mt is a local martingale with respectto P then

M = M − 〈M,L〉

is a local martingale with respect to Q.

Proof. We have checked that if MDt is a (local) martingale with respect to P ,then M is a (local) martingale with respect to Q.

Now by Ito’s formula,

d(MtDt) = dMtDt + MtdDt + 〈dM, dD〉t= DtdMt −Dtd〈M,L〉t + MtdDt + 〈dM, dD〉t,

because Dt is a martingale and so its variation with respect to a finite variationprocess is 〈〈M,L〉, D〉 = 0.

We can also compute

dDt = d(eLt−12 〈L,L〉t) = D

(dLt −

1

2d〈L,L〉t +

1

2d〈L,L〉t

)= D(dLt).

Therefore

d(MtDt) = DtdMt −Dd〈M,L〉t + MtDt +Dt〈dM, dL〉t = DtdMt + MtdDt.

Because Mt and Dt are martingales, MtDt is a martingale with respect to P .

26.1 Cameron–Martin formula

For example, take Lt∫ t

0b(Bs, s)dBs by setting

Qt = Pte∫ t0b(Bs,s)dBs− 1

2

∫ t0b2(Bs,s)ds.

If Mt = Bt, then

M = Bt −⟨Bt,

∫ t

0

b(Bs, s)dBs

⟩= Bt −

∫ t

0

b(Bs, s)ds

is a Brownian motion with respect to Q.If b(Bs, s) = g(s), then

D∞ = e∫∞0g(s)dBs− 1

2

∫∞0g2(s)ds.

If we write h(t) =∫ t

0g(s)ds, then for any function Φ on C(R+,R),

EW [Φ(B + h)] = EQ[Φ((B − h) + h)] = EQ[Φ(B)] = EW [D∞Φ(B)]

Math 288 Notes 57

where W is the Wiener measure with respect to P . This is the Cameron–Martin formula.

Intuitively, this is true because, in the case when Φ is given by the simplediscrete form F (B(t1))G(B(t2)),

EW [F (B(t1) + h(t1))G(B(t2) + h(t2))]

=

∫e−

x2

2t1 e− (x−y)2

2(t2−t1)F (x+ h(t1))G(y + h(t2))

=

∫F (x)G(y)e

− x2

2t1− (x−y)2

2(t2−t1)+h(t1)xt1

+(h(t1)−h(t2))(x−y)

(t2−t1)− 1

2 (h(t1)2

t1+

(h(t2)−h(t1))2

t2−t1)2

.

Here the last three terms correspond to D∞.There are some applications of Ito calculus in solving differential equations

involving processes.

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

u(x, t) = Ex[φ(xt)]

for some xt.

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

u(x) = Ex[φ(xT )], T = inft : xt /∈ D.

3. Feynman–Kac formula.

Let us prove the first statement. Let v(s, x) = u(t− s, x). Then

∂sv +∆

2v = 0.

Now Ito’s formula tells us

Ex[v(t,X(t))] = Ex[v(0, x0)]+E∫ t

0

(∂v∂s

+∆v

2

)ds+E[martingale] = v(0, x) = u(t, x).

So Ex[φ(X(t))] = u(t, x).

Math 288 Notes 58

27 April 5, 2017

27.1 Application to the Dirichlet problem

Let us look at the Dirichlet problem, given by

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

Consider the Brownian motion Bt starting from x. Then

u(Bt) = u(B0) +

∫ t

0

∆

2u(Bs)ds+ (martingale term) = u(B0) + (martingale).

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

Ex[u(Bt∧T )] = u(x).

Let us assume that P (T <∞) = 1. This is true if, in particular, D is bounded.Then

u(x) = limt→∞

Ex[Bt∧T ] = Ex[BT ] = Ex[g(BT )].

Let T(x, dy) be the probability measure of the Brownian motion starting fromx to exit at y. Then

u(x) =

∫g(y)T(x, dy). (∗)

This arguments shows that if u solves the equation, it has to be given bythis integral. But when does (∗) actually solve the problem? You need somecondition on D. There is something called the exterior cone condition thatimplies the existence of the Dirichlet problem. The same condition implies that(∗) solves the equation.

Consider the Dirichlet problem given by

g(x) =

0 if |x| = R,

1 if |x| = ε.

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

uRε (x) =logR− log|x|logR− log ε

,

and for d = 3,

uRε (x) =

1R −

1|x|

1R −

1ε

.

This has the interpretation that u(x) is the probability of the Brownian motionstarting from x exits ε ≤ |x| ≤ R from Sε.

Math 288 Notes 59

Taking R→∞, we get

uRε (x)→

1 if d = 2

< 1 if d ≥ 3.

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

27.2 Stochastic differential equations

We want to look at equations of the form

dXt = σ(t,Xt)dBt + b(t,Xt)dt,

with respect to a filtration Ft. Something being a solution to the equationE(σ, b) means

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

(ii) Bt is a Brownian motion with respect to Ft,(iii) there exists Xt a continuous progressively measurable with respect to Ft

such that

Xt =

∫ t

0

σ(s,Xs)dBs +

∫ t

0

b(s,Xs)ds+X0.

Definition 27.1. A solution X is called a strong solution if Ft is adaptedwith respect to the canonical filtration of the Brownian motion.

Definition 27.2. A equation is pathwise unique if any two solutions X andX ′ for the same Bt has X = X ′ almost surely.

Theorem 27.3. If σ and b are uniformly Lipschitz then the SDE has a strongsolution and is pathwise unique.

Proof. Iterate this process, i.e., let X0t = X and

Xnt =

∫ t

0

σ(s,Xn−1s )dBs +

∫ t

0

b(s,Xn−1s )ds+X.

We want to check that this iteration converges. To show this, define

gn(t) = E[

sup0≤s≤t

|Xns −Xn−1

s |2].

We claim that

gn(t) ≤ C ′T (CT )n−1 tn−1

(n− 1)!.

Math 288 Notes 60

We have

E[

sup0≤s≤t

|Xn+1s −Xn

s |2]≤ 2E

[sup

0≤s≤t

∣∣∣∣∫ t

0

(σ(s,Xns )− σ(s,Xn−1

s ))dBs

∣∣∣∣2]+ 2E

[sup

0≤s≤t

∣∣∣∣∫ t

0

(b(s,Xns )− b(s,Xn−1

s ))ds

∣∣∣∣2].Using the Lipschitz condition, we are going to bound this, next time.

Math 288 Notes 61

28 April 7, 2017

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

28.1 Existence and uniqueness in the Lipschitz case

Theorem 28.1. Consider the stochastic differential equation

dX = σ(t,Xt)dBt + b(t,Xt)dt, (∗)

where σ(t, x) and b(t, x) are uniformly Lipschitz with respect to x. Then thereexists a pathwise unique solution to (∗), i.e., there exists a (Ω,F , P ) with con-tinuous sample paths such that

Xt = X +

∫ t

0

σ(s,Xs)dBs +

∫ t

0

b(s,Xs)ds

almost surely.

Proof of existence. Let us define, X0t = X and

Xnt = X +

∫ t

0

σ(s,Xn−1s )dBs +

∫ t

0

b(s,Xn−1s )ds.

Let us definegn(T ) = E

[sup

0≤s≤t|Xn

s −Xn−1s |2

].

For fixed T and t ∈ [0, T ], we have

gn+1(t) = E[

sup0≤s≤t

|Xn+1s −Xn

s |2]≤ 2E

[sup

0≤s≤t

(∫ s

0

(σ(Xnu )− σ(Xn−1

u ))dBu

)2]+ 2E

[sup

0≤s≤t

(∫ s

0

(b(Xnu )− b(Xn−1

u ))du)2]

≤ CE[∫ t

0

(σ(Xnu )2 − σ(Xn−1

u ))2du+ t

∫ t

0

(b(Xnu )− b(Xn−1

u ))2du]

≤ CTE[∫ t

0

sup0≤s≤u

|Xns −Xn−1

s |2du]≤ CT

∫ t

0

gn(u)du,

by the Burkholder–Davis–Gundy inequality and the Lipschitz condition. Fromthis inequality, you will inductively get

gn+1(t) ≤ (CT )ntn

n!.

That is, for any T fixed,∑∞n=1 gn(T )1/2 <∞. Hence the process converges and

it is a solution.

Math 288 Notes 62

Proof of uniqueness. This actually just follows from the same arguments. If wedefine

g(t) = E[

sup0≤s≤t

|Xs − Ys|2],

then

g(t) ≤ C∫ t

0

g(s)ds.

Then by Gronwall’s inequality, g = 0.

Let us apply Ito’s formula to a solution Xt. We get

f(Xt) = f(X0) +

∫ t

0

∇fsdXs +1

2

∫ t

0

f ′′(Xs)d〈X,X〉s

= f(X0) +

∫ t

0

(∇fs)σ(s,Xs)dBs +

∫ t

0

(∇fs)2b(s,Xs)ds

+1

2

∫ t

0

f ′′(Xs)σ2(s,Xs)ds

= f(X0) +

∫ t

0

(1

2σ2∆ + b∇

)f(Xs)ds+ (martingale).

We call the operator

L =1

2σ2∆ + b∇

the generator of the process.The Brownian motion is a solution to ∂tu = 1

2∆u. This is illustrated byIto’s formula

f(Bt) = f(B0) +

∫ t

0

1

2∆f(Bs)ds+M.

Likewise, to solve the equation ∂tu = Lu for L = 12σ

2∆ + b∇, you can solve theequation

ds = σdB + ddt.

28.2 Kolomogorov forward and backward equations

Theorem 28.2. Let P (s, x; t, y) be a transition probability density such that

Ex,s[h(x(t))] =

∫p(s, x; t, y)h(y)dy.

Then

∂sp(s, x; t, y) + Lxp(s, x; t, y) = 0,

∂tp(s, x; t, y) = L∗yp(s, x; t, y).

Math 288 Notes 63

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

f(x, t) =

∫p(0, z; t, x)f0(z)dz.

Then we get ∫f(t, x)h(x)dx =

∫f0(x)Ex[h(x(t))]dx.

The derivative with respect to t gives

∂t

∫f(t, x)h(x)dx = ∂t

∫f0(x)Ex[h(x(t))]dx

= ∂t

∫f0(x)Ex

[h(x(0)) +

∫ t

0

(Lh)(x(s))ds]

=

∫f0(x)Ex(Lh)(x(t)) =

∫(Lh)(x)f(t, x)dx

=

∫h(x)(L∗f)(t, x)dx.

This gives the forward equation, and its dual is the backward equation.

Math 288 Notes 64

29 April 10, 2017

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

dXt = σ(Xt)dBt + b(Xt)dt,

let f(t, x) be the probability density of the solution Xt. Then

∂tf(t, x) = L∗f(t, x), where L =σ(x)2

2∆ + b(x)∇.

The Kolmogorov equations say that, if the transition probability is p(s, x; t, y),

Ex;s[h(Xt)] =

∫p(s, x; t, y)h(y)dy,

then(∂s + Lx)p(s, x; t, y) = 0, (∂t − L∗y)p(s, x; t, y) = 0.

Proof. Recall that by Ito’s formula,

h(Xt) = h(Xs) +

∫ t

0

(Lh)(Xs)ds+ (martingale).

So

∂t

∫f(t, x)h(x) = ∂t

∫f0(x)Ex(h(Xt)) =

∫f0(x)Ex[(Lh)(Xt)]

=

∫f(t, x)(Lh)(x)dx =

∫(L∗f)(t, x)h(x)dx.

More generally, if f(t, x) is the density relative to µ, then

∂t

∫f(t, x)h(x)µ(dx) =

∫f(t, x)(Lh)(x)µ(dx) =

∫(L∗µ)f(t, x)h(x)µ(dx)

and so∂tf = L∗µf. (∗)

Definition 29.1. µ is called an invariant measure of the process if L∗µ1 = 0,i.e., 1 is a solution to (∗).

In other words, µ is invariant if and only if∫(Lf)dµ = 0

for any f with compact support.

Definition 29.2. µ is called a reversible measure of the process if L∗µ = L,i.e., ∫

f(Lh)dµ =

∫(Lf)hdµ.

In this case,

−∫h(Lh)dµ = Dµ(h)

is called the Dirichlet form of the process.

Math 288 Notes 65

29.1 Girsanov’s formula as an SDE

Girsanov’s formula says that if

dQ

dP

∣∣∣[0,t]

= e∫ t0b(Xs)dBs− 1

2

∫ t0b2(Xs)ds

and M is a local martingale with respect to P , then

M = M −⟨M,

∫ t

0

b(Xs)dBs

⟩is a local martingale with respect to Q.

Take M = B with respect to P . Then this implies that

βt = M = Bt −⟨Bt,

∫ t

0

b(Bs)dBs

⟩= Bt −

∫ t

0

b(Bs)ds

is a martingale with respect to Q. We note that 〈M, M〉t = t. Therefore M isin fact a Brownian motion with respect to Q. But Bt is no longer a Brownianmotion in Q. So we change notation and write Xt instead of Bt. Then

dXt = dβt + b(Xt)dt, (†)

i.e., Xt is the solution to the SDE (†), i.e., the probability measure Q is thesolution to (†).

This means that when you have a SDE that looks like dXt = σdBt + bdt,the really important term is σdBt, since the bdt comes from Girsanov’s formulaby changing the probability measure.

We will look at three examples of SDEs:

(1) Ornsetin–Uhlenbeck process

(2) Geometric Brownian motion - Black–Scholes formula

(3) Bessel process

29.2 Ornstein–Uhlenbeck process

Let us first look at the Ornstein–Uhlenbeck process:

dXt = dBt − λXtdt,

where the negative coefficient means that we are pushing back the process intothe origin. If we define Xt = e−λtYt, then

dYt = d(eλtXt) = λYtdt+ eλtdBt − eλtλXtdt

and so dYt = eλtdBt. That is,

Xt = e−λtYt = e−λtX0 +

∫ t

0

e−λ(t−s)dBs,

Math 288 Notes 66

and this is the exact solution.The generator is

L =1

2

∂2

∂x2− λx ∂

∂x,

and you can check that the Gaussian measure µ = dλe−cλx2

is an invariantmeasure.

29.3 Geometric Brownian motion

Let us now look at the geometric Brownian motion:

dXt = σXtdBt + rXtdt.

If we again make a substitution Xt = ertYt, then

dYt = −rYtdt+ e−rtrXtdt+ e−rtσXtdBt = σYtdBt.

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

dYt = f ′(Bt, t)dBt +f ′′(Bt, t)

2dt+ (∂tf)(Bt, t)dt.

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

Yt = f(Bt, t) = eσBt − σ2

2t.

Therefore the exact solution is given by

Xt = erteσBt−σ2

2 t +X0.

Math 288 Notes 67

30 April 12, 2017

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

dXt = σXtdBt + γXtdt.

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

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

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

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

= (atµSt + brβt)dt+ atσStdBt.

Suppose Vt = f(t, St) is a function of t and St. Then by Ito’s formula,

dVt = ∂tf +1

2f ′′d〈S, S〉t + f ′(t, St)dSt

=[∂tf +

1

2σ2S2f ′′ + µStf

′]dt+ f ′σStdBt.

Comparing this with the other equation gives

at = f ′(t, St), bt =ft + 1

2f′′σ2St

rβt.

Using the formula f(t, St) = atSt + btβt gives

ft +1

2σ2x2fxx(t, x) + fx(t, x)x− rf(t, x) = 0.

This is the Black–Scholes equation.

30.1 Bessel process

Let us now look at the Bessel process given by

dXt = 2√XtdBt +mdt, m > 0.

If we look at the d-dimensional Brownian motion βt = (β1, . . . , βd) and defineXt = β2

t = (β1)2 + · · ·+ (βd)2, then we can check that dXt = 2√XtdBt + ddt.

So m = d is exactly the process of |βt|2.Let us define

Tr = inft ≥ 0 : Xt = r,

where we assume x ≤ r. We now claim that P (Tr =∞) = 0.

Math 288 Notes 68

Suppose that ε < x < r and define

Mt =

X

1−m/2t m 6= 2

logXt m = 2.

Then Ito’s calculus gives

dMt∧Tε =(

1− m

2

)X−m/2t dXt∧Tε +

1

2

(1− m

2

)(−m

2

)X−m/2−1d〈X,X〉t∧Tε

=[(

1− m

2

)mX

−m/2t∧Tε +

1

2

(1− m

2

)(−m

2

)X−m/2t∧Tε

]dt

+ 2(

1− m

2

)X−m/2+1/2t∧Tε dBt∧Tε = 2

(1− m

2

)X−m/2+1/2t∧Tε dBt∧Tε .

Thus

Mt∧Tε = 2(

1− m

2

)∫ t∧Tε

0

X−m/2+1/2s dBs.

This is a martingale for ε fixed, and just a local martingale if ε→ 0. By optionalstopping, we have E[Mt∧Tε∧TA ] = x1−m/2 and if you solve the equation, we get

P (Tε < TA) =

A1−m/2 − x1−m/2

A1−m/2 − ε1−m/2m 6= 2

logA− log x

logA− log εm = 2.

This shows that for m ≥ 2, as ε → 0, the probability that the Bessel processreach 0 is zero.

Math 288 Notes 69

31 April 14, 2017

Last time we talked about the Bessel process. This is given by

dXt = 2√XtdBt +mdt.

An example, when m = d is an integer, is Xt = |Bdt |2. If we define Mt = X1−m/2t

thendMt =

(1− m

2

)X−m/2t 2

√XtdBt.

Then Mt is a local martingale.In the case of m = 3, we would have Mt = |Bt|−1, and so d|Bt|−1 =

C|Bt|−2dBt. Then

|Bt|−1 =

∫ t

0

|Bs|−2dBs

is a local martingale.Recall that E(

∫ t0|Bs|−2dBs)

2 <∞ then |Bt|−1 would be a martingale. How-ever, you can check that

E[∫ t

0

|Bs|−4ds]

=∞.

Indeed, |Bt|−1 is a local martingale but not a martingale. Likewise, in d = 2,Mt = log|Bt| is a local martingale but not a martingale.

31.1 Maximum principle

Theorem 31.1. Suppose you want to solve the equation

∂tu+1

2∆u+ g = 0, u(T, x) = f(x)

with boundary condition u(t, x) = k(t, x) for x ∈ ∂Ω and T > 0. Or moregenerally, consider the equation ∂tu + Lu + g = 0 (in this case, we use thestochastic process generated by L instead of the Brownian motion). Then

u(0, x) = Ex[f(x, T ))1(τ≥T ) + k(τ, x(τ))1(τ<T ) +

∫ τ∧T

0

g(s, x(s))].

Proof. From Ito’s formula, for a process X,

u(t,X(t)) =

∫ t

0

(∂tu+

∆

2u)ds+ (martingale) + u(0, x).

If we define τ = τ ∧ T , then

Ex[u(τ ∧ T,X(τ ∧ T ))] + Ex∫ τ∧T

0

g(s,Xs)ds = u(0, x).

This proves the theorem.

Math 288 Notes 70

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

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

31.2 2-dimensional Brownian motion

Consider a 2-dimensional Brownian motion Bt = Xt + iYt.

Theorem 31.3 (Conformal invariance, Theorem 7.8). Suppose Φ is analytic.Then Φ(B) is a Brownian motion up to a time change.

Proof. Let us compute dΦ(B). We have

dΦ(B) = d(g(Xt, Yt) + ik(Xt, Yt)) =∂g

∂xdXt +

∂g

∂ydYt + i

(∂k∂xdXt +

∂k

∂ydYt

).

The quadratic variation of one component is

〈dg, dg〉 =[(∂g∂x

)2

+(∂g∂y

)2]dt.

So g(Xt, Yt) is a Brownian motion with time change. The similar thing holdsfor k. Moreover,

〈dg, dk〉 =⟨∂g∂xdX +

∂g

∂ydY,

∂k

∂xdX +

∂k

∂tdY⟩

=(∂g∂x

∂k

∂x+∂g

∂y

∂k

∂y

)dt = 0

from the Cauchy–Riemann equations.

Math 288 Notes 71

32 April 17, 2017

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

32.1 More on 2-dimensional Brownian motion

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

Bt = exp(βHt + iγHt)

where Ht =∫ t

0ds/|Bs|2 and β and γ are independent Brownian motions.

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

d logBt =dXt + idYtXt + iYt

− 1

2

( 1

Xt + iYt

)2

(dXt + idYt)2 =

dXt + idYtXt + iYt

.

Then we get

dβt =XdX + Y dY

(X2 + Y 2), dγt =

XdY − Y dXX2 + Y 2

.

Now we have

〈dβt, dβt〉 =(XdX + Y dY )2

(X2 + Y 2)2=

(X2 + Y 2)dt

(X2 + Y 2)2=

dt

(X2 + Y 2),

〈dγt, dγt〉 = · · · = dt

(X2 + Y 2),

〈dβt, dγt〉 =(XdX + yDY )(XdY − Y dX)

(X2 + Y 2)2= 0.

What we want is actually βHt instead of βt. If β is a Brownian motion, then

〈dβHt , dβHt〉 = dHt =1

X2t + Y 2

t

dt

and so we see that βHt for β a Brownian motion works.

From this we get, we get

dHt =1

|Bt|2dt =

1

exp(2βHt)dt

and soexp(2βHt)dHt = dt.

Lemma 32.2. Let T(λ)1 = inft ≥ 0;β

(λ)t = 1, where B

(λ)t = λ−1βλ2t is the

rescaled Brownian motion. Then

4

(log t)2Ht − T (log t/2)

1 → 0

in probability as t→∞.

Math 288 Notes 72

Proof. From exp(2βHt)dHt = dt, we have

infs

∫ s

0

e2βudu = t

= Ht.

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

P (Ht > λ2tT

λt1+ε)→ 0.

Then

P (Ht > λ2tT1+ελt) = P

[∫ λ2tT

(λt)1+ε

0

e2βudu < t

]= P

[log λtλt

+1

2λtlog

∫ T(λt)1+ε

0

e2λtβ(λt)v dv < 1

]≤ P

[1

2λtlog

∫ T1+ε

0

e2λtβvdv < 1

]= P

(max

0≤v≤T1+ε

βv < 1)

= 0

after the rescaling u = λ2t v.

Theorem 32.3. If Bs is a 2-dimensional Brownian motion starting from anyz 6= 0,

limt→∞

P(

inf0≤s≤t

|Bs| ≤ t−a/2)

=1

1 + a.

Proof. Due to scaling, we as may as well assume that z = 1. We have

log(

inf0≤s≤t

|Bs|)

= inf0≤s≤t

βHs = inf0≤u≤Ht

βu

by monotonicity. Now we have

2

log tlog(

inf0≤s≤t

|Bs|)

=1

λtinf

0≤u≤Htβu = inf

0≤s≤λ−2t Ht

β(λt)s .

By the lemma, λ−2t Ht converges to T

(λt)1 and also

inf0≤s≤T (λt)

1

β(λt)s = inf

0≤s≤T1

βs.

We will continue next time.

Math 288 Notes 73

33 April 19, 2017

Last time we showed that a 2-dimensional Brownian motion can be expressedas

Bt = eβHt+iγHt , Ht =

∫ t

0

1

|Bs|2ds,

where β and γ are independent Brownian motions.

Lemma 33.1 (7.21). In probability,

4

(log t)2Ht − T (log t/2)

1 → 0

as t→∞, where T(λ)1 = infs ≥ 0, β

(λ)t = 1.

Proposition 33.2 (7.22). limt→∞

P(

inf0≤s≤t

|Bs| ≤ t−a/2)

=1

1 + a.

Proof. Because Ht is monotone, we have

log(

inf0≤s≤t

|Bs|)

= inf0≤s≤t

βHs = inf0≤u≤Ht

βu.

Then

2

log tlog(

inf0≤s≤t

|Bs|)

=1

λtinf

0≤u≤Htβu = inf

0≤v≤Ht/λ2t

β(λt)v = inf

0≤v≤T (λt)1

β(λt)v

after the rescaling u = λ2t v. Then the probability is

P( 1

λtlog inf

0≤s≤t|Bs| ≤ −a

)= P

(inf

0≤v≤T (λt)1

β(λt)v ≤ −a

)= P

(inf

0≤v≤T1

βv ≤ −a)

= P (T−a < T1) =1

1 + a

because of the property of the 1-dimensional Brownian motion. (You can seethis from solving the Laplace equation with u(−a) = 1 and u(1) = 0.)

33.1 Feynman–Kac formula

Theorem 33.3. Consider an equation

∂tu+1

2∆u+ V (u) = 0

with the boundary condition u(T, x) = f(x). Then

u(0, x) = Ex[exp

(∫ t

0

V (x(s))ds

)f(x(T ))

].

Math 288 Notes 74

Proof. Take a u solving the equation. Then

d

[u(t, x(t)) exp

(∫ t

0

V (x(s))ds

)]=(∂tu+

1

2∆u)

exp

(∫ t

0

V (x(s))ds

)dt

+∇xu(t, x(t))dx(t) + u exp

(∫ t

0

V (x(s))ds

)V (x(t))dt

+ du d exp

(∫ t

0

V (x(s))ds

)= (∂tu+

1

2∆u+ V u) exp

(∫ t

0

V (x(s))ds

)dt+ (martingale)

is a martingale. So we get

u(0, x) = Ex[exp

(∫ T

0

V (x(s))ds

)f(x(T ))

].

Here is an intuitive proof. The equation can be written as ∂tu = ( 12∆+V )(u)

and so we can writeu = et(

12 ∆+V )u0.

There is also the Trotter product formula

eA+B = limn→∞

(eA/neB/n)n.

Then our solution can be thought of as

e∆2tn e

tnV e

∆2tn e

tnV · · · = 1√

2π · · ·e−

(x0−x1)2

2(t/n) etnV (x1)e0

(x1−x2)2

2(t/n) etnV (x2) · · · .

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

etn (V (x1)+···+V (xn)) → e

∫ t0V (x(s))ds

under the limit n→∞.

33.2 Girsanov formula again

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

dX = bdt+ dβ.

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

dQ

dP= exp

(∫ t

0

b(s,X(s))dx(s)− 1

2

∫ t

0

b2ds

).

Math 288 Notes 75

Let us define

Z(t) =

∫ t

0

b(s,X(s))dX(s)

and consider the object

d exp

(λX(t) + Z(t)− 1

2

∫ t

0

(λ2 + 2λb(s,X(s)) + b2)ds

).

This is going to be a martingale plus a dt term. Our claim is that the dt termvanish. We check that Ito calculus gives

λdX − 1

2(λ2 + 2λb+ b2)dt+ dZ + λdXdZ +

1

2λ2dXdX +

1

2dZdZ.

Here λdXdZ = bdt and 12λ

2dXdX = λ2

2 dt and 12dZdZ = b2

2 dt. So the dt termindeed vanishes. Everything follows from this.

Math 288 Notes 76

34 April 21, 2017

The Girsanov formula says that

dQ

dP= exp

(∫ t

0

b(s,X(s))dX(s)− 1

2

∫ t

0

b2(s,X(s))ds

),

where Q solves dX = b(t,X(t))dt = dβ.The idea is that the Brownian motion measures something like

(const)e−(x2−x1)2

2t/n− (x3−x2)2

2t/n−···.

Here the discrete difference Xt −Xt+1 is like dβ(t) = dX − b(t,X(t))dt and sothis quantity is like

exp(− 1

2t/n

[x2−x1−b(t1, x1)

t

n

]2−· · ·

)= exp

(− (x1 − x2)2

2t/n+b(t1, x1)(x2−x1)−· · ·

).

Here∑j b(tj , xj)(xj+1 − xj) is like

∫ t0b(s, x(s))ds. So this is why we get Gir-

sanov’s formula.

34.1 Kipnis–Varadhan cutoff lemma

Theorem 34.1. Suppose µ is a reversible measure with respect to a processwith generator L. Let D(f) = −

∫fLfdµ be the Dirichlet form. Then

Pµ(

sup0≤s≤t

|f(X(s))| ≥ `)≤ 1

`

√‖f‖2L2µ +D(f).

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

du(t, x) = (∂tu+ Lu)dt+ (martingale).

Now consider the stopping time τ = infs ≥ 0, f(X(s)) = `. Then

u(t ∧ τ, x) =

∫ t∧τ

0

−g(s,X(s))ds+ u(0, x) + (martingale).

Suppose a function φλ solves the equation(λ− L)φ = 0 x ∈ G = y : f(y) < `,φλ = 1 x ∈ ∂G.

If we take u(t, x) = φλ(x)e−λt then

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

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

P (τ < t) ≤ eλtEµφλ(x) ≤ eλt√Eµ[φλ(x)2]

Math 288 Notes 77

by Chebyshev’s inequality.But recall that φλ is an eigenfunction of L. But note that for any h with

h = 1 on ∂G, we have 〈h, (λ− L)h〉µ ≥ 〈φλ, (λ− L)φλ〉µ and so

‖φλ‖2L2 ≤ ‖h‖22 +1

λD(h).

Therefore

P (τ < t) ≤ eλt√‖h‖22 +

1

λD(h).

Take λ = 1/t and h = `−1(|f | ∧ `) gives the theorem.

Math 288 Notes 78

35 April 24, 2017

Let µ be a reversible (∫gLfdµ =

∫(Lg)fdµ) measure of a process with Dirichlet

form D = −∫fLfdµ. Then we want to show

P(

sup0≤s≤t

|f(X(s))| ≥ `)≤ c

`

√‖f‖2L2

µ+ tD(f).

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

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

This being a ground state means that λ is minimal. So φλ is nonnegative. (Thefirst eigenfunction of an elliptic problem is nonnegative.)

If we define u(t, x) = φλ(x)e−λt then ∂tu + Lu = 0. Then Ito’s formulaimplies that

u(0, x) = −Ex[∫ t

0

(∂s + L)u(s, x)ds

]+ Ex[u(t, x)] = Ex[u(t, x)].

So

φλ(x) = Ex[u(t ∧ τ, x)] = Ex[u(τ, x)1(τ≤t)] + Ex[u(t, x)1(τ>t)]

= Ex[e−λτ1(τ≤t) + e−λtφλ(x)1(τ>t)].

Now we have, with respect to µ,

e−λtPµ(τ ≤ t) ≤ Eµ[e−λτ1(τ≤t) ≤ Eµφλ(x) ≤ (Eµ(φλ(x)2))1/2.

Since φλ is the minimizer, we have

〈φλ, (λ− L)φλ〉 ≤ 〈h, (λ− L)h〉

for any h such that h(x) = 1 on ∂G. This implies that λ‖φλ‖2 + D(φλ) ≤λ‖h‖2 +D(h). If we let h = 1

` min(|f(x)|, `), we have

‖φλ‖22 ≤ ‖h‖22 +1

λD(h) =

1

`2‖f‖22 +

1

`2λD(f).

Plugging this into the previous formula gives

Pµ(τ ≤ t) ≤ eλt

`

√‖f‖22 +

1

λD(f) ≤ e

`

√‖f‖22 + tD(f).

35.1 Final review

Here are the things we have done so far:

• construction of Brownian motions (Gaussian processes, Levy’s construc-tion)

Math 288 Notes 79

• martingales (discrete)

• local martingales (continuous time); roughly speaking, M is a local mar-tingale if it can be appproximated by a martingale

• a nonnegative local martingale is a supermartingale

• construction of quadratic variation 〈M,M〉t• if E[〈M,M〉t] <∞ for each t then M is a martingale (Theorem 4.13)

• stochastic integration∫HdM , when

∫Hsd〈M,M〉 <∞ almost surely

• if E[∫Hsd〈M,M〉∞] <∞ then it is a marginale

• Ito’s formula (in general, we only know that∫· · · dM is a local martingale)

• Burkholder–Davis–Gundy inequality

E[(M∗T )p] ≤ CpE[〈M,M〉p/2T ].

• Levi’s theorem for characterizing Brownian motions, 〈M,M〉t = t

• dominated convergence,∫ t

0Hns dXs →

∫ t0HsdXs in probability if Hn

s →Hs and |Hn

s | ≤ Ks and∫ t

0K2sd〈X,X〉s <∞ (Proposition 5.18)

• existence and uniqueness of dX = bdt+ σdβ for σ and β Lipschitz

• applications of Ito’s formula; dF = ∇FdM + 12∇

2Fd〈M,M〉, check that∇FdM is a martingale, and then take expectation

• Feynman–Kac formaula, Girnasov, Dirichlet problem, Kipnis–Varadhan,. . .

You will not remember the details, and I will not remember the details. Butit is important to have the big picture.

Index

adapted variable, 21

backward martingale, 32Bessel process, 67Black–Scholes model, 67Blumenthal’s zero-one law, 17bracket, 39Brownian motion, 11, 15Burkholder–Davis–Gundy

inequality, 51

Cameron–Martin formula, 57conditional expectation, 8covariant matrix, 7

Dirichlet form, 64Doob inequality, 23Doob upcrossing inequality, 24, 25

elementary process, 41

filtration, 20finite variance process, 32

Gaussian process, 8Gaussian random variable, 5Gaussian space, 8Gaussian vector, 6Gaussian white noise, 10generator of a process, 62geometric Brownian motion, 66Girsanov’s theorem, 55

indistinguishable, 13invariant measure, 64Ito’s formula, 48

Kipnis–Varadhan cutoff lemma, 76Kolmogorov’s theorem, 13

Levi’s theorem, 27Levi’s theorem on local

martingales, 50local martingale, 33

martingale, 22mean, 7moment generating function, 5

optional stopping theorem, 23Ornstein–Uhlenbeck process, 65

pathwise uniqueness, 59predictable, 30progressive measurability, 21

quadratic variation, 34

reflection principle, 19reversible measure, 64right continuous, 20

semimartingale, 39stopping time, 18, 21strong Markov property, 18strong solution, 59submartingale, 22submartingale decomposition, 30

uniform integrability, 6

Wiener measure, 16

80

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