Math 288 - Probability Theory and Stochastic Process€¦ · Math 288 - Probability Theory and...
Transcript of 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
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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
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34 April 21, 2017 7634.1 Kipnis–Varadhan cutoff lemma . . . . . . . . . . . . . . . . . . . 76
35 April 24, 2017 7835.1 Final review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78
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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
11 February 15, 2017
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
12 February 17, 2017
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
13 February 22, 2017
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
14 February 24, 2017
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
15 February 27, 2017
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