Stochastic Analysis

Prof. Dr. Nina Gantert

Lecture at TUM in WS 2011/2012

June 13, 2012

Produced by Leopold von Bonhorst and Nina Gantert

Contents

1 Definition and Construction of Brownian Motion 3

2 Some properties of Brownian Motion 7

3 The Cameron-Martin Theorem and the Paley-Wiener stochastic integral 13

4 Brownian Motion as a continuous martingale 17

5 Stochastic integrals with respect to Brownian Motion 19

6 Ito’s formula and examples 24

7 Pathwise stochastic integration with respect to continuous semimartingales 27

8 Cross-variation and Ito’s product rule 32

9 Stochastic Differential Equations 35

10 Girsanov transforms 38

2

1 Definition and Construction of Brownian Motion

1.1 Historic origin

Brown (1827): Movement of a pollen in a liquid

Bachelier (1900): Model for stock market fluctuations

Einstein (1905): Motion of a particle

1.2 Heuristic description with symmetric random walks

Y1, Y2, ... iid with P [Yi = 1] = 12= P [Yi = −1],

Sk =∑k

i=1 Yi, k = 1, 2, ..., S0 = 0. Fix N ∈ N

Rescale the process to the time interval [0, 1]:

X kN=

1√NSk, k = 0, 1, 2 . . . , N

Then,

(i) X0 = 0.

(ii) For 0 ≤ t0 < t1 < t2 < ... < tm ≤ 1, ti = kiN, ki ∈ 0, 1, ..., N, Xti − Xti−1

are

independent and E[Xti −Xti−1] = 0 and

V ar(Xti −Xti−1) = V ar(

1√N

ki∑

j=ki−1+1

Yj) =1

N(ki − ki−1).

Due to the CLT, the laws of (Xti −Xti−1) converge to N(0, ti − ti−1) for N → ∞

(and kiN

−−−→N→∞

ti ∈ [0, 1]).

This motivates the following definition:

1.3 Basic Definitions

Definition 1.1 Brownian Motion (BM) is a stochastic process Bt(ω), (0 ≤ t ≤ 1) on

a probability space (Ω,A, P ) such that

(i) B0 = 0 P-a.s.

(ii) For 0 ≤ t0 < t1 < t2 < ... < tn ≤ 1, the increments Bti − Bti−1, 1 ≤ i ≤ n , are

independent with law N(0, ti − ti−1)

(iii) t → Bt(ω) is continuous for P-a.a.ω.

Definition 1.2 Let (Bt)0≤t≤1 be a BM on the probability space (Ω,A, P ).

Then, the image measure of P under the map

Ω → C[0, 1]

3

ω → (Bt(ω))0≤t≤1

is the Wiener measure.

The Wiener measure is a prob. measure on (C[0, 1],F), with F = σ(gt : 0 ≤ t ≤ 1),where gt : C[0, 1] → R, gt(x) = x(t) (x ∈ C[0, 1]).

Interpretation:

Def. 1.1: t → Bt(ω) is a stochastic evolution in time.

Def. 1.2: (Bt)0≤t≤1 is a random variable with values in C[0, 1].

Theorem 1.3 Brownian motion exists.

There are different proofs of Theorem 1.3.

We give here a proof, which constructs ”linear interpolations” on the sets Dn = k2n

: 0 ≤k ≤ 2n. We follow the proof of Theorem 1.3. in [4].

Proof: Let D =⋃

n Dn and let (Ω,A, P ) be a probability space such that Zt, t ∈ D are

iid random variables on (Ω,A, P ), with law N(0, 1).

Let B0 := 0 and B1 := Z1. For each n ∈ N, we define the random variables Bs, s ∈ Dn

such that:

(1) For r < s < t, r, s, t ∈ Dn, Bt − Bs is independent of Bs − Br, and Bt − Bs has the

law N(0, t− s)

(2) The vectors (Bs, s ∈ Dn) and (Zt, t ∈ D \Dn) are independent.

For D0 = 0, 1, we are done.

Proceeding inductively, assume that we followed the construction for some n−1. We then

define Bs for s ∈ Dn \Dn−1 by

Bs =1

2(Bs− 1

2n+Bs+ 1

2n) +

1

2n+12

Zs.

The first term is the linear interpolation of B at the neighboring points of s in Dn−1.

Therefore, Bs is independent of (Zt, t ∈ D \Dn) and (2) is satisfied.

Moreover, since1

2(Bs+ 1

2n− Bs− 1

2n)

4

depends only on (Zt, t ∈ Dn−1), it is independent of 1

2n+12Zs. By induction assump-

tion, both terms have law N(0, 12n+1 ). Hence, their sums Bs − Bs− 1

2nand their difference

Bs+ 12n

− Bs are iid with law N(0, 12n).

Exercise:

X and Y iid random variables with law N(0, σ2)

⇒ X + Y,X − Y are iid random variables with law N(0, 2σ2).

To see that all increments Bs − Bs− 12n, s ∈ Dn \ 0 are independent, it suffices to show

that they are pairwise independent, since the vector of increments is Gaussian. We saw

that Bs−Bs− 12n, Bs+ 1

2n−Bs (with s ∈ Dn \Dn−1) are independent. The other possibility

is that the increments are over intervals separated by some s ∈ Dn−1. Choose s ∈ Dj

with this property and j minimal, so that the two intervals are contained in [s − 12j, s]

and [s, s+ 12j].

By induction hypothesis, the increments over these two intervals of length 12j

are inde-

pendent, and the increments over the intervals of lengths 12n

are constructed from the

independent increments Bs − Bs− 1

2jand Bs+ 1

2j− Bs, respectively, using disjoint sets of

random variables (Zt, t ∈ Dn).

Hence they are independent ⇒ (1) is satisfied. This completes the induction.

Now, we interpolate between the dyadic points. More precisely, let

f0(t) =

Z1 t = 1

0 t = 0

linear in between.

and for each n ≥ 1,

fn(t) =

2−n+12 Zt t ∈ Dn \Dn−1

0 t ∈ Dn−1

linear between consecutive points in Dn

f0, f1, f2, ...are continuous functions and, ∀n and s ∈ Dn

Bs :=

n∑

j=0

fj(s) =

∞∑

j=0

fj(s). (1.1)

5

We prove (1.1) by induction. (1.1) holds for n = 0. Suppose it holds for n − 1. Let

s ∈ Dn \Dn−1. Since for 0 ≤ j ≤ n− 1, the function fj is linear on [s− 12n, s+ 1

2n] we get

n−1∑

j=0

fj(s) =

n−1∑

j=1

fj(s− 12n) + fj(s+

12n)

2=

1

2(Bs− 1

2n+Bs+ 1

2n).

Since fn(s) =1

2n+12Zs, this gives (1.1).

Since Zsd= N(0, 1), we have for c > 1, and n large enough,

P [|Z1| ≥ c√n] ≤ e−

c2n2

(since∫∞x

e−u2

2 du ≤ 1xe−

x2

2 , Proof: exercise).

⇒ the series∞∑

n=0

P [∃s ∈ Dnwith|Zs| ≥ c√n] ≤

∞∑

n=0

∑

s∈Dn

P [|Zs| ≥ c√n] ≤

∞∑

n=0

(2n + 1)e−c2n2

converges if c >√2 log 2. Fix c >

√2 log 2.

Apply the Borel-Cantelli lemma :

∃N0(ω) < ∞ s.t. for n ≥ N0(ω), and s ∈ Dn we have |Zs| < c√n

⇒For n ≥ N0(ω), ||fn||∞ < c√n 1

2n2

⇒For P -a.a. ω, the sequence Bmt =

∑mn=0 fn(t) converges uniformly in t ∈ [0, 1] for

m → ∞⇒ Bt := lim

m→∞Bm

t is continuous in t.

We check that the increments of B have the right finite-dimensional distributions:

Assume 0 ≤ t1 < t2 < ... < tn ≤ 1. Then, we find 0 ≤ t1,k ≤ t2,k ≤ ... ≤ tn,k ≤ 1 with

ti,k ∈ D and limk→∞

ti,k = ti and since t → Bt is continuous for P -a.a. ω, Bti+1− Bti =

limk→∞

(Bti+1,k− Bti,k) P-a.s.

Since

limk→∞

E[Bti+1,k− Bti,k ] = 0 and

limk→∞

Cov(Bti+1,k− Bti,k , Btj+1,k

−Btj,k) = limk→∞

Ii=j(ti+1,k − ti,k) = Ii=j(ti+1 − ti),

the increments Bti+1−Bti , i = 1, 2, ..., n, are independent Gaussian random variables with

mean 0 and the variance ti+1 − ti, using Lemma 1.4 (s. below)

Lemma 1.4 (Xn)n∈N sequence of Gaussian random vectors and limn→∞

Xn = X P-a.s..

If b := limn→∞

E[Xn] and C := limn→∞

Cov(Xn) exist, then X is Gaussian with mean b and

Covariance Matrix C.

Proof: See [4], Prop. 12.15. -

6

Definition 1.5 A stochastic process (Bt)t≥0 on some prob. space (Ω,A, P ) is a Brownian

Motion if:

(i) B0 = 0 P-a.s.

(ii) For 0 ≤ t0 < t1 < ... < tn, the increments Bti − Bti−1are independent with law

N(0, ti − ti−1)

(iii) t → Bt(ω) is continuous for P-a.a. ω.

We obtain (Bt)t≥0 from a sequence of iid BMs (Bt)0≤t≤1 as follows:

Bt = B⌊t⌋t−⌊t⌋ +

⌊t⌋−1∑

i=0

Bi1 (t ≥ 0)

i.e. by glueing the paths (Bit)0≤t≤1 together.

Then (Bt)t≥0 is a BM. Proof: exercise

Definition 1.6 A stochastic process (Vt)t≥0 is a Gaussian process if for all t1 < t2 < ... <

tn, the vector (Vt1 , ..., Vtn) is a Gaussian random vector.

(Bt)t≥0 is a Gaussian process.

Proof: See exercises.

2 Some properties of Brownian Motion

The paths of BM are random fractals in the follows sense:

Lemma 2.1 (Scaling invariance)

Let (Bt)t≥0 be a BM and let a > 0. Then the process (Xt)t≥0 given by Xt =1aBa2t (t ≥ 0)

is also a BM.

7

Proof: Independence and stationarity of the increments and continuity of the paths per-

sist under the scaling. It remains to show that Xt − Xs = 1a(Ba2t − Ba2s) has the

law N(0, t − s). But Xt − Xs is a Gaussian RV with the expectation 0 and variance1a2E[(Ba2t −Ba2s)

2] = 1a2a2(t− s) = t− s. -

Example 2.2 (1) Let a < 0 < b and consider Ta,b with

Ta,b := inft ≥ 0 : Bt ∈ a, b

the first exit time of BM from the intervall (a, b).

Then, with Xt =1|a|Ba2t,

E[Ta,b] = a2E[inft ≥ 0 : Xt ∈ −1,b

|a|] = a2E[T−1, b|a|]

In particular E[Tb,b] = b2E[T−1,1] = const · b2.

(2) Ruin probabilities:

P [(Bt)t≥0 exits (a, b) at a] = P [(Xt)t≥0 exits (−1,b

|a|) at − 1]

depends only on the ratio b|a| .

Theorem 2.3 (Time inversion)

Let (Bt)t≥0 be a BM. Then the process (Xt)t≥0 given by

Xt =

0 t = 0

tB 1t

t > 0

is again a BM.

Proof: Recall that (Bt1 , ..., Btn), 0 ≤ t1 < t2 < ... ≤ tn are Gaussian random vectors and

are therefore characterized by their expectations and their Covariances

Cov(Bti , Btj ) = ti ∧ tj (2.1)

Proof of (2.1): Let ti < tj . Then

E[BtiBtj ] = E[Bti(Btj − Bti)] + E[B2ti] = 0 + ti = ti

8

(Xt)t≥0 is also a Gaussian process (check!) and the Gaussian random vectors (Xt1 , ..., Xtn)

have expectations E[Xti ] = 0, 1 ≤ i ≤ n. For t > 0, h ≥ 0, the Covariance of Xt and Xt+h

is given by

Cov(Xt, Xt+h) = Cov(tB 1t, (t+ h)B 1

t+h) = t(t + h)Cov(B 1

t, B 1

t+h) = t

Hence, the laws of all the finite vectors (Xt1 , ..., Xtn), 0 ≤ t1 < t2 < ... ≤ tn, are the same

as for BM. The paths t 7→ Xt are clearly continuous for all t > 0 (for P-a.a. ω).

For t = 0, we use the following two facts:

(1) Since Q is countable, (Xt, t ≥ 0, t ∈ Q) has the same law as (Bt, t ≥ 0, t ∈ Q)

⇒ limtց0,t∈Q

Xt = 0 P-a.s.

(2) Q ∩ (0,∞) is dense in (0,∞) and (Xt)t≥0 is continuous on (0,∞) (for P-a.a. ω) so

that 0 = limtց0,t∈Q

Xt = limt→0

Xt = 0 P-a.s.

-

Example 2.4 (Ornstein-Uhlenbeck-process)

Let (Bt)t≥0 be a BM, and set Xt = e−tBe2t , (t ∈ R).

Then Xtd= N(0, 1) ∀t

Proof: Xt is a Gaussian RV with E[Xt] = 0,Var(Xt) = e−2te2t = 1 -

Further (X−t)t∈R has the same law as (Xt)t∈R.

Proof: Set Xt = X−t, (t ∈ R) and

Bt =

tB 1

tt > 0

0 t = 0.

Then Xt = etBe−2t = etBe2te−2t = e−tBe2t .

Since (Bt)t≥0 is a BM, (Xt)t∈Rd= (Xt)t∈R. -

(Xt)t∈R is a Gaussian process with E[Xt] = 0, ∀t and

Cov(Xs, Xt) = E[Xs, Xt] = e−(s+t)E[Be2sBe2t ](2.1)= e−(s+t)e2(s∧t) = e−|t−s|.

Later we will see that ( 1√2Xt)t≥0 is a (weak) solution of the stochastic differential equation

dXt = dBt −Xtdt.

Corollary 2.5 (Law of large numbers)

(Bt)t≥0 BM. Then,

limt→∞

1

tBt = 0 P-a.s.

9

Remark:

limn→∞

1

nBn = 0 P-a.s.

since Bn =∑n

i=1(Bi −Bi−1) =∑n

i=1 Yi, (Yi)i≥1 iid with law N(0, 1).

Proof: Let

Xt =

tB 1

tt > 0

0 t = 0.

Then limt→∞

1tBt = lim

t→∞X 1

t= X0 = 0 P-a.s. -

Remark 2.6 The law of the iterated logarithm says that

lim supt→∞

Bt√2t log log t

= 1 P-a.s. (2.2)

lim inft→∞

Bt√2t log log t

= −1 P-a.s. (2.3)

In particular,

lim supt→∞

Bt√t= +∞, lim inf

t→∞

Bt√t= −∞ P-a.s.

Theorem 2.7 (Paley, Wiener, Zygmund)

Let (Bt)t≥0 be a BM. Then

P [ω : t 7→ Bt(ω)is nowhere differentiable] = 1

Proof: Assume that there is t0 ∈ [0, 1] such that t 7→ Bt(ω) is diff. in t0. Then, there is a

constant M < ∞ such that

sups∈[0,1]

∣∣∣∣Bt0+s − Bt0

s

∣∣∣∣ ≤ M (2.4)

If t0 ∈[k−12n

, k2n

]for some n > 2, k ≤ 2n, then we have for 1 ≤ j ≤ n,

∣∣∣∣Bk+j

2n− Bk+j−1

2n

∣∣∣∣ ≤ M(2j + 1)1

2n(2.5)

Proof of (2.5):

∣∣∣∣Bk+j

2n− Bk+j−1

2n

∣∣∣∣ ≤∣∣∣∣Bk+j

2n−Bt0

∣∣∣∣+∣∣∣∣Bt0 −

Bk+j−1

2n

∣∣∣∣

≤ Mj + 1

2n+M

j

2n

≤ M(2j + 1)1

2n.

10

Let An,k ⊆ C[0, 1] be the collection of functions satisfying (2.5) for j = 1, 2, 3.

Claim:

P [An,k] ≤ P [|B1| ≤7M√2n

]3 (2.6)

Proof of (2.6):

Bk+j

2n− Bk+j−1

2nd= N(0,

1

2n)

d=

1√2n

B1

andBk+j

2n− Bk+j−1

2n, j = 1, 2, 3 are independent. Further,

P

[|B1| ≤

7M√2n

]3≤(7M√2n

)3

.

Hence,

P

[2n⋃

k=1

An,k

]≤ 2n

(7M√2n

)3

=(7M)3√

2n

⇒∞∑

n=2

P

[2n⋃

k=1

An,k

]< ∞.

Therefore, using the Borel-Cantelli lemma,

P [(2.4) holds for some t0 ∈ [0, 1]] ≤ P

[2n⋃

k=1

An,k happens for infinitely n

]

= P

[ ∞⋂

m=2

∞⋃

n=m

2n⋃

k=1

An,k

]

= 0.

-

Corollary 2.8 For P-a.a. ω, the function t 7→ Bt(ω) is not of bounded variation on any

interval.

Recall that for g which is right-continuous on [a, b], we set

Vg[a, b] = sup

m∑

k=1

|g(tk)− g(tk−1)| : a ≤ t0 < t1 < ... < tm ≤ b,m ∈ N

.

Vg[a, b] is the variation of g on [a, b]. We say that g is of bounded variation (BV) on [a, b]

if Vg[a, b] < ∞.

Example:

If t 7→ g(t) is increasing on [a, b], g is BV on [a, b], and Vg[a, b] = g(b)− g(a).

Lemma 2.9 Assume g is BV on [a, b] and right-continuous ⇒ ∃g1, g2 : [a, b] → R, g1, g2

increasing and right-continuous such that

g = g1 − g2 (2.7)

11

Proof: See literature.

Proof of Corollary 2.8: A theorem by Lebesgue says that an increasing function g : [a, b] →R is differentiable for λ-a.a. s ∈ [a, b]. -

Remark 2.10 Let g be right-continuous and increasing on [a, b]. Then, g defines a

measure νg on [a, b] by

νg((a1, b1]) = g(b1)− g(a1), a ≤ a1 < b1 ≤ b.

If g is BV on [a, b], and f ∈ C[a, b] we can define

∫ b

a

f(s)dg(s) :=

∫ b

a

f(s)νg1(ds)−∫ b

a

f(s)νg2(ds)

(with g1, g2 from (2.7)). For a BM (Bt)t≥0, this procedure can not be applied - nevertheless,

we will be able to define∫ b

af(s)dBs.

Remark 2.11 A continuous function g which is BV on [0,1] has quadratic variation

0, i.e. for En ⊆ [0, 1], En = 0, t1, ..., tn, 1, (0 ≤ t1 < ... < tn ≤ 1) we have

∑

ti∈En

(g(ti+1)− g(ti))2 ≤ max

ti∈En

|g(ti+1)− g(ti)| ·∑

ti∈En

|g(ti+1)− g(ti)| −→n→∞

0

if s(En) −→n→∞

0, where s(En) := supti∈En|ti+1 − ti|. is the mesh of En.

Theorem 2.12 (Bt)t≥0 BM. Let (En) be a sequence of partitions with s(En) −→n→∞

0, En ⊆En+1 ⊆ En+2 ⊆ ... Then, ∀t > 0,

V nt :=

∑

ti∈En,

ti+1≤t

(Bti+1− Bti)

2 −→n→∞

t P-a.s. and in L2.

Proof:

(1) Convergence in L2:

E[V nt ] =

∑

ti∈En,

ti+1≤t

(ti+1 − ti) −→n→∞

t.

Using the independence of the increments,

Var(V nt ) =

∑

ti∈En,

ti+1≤t

Var((Bti+1− Bti)

2) =(∗)

∑

ti∈En,

ti+1≤t

2(ti+1 − ti)2

⇒ Var(V nt ) −→

n→∞0, see Remark 2.11

Proof of (*): Yd= N(0, σ2)

⇒ Var(Y 2) = E[Y 4]− E[Y 2]2 = 3σ4 − σ4 = 2σ4.

12

(2) We first show P-a.s. convergence along dyadic partitions En = 0, 12n, 22n, 32n, ..., 1.

For these partitions,

Var(V nt ) = 2 · 2n

(1

2n

)2

=2

2n

⇒∑∞

n=1Var(Vnt ) < ∞.

Using Lemma 2.13 below we conclude that V nt −→

n→∞t P-a.s..

(3) For general sequence of partitions, use (2) and an approximation argument. -

Lemma 2.13 Y1, Y2, ... RVs with∑∞

n=1Var(Yn) < ∞. Then, (Yn −E[Yn]) −→n→∞

0 P-a.s..

Proof: See exercises.

3 The Cameron-Martin Theorem and the Paley-Wiener

stochastic integral

We know several facts about paths of BM, for instance:

P [∃t0 ∈ [0, 1] s.t. t → Bt differentiable in t0] = 0

Do these properties remain true for (Bt + ct)0≤t≤1 or, more generally, for (Bt + h(t))0≤t≤1

where h ∈ C[0, 1]?

We denote by

H =

h ∈ C[0, 1] : there is f ∈ L2[0, 1] s.t. h(t) =

∫ t

0

f(s)ds, 0 ≤ t ≤ 1

the Cameron-Martin space (or Dirichlet space).

Given h ∈ H , f is uniquely determined as an element of L2[0, 1] and we write

h′ = f

Example:

h′ = f is true λ-a.s. (λ = Lebesgue measure on [0, 1]).

13

Recall that for two measures µ, ν on (Ω,A) we write µ ⊥ ν and say ”µ and ν are singular”

if there is A ∈ A with µ(A) = 0, ν(Ac) = 0.

We write ν ≪ µ and say ”ν is absolutely continuous w.r.t. µ” if, ∀A ∈ A, µ(A) = 0 ⇒ν(A) = 0.

We write ν ≈ µ and say ”ν and µ are equivalent” if ν ≪ µ and µ ≪ ν.

Let µ be Wiener measure on (C[0, 1],F) and µh be the law of (Bt + h(t))0≤t≤1.

Theorem 3.1 Assume h ∈ C[0, 1] and h(0) = 0.

(i) If h /∈ H then µh ⊥ µ.

(ii) If h ∈ H then µh ≈ µ.

For the proof, we will need the following quantity:

Qn(h) := 2n2n∑

j=1

(h

(j

2n

)− h

(j − 1

2n

))2

(n = 1, 2, ...) (3.1)

Lemma 3.2 Qn(h), n = 1, 2, ... is an increasing sequence and

h ∈ H ⇔ supn

Qn(h) < ∞.

Moreover, if h ∈ H, then

Qn(h) −→n→∞

∫ 1

0

h′(s)2ds =

∫ 1

0

f(s)2ds.

Proof: The general inequality (a+ b)2 ≤ 2a2 + 2b2 gives

[h

(j

2n

)− h

(j − 1

2n

)]2≤ 2

[h

(2j − 1

2n+1

)− h

(j − 1

2n

)]2+ 2

[h

(j

2n

)− h

(2j − 1

2n+1

)]2

Summing this inequality over j ∈ 1, 2, ..., n gives Qn(h) ≤ Qn+1(h) ⇒ Qn(h) is increas-

ing in n. For h ∈ H with h′ = f , we have, using Jensen’s inequality

Qn(h) = 2n2n∑

j=1

(∫ j

2n

j−12n

f(s)ds

)2

≤2n∑

j=1

∫ j

2n

j−12n

f(s)2ds =

∫ 1

0

f(s)2ds

Hence, h ∈ H ⇒ supnQn(h) < ∞.

Proof of ”⇒”: Let t ∼ U([0, 1]), then there exists a sequence of intervals In(t) = [an, bn] =[kn−12n

, kn2n

]s.t. t ∈ In(t), ∀n. Given I1(t), ...In(t), the interval In+1(t) is, with probability

12the left or the right half of In(t). Let Mn = Mn(t) = 2n(h(bn))− h(an)), then (Mn)n≥1

is a martingale w.r.t. σ(In(t) (on ([0, 1], B[0,1], λ)). Furthermore:

E[M2n ] = 22n

2n∑

k=1

(h

(k

2n

)− h

(k − 1

2n

))2

· 1

2n= Qn(h)

If supn Qn(h) < ∞, (Mn)n≥1 is a martingale which is bounded in L2, i.e. supn E[M2n] < ∞.

We prove later:

14

Lemma 3.3 Assume (Mn)n≥1 is a martingale on (Ω,A, P ) and (Mn)n≥1 is bounded in

L2. Then there exists a RV X s.t.

Mn −→n→∞

X P-a.s. and in L2

By Lemma 3.3, Mn → X a.s. and in L2. Let

g(s) :=

∫ s

0

X(t)dt

For j,m fixed, we have ∀n:

h

(j

2m

)=

∫ j

2m

0

Mn(t)dt −→n→∞

∫ j

2m

0

X(t)dt

⇒ h(

j2m

)= g

(j2m

)∀j,m and by continuity, g(s) = h(s) ∀s ∈ [0, 1] and h′(s) =

X(s) λ-a.s.

⇒ h ∈ H and

Qn(h) = E[M2n ] −→

n→∞E[X2] =

∫ 1

0

(h′(t))2dt.

-

Proof of Lemma 3.3: Mn is bounded in L1 and by the martingale convergence theorem

(Theorem 14.13 in Probability Theory lecture notes) Mn −→n→∞

X P-a.s. and X ∈ L1. We

have for m > n:

E[(Mm −Mn)2] = E[M2

m]− E[M2n] (3.2)

since

E[MmMn] = E[E[MmMn|An]] = E[MnE[Mm|An]] = E[M2n ]

Fatou’s Lemma implies from (3.2)

E[(X −Mn)2] ≤ E

[lim infm→∞

(Mm −Mn)2]≤ lim inf

m→∞E[M2

m]− E[M2n]

The last expression tends to 0 a.s. for n → ∞, since E[M2n ] is increasing in n:

E[M2n]

(3.2)=

n∑

k=2

E[(Mk −Mk−1)2] + E[M2

1 ].

-

Lemma 3.4 (The Paley-Wiener stochastic integral)

Let (Bt)t≥0 be a BM and h ∈ H. Then

ξn := 2n∑

j=1

2n(h

(j

2n

)− h

(j − 1

2n

))(B j

2n− B j−1

2n)

converges a.s. and in L2. The limit is denoted by∫ 1

0h′dB.

15

Proof: Recall from the construction of BM that

B 2j−12n

=1

2

(B 2j−2

2n+B 2j

2n

)+ σnZ 2j−1

2n

where σn = 2−(n+1)/2 and Zt are iid standard normal.

Therefore:

ξn − ξn−1 = 2nσn

2n−1∑

j=1

(2h

(2j − 1

2n

)− h

(2j − 2

2n

)− h

(2j

2n

))Z 2j−1

2n

which implies that (ξn)n≥1 is a martingale.

(E[ξn|An−1] = E[ξn − ξn−1|An−1] + E[ξn−1|An−1] = E[ξn−1|An−1])

E[ξ2n] = 22n2n∑

j=1

(h

(j

2n

)− h

(j − 1

2n

))2

E

[(B j

2n− B j−1

2n

)2]= Qn(h).

Hence, for h ∈ H , the convergence follows from Lemma 3.3. -

Proof of the Cameron-Martin Theorem: Let µn and µh,n denote the finite-dimensional

distributions on the set Dn. Then the Radon-Nikodym derivativedµh,n

dµnis the ratio of the

two Lebesgue densities. For x ∈ C[0, 1] and ∆jx = ∆(n)j x = x

(j2n

)− x

(j−12n

),

dµh,n

dµn(x) =

2n∏

j=1

exp

(−(∆jx−∆jh)

2

21−n

)exp

((∆jx)

2

21−n

)= exp(−Hn(x))

with Hn(x) = 2n−1∑n

j=1((∆jh)2 − 2∆jx∆jh).

By Theorem 14.5 in the Probability Theory lecture notes, exp(−Hn(x)) is a martingale

under µ, since it is non-negative it converges a.s. to a finite RV X . We show later:

µh(A) =

∫

A

Xdµ+ µh(A ∩ X = ∞)

for A ∈ F . This implies:

µ(X = 0) = 1 ⇒ µ ⊥ µh

µ(X > 0) = 1 ⇒ µ ≪ µh

We have Eµ[Hn] =∫Hn(x)µ(dx) =

12Qn(h) and Varµ(Hn) = 22n−2·4·

∑2n

j=1(∆jh)2Varµ(∆jx) =

Qn(h). By Chebyshev’s inequality, we get

Pµ

(Hn ≤ 1

4Qn(h)

)= Pµ

(1

2Qn(h)−Hn ≥ 1

4Qn(h)

)≤ Qn(h)(

14Qn(h)

)2 =16

Qn(h).

Now, if h /∈ H , then by Lemma 3.2, Hn → ∞ and x = 0 µ-a.s..

For the converse, suppose h ∈ H . By Lemma 3.4 (and the second part of Lemma 3.2),

H(x) −→n→∞

1

2||h′||22 −

∫ 1

0

h′dB.

Therefore x > 0 µ-a.s. and µ ≪ µh. Finally note that µh ≪ µ ⇔ µ ≪ µ−h. -

16

4 Brownian Motion as a continuous martingale

Definition 4.1 Consider a probability space (Ω,F , P ).

(i) A filtration is a sequence of σ-fields (Ft)t≥0 with Fs ⊆ Ft ⊆ F , ∀s < t.

(ii) A stochastic process (Xt)t≥0 on (Ω,F , P ) is adapted to (Ft)t≥0 ifXt is Ft-measurable,

∀t ≥ 0.

Suppose (Xt)t≥0 is a stochastic process on (Ω,F , P ). Then we can define a filtration

(Ft)t≥0 by taking Ft := σ(Xs, 0 ≤ s ≤ t), i.e. Ft is the σ-field generated by Xs, 0 ≤s ≤ t. Then, (Xt)t≥0 is adapted to (Ft)t≥0.

Definition 4.2 A real-valued stochastic process (Xt)t≥0 is a martingale with respect to

a filtration (Ft)t≥0 if it is adapted to (Ft)t≥0 and

E[|Xt|] < ∞ ∀t ≥ 0 (4.1)

and, for 0 ≤ s ≤ t

E[Xt|Fs] = Xs P-a.s. (4.2)

The process (Xt)t≥0 is a submartingale with respect to a filtration (Ft)t≥0 if it is adapted

to (Ft)t≥0, (4.1) holds and for 0 ≤ s ≤ t

E[Xt|Fs] ≥ Xs P-a.s. (4.3)

and it is a supermartingale with respect to a filtration (Ft)t≥0 if it is adapted to (Ft)t≥0,

(4.1) holds and, for 0 ≤ s ≤ t

E[Xt|Fs] ≤ Xs P-a.s. (4.4)

Remark 4.3 If (Xt)t≥0 is a martingale with respect to (Ft)t≥0, |Xt| is in general not a

martingale but a submartingale. More generally, if (Xt)t≥0 is a martingale with respect

to (Ft)t≥0 and f : R → R is a convex function such that E[|f(Xt)|] < ∞, ∀t ≥ 0, then

f(Xt)t≥0 is a submartingale with respect to (Ft)t≥0.

Proof:

E[f(Xt)|Fs]Jensen≥ f(E[Xt|Fs]) = f(Xs) P-a.s.

hence (4.2) holds. -

Remark 4.4 Let (Bt)t≥0 be a BM and Ft = σ(Xs, 0 ≤ s ≤ t). Then (Bt)t≥0 is a

martingale with respect to (Ft)t≥0.

Proof:

E[Bt|Fs] = E[Bt − Bs|Fs] + E[Bs|Fs] = 0 +Bs P-a.s.

(See exercise 3.3 (ii): Bt − Bs is independent of Fs, hence E[Bt − Bs|Fs] = 0.) -

17

Definition 4.5 Let (Ω,F , P ) be a probability space with a filtration (Ft)t≥0. A RV T

with values in [0,∞] is a stopping time with respect to (Ft)t≥0 if

T ≤ t ∈ Ft, ∀t ≥ 0.

Example 4.6 Let (Bt)t≥0 be a BM and Ft = σ(Xs, 0 ≤ s ≤ t), t ≥ 0.

(i) Let y 6= 0 and T = inft ≥ 0 : Bt = y. Then, T is a stopping time with respect to

(Ft)t≥0.

Proof:

T ≤ t =

∞⋂

n=1

⋃

s:s∈Q∩(0,t)Bs ∈ U(y,

1

n) ∈ Ft, where

U(y,1

n) := z ∈ R : |z − y| < 1

n

(ii) Let I = (a, b), 0 < a < b and T = inft ≥ 0 : Bt ∈ I. Then, T is not a stopping

time because T ≤ t /∈ Ft.

Proof: See [4].

Definition 4.7 Assume T is a stopping time with respect to the filtration (Ft)t≥0. Define

FT := A ∈ F : A ∩ T ≤ t ∈ Ft, ∀t ≥ 0.

FT is called the σ-field of events observable until time T .

A martingale (Xt)t≥0 is a continuous martingale if P [t → Xt(ω) is continuous] = 1.

Theorem 4.8 (Optional stopping)

Suppose (Xt)t≥0 is a continuous martingale and 0 ≤ S ≤ T stopping times. If the process

(Xt∧T )t≥0 is dominated by an integrable RV X, i.e. |Xt∧T | ≤ X, ∀t ≥ 0 a.s. and E[|X|] <∞, then E[XT |FS] = XS P-a.s.

Proof: This can be derived from the result for martingales in discrete time (see Theorem

14.9 in the Probability Theory lecture notes for the case S = 0). See [4] for details.

Theorem 4.9 (Wald’s Lemma for BM)

Let (Bt)t≥0 be a BM and T a stopping time such that either

(a) E[T ] < ∞ or

(b) (Bt∧T )t≥0 is dominated by an integrable RV.

Then, we have E[BT ] = 0.

Remark 4.10 One does need a condition on T , as the following example shows:

Let T = inft : Bt = 1. (Then T < ∞ P-a.s., see exercise 5.2). Clearly, E[BT ] = 1. We

conclude from Theorem 4.9 that E[T ] = ∞ and that (Bt∧T )t≥0 is not dominated by an

integrable RV.

18

Proof of Theorem 4.9: We show that (a) implies (b). Suppose E[T ] < ∞, and define

Mk = max0≤t≤1 |Bt+k − Bk| and M =∑⌊T ⌋

k=1Mk. Then

E[M ] = E[

⌊T ⌋∑

k=1

Mk]

=∞∑

k=1

E[IT>k−1Mk]

=

∞∑

k=1

P [T > k − 1]E[Mk]

≤ E[M0]E[T + 1]

But E[M0] = E[max0≤t≤1 |Bt|] < ∞ (Proof: exercise).

If (b) is satisfied, we can apply the optional stopping theorem (Theorem 4.8) with S = 0,

giving E[XT |F0] = X0, P-a.s. which yields that E[BT ] = 0. -

5 Stochastic integrals with respect to Brownian Motion

Let (Bt)t≥0 be a BM on some probability space (Ω,F , P ) and Ft the completion of

σ(Bs, s ≤ t) (see Theorem 1.32 in the Probability Theory lecture notes). Then, (Bt)t≥0

is adapted to (Ft)t≥0 .

Definition 5.1 A process Xt(ω) : t ≥ 0, ω ∈ Ω is progressively measurable if for

each t ≥ 0 the mapping X : [0, t] × Ω → R is measurable with respect to the σ-field

B[0,t] ⊗Ft.

Lemma 5.2 Any process (Xt)t≥0 which is adapted and either right-continuous or left-

continuous is also progressively measurable.

Proof: Assume (Xt)t≥0 is right-continuous. Fix t > 0. For n ∈ N, 0 ≤ s ≤ t define

X(n)0 (ω) = X0(ω)

and

X(n)s (ω) = X (k+1)t

2n(ω) for

kt

2n< s ≤ (k + 1)t

2nk = 0, 1, 2, ..., 2n − 1.

The mapping (s, ω) → X(n)s (ω) is B[0,t] ⊗Ft-measurable.

By right-continuity, we have limn→∞

X(n)s (ω) = Xs(ω) for all s ∈ [0, t] and ω ∈ Ω, hence the

limit mapping (s, ω) → Xs(ω) is also B[0,t] ⊗ Ft-measurable.

The left-continuous case is analogous.

-

19

We construct the integral by starting with ”simple” progressively measurable processes

Ht(ω) and then proceeding to more complicated ones. Consider first step processes

Ht(ω) : t ≥ 0, ω ∈ Ω of the form

Ht(ω) =k∑

j=1

Aj(ω)I(tj ,tj+1](t) for 0 ≤ t1 < ... < tk+1

where Aj is Ftj -measurable, 1 ≤ j ≤ k.

We define the integral as

∫ ∞

0

HsdBs :=

k∑

j=1

Aj(Btj+1− Btj )

Now letH be a progressively measurable process satisfying E[∫∞

0H2

sds]< ∞. Suppose H

can be approximated by a sequence of progressively measurable step processes H(n), n ≥ 1,

then we define ∫ ∞

0

HsdBs := limn→∞

∫ ∞

0

H(n)s dBs . (5.1)

More precisely, let ||H||22 := E[∫∞

0H2

sds]. We will show that

(1) Every progressively measurable H satisfying E[∫∞

0H2

sds]< ∞ can be approximated

in the || · ||2 - norm by progressively measurable step processes.

(2) For each approximating sequence, the limit in (5.1) exists in the L2-sense.

(3) This limit does not depend on the approximating sequence of step processes.

We start with (1):

Lemma 5.3 For every progressively measurable process Hs(ω) : s ≥ 0, ω ∈ Ω satisfying

E[∫∞

0H2

sds]< ∞ there exists a sequence (H(n))n∈N of progressively measurable step

processes such that limn→∞

||H(n) −H||2 = 0.

Proof: We approximate the progressively measurable process successively by

• a bounded progressively measurable process

• a bounded, almost surely continuous progressively measurable process

• a progressively measurable step process.

Let H = Hs(ω), s ≥ 0, ω ∈ Ω be a progressively measurable process with ||H||2 < ∞.

First define

H(n)s (ω) =

Hs(ω) s ≤ n

0 otherwise.

20

Clearly, limn→∞

||H(n) −H||2 = 0.

Second, approximate any progressively measurable process H on a finite interval by trun-

cating its values, i.e. define H(n) by

H(n)s = (Hs(ω) ∧ n) ∨ (−n).

Clearly, H(n) is progressively measurable and ||H(n) −H||2 → 0.

Now we approximate any uniformly bounded progressively measurable H by bounded,

almost surely continuous progressively measurable processes.

Let h = 1nand

H(n)s (ω) =

1

h

∫ s

s−h

Ht(ω)dt

(where we set Hs(ω) = H0(ω) for s < 0). H(n) is again progressively measurable (since

we only take averages over the past). H(n) is almost surely continuous. Further, ∀ω ∈ Ω

and almost every (with respect to Lebesgue measure) s ∈ [0, t],

limh→0

1

h

∫ s

s−h

Ht(ω)dt = Hs(ω).

Since H is uniformly bounded, we obtain that limn→∞

||H(n) −H||2 = 0.

Finally, a bounded, amost surely continuous, progressively measurable process H can be

approximated by a sequence of progressively measurable step processes H(n) by taking

H(n)s = H

(j

n, ω

)for

j

n≤ s ≤ j + 1

n.

The process H(n) are again progressively measurable and one easily sees that

limn→∞

||H(n) −H||2 = 0.

This completes the proof of Lemma 5.3. -

Lemma 5.4 Let H be a progressively measurable step process and E[∫∞

0H2

sds]< ∞ .

Then

E

[(∫ ∞

0

HsdBs

)2]= E

[∫ ∞

0

H2sds

]

Proof: Let H =∑k

i=1AiI(ai,ai+1] be a progressively measurable step process. Then

E

[(∫ ∞

0

HsdBs

)2]= E

[k∑

i,j=1

AiAj(Bai+1−Bai)(Baj+1

−Baj )

]

=

k∑

i=1

E[A2i (Bai+1

− Bai)2] + 2

k∑

i=1

k∑

j=i+1

E[AiAj(Bai+1−Bai) ·E[(Baj+1

− Baj )|Faj ]]

=

k∑

i=1

E[A2i ]E[(Bai+1

−Bai)2] = E

[∫ ∞

0

H2sds

]

-

21

Corollary 5.5 Suppose (H(n))n∈N is a sequence of progressively measurable step processes

such that

E[

∫ ∞

0

(H(n)s −H(m)

s )2ds] −→n,m→∞

0.

Then

E[(

∫ ∞

0

(H(n)s −H(m)

s )dBs)2] −→

n,m→∞0.

Proof: Because the difference of two progressively measurable step processes is again a

progressively measurable step process, Lemma 5.4 can be applied to H(n) − H(m) and

yields the claim. -

We showed (1). The following theorem addresses (2) and (3).

Theorem 5.6 Suppose (H(n))n∈N is a sequence of progressively measurable step processes

and H a progressively measurable process such that

limn→∞

E[

∫ ∞

0

(H(n)s −Hs)

2ds] = 0

Then

limn→∞

∫ ∞

0

H(n)s dBs =:

∫ ∞

0

HsdBs

exists as a limit in the L2-sense and does not depend on the choice of (H(n))n∈N. Moreover,

we have

E[

∫ ∞

0

(H(n)s −H(m)

s )2dBs] −→n,m→∞

0 (5.2)

Proof: By the triangle inequality (H(n))n∈N satisfies the assumptions of Corollary 5.5 and

hence(∫∞

0H

(n)s dBs

)n∈N

is a Cauchy sequence in L2. Since L2 is complete, the limit exists

and does not depend on the choice of the approximating sequence. Finally, (5.2) follows

from Lemma 5.4, applied to H(n), taking the limit as n → ∞. -

This completes the construction of the stochastic integral∫∞0

H2sdBs for progressively

measurable processes H with E[∫∞

0H2

sds]< ∞.

Remark 5.7 If the sequence of step processes in Theorem 5.6 is chosen such that

∞∑

n=1

E

[∫ ∞

0

(H(n)s −Hs)

2ds

]< ∞,

then by (5.2) we get∞∑

n=1

E

[(∫ ∞

0

(H(n)s −Hs)dBs

)2]< ∞

and therefore, almost surely,

∞∑

n=1

(∫ ∞

0

H(n)s Bs −

∫ ∞

0

HsdBs

)2

< ∞.

22

This implies that, almost surely,

limn→∞

∫ ∞

0

H(n)s dBs =

∫ ∞

0

HsdBs.

We now want to describe the stochastic integral as a process in time. We will see that it

will be a continuous martingale.

Definition 5.8 Suppose H = Hs(ω) : s ≥ 0, ω ∈ Ω is progressively measurable with

E[∫∞

0H2

sds]< ∞. Define the progressively measurable process H t

s(ω), s ≥ 0, ω ∈ Ω

H ts(ω) := Hs(ω)Is≤t.

Then, the stochastic integral of H up to time t is defined∫ t

0

HsdBs :=

∫ ∞

0

H tsdBs.

Remark 5.9 We have seen already in Lemma 3.4 that for g ∈ L2[0, 1] we can define∫ 1

0g(s)dBs. Provided that both integrals exist, the Paley-Wiener integral from Lemma

3.4 agrees with the stochastic integral just defined.

Proof: See exercises.

Definition 5.10 A stochastic process (Xt)t≥0 is a modification of a stochastic process

(Yt)t≥0 if, for every t ≥ 0, we have P [Xt = Yt] = 1.

Theorem 5.11 Assume that (Hs(ω))s≥0 is progressively measurable and E[∫ t

0Hs(ω)

2ds]<

∞, ∀t ≥ 0. Then there exists a modification (Mt)t≥0 of(∫ t

0HsdBs

)t≥0

such that

P [t 7→ Mt(ω) is continuous] = 1.

Further, (Mt)t≥0 is a martingale and hence

E

[∫ t

0

HsdBs

]= 0 ∀t ≥ 0.

Proof: Fix t0 ∈ N and let H(n) be a sequence of progressively measurable step processes

such that

||H(n) −H t0 ||2 −→n→∞

0

⇒ E

[(∫ ∞

0

(H(n)s −H t0

s )dBs

)2]−→n→∞

0.

For s ≤ t the random variable∫ s

0H

(n)u dBu is Fs-measurable and E[

∫ t

sH

(n)u dBu|Fs] = 0

(proof: exercise!) which implies that the process(∫ t

0H

(n)u dBu

)0≤t≤t0

is a martingale, ∀n.By Doob’s maximal inequality, see below, for p = 2,

E

[sup

0≤t≤t0

(∫ t

0

H(n)s dBs −

∫ t

0

H(m)s dBs

)2]≤ 4E

[(∫ t0

0

(H(n)s −H(m)

s )dBs

)2]−→n→∞

0.

23

This implies that M(n)t :=

∫ t

0H

(n)s dBs, 0 ≤ t ≤ t0, n = 0, 1, 2, . . . defines a Cauchy

sequence in the space of continuous functions on [0, t0] (equipped with the supremum

norm). We denote the limit of this Cauchy sequence by (Mt)0≤t≤t0 . Hence, the process

(Mt)0≤t≤t0 is almost surely a uniform limit of continuous processes and therefore almost

surely continuous. Due to Theorem 5.6,

P

[Mt =

∫ t

0

HsdBs

]= 1.

For fixed t ∈ [0, t0], the random variable∫ t

0HsdBs is the limit (in L2) of

∫ t

0H

(n)s dBs, hence

it is Ft-measurable, and∫ t0t

HsdBs has conditional expectation E[∫ t0t

HsdBs|Ft] = 0.

Therefore,∫ t

0HsdBs is a conditional expectation of Mt0 , given Ft, i.e.

Mt = E

[∫ t0

0

HsdBs|Ft

].

Therefore, (Mt)0≤t≤t0 is a martingale, as a process of successive predictions, (see (14.4) in

Probability Theory lecture notes). -

Doob’s maximal inequality

Suppose (Xt)t≥0 is a continuous martingale and p > 1. Then, for any t ≥ 0

E[ sup0≤s≤t

|Xs|p] ≤(

p

p− 1

)p

E[|Xt|p].

Proof: See literature.

6 Ito’s formula and examples

Let f ∈ C1(R) (f continuously differentiable) and x : [0,∞) → R x continuous and BV

on [0, t]. Then

f(x(t))− f(x(0)) =

∫ t

0

f ′(x(s))dx(s)

Example:

x(s) = s, ∀s ≥ 0. Then f(t)− f(0) =∫ t

0f ′(s)ds. Ito’s formula gives an analogue for the

case when x is replaced by a BM Bt. The crucial difference is that the second derivative

of f is needed.

Theorem 6.1 (Ito’s formula I)

Let f : R → R be twice continuously differentiable such that E[∫ t

0(f ′(Bs))

2ds] < ∞ for

some t > 0, where (Bt)t≥0 is a BM. Then, almost surely for all s ∈ [0, t]

f(Bs)− f(B0) =

∫ s

0

f ′(Bu)dBu +1

2

∫ s

0

f ′′(Bu)du.

24

Example 6.2 Let f(x) = x2. We have E[∫ s

0B2

udu] =∫ s

0udu < ∞, ∀s > 0. Hence

B2s = 2

∫ s

0

BudBu + s ⇒ B2s − s = 2

∫ s

0

BudBu.

We conclude from Theorem 5.11 that (B2s − s)s≥0 is a martingale.

Example 6.3 Let f(x) = x3. We have E[∫ s

0B4

udu] =∫ s

0E[B4

u]du < ∞, ∀s > 0. Hence

B3s = 3

∫ s

0

B2udBu + 3

∫ s

0

Budu .

Let

Ms = B3s − 3

∫ s

0

Budu, s ≥ 0.

We conclude from Theorem 5.11 that (Ms)s≥0 is a martingale.

To prove Theorem. 6.1, we will need the following.

Theorem 6.4 Let f : R → R is continuous, t > 0, and 0 = t(n)1 < ... < t

(n)n = t are

partitions such that their mesh

max1≤i≤n−1

|t(n)i+1 − t(n)i | → 0.

Thenn−1∑

j=1

f(Bt(n)j

)(Bt(n)j+1

−Bt(n)j

)2 →∫ t

0

f(Bs)ds

Proof: For f ≡ 1, see Theorem 2.12. For general f , see Theorem 7.12 in Morter/Peres.

Proof of Theorem 6.1: We write w(δ,M) for the modulus of continuity of f ′′ on [−M,M ]:

w(δ,M) = sups,t∈[−M,M ],|s−t|<δ

|f ′′(s)− f ′′(t)|

Using Taylor’s formula, for any x, y ∈ [−M,M ], x < y with |x− y| < δ,

|f(y)− f(x)− f ′(x)(y − x) +1

2f ′′(x)(y − x)2| ≤ w(δ,M)(y − x)2.

Take a sequence 0 = t(n)1 < ... < t

(n)n = t. We write 0 = t1 < ... < tn = t for simplicity.

With

δB := max1≤i≤n−1

|Bti+1−Bti |

and

MB := max0≤s≤t

|Bs|,

we get

|n−1∑

i=1

(f(Bti+1)− f(Bti))−

n∑

i=1

f ′(Bti)(Bti+1−Bti)−

n−1∑

i=1

1

2f ′′(Bti)(Bti+1 −Bti)

2|

25

≤ w(δB,MB)

n−1∑

i=1

(Bti+1−Bti)

2

Now,∑n−1

i=1 f(Bti+1)− f(Bti) = f(Bt)− f(B0), and there is a sequence of partitions with

mesh going to 0 s.t.

n−1∑

i=1

f ′(Bti)(Bti+1− Bti) →

∫ t

0

f ′(Bs)dBs P − a.s.

n−1∑

i=1

f ′′(Bti)(Bti+1− Bti)

2 →∫ t

0

f ′′(Bs)ds P − a.s.

n−1∑

i=1

(Bti+1− Bti)

2 → t P − a.s.

By continuity of the Brownian path, w(δB,MB) converges almost surely to 0.

This proves Ito’s formula for fixed t, or indeed almost surely for all s ∈ Q∩ [0, t]. Since all

the terms in Ito’s formula are continuous almost surely, we get the result simultaneously

for all s ∈ [0, t]. -

We next state Ito’s formula for functions f which can depend also on time.

Theorem 6.5 (Ito’s formula II)

Let f : R×R → R, (x, t) 7→ f(x, t) be twice continuously differentiable in the x-coordinate

and once continuously differentiable in the t-coordinate. Assume that

E[∫ t

0(∂xf(Bs, s))

2ds] < ∞ for some t > 0. Then, almost surely for all s ∈ [0, t]:

f(Bs, s)− f(B0, 0) =

∫ s

0

∂xf(Bu, u)dBu +

∫ s

0

dtf(Bu, u)du+1

2

∫ s

0

∂xxf(Bu, u)du

Proof: See Morter/Peres.

Example 6.6 Fix α > 0.

f(Bt, t) = eαBt− 12α2t = Mt (M0 = 1)

Then

f(Bs, s)− 1 =

∫ s

0

αMudBu +

∫ s

0

(−1

2α2)Mudu+

1

2

∫ s

0

α2Mudu

⇒ Ms −M0 =

∫ s

0

αMudBu.

We conclude from Theorem 5.11 that (Mt)t≥0 is a martingale (details: exercise). M solves

the stochastic differential equation

”dMs = αMsdBs, M0 = 1”

which can be written in integral form

Ms = 1 +

∫ s

0

MudBu, s ≥ 0.

26

Definition 6.7 (Bt)t≥0 = (B(1)t , ..., B

(d)t )t≥0 is a d-dim. BM if (B

(1)t )t≥0, ..., (B

(d)t )t≥0 are

iid one-dimensional BMs. For Hs = (H(1)s , ..., H

(d)s ) we write

∫ t

0

HsdBs =d∑

i=1

∫ t

0

H(i)s dB(i)

s .

Theorem 6.8 (Multidimensional Ito formula)

Let (Bt)t≥0 be a d-dim. BM and f : Rd+1 → R be such that the partial derivatives ∂if

and ∂jkf exist for all 1 ≤ i ≤ d+ 1, 1 ≤ j, k ≤ d and are continuous. If for some t > 0

E

[∫ t

0

|∇xf(Bs, s)|2ds]< ∞

where

∇xf = (∂1f, ..., ∂df)

then, almost surely, for all 0 ≤ s ≤ t

f(Bs, s)− f(B0, 0) =

∫ s

0

∇xf(Bu, u)dBu +

∫ s

0

∂d+1f(Bu, u)du+1

2

∫ s

0

xf(Bu, u)du

where xf =∑d

j=1 ∂jjf .

7 Pathwise stochastic integration with respect to

continuous semimartingales

We saw that for a continuous process H with E[∫ t

0H2

sds] < ∞,∫ t

0HsdBs can be defined

as an almost sure limit, see Remark 5.7.

Can we replace (Bt)t≥0 with another continuous martingale?

Definition 7.1 Let En = 0 = t(n)1 < ... < t

(n)n be a sequence of partitions with

s(En) = sup |t(n)i+1 − t(n)i | → 0.

Then, the function Xt has continuous quadratic variation along the sequence En if

〈X〉t = limn→∞

∑

ti∈En

ti+1≤t

(Xtj+1−Xtj )

2 (7.1)

exists P-a.s.

Remark 7.2 〈X〉t is increasing and continuous, hence it defines a measure ν on (R,B)given by ν((a, b]) = 〈X〉b − 〈X〉a.In particular, if f : R → R is continuous,

∫ t

0f(s)d〈X〉s is well-defined.

In analogy to Theorem 6.4, we have

27

Theorem 7.3 Assume that Xt has continuous quadratic variation 〈X〉t and f : R → R

is continuous. Then, for t > 0,

∑

ti∈En

ti+1≤t

f(Xti)(Xtj+1−Xtj )

2 →∫ t

0

f(Xs)d〈X〉s

Proof: Clear for f ≡ 1 due to (7.1). Rest see literature.

Note that the assumption “Xt has continuous quadratic variation” can only be satisfied

for continuous processes.

Theorem 7.4 (Ito’s formula for (deterministic) functions with continuous quadratic

variation)

Assume that the function Xt has continuous quadratic variation 〈X〉t and let f : R → R

be twice continuous differentiable. Then,

f(Xt)− f(X0) =

∫ t

0

f ′(Xu)dXu +1

2

∫ t

0

f ′′(Xu)d〈X〉u (7.2)

where ∫ t

0

f ′(Xu)dXu = limn→∞

∑

ti∈En

ti+1≤t

f ′(Xti)(Xti+1−Xti) .

Sketch of proof: As in the proof of Theorem 6.1, we have

|∑

ti∈En

ti+1≤t

f(Xti+1)− f(Xti)−

∑

ti∈En

ti+1≤t

f ′(Xti)(Xti+1−Xti)−

∑

ti∈En

ti+1≤t

1

2f ′(Xti)(Xti+1

−Xti)2|

≤ w(δX ,MX)∑

ti∈En

ti+1≤t

(Xti+1−Xti)

2

where

δX = maxti∈En

ti+1≤t

|Xti+1−Xti |

and

MX = max0≤s≤t

|Xs|

and

w(δ,M) = sups,t∈[−M,M ]

|s−t|<δ

|f ′′(s)− f ′′(t)|.

Now, ∑

ti∈En

ti+1≤t

(f(Xti+1)− f(Xti)) → f(Xt)− f(X0)

and ∑

ti∈En

ti+1≤t

f ′′(Xti)(Xti+1−Xti)

2 →∫ t

0

f ′′(Xs)d〈X〉s

28

and ∑

ti∈En

ti+1≤t

(Xti+1−Xti)

2 → 〈X〉t.

Moreover, w(δX ,MX) −→δ→0

0. Hence,

∑

ti∈En

ti+1≤t

f ′(Xti)(Xti+1−Xti)

has to converge as well and (7.2) holds.

Remark 7.5 (1) Theorem 7.4 gives a ”pathwise” version of Ito’s formula, ”without prob-

ability”.

(2) If X is BV on [0, t], 〈X〉t = 0 and (7.2) becomes

f(Xt)− f(X0) =

∫ t

0

f ′(Xu)dXu.

Short notation:

df(X) = f ′(X)dX “classical differential”

If X has continuous quadratic variation 〈X〉 and 〈X〉t 6= 0, then

df(X) = f ′(X)dX +1

2f ′′(X)d〈X〉 “Ito differential”

We have defined∫ t

0f ′(Xu)dXu for all continuous Xt with continuous quadratic variation

〈X〉t. Which stochastic processes Xt have a.s. continuous quadratic variation 〈X〉t?

Lemma 7.6 (i) Assume Xt has continuous quadratic variation 〈X〉t. Then, if f is

continuously differentable, f(Xt) has continuous quadratic variation

〈f(X)〉t =∫ t

0

f ′(Xs)2d〈X〉s

(ii) Let Xt = Mt+At, t ≥ 0, where Mt has quadratic variation 〈M〉t and At has quadratic

variation 〈A〉t = 0, then Xt has quadratic variation 〈X〉t and 〈X〉t = 〈M〉t.

(iii) Let f ∈ C1(R) and assume Xt has continuous quadratic variation 〈X〉t. Then,

Mt =∫ t

0f(Xs)dXs has continuous quadratic variation

〈M〉t =∫ t

0

f(Xs)2d〈X〉s.

(iv) Let f ∈ C1(R2) and assume Xt has continuous quadratic variation 〈X〉t. Then,

g(t) = f(Xt, t) has continuous quadratic variation

〈g〉t =∫ t

0

(∂f

∂x(Xs, s)

)2

d〈X〉s.

29

Examples:

1. (Bt)t≥0 BM, α > 0, Zt = eαBt , t ≥ 0. Then, Lemma 7.4 (i) implies that 〈Z〉t =∫ t

0α2e2αBsds =

∫ t

0α2Z2

sds P-a.s. Note that 〈Z〉t is random (whereas 〈B〉t = t, ∀t P-a.s.).2. (Bt)t≥0 BM, α > 0,Mt = eαBt− 1

2α2t, t ≥ 0. Then, we know thatMt = 1+

∫ t

0αMsdBs, t ≥

0. Hence, Lemma 7.4 (iv) implies that 〈M〉t =∫ t

0α2M2

s ds P-a.s. Note that 〈M〉t is ran-dom (whereas 〈B〉t = t, ∀t P-a.s.).

Proof of Lemma 7.4:

(i) iX := Xti+1−Xti . Then

f(Xti+1)− f(Xti) = f(Xti)iX +Ri

and

|Ri| ≤ sups,u∈[ti,ti+1]

|f ′(Xs)− f ′(Xu)| · iX

Hence,

∑

ti∈En

ti+1≤t

(f(Xti+1)− f(Xti))

2 =∑

ti∈En

ti+1≤t

f ′(Xti)2(iX)2 +

∑

ti∈En

ti+1≤t

R2i + 2

∑

ti∈En

ti+1≤t

f ′(Xti)iX ·Ri

But ∑

ti∈En

ti+1≤t

f ′(Xti)2(iX)2 →

∫ t

0

f ′(Xs)2d〈X〉s

due to Theorem 7.3.

Rest: Exercise.

(ii) iM = Mti+1−Mti , iA = Ati+1

− Ati . Then

(iX)2 = (iM)2 + (iA)2 + 2iMiA.

Hence

∑

i:ti∈En

ti+1≤t

(iX)2 =∑

i:ti∈En

ti+1≤t

(iM)2 +∑

i:ti∈En

ti+1≤t

(iA)2 + 2

∑

i:ti∈En

ti+1≤t

iMiA → 〈M〉t,

since ∑

i:ti∈En

ti+1≤t

(iA)2 → 0

and ∑

i:ti∈En

ti+1≤t

iMiA → 0

(details: exercise).

30

(iii) Due to (7.2), taking g such that g′ = f ,

Mt = g(Xt)− g(X0)−1

2

∫ t

0

g′′(Xs)d〈X〉s.

But∫ t

0g′′(Xs)d〈X〉s is continuous and BV as a function of t. Using (ii) and (i),

〈M〉t = 〈g(X)〉t =∫ t

0

g′(Xs)2d〈X〉s =

∫ t

0

f(Xs)2d〈X〉s

(iv) See [5], Remark 1.3.22.

-

Theorem 7.7 If (Xt)t≥0 is a continuous martingale with X0 = 0 and E[X2t ] < ∞,

∀t there exists a unique process 〈X〉t with 〈X〉0 = 0 which is continuous, adapted and

increasing, such X2t − 〈X〉t is a martingale. Moreover, Xt has quadratic variation 〈X〉t.

Proof: If (Bt)t≥0 is a BM and (Ht)t≥0 progressively measurable and Xt =∫ t

0HsdBs, then

〈X〉t =∫ t

0H2

sds P-a.s. (see Lemma 5.4 and Lemma 7.4(i).)

General case: See literature. -

〈X〉t is called quadratic variation or bracket or compensator of (Xt).

The first part of Theorem 7.6 has a discrete time analogue.

A stochastic process (Yn) is previsible w.r.t. a filtration (An) if Yn is An−1-measurable

∀n.

Theorem 7.8 (Doob decomposition)

Suppose (Xn)n=0,1,2,... is a martingale w.r.t. the filtration (An) and E[X2n] < ∞, ∀n.

Then, there is a unique previsible increasing process (An) with A0 = 0 such that (X2n −

An)n=0,1,2,... is a martingale w.r.t. (An).

Proof: Let A0 = 0 and define, for n ≥ 1, An = An−1 +E[X2n|An−1]−X2

n−1. Clearly, An is

An−1-measurable ∀n. Since (X2n)n=0,1,2,... is a submartingale w.r.t. An, An is increasing.

Further, E[X2n −An|An−1] = E[X2

n−1 −An−1|An−1] = X2n−1 −An−1 ⇒ (X2

n −An)n=0,1,2,...

is a martingale w.r.t. (An).

To prove the uniqueness, assume that (An) and (Bn) both fulfill the requirements. Then

An − Bn is a previsible martingale starting at 0, and we infer that An − Bn = E[An −Bn|An−1], hence, by induction, An = Bn ∀n. -

Definition 7.9 A continuous semimartingale w.r.t. a filtration (Ft)t≥0 is a process

(Xt)t≥0 which is adapted to (Ft)t≥0 and which has a decomposition Xt = X0 + Mt +

At, ∀t ≥ 0 P-a.s. where (Mt)t≥0 is a continuous martingale and (At)t≥0 is a continuous

adapted process which is BV (on each interval [0, t]).

If E[M2t ] < ∞, ∀t, we define for f ∈ C1(R)

∫ t

0

f(s)dXs :=

∫ t

0

f(s)dMs +

∫ t

0

f(s)dAs

31

8 Cross-variation and Ito’s product rule

Definition 8.1 The cross-variation 〈X, Y 〉 is given by

〈X, Y 〉t = limn→∞

∑

ti∈En

ti+1≤t

(Xti+ −Xti)(Yti+ − Yti)

(provided that the limit exists).

Clearly, 〈X,X〉 = 〈X〉.

Lemma 8.2 The following statements are equivalent

(i) 〈X, Y 〉 exists and t → 〈X, Y 〉t is continuous.

(ii) 〈X + Y 〉 exists and t → 〈X + Y 〉t is continuous.

If (i) and (ii) hold, then

〈X, Y 〉 = 1

2(〈X + Y 〉 − 〈X〉 − 〈Y 〉) (8.1)

In particular, in this case∫ t

0g(s)d〈X, Y 〉s is well-defined for g ∈ C[0,∞).

Proof:

(Xti+1−Xti)(Yti+1

− Yti) =

1

2

(((Xti+1

+ Yti+1)− (Xti + Yti)

)2 − (Xti+1−Xti)

2 − (Yti+1− Yti)

2).

-

Example 8.3 Let (Xt)t≥0 and (Yt)t≥0 be independent BMs.

Claim:

For P-a.a. ω

〈X(ω), Y (ω)〉t = 0 ∀t

Proof:

〈X, Y 〉 = 1

2(〈X + Y 〉 − 〈X〉 − 〈Y 〉)

and we have 〈X〉t = t, ∀t, and 〈Y 〉t = t, ∀t P-a.s. Suffices to show 〈X + Y 〉 = 2t ∀t P-a.s.Zt :=

1√2(Xt + Yt), t ≥ 0 is again a BM (Proof: exercise). Therefore, 〈Z〉t = t ∀t P-a.s.

implying that 〈X + Y 〉 = 2t, ∀t P-a.s. -

Lemma 8.4 X, 〈X〉 continuous as before, f, g ∈ C1(R),

Yt =

∫ t

0

f(Xs)dXs, Zt =

∫ t

0

g(Xs)dXs

Then

〈Y, Z〉t =∫ t

0

f(Xs)g(Xs)d〈X〉s (8.2)

32

Proof: Define F and G by F ′ = f , G′ = g, F (0) = G(0) = 0.

Then, Yt = F (Xt) + At where

At = −F (X0)−1

2

∫ t

0

F ′′(Xs)d〈X〉s

and Zt = G(Xt)− Bt where

Bt = −G(X0)−1

2

∫ t

0

G′′(Xs)d〈X〉s.

At, Bt are continuous with quadratic variation 0. Hence,

〈Y, Z〉t =1

2(〈Y + Z〉t − 〈Y 〉t − 〈Z〉t)

=1

2

(∫ t

0

(f(Xs) + g(Xs))2d〈X〉s −

∫ t

0

f(Xs)2d〈X〉s −

∫ t

0

g(Xs)2d〈X〉s

)

=

∫ t

0

f(Xs)g(Xs)d〈X〉s

-

Theorem 8.5 (Ito’s formula in d dimensions)

X = (X(1), ..., X(d)) : [0,∞) → Rd where X(i), 1 ≤ i ≤ d are continuous with continuous

quadratic variation 〈X(i)〉t and continuous cross-variations 〈X(i), X(k)〉t, 1 ≤ i, k ≤ d.

Let f ∈ C2(Rd). Then,

f(Xt)− f(X0) =

∫ t

0

(∇f, dXs) +1

2

∫ t

0

d∑

i,k=1

∂2f

∂xi∂xk(Xs)d〈X(i), X(k)〉s

where ∫ t

0

(∇f, dXs) =d∑

k=1

∫ t

0

∂f

∂xi

(Xs)dX(i)s .

Proof: Analogous to the proof of Theorem 6.1: Taylor formula forf(Xti+1)− f(Xti). See

literature. -

Note that Theorem 6.8 follows, using Example 8.3.

Corollary 8.6 (Ito’s product rule)

Assume that X, Y, 〈X〉, 〈Y 〉 and 〈X, Y 〉 are continuous. Then, ∀t > 0

Xt · Yt = X0 · Y0 +

∫ t

0

YsdXs +

∫ t

0

XsdYs + 〈X, Y 〉t (8.3)

Short notation:

d(X · Y ) = Y dX +XdY + d〈X, Y 〉 (8.4)

Proof: Apply Theorem 8.5 with d = 2, f(x, y) = x · y. -

33

Example 8.7 (Ornstein-Uhlenbeck process)

Let α > 0 and (Bt)t≥0 a BM, x0 ∈ R. Then, we say that

Xt = e−αtx0 + e−αt

∫ t

0

eαsdBs, t ≥ 0 (8.5)

is an Ornstein-Uhlenbeck process with parameter α and starting point x0.

Claim:

(Xt)t≥0 solves the stochastic differential equationdXt = dBt − αXtdt

X0 = x0

i.e.

Xt = Bt −

∫ t

0αXsds

X0 = x0

Proof:

Xt = e−αtx0 + e−αt

∫ t

0

eαsdBs

dXt = −αx0e−αtdt− α(e−αt

∫ t

0

eαsdBs)dt+ e−αteαtdBt

= −αXtdt+ dBt ,

where we applied Ito’s product rule to e−αt∫ t

0eαsdBs. -

Claim: If X0d= N(0, 1

2α), X0 independent of (Bt)t≥0, then

Xt = e−αtX0 + e−αt

∫ t

0

eαsdBs

is a Gaussian process with E[Xt] = 0, ∀t and Cov(Xs, Xt) =12αe−α|t−s|. Hence, for any t,

E[X2t ] =

12α

⇒ Xtd= N(0, 1

2α).

In particular, for the process Zt = e−tBe2t , t ≥ 0, in Example 2.4, ( 1√2Zt)t≥0 is an Ornstein-

Uhlenbeck process with α = 1 and X0 = N(0, 12).

Proof: Clearly, E[Xt] = 0, ∀t. Assume s ≤ t. Then,

E[XsXt] = E[X20 ]e

−αte−αs + 0 + 0 + e−α(s+t)E[

∫ t

0

eαudBu

∫ s

0

eαvdBv]

For any martingale (Mt)t≥0, E[MtMs] = E[M2s ] for s ≤ t.(Proof: exercise).

Hence, applying this to the martingaleMt =∫ t

0eαudBu,

E[XsXt] =1

2αe−α(t+s) + e−α(t+s)E[(

∫ s

0

eαvdBv)2]

=1

2αe−α(t+s) + e−α(t+s)

∫ s

0

e2αvdv

=1

2αe−α(t+s) + e−α(t+s) 1

2α(e2αs − 1)

=1

2αe−α(t−s) .

-

34

9 Stochastic Differential Equations

In this chapter we want to study stochastic differential equations (SDE) of the form:

X0 = ξ

dXt = σ(t, Xt) · dBt + b(t, Xt)dt (9.1)

(Xt)t≥0 = (X(1)t , X

(2)t , ..., X

(n)t )t≥0 is an unknown R-valued process,

(Bt)t≥0 = (B(1)t , B

(2)t , ..., B

(m)t )t≥0 is an m-dimensional Brownian motion and b(t, x) and

σ(t, x) are measurable functions of (t, x) ∈ R+ ×Rn. The drift vector b(t, x) is Rn-valued

and the dispersion matrix σ(t, x) is an n × m-matrix valued. Further ξ is a random

variable with values in Rn which is independent of (Bt)t≥0.

Definition 9.1 We say the SDE (9.1) has the strong solution (Xt)t≥0 if the following

conditions hold:

(i) (Xt)t≥0 is adapted to the filtration (F ξt )t≥0 where F ξ

t is the completion of σ(ξ, Bs :

s ≤ t) for t ≥ 0.

(ii) (Xt)t≥0 satisfies the following integral equation:

X(i)t = ξ(i) +

m∑

j=1

∫ t

0

σijdB(j)s +

∫ t

0

bi(s,Xs)ds (9.2)

for t ≥ 0, i = 1, 2, ...n.

Remark 9.2 (1) Processes which fulfill the integral equations in (9.2) and are defined on

a possibly enlarged probability space, but do not have to be adapted to the filtration

(F ξt )t≥0, are called weak solutions.

(2) The second property of a solution of a SDE includes the requirement that the integrals

in (9.2) are well–defined.

Example 9.3 For α, β ∈ R consider the SDE

X0 = 1

dXt = αXtdBt + βXtdt

for a one-dimensional BM (Bt)t≥0 . This SDE has the (unique) strong solution Xt =

exp(αBt + (β − α2

2)t).

Proof: exercise

Remark

The SDE from Example 9.3 is used for the Black-Scholes model.

35

Remark 9.4 In the following Theorem we use

||x||2 =n∑

i=1

(xi)2 for x = (x1, ..., xn) ∈ Rn ,

||σ||2 =n∑

i=1

m∑

j=1

(σij)2 for σ = (σij)ij ∈ Rn×m .

Theorem 9.5 (Existence and uniqueness)

Assume that b : R+ × Rn and σ : R+ × Rn → Rn×m are measurable and satisfy the

Lipschitz-condition

||σ(t, x)− σ(t, y)||+ ||b(t, x)− b(t, y)|| ≤ K · ||x− y|| (9.3)

as well as the growth condition

||σ(t, x)||2 + ||b(t, x)||2 ≤ K2 · (1 + ||x||2) (9.4)

for a constant K > 0, for all t ≥ 0 and x, y ∈ Rn. Let ξ be a random variable with values

in Rn which is independent of the m-dimensional BM (Bt)t≥0 such that E[||ξ||2] < ∞.

Then the SDE (9.1) has a unique, continuous, strong solution (Xt)t≥0 such that

E

[∫ T

0

||Xt||2dt]< ∞ ∀T > 0 (9.5)

Remark 9.6 Uniqueness in Theorem 9.5 means that for two continuous solutions (Xt)t≥0,

(X ′t)t≥0 of the SDE (9.1) which fulfill the properties of Theorem 9.5 we have

P (Xt = X ′t ∀t ≥ 0) = 1.

Example 9.7 To illustrate that we need some conditions like (9.3) and (9.4) let us look

at the following examples from ODEs:

dXt

dt= (Xt)

2, X0 = 1

corresponding to b(x) = x2 (which does not satisfy (9.4)) has the unique solution

Xt =1

1− tfor 0 ≤ t < 1.

Another example:dXt

dt= 3(Xt)

2/3, X0 = 0

where b(x) = 3x2/3 does not satisfy (9.3) at x = 0 and

Xt =

0, for t ≤ a,

(t− a)3, for t > a.

are solutions for all a > 0.

36

Sketch of the proof of Theorem 9.5:

Uniqueness:

Uses the following lemma

Lemma 9.8 Gronwall inequality

Let f : [0, T ] → R be integrable and A ∈ R, C > 0 such that

f(t) ≤ A+ C ·∫ t

0

f(s)ds ∀t ∈ [0, T ],

Then

f(t) ≤ A+ eCt ∀t ∈ [0, T ] .

Proof: see literature.

Consider two continuous solutions (Xt)t≥0, (X′t)t≥0 of the SDE (9.1) which fulfill condition

(9.5)

Xt −X ′t =

∫ t

0

(σ(s,Xs)− σ(s,X ′s)) · dBs +

∫ t

0

(b(s,Xs)− b(s,X ′s))ds

Therefore (using (a + b)2 ≤ 2a2 + 2b2),

||Xt −X ′t|| ≤ 2||

∫ t

0

(σ(s,Xs)− σ(s,X ′s)) · dBs||2 + 2||

∫ t

0

(b(s,Xs)− b(s,X ′s))ds||2

A short calculation for the first term (see Theorem 5.6 for n = m = 1) and an application

of the Cauchy-Schwarz inequality shows

E[||Xt−X ′t||2] ≤ 2

∫ t

0

E[||(σ(s,Xs)−σ(s,X ′s))||2]dBs+2t

∫ t

0

E[||(b(s,Xs)−b(s,X ′s))||2]ds

Therefore we have for f(t) := E[||Xt − X ′t||2] and C := 2(T + 1)K2 (for T > 0) due to

condition (9.3)

f(t) ≤ C ·∫ t

0

f(s)ds for 0 ≤ t ≤ T

and the Gronwall inequality implies f ≡ 0. We can conclude due to the continuity of

(Xt)t≥0, (X′t)t≥0 that we have

P (Xt = X ′t ∀t ≥ 0) = 1.

(using the fact that Q+ is countable and dense in R+).

Existence: We define for n ∈ N

X(n+1)t := ξ +

∫ t

0

σ(s,X(n)s ) · dBs +

∫ t

0

b(s,X(n)s )ds for t ≥ 0

inductively where X0t := ξ. Using the growth condition we can show inductively that for

T > 0 ∫ T

0

E[||X(n)s ||2]ds < ∞ ,

i.e. the stochastic integral is well–defined in every step of the recursion. One can show

that (X(n))n∈N0 converges a.s. uniformly on [0, T ] for every T > 0 and the limit solves

the SDE (9.1) and has the properties of Theorem 9.5 (more details can be found in the

literature). -

37

10 Girsanov transforms

Goal:

Construction of a stochastic process (Xt)t≥0 with

dXt = b(Xt, t)dt+ dBt

X0 = x0

where (Bt)t≥0 is a BM, i.e.

Xt = x0 +

∫ t

0

b(Xs, s)ds+Bt (10.1)

Interpretation:

Deterministic process Xt withddtXt = b(Xt, t) with additional ”noise” (Bt)t≥0.

Possible strategies:

1) Construction of a ”strong” solution:

For a given BM (Bt)t≥0 on (Ω,A, P ), solve (10.1).

Example 10.1 (Ornstein-Uhlenbeck process)

dXt = dBt − αXtdt

X0 = x0

has the strong solution

Xt = e−αt(x0 +

∫ t

0

eαsdBs) , (10.2)

see Example 8.8.

Drawback: a strong solution does not always exist.

2) Construction of a ”weak” solution:

Find a BM (Bt)t≥0 and a process (Xt)t≥0 on some probability space (Ω,A, P ) such

that (10.1) holds, i.e. find (Xt)t≥0 such that

Bt = Xt − x0 −∫ t

0

b(Xs, s)ds

is a BM.

General Girsanov transformation

(Ω,A, P ) probability space, (Ft)t≥0 filtration, P probability measure on (Ω,A). Assume

that for all t, P |Ft≪ P |Ft

.

Then, there are Radon-Nikodym derivatives

Zt :=dP

dP|Ft

=dP |Ft

dP |Ft

(10.3)

38

and (Zt)t≥0 is a martingale with respect to (Ft)t≥0 and P (see Probability Theory lecture

notes, 14.1.3.)

We always assume that (Zt)t≥0 has continuous paths and that

inft ≥ 0 : Zt(ω) = 0 = ∞ P -a.s.

Definition 10.2 (Mt)t≥0 is a local martingale (up to ∞) if there is a sequence of

stopping times T1 ≤ T2 ≤ ... such that

i) supn

Tn = ∞ P -a.s.

ii) (Mt∧Tn)t≥0 is a martingale, ∀n.

Each martingale is a local martingale but there are local martingales which are not mar-

tingales.

We will use the following important fact:

If M is a continuous local martingale and (Yt)t≥0 a continuous adapted process, then∫ t

0YsdMs, t ≥ 0 is again a continuous local martingale. See the literature, for instance

[5], Proposition 1.4.29.

Lemma 10.3 In the above setup, with Zt :=dPdP

|Ft, the following holds.

(i) For s ≤ t and a function gt which is Ft-measurable and bounded, we have

E[gt|Fs] =1

ZsE[gtZt|Fs] P -a.s.

where E denotes expectation with respect to P .

(ii) For M = (Mt)t≥0 continuous and adapted, the following two statements are equiva-

lent:

a) (Mt)t≥0 is a local martingale with respect to P .

b) (MtZt)t≥0 is a local martingale with respect to P .

Proof:

(i) Assume gs is Fs-measurable and bounded.

E[gsgt] = E[gsgtZt]

= E[gsE[gtZt|Fs]]

= E

[gsZsE[gtZt|Fs]

1

Zs

]

= E

[gs

1

ZsE[gtZt|Fs]

]

⇒ E[gt|Fs] =1

ZsE[gtZt|Fs]

39

(ii) Assume that (MtZt)t≥0 is a martingale with respect to P . Then

E[Mt|Fs] =(i)

1

ZsE[MtZt|Fs] =

1

ZsMsZs = Ms, P-a.s.

⇒ (Mt)t≥0 is a martingale with respect to P , hence b) ⇒ a) for martingales.

Rest of the proof: Exercise. -

Theorem 10.4 Assume that (Mt) is a continuous local martingale with respect to P and

Zt is defined as in (10.3). Then,

Mt = Mt −∫ t

0

1

Zsd〈M,Z〉s

is a continuous local martingale with respect to P .

Short notation:

dM = dM +1

Zd〈M,Z〉

Proof: Due to Lemma 10.1, it suffices to show that (MtZt)t≥0 is a continuous local mar-

tingale with respect to P .

Let At =∫ t

01Zsd〈M,Z〉s. Then,

MtZt = Zt(Mt−At)Ito’s product rule

= Z0M0+

∫ t

0

(Ms−As)dZs−∫ t

0

ZsdAs+ 〈Z,M〉t (10.4)

The definition of A implies that

∫ t

0

ZsdAs = 〈Z,M〉t .

Due to (10.4), MtZt − M0Z0 is a stochastic integral, hence again a continuous local

martingale with respect to P . -

Remark 10.5

logZt = logZ0 +

∫ t

0

1

ZsdZs −

∫ t

0

1

Z2s

d〈Z〉s

Z continuous local martingale ⇒ Yt =∫ t

01ZsdZs is a continuous local martingale with

〈Y 〉 =∫ t

01Z2sd〈Z〉s, hence

logZt = logZ0 + Yt −1

2〈Y 〉t ⇒ Zt = Z0e

Yt− 12〈Y 〉t

and Z solves dZ = ZdY .

Using 〈M,Z〉t =∫ t

0Zsd〈M,Y 〉s, see Lemma 10.6 below we get from Theorem 10.3 that

dM = dM + d〈M,Y 〉 (10.5)

40

Lemma 10.6 (Generalization of Lemma 8.4)

Asume that M and Y are local martingales with continuous quadratic variation 〈M〉 and〈Y 〉 and f, g ∈ C2(R2). Let

Vt =

∫ t

0

f(Ms, s)dMs, Rt =

∫ t

0

g(Ys, s)dYs .

Then,

〈V,R〉t =∫ t

0

f(Ms, s)g(Ys, s)d〈M,Y 〉s.

Proof: See literature. For M = Y , we recover Lemma 8.4.

Now, consider

Zt =

∫ t

0

ZsdYs, Mt =

∫ t

0

1dMs

Lemma 10.6 yields

〈M,Z〉t =∫ t

0

Zsd〈M,Y 〉s.

Girsanov transform on Wiener space

Let (Xt)0≤t≤1 be a BM with respect to P .

Theorem 10.7 (Girsanov)

Assume (bt)0≤t≤1 is progressively measurable and adapted and E[∫ 1

0b2sds] < ∞. Let Z1 :=

e∫ 10 bsdXs− 1

2

∫ 10 b2sds and assume

E[Z1] = 1 (10.6)

Define P by dPdP

|F1 = Z1.

Then, Bt = Xt −∫ t

0bsds, 0 ≤ t ≤ 1 is a BM with respect to P .

For the proof, we will need

Theorem 10.8 (Levy’s characterization of BM)

Assume (Bt) is a continuous local martingale with quadratic variation 〈B〉t = t. Then,

(Bt)t≥0 is a BM.

Proof: We use characteristic functions:

A probability measure µ on R is characterized by its Fourier transform:

µ(u) =

∫eiuxµ(dx) .

(Example: µ = N(0, σ2), µ(u) = e−12σ2u2

).

We show that Bt − Bs is independent of Fs, with law N(0, t − s). Ito’s formula for

f(x) = eiux (verify by separating real and imaginary parts) yields

eiuXt − eiuXs =

∫ t

s

iueiuXrdXr +1

2

∫ t

s

−u2eiuXrdr

41

Divide by eiuXs , take conditional expectation w.r.t Fs

⇒ E[eiu(Xt−Xs)|Fs]− 1 = E

[∫ t

s

iueiu(Xr−Xs)dXr |Fs

]− 1

2E

[∫ t

s

u2eiu(Xr−Xs)dr|Fs

]

Take A ∈ Fs. Then,

E[eiu(Xt−Xs)IA]− P [A] = −1

2u2

∫ t

s

E[eiu(Xr−Xs)IA]dr

Let g(t) := E[eiu(Xt−Xs)IA]. Then, g(t) − g(s) = −12u2

t∫s

g(r)dr which implies g(t) =

g(s) exp(−12u2(t − s)). Hence E[eiu(Xt−Xs)IA] = P [A]e−

12u2(t−s) = P [A]µ(u) where µ =

N(0, t− s), ∀A ∈ Fs. This implies that Xt −Xs is independent of Fs with law N(0, t −s). -

Proof of Thm. 10.7: We have dPdP

|Ft= Zt = e

∫ t

0bsdXs− 1

2

∫ t

0b2sds hence Zt = eYt− 1

2〈Y 〉t with

Yt =∫ t

0bsdXs. Due to (10.4), dXt = dBt + d〈X,

∫ ·0bsdXs〉t where (Bt)t≥0 is a continuous

local martingale w.r.t. P .

Since 〈X,∫ ·0bsdXs〉t =

∫ t

0bsds (Lemma 8.4) we have dXt = dBt + btdt. Hence, P -a.s.

〈Bt〉 = t, Xt = Bt +∫ t

0bsds. ⇒ P -a.s. 〈B〉t = t, ∀t since P ≪ P .

Hence, Levy’s characterization of BM implies that (Bt)t≥0 is a BM w.r.t. P . -

Remark 10.9 Thm. 10.5 also applies to bt = b(Xu, u ≤ t) and yields a weak solution

of dXt = dBt − btdt, i.e. bt can depend on the whole ”history” up to time t. (10.5) and

(10.6) are only weak regularity assumptions.

42

Acknowledgements: We thank Michael Kochler and Yaroslav Yevmenenko-Shul’ts for

many corrections.

References

[1] Ioannis Karatzas and Steven Shreve: Brownian motion and stochastic cal-

culus, Springer, 2007.

[2] Achim Klenke: Probability Theory, Springer, 2008.

[3] Thomas Liggett: Continuous time Markov processes: An introduction,

American Mathematical Society, 2010.

[4] Peter Morters and Yuval Peres: Brownian motion, Cambridge University

Press, 2010.

[5] Michael Rockner: Introduction to stochastic analysis. Available online at

http://www.math.uni-bielefeld.de/∼roeckner/teaching1112/stochana.pdf

43