f (x) = xk . Then f (x) = kxk’1 and f (x) = k(k ’ 1)xk’2 . We then have

t t

Wtk k k’1 k’2

1

k(k ’ 1)Ws d W

= W0 + kWs dWs + s

2

0 0

t t

k(k ’ 1)

k’1 k’2

= kWs dWs + Ws ds.

2

0 0

t

When k = 3, this says Wt3 ’ 3 0 Ws ds is a stochastic integral with respect to a Brownian

motion, and hence a martingale.

For a semimartingale Xt = Mt +At we set X t = M t . Given two semimartingales

X, Y , we de¬ne

X, Y t = 1 [ X + Y t ’ X t ’ Y t ].

2

The following is known as Ito™s product formula. It may also be viewed as an

integration by parts formula.

Proposition 13.2. If Xt and Yt are semimartingales,

t t

Xt Yt = X0 Y0 + Xs dYs + Ys dXs + X, Y t .

0 0

Proof. Applying Ito™s formula with f (x) = x2 to Xt + Yt , we obtain

t

2 2

(Xt + Yt ) = (X0 + Y0 ) + 2 (Xs + Ys )(dXs + dYs ) + X + Y t .

0

Applying Ito™s formula with f (x) = x2 to X and to Y , then

t

2 2

Xt = X0 +2 Xs dXs + X t

0

and

t

Yt2 Y02

= +2 Ys dYs + Y t .

0

Then some algebra and the fact that

Xt Yt = 2 [(Xt + Yt )2 ’ Xt ’ Yt2 ]

2

1

yields the formula.

1 d

There is a multidimensional version of Ito™s formula: if Xt = (Xt , . . . , Xt ) is a

vector, each component of which is a semimartingale, and f ∈ C 2 , then

d d

t t

‚2f

‚f i i j

1

f (Xt ) ’ f (X0 ) = (Xs )dXs + 2 (Xs )d X , X s .

2

‚xi ‚xi

0 0

i=1 i,j=1

The following application of Ito™s formula, known as L´vy™s theorem, is important.

e

61

Theorem 13.3. Suppose Mt is a continuous martingale with M = t. Then Mt is a

t

Brownian motion.

Before proving this, recall from undergraduate probability that the moment generating

function of a r.v. X is de¬ned by mX (a) = E eaX and that if two random variables have

the same moment generating function, they have the same law. This is also true if we

replace a by iu. In this case we have •X (u) = E eiuX and •X is called the characteristic

function of X. The reason for looking at the characteristic function is that •X always

exists, whereas mX (a) might be in¬nite. The one special case we will need is that if X is

2

a normal r.v. with mean 0 and variance t, then •X (u) = e’u t/2 . This follows from the

formula for mX (a) with a replaced by iu (this can be justi¬ed rigorously).

Proof. We will prove that Mt is a N (0, t); for the remainder of the proof see Note 1.

We apply Ito™s formula with f (x) = eiux . Then

t t

iuMt iuMs

(’u2 )eiuMs d M s .

1

e =1+ iue dMs + 2

0 0

Taking expectations and using M s = s and the fact that a stochastic integral is a

martingale, hence has 0 expectation, we have

t

u2

iuMt

eiuMs ds.

=1’

Ee

2 0

Let J(t) = E eiuMt . The equation can be rewritten

t

u2

J(t) = 1 ’ J(s)ds.

2 0

So J (t) = ’ 1 u2 J(t) with J(0) = 1. The solution to this elementary ODE is J(t) =

2

’u2 t/2

e . Since

2

E eiuMt = e’u t/2 ,

then by our remarks above the law of Mt must be that of a N (0, t), which shows that Mt

is a mean 0 variance t normal r.v.

Note 1. If A ∈ Fs and we do the same argument with Mt replaced by Ms+t ’ Ms , we have

t t

iu(Ms+t ’Ms ) iu(Ms+r ’Ms )

(’u2 )eiu(Ms+r ’Ms ) d M r .

1

e =1+ iue dMr + 2

0 0

Multiply this by 1A and take expectations. Since a stochastic integral is a martingale, the

stochastic integral term again has expectation 0. If we let K(t) = E [eiu(Mt+s ’Mt ) ; A], we

1

now arrive at K (t) = ’ 2 u2 K(t) with K(0) = P(A), so

2

K(t) = P(A)e’u t/2

.

62

Therefore

E eiu(Mt+s ’Ms ) ; A = E eiu(Mt+s ’Ms ) P(A). (13.2)

If f is a nice function and f is its Fourier transform, replace u in the above by ’u, multiply

by f (u), and integrate over u. (To do the integral, we approximate the integral by a Riemann

sum and then take limits.) We then have

E [f (Ms+t ’ Ms ); A] = E [f ((Ms+t ’ Ms )]P(A).

By taking limits we have this for f = 1B , so

P(Ms+t ’ Ms ∈ B, A) = P(Ms+t ’ Ms ∈ B)P(A).

This implies that Ms+t ’ Ms is independent of Fs .

Note Var (Mt ’ Ms ) = t ’ s; take A = „¦ in (13.2).

63

14. The Girsanov theorem.

Suppose P is a probability and

dXt = dWt + µ(Xt )dt,

where Wt is a Brownian motion. This is short hand for

t

Xt = X0 + Wt + µ(Xs )ds. (14.1)

0

Let

t t

µ(Xs )2 ds/2 .

Mt = exp ’ µ(Xs )dWs ’ (14.2)

0 0

Then as we have seen before, by Ito™s formula, Mt is a martingale. This calculation is

reviewed in Note 1. We also observe that M0 = 1.

Now let us de¬ne a new probability by setting

Q(A) = E [Mt ; A] (14.3)

if A ∈ Ft . We had better be sure this Q is well de¬ned. If A ∈ Fs ‚ Ft , then E [Mt ; A] =

E [Ms ; A] because Mt is a martingale. We also check that Q(„¦) = E [Mt ; „¦] = E Mt . This

is equal to E M0 = 1, since Mt is a martingale.

What the Girsanov theorem says is

Theorem 14.1. Under Q, Xt is a Brownian motion.

Under P, Wt is a Brownian motion and Xt is not. Under Q, the process Wt is no

longer a Brownian motion.

In order for a process Xt to be a Brownian motion, we need at a minimum that Xt

is mean zero and variance t. To de¬ne mean and variance, we need a probability. Therefore

a process might be a Brownian motion with respect to one probability and not another.

Most of the other parts of the de¬nition of being a Brownian motion also depend on the

probability.

Similarly, to be a martingale, we need conditional expectations, and the conditional

expectation of a random variable depends on what probability is being used.

There is a more general version of the Girsanov theorem.

Theorem 14.2. If Xt is a martingale under P, then under Q the process Xt ’ Dt is a

martingale where

t

1

Dt = d X, M s .

Ms

0

64

X is the same under both P and Q.

t

Let us see how Theorem 14.1 can be used. Let St be the stock price, and suppose

dSt = σSt dWt + mSt dt.

(So in the above formulation, µ(x) = m for all x.) De¬ne

2

/2σ 2 )t

Mt = e(’m/σ)(Wt )’(m .

Then from (13.1) Mt is a martingale and

t

m

’

Mt = 1 + Ms dWs .

σ

0

Let Xt = Wt . Then

t t

m m

’ Ms ds = ’

X, M = Ms ds.

t

σ σ

0 0

Therefore

t t

1 m

=’ ds = ’(m/σ)t.

d X, M s

Ms σ

0 0

De¬ne Q by (14.3). By Theorem 14.2, under Q the process Wt = Wt + (m/σ)t is a

martingale. Hence

dSt = σSt (dWt + (m/σ)dt) = σSt dWt ,

or

t

St = S0 + σSs dWs

0

is a martingale. So we have found a probability under which the asset price is a martingale.

This means that Q is the risk-neutral probability, which we have been calling P.

Let us give another example of the use of the Girsanov theorem. Suppose Xt =

Wt + µt, where µ is a constant. We want to compute the probability that Xt exceeds the

level a by time t0 .

We ¬rst need the probability that a Brownian motion crosses a level a by time t0 .

If At = sups¤t Wt , (note we are not looking at |Wt |), we have

d

P(At > a, c ¤ Wt ¤ d) = •(t, a, x), (14.4)

c

where 2

√ 1 e’x /2t x≥a

2πt

•(t, a, x) = 2

√ 1 e’(2a’x) /2t x < a.

2πt

65

This is called the re¬‚ection principle, and the name is due to the derivation, given in Note

2. Sometimes one says

P(Wt = x, At > a) = P(Wt = 2a ’ x), x < a,