3The stochastic integral
III Stochastic Calculus and Applications
3.7 Girsanov’s theorem
Girsanov’s theorem tells us what happens to our (semi)-martingales when we
change the measure of our space. We first look at a simple example when we
perform a shift.
Example.
Let
X ∼ N
(0
, C
) be an
n
-dimensional centered Gaussian with
positive definite covariance
C
= (
C
ij
)
n
i,j=1
. Put
M
=
C
−1
. Then for any
function f , we have
Ef(X) =
det
M
2π
1/2
Z
R
n
f(x)e
−
1
2
(x,Mx)
dx.
Now fix an a ∈ R
n
. The distribution of X + a then satisfies
Ef(X + a) =
det
M
2π
1/2
Z
R
n
f(x)e
−
1
2
(x−a,M(x−a))
dx = E[Zf (X)],
where
Z = Z(x) = e
−
1
2
(a,Ma)+(x,Ma)
.
Thus, if P denotes the distribution of X, then the measure Q with
dQ
dP
= Z
is that of N(a, C) vector.
Example.
We can extend the above example to Brownian motion. Let
B
be a
Brownian motion with
B
0
= 0, and
h
: [0
, ∞
)
→ R
a deterministic function. We
then want to understand the distribution of B
t
+ h.
Fix a finite sequence of times 0 =
t
0
< t
1
< ··· < t
n
. Then we know that
(
B
t
i
)
n
i=1
is a centered Gaussian random variable. Thus, if
f
(
B
) =
f
(
B
t
1
, . . . , B
t
n
)
is a function, then
E(f(B)) = c ·
Z
R
n
f(x)e
−
1
2
P
n
i=1
(x
i
−x
i−1
)
2
t
i
−t
i−1
dx
1
···dx
n
.
Thus, after a shift, we get
E(f(B + h)) = E(Zf(B)),
Z = exp
−
1
2
n
X
i=1
(h
t
i
− h
t
i−1
)
2
t
i
− t
i−1
+
n
X
i=1
(h
t
i
− h
t
i−1
)(B
t
i
− B
t
i−1
)
t
i
− t
i−1
!
.
In general, we are interested in what happens when we change the measure
by an exponential:
Definition
(Stochastic exponential)
.
Let
M
be a continuous local martingale.
Then the stochastic exponential (or Dol´eans–Dade exponential) of M is
E(M )
t
= e
M
t
−
1
2
hMi
t
The point of introducing that quadratic variation term is
Proposition.
Let
M
be a continuous local martingale with
M
0
= 0. Then
E(M ) = Z satisfies
dZ
t
= Z
t
dM,
i.e.
Z
t
= 1 +
Z
t
0
Z
s
dM
s
.
In particular,
E
(
M
) is a continuous local martingale. Moreover, if
hMi
is
uniformly bounded, then E(M) is a uniformly integrable martingale.
There is a more general condition for the final property, namely Novikov’s
condition, but we will not go into that.
Proof. By Itˆo’s formula with X = M −
1
2
hMi, we have
dZ
t
= Z
t
d
M
t
−
1
2
dhMi
t
+
1
2
Z
t
dhMi
t
= Z
t
dM
t
.
Since
M
is a continuous local martingale, so is
R
Z
s
d
M
s
. So
Z
is a continuous
local martingale.
Now suppose hMi
∞
≤ b < ∞. Then
P
sup
t≥0
M
t
≥ a
= P
sup
t≥0
M
t
≥ a, hM i
∞
≤ b
≤ e
−a
2
/2b
,
where the final equality is an exercise on the third example sheet, which is true
for general continuous local martingales. So we get
E
exp
sup
t
M
t
=
Z
∞
0
P(exp(sup M
t
) ≥ λ) dλ
=
Z
∞
0
P(sup M
t
≥ log λ) dλ
≤ 1 +
Z
∞
1
e
−(log λ)
2
/2b
dλ < ∞.
Since hM i ≥ 0, we know that
sup
t≥0
E(M )
t
≤ exp (sup M
t
) ,
So E(M ) is a uniformly integrable martingale.
Theorem
(Girsanov’s theorem)
.
Let
M
be a continuous local martingale with
M
0
= 0. Suppose that
E
(
M
) is a uniformly integrable martingale. Define a new
probability measure
dQ
dP
= E(M )
∞
Let
X
be a continuous local martingale with respect to
P
. Then
X −hX, M i
is
a continuous local martingale with respect to Q.
Proof. Define the stopping time
T
n
= inf{t ≥ 0 : |X
t
− hX, M i
t
| ≥ n},
and
P
(
T
n
→ ∞
) = 1 by continuity. Since
Q
is absolutely continuous with
respect to
P
, we know that
Q
(
T
n
→ ∞
) = 1. Thus it suffices to show that
X
T
n
− hX
T
n
, Mi is a continuous martingale for any n. Let
Y = X
T
n
− hX
T
n
, Mi, Z = E(M ).
Claim. ZY is a continuous local martingale with respect to P.
We use the product rule to compute
d(ZY ) = Y
t
dZ
t
+ Z
t
dY
t
+ dhY, Zi
t
= Y Z
t
dM
t
+ Z
t
(dX
T
n
− dhX
T
n
, Mi
t
) + Z
t
dhM, X
T
n
i
= Y Z
t
dM
t
+ Z
t
dX
T
n
So we see that
ZY
is a stochastic integral with respect to a continuous local
martingale. Thus ZY is a continuous local martingale.
Claim. ZY is uniformly integrable.
Since
Z
is a uniformly integrable martingale,
{Z
T
:
T is a stopping time}
is
uniformly integrable. Since
Y
is bounded,
{Z
T
Y
T
:
T is a stopping time}
is also
uniformly integrable. So ZY is a true martingale (with respect to P).
Claim. Y is a martingale with respect to Q.
We have
E
Q
(Y
t
− Y
s
| F
s
) = E
P
(Z
∞
Y
t
− Z
∞
Y
s
| F
s
)
= E
P
(Z
t
Y
t
− Z
s
Y
s
| F
s
) = 0.
Note that the quadratic variation does not change since
hX −hX, M ii = hXi
t
= lim
n→∞
b2
n
tc
X
i=1
(X
i2
−n
− X
(i−1)2
−n
)
2
a.s.
along a subsequence.