3The stochastic integral
III Stochastic Calculus and Applications
3.2 Itˆo isometry
We now try to extend the above definition to something more general than
simple processes.
Definition
(
L
2
(
M
))
.
Let
M ∈ M
2
c
. Define
L
2
(
M
) to be the space of (equiva-
lence classes of) previsible H : Ω × [0, ∞) → R such that
kHk
L
2
(M)
= kHk
M
= E
Z
∞
0
H
2
s
dhMi
s
1/2
< ∞.
For H, K ∈ L
2
(M), we set
(H, K)
L
2
(M)
= E
Z
∞
0
H
s
K
s
dhMi
s
.
In fact,
L
2
(
M
) is equal to
L
2
(Ω
×
[0
, ∞
)
, P,
d
P
d
hMi
), where
P
is the
previsible σ-algebra, and in particular is a Hilbert space.
Proposition. Let M ∈ M
2
c
. Then E is dense in L
2
(M).
Proof.
Since
L
2
(
M
) is a Hilbert space, it suffices to show that if (
K, H
) = 0 for
all H ∈ E, then K = 0.
So assume that (K, H) = 0 for all H ∈ E and
X
t
=
Z
t
0
K
s
dhMi
s
,
Then
X
is a well-defined finite variation process, and
X
t
∈ L
1
for all
t
. It
suffices to show that
X
t
= 0 for all
t
, and we shall show that
X
t
is a continuous
martingale.
Let 0
≤ s < t
and
F ∈ F
s
bounded. We let
H
=
F
1
(s,t]
∈ E
. By assumption,
we know
0 = (K, H) = E
F
Z
t
s
K
u
dhMi
u
= E(F (X
t
− X
S
)).
Since this holds for all F
s
measurable F , we have shown that
E(X
t
| F
s
) = X
s
.
So X is a (continuous) martingale, and we are done.
Theorem. Let M ∈ M
2
c
. Then
(i)
The map
H ∈ E 7→ H ·M ∈ M
2
c
extends uniquely to an isometry
L
2
(
M
)
→
M
2
c
, called the Itˆo isometry.
(ii) For H ∈ L
2
(M), H · M is the unique martingale in M
2
c
such that
hH · M, Ni = H · hM, N i
for all
N ∈ M
2
c
, where the integral on the LHS is the stochastic integral
(as above) and the RHS is the finite variation integral.
(iii) If T is a stopping time, then (1
[0,T ]
H) · M = (H · M )
T
= H · M
T
.
Definition
(Stochastic integral)
. H · M
is the stochastic integral of
H
with
respect to M and we also write
(H · M)
t
=
Z
t
0
H
s
dM
s
.
It is important that the integral of martingale is still a martingale. After
proving Itˆo’s formula, we will use this fact to show that a lot of things are in
fact martingales in a rather systematic manner. For example, it will be rather
effortless to show that
B
2
t
− t
is a martingale when
B
t
is a standard Brownian
motion.
Proof.
(i)
We have already shown that this map is an isometry when restricted to
E
.
So extend by completeness of M
2
c
and denseness of E.
(ii)
Again the equation to show is known for simple
H
, and we want to show
it is preserved under taking limits. Suppose
H
n
→ H
in
L
2
(
M
) with
H
n
∈ L
2
(M). Then H
n
· M → H · M in M
2
c
. We want to show that
hH · M, Ni
∞
= lim
n→∞
hH
n
· M, Ni
∞
in L
1
.
H · hM, Ni = lim
n→∞
H
n
· hM, Ni in L
1
.
for all N ∈ M
2
c
.
To show the first holds, we use the Kunita–Watanabe inequality to get
E|hH · M − H
n
· M, Ni
∞
| ≤ E (hH · M − H
n
· Mi
∞
)
1/2
(EhNi
∞
)
1/2
,
and the first factor is
kH ·M −H
n
·Mk
M
2
→
0, while the second is finite
since N ∈ M
2
c
. The second follows from
E |((H − H
n
) · hM, Ni)
∞
| ≤ kH − H
n
k
L
2
(M)
kNk
M
2
→ 0.
So we know that
hH · M, N i
∞
= (
H · hM, N i
)
∞
. We can then replace
N
by the stopped process N
t
to get hH · M, Ni
t
= (H · hM, Ni)
t
.
To see uniqueness, suppose
X ∈ M
2
c
is another such martingale. Then we
have
hX − H · M, Ni
= 0 for all
N
. Take
N
=
X − H · M
, and then we
are done.
(iii) For N ∈ M
2
c
, we have
h(H · M)
T
, Ni
t
= hH · M, Ni
t∧T
= H · hM, Ni
t∧T
= (H1
[0,T ]
· hM, Ni)
t
for every N. So we have shown that
(H · M)
T
= (1
[0,T ]
H · M)
by (ii). To prove the second equality, we have
hH ·M
T
, Ni
t
= H ·hM
T
, Ni
t
= H ·hM, N i
t∧T
= ((H1
[0,T ]
·hM, N i)
t
.
Note that (ii) can be written as
*
Z
(−)
0
H
s
dM
s
, N
+
t
=
Z
t
0
H
s
dhM, Ni
s
.
Corollary.
hH · M, K · N i = H · (K · hM, Ni) = (HK) · hM, N i.
In other words,
*
Z
(−)
0
H
s
dM
s
,
Z
(−)
0
K
s
dN
s
+
t
=
Z
t
0
H
s
K
s
dhM, Ni
s
.
Corollary.
Since
H · M
and (
H · M
)(
K · N
)
− hH · M, K · N i
are martingales
starting at 0, we have
E
Z
t
0
H dM
s
= 0
E
Z
t
0
H
s
dM
s
Z
t
0
K
s
dN
s
=
Z
t
0
H
s
K
s
dhM, Ni
s
.
Corollary.
Let
H ∈ L
2
(
M
), then
HK ∈ L
2
(
M
) iff
K ∈ L
2
(
H · M
), in which
case
(KH) · M = K · (H · M).
Proof. We have
E
Z
∞
0
K
2
s
H
2
s
dhM
s
i
= E
Z
∞
0
K
2
s
hH · Mi
s
,
so kKk
L
2
(H·M)
= kHKk
L
2
(M)
. For N ∈ M
2
c
, we have
h(KH) ·M, N i
t
= (KH ·hM, N i)
t
= (K ·(H ·hM, N i))
t
= (K ·hH ·M, N i)
t
.