3The stochastic integral
III Stochastic Calculus and Applications
3.1 Simple processes
We now have all the background required to define the stochastic integral, and
we can start constructing it. As in the case of the Lebesgue integral, we first
define it for simple processes, and then extend to general processes by taking a
limit. Recall that we have
Definition
(Simple process)
.
The space of simple processes
E
consists of func-
tions H : Ω × [0, ∞) → R that can be written as
H
t
(ω) =
n
X
i=1
H
i−1
(ω)1
(t
i−1
,t
i
]
(t)
for some 0 ≤ t
0
≤ t
1
≤ ··· ≤ t
n
and bounded random variables H
i
∈ F
t
i
.
Definition (H ·M). For M ∈ M
2
and H ∈ E, we set
Z
t
0
H dM = (H ·M)
t
=
n
X
i=1
H
i−1
(M
t
i
∧t
− M
t
i−1
∧t
).
If
M
were of finite variation, then this is the same as what we have previously
seen.
Recall that for the Lebesgue integral, extending this definition to general
functions required results like monotone convergence. Here we need some similar
results that put bounds on how large the integral can be. In fact, we get
something better than a bound.
Proposition. If M ∈ M
2
c
and H ∈ E, then H · M ∈ M
2
c
and
kH · Mk
2
M
2
= E
Z
∞
0
H
2
s
dhMi
s
. (∗)
Proof.
We first show that
H · M ∈ M
2
c
. By linearity, we only have to check it
for
X
i
t
= H
i−1
(M
t
i
∧t
− M
t
i−1
∧t
)
We have to check that
E
(
X
i
t
| F
s
) = 0 for all
t > s
, and the only non-trivial case
is when t > t
i−1
.
E(X
i
t
| F
s
) = H
i−1
E(M
t
i
∧t
− M
t
i−1
∧t
| F
s
) = 0.
We also check that
kX
i
k
M
2
≤ 2kHk
∞
kMk
M
2
.
So it is bounded. So H · M ∈ M
2
c
.
To prove (∗), we note that the X
i
are orthogonal and that
hX
i
i
t
= H
2
i−1
(hMi
t
i
∧t
− hMi
t
i−1
∧t
).
So we have
hH ·M, H ·Mi =
X
hX
i
, X
i
i =
X
H
2
i−1
(hMi
t
i
∧t
−hMi
t
i−1
∧t
) =
Z
t
0
H
2
s
dhMi
s
.
In particular,
kH · Mk
2
M
2
= EhH · Mi
∞
= E
Z
∞
0
H
2
s
dhMi
s
.
Proposition. Let M ∈ M
2
c
and H ∈ E. Then
hH · M, N i = H · hM, Ni
for all N ∈ M
2
.
In other words, the stochastic integral commutes with the bracket.
Proof. Write H · M =
P
X
i
=
P
H
i−1
(M
t
i
∧t
− M
t
i−1
∧t
) as before. Then
hX
i
, Ni
t
= H
i−1
hM
t
i
∧t
− M
t
i−1
∧t
, Ni = H
i−1
(hM, N i
t
i
∧t
− hM, N i
t
i−1
∧t
).