Chapter 66
The Integral∫ t
0 (Y,dM)HFirst the integral is defined for elementary functions.
Definition 66.0.1 Let an elementary function be one which is of the form
m−1
∑i=0
YiX(ti,ti+1] (t)
where Yi is Fti measurable with values in H a separable real Hilbert space for 0 = t0 <t1 < · · ·< tm = T.
Definition 66.0.2 Now let M be a H valued continuous local martingale, M (0) = 0. Thenfor Y a simple function as above,∫ t
0(Y,dM)≡
m−1
∑i=0
(Yi,M (t ∧ ti+1)−M (t ∧ ti))H
Assumption 66.0.3 We will always assume that d [M] is absolutely continuous with respectto Lebesgue measure. Thus d [M] = kdt where k ≥ 0 and is in L1 ([0,T ]×Ω). This is doneto avoid technical questions related to whether t→
∫ t0 d [M] is continuous and also to make
it easier to get examples of a certain class of functions.
This includes the usual stochastic integral M (t) =∫ t
0 ΦdW where [M] (t) =∫ t
0 ∥Φ∥2L2
dsso d [M] = ∥Φ∥2 dt.
Next is to consider how this relates to stopping times which have values in the {ti}. Letτ be a stopping time which takes the values {ti}m
i=0. Then∫ t∧τ
0(Y,dM)≡
m−1
∑i=0
(Yi,M (t ∧ ti+1∧ τ)−M (t ∧ ti∧ τ))H (66.0.1)
Now consider X[0,τ]Y. Is it also an elementary function?
X[0,τ]Y =m−1
∑i=0
X[0,τ] (t)YiX(ti,ti+1] (t)
To get the ith term to be non zero, you must have τ ≥ t and t ∈ (ti, ti+1]. Thus it must be thecase that τ > ti. Also, if τ > ti and t ∈ (ti, ti+1], then τ ≥ ti+1 because τ has only the valuesti. Hence also τ ≥ t. Thus the above sum reduces to
m−1
∑i=0
X[τ>ti] (ω)YiX(ti,ti+1] (t)
This shows that X[0,τ]Y is of the right sort, the sum of Fti measurable functions timesX(ti,ti+1] (t). Thus from the definition of this funny integral,
∫ t
0
(X[0,τ]Y,dM
)≡
m−1
∑i=0
Fti︷ ︸︸ ︷
X[τ>ti] (ω)Yi,M (t ∧ ti+1)−M (t ∧ ti)
H
(66.0.2)
2273