64.5. HILBERT SPACE VALUED WIENER PROCESSES 2203
Thus (h,W (t)−W (s)) is normally distributed with mean 0 and variance (t− s)(Qh,h). Itis obvious from the definition that W (0) = 0. Also to find the covariance, consider
E ((h1,W (t)−W (s))(h2,W (t)−W (s))) ,
and use 64.5.23 to obtain this is equal to
E
(∞
∑k=1
√λ k (ψk (t)−ψk (s))(h1,ek)
∞
∑j=1
√λ j
(ψ j (t)−ψ j (s)
)(h2,e j)
)
= limn→∞
E
(n
∑k=1
√λ k (ψk (t)−ψk (s))(h1,ek)
n
∑j=1
√λ j
(ψ j (t)−ψ j (s)
)(h2,e j)
)
= limn→∞
(t− s)n
∑k=1
λ k (h1,ek)(h2,e j) = (t− s)(Qh1,h2)
(Recall Q≡ ∑k λ kek⊗ ek.)Next I show the increments are independent. Let N be the subsequence defined above
and let W N (t) be given by the appropriate partial sum and let{
h j}m
j=1 be a finite list ofvectors of U . Then from the independence properties of ψ j explained above,
E
(exp
m
∑j=1
i(h j,W N (t j)−W N (t j−1
))U
)
E
(exp
m
∑j=1
i
(h j,
N
∑k=1
√λ kek
(ψk (t j)−ψk
(t j−1
)))U
)
= E
(exp
m
∑j=1
N
∑k=1
i√
λ k (h j,ek)U(ψk (t j)−ψk
(t j−1
)))
= E
(∏j,k
exp(
i√
λ k (h j,ek)U(ψk (t j)−ψk
(t j−1
))))
= ∏j,k
E(
exp(
i√
λ k (h j,ek)U(ψk (t j)−ψk
(t j−1
))))This can be done because of the independence of the random variables{
ψk (t j)−ψk(t j−1
)}j,k .
Thus the above equals
=m
∏j=1
exp
(−1
2
N
∑k=1
λ k (h j,ek)2U
(t j− t j−1
))