2206 CHAPTER 64. WIENER PROCESSES
and consider E(eiP). This equals
eiP = E
(exp
(i
n
∑q=1
m
∑j=1
sq j
((W (tq) ,ek j
)U−(
W(tq−1
),ek j
)U
)))(64.5.30)
= E
(exp
(i
n
∑q=1
((W (tq) ,
m
∑j=1
sq jek j
)U
−
(W(tq−1
),
m
∑j=1
sq jek j
)U
)))
= E
(n
∏q=1
exp
(i
(W (tq)−W
(tq−1
),
m
∑j=1
sq jek j
)U
))Now recall that by assumption the increments W (t)−W (s) are independent. Therefore,the above equals
n
∏q=1
E
(exp
(i
(W (tq)−W
(tq−1
),
m
∑j=1
sq jek j
)U
))
Recall that by assumption (W (t)−W (s) ,h)U is normally distributed with variance givenby the expresson (t− s)(Qh,h) and mean 0. Therefore, the above equals
n
∏q=1
exp
(−1
2(tq− tq−1
)(Q
m
∑j=1
sq jek j ,m
∑j=1
sq jek j
))
=n
∏q=1
exp
(−1
2(tq− tq−1
) m
∑j=1
s2q jλ k j
)
exp
(−1
2
n
∑q=1
m
∑j=1
(tq− tq−1
)s2
q jλ k j
)(64.5.31)
Alson
∏q=1
n
∏j=1
E(
exp(
isq j
((W (tq) ,ek j
)U−(
W(tq−1
),ek j
)U
)))(64.5.32)
=n
∏q=1
n
∏j=1
E(
exp(
isq j
((W (tq)−W
(tq−1
),ek j
)U
)))=
n
∏q=1
n
∏j=1
exp(−1
2(tq− tq−1
)s2
q j
(Qek j ,ek j
))
=n
∏q=1
n
∏j=1
exp(−1
2(tq− tq−1
)s2
q jλ k j
)
= exp
(−1
2
n
∑q=1
m
∑j=1
(tq− tq−1
)s2
q jλ k j
)