9.10. The key observation is that B is independent of W (1). Indeed, note that for all 0 < t1 < . . . < tk < 1,
(B(t1 ), . . . , B(tk ), W (1)) is a centered normal random variable. Therefore, it suffices to show that
E[B(ti )W (1)] = 0, but this is easy to check. Now,
*
*
$
$ # Pm
$
# Pm
# Pm
*
*
E ei j=1 uj W (tj ) * |W (1)| ≤ ε = E ei j=1 uj B(tj ) E ei j=1 uj W (1) * |W (1)| ≤ ε .
As for the second term, conditional on |W (1)| ≤ ε,

*
 * 
*
*
m
m
m
"
! Pm
*
*
i j=1 uj W (1)
= sin 
|uj |** ≤ ε
|uj |,
Im e
uj W (1) ≤ **sin ε
*
*
j=1
j=1
j=1
for all ε ∈ (0, 1/
This proves that
0m
j=1
|uj |). This is because sin is increasing on (0, 1), and | sin x| = sin x ≤ x there.
m
*
! # Pm
$"
i j=1 uj W (1) *
Im E e
|uj | → 0
* |W (1)| ≤ ε ≤ ε
j=1
Similarly, Re(· · · ) → 1. This proves the result.
as ε → 0.