Originally Posted by
elenaas Thanks for your help,
I solved a) and c) but now I have difficulty to prove that
which implies b).
What do you think?
From a), you have $\displaystyle E[e^{iuB_t}]=e^{-\frac{u^2 t}{2}}$, hence $\displaystyle E[\sum_{n=0}^\infty \frac{(iuB_t)^n}{n!}]=\sum_{k=0}^\infty \frac{(-u^2 t)^k}{2^k k!}$ by using the power series expansion of the exponentials. Using the bounded convergence theorem, you can justify that $\displaystyle E[\sum_{n=0}^\infty \frac{(iuB_t)^n}{n!}]=\sum_{n=0}^\infty E [\frac{(iuB_t)^n}{n!}]$. So that you get, for all real $\displaystyle u$,
$\displaystyle \sum_{n=0}^\infty \frac{i^n E[B_t^n]}{n!}u^n = \sum_{k=0}^\infty \frac{(-1)^k t^k}{2^k k!}u^{2k}$.
Identifying coefficients gives $\displaystyle E[B_t^n]=0$ if $\displaystyle n$ is odd, and your formula if $\displaystyle n=2k$.