1. ## characteristic function

Let $W_t$ and $B_t$ be independent standard Brownian Motions, $t \in [0,T], T< \infty$

find the characteristic function of
$\phi(u) = Eexp[iuW_tB_T]$
where u is a real variable, $i = \sqrt{-1}$

well i know that im supposed to integrate this, but i have no idea how to do it.. should i just:

$\int_0^\infty exp[iuW_tB_T] \times \frac{1}{\sigma\sqrt{2\pi}} exp[\frac{(y-\mu)^2}{2\sigma^2}] dW_tB_T$ ?

is this the right way to answer it?

2. Originally Posted by Pengu
Let $W_t$ and $B_t$ be independent standard Brownian Motions, $t \in [0,T], T< \infty$

find the characteristic function of
$\phi(u) = Eexp[iuW_tB_T]$
where u is a real variable, $i = \sqrt{-1}$

well i know that im supposed to integrate this, but i have no idea how to do it.. should i just:

$\int_0^\infty exp[iuW_tB_T] \times \frac{1}{\sigma\sqrt{2\pi}} exp[\frac{(y-\mu)^2}{2\sigma^2}] dW_tB_T$ ?

is this the right way to answer it?
You should have a look here.

3. so is this right...

$\int_0^\infty exp[-u^2x^2/2] \times \frac{1}{\sqrt{2\pi T}} exp[\frac{(x)^2}{2T}] dx$

is that right now??

4. ## lol

is that you henry?? LOL

are you sure variance of a standard brownian motion is 1....thats the correct answer if the s.b.m~N(0,1)

5. Originally Posted by Pengu
so is this right...

$\int_0^\infty exp[-u^2x^2/2] \times \frac{1}{\sqrt{2\pi T}} exp[\frac{(x)^2}{2T}] dx$

is that right now??
Yes, except that you forgot $t$, a minus sign, and you should integrate on $\mathbb{R}$. I.e.: $E[e^{iuB_t W_T}]=\int e^{-\frac{u^2 x^2 t}{2}} \frac{e^{-\frac{x^2}{2T}}}{\sqrt{2\pi T}} dx$. I let you find where you did a mistake.

Now you can compute this integral. Remember that the density of a Gaussian integrates to 1: $\int e^{-\frac{x^2}{2v}}dx = \sqrt{2\pi v}$ for any $v>0$. Apply this here for the suitable $v$ and conclude.