stochastic processes

**elenaas** · September 23rd 2009, 02:02 PM

let

be Brownian motion on R,

. Put

a) Prove

for all

b) Use power series expansion of the exponential function on both sides, compare the terms with the same power of u and deduse that

c) Prove that

for all functions f such that the integral of the right converges.

**Laurent** · September 23rd 2009, 02:35 PM

Originally Posted by elenaas

let

be Brownian motion on R,

. Put

a) Prove

for all

b) Use power series expansion of the exponential function on both sides, compare the terms with the same power of u and deduse that

c) Prove that

for all functions f such that the integral of the right converges.

a) and c) are easy consequences of the fact that $B_t$ is a Gaussian random variable of mean 0 and variance $t$ . As for b), you already have a hint given, you should specify what causes you difficulty.

**elenaas** · September 24th 2009, 02:21 AM

I have a problem with a)
if

is a random variable with density

then for any (well behaved) function

:

From that:

So I have a problem to calculate this integral

I tried with integration by parts and got more complicated integral.
So what do you think?

**Laurent** · September 24th 2009, 09:12 AM

Originally Posted by elenaas

I have a problem with a)
if

is a random variable with density

then for any (well behaved) function

:

From that:

So I have a problem to calculate this integral

I tried with integration by parts and got more complicated integral.
So what do you think?

I think your teacher assumes you already know the characteristic function of a Gaussian random variable: if $X\sim \mathcal{N}(\mu,\sigma^2)$ , then $\Phi_X(t)=E[e^{itX}]=e^{it\mu-t^2\sigma^2/2}$ . This is well-known but not quite easy to prove, you can't find it by direct computation. One way is to see that, if $X\sim \mathcal{N}(0,\sigma^2)$ and $f(t)=E[e^{itX}]$ , then $f'(t)=-\sigma^2 t f(t)$ (you need to differentiate and then integrate by part), and $f(0)=1$ , hence $f(t)=e^{-\sigma^2 t^2/2}$ . The case $\mathcal{N}(\mu,\sigma^2)$ follows. But, once again: you can't be expected to invent that, this is the kind of thing one is supposed to know (not the proof, but the formula).

**elenaas** · September 27th 2009, 11:04 AM

Thanks for your help,
I solved a) and c) but now I have difficulty to prove that $E\left [ B_{t}^{2k} \right ]=\frac{\left ( 2k \right )!}{2^{k}k!}t^{k};k\in N$
which implies b).
What do you think?

**Laurent** · September 27th 2009, 12:10 PM

Originally Posted by elenaas

Thanks for your help,
I solved a) and c) but now I have difficulty to prove that $E\left [ B_{t}^{2k} \right ]=\frac{\left ( 2k \right )!}{2^{k}k!}t^{k};k\in N$
which implies b).
What do you think?

From a), you have $E[e^{iuB_t}]=e^{-\frac{u^2 t}{2}}$ , hence $E[\sum_{n=0}^\infty \frac{(iuB_t)^n}{n!}]=\sum_{k=0}^\infty \frac{(-u^2 t)^k}{2^k k!}$ by using the power series expansion of the exponentials. Using the bounded convergence theorem, you can justify that $E[\sum_{n=0}^\infty \frac{(iuB_t)^n}{n!}]=\sum_{n=0}^\infty E [\frac{(iuB_t)^n}{n!}]$ . So that you get, for all real $u$ ,

$\sum_{n=0}^\infty \frac{i^n E[B_t^n]}{n!}u^n = \sum_{k=0}^\infty \frac{(-1)^k t^k}{2^k k!}u^{2k}$ .

Identifying coefficients gives $E[B_t^n]=0$ if $n$ is odd, and your formula if $n=2k$ .

**elenaas** · September 28th 2009, 06:55 AM

Thank you!
I really appreciate your help.

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