5
$\begingroup$

Let $W_t$ be a standard one dimension Brownian Motion with $W_0=0$ and $X_t=\int_0^t{W_sds}$. With the help of ito formula, we could get $E[(X_t)^2]=\frac{1}{3}t^3$ $E[(X_t)^3]=0$

When I try to employ the same method to calculate the general case $E[(X_t)^n]$, I got stuck. I guess $X_t$ should be normal distribution since it could be the limit of the following $\lim_{n\rightarrow \infty}{\sum_{i=0}^{n-1}{W_{t_i}(t_{i+1}-t_i)}},$

where $ W_{t_i}\sim norm(0,\sqrt{\frac{t_i}{n}}).$

If it is true, the problem would be trivial.

Update: Thanks for all the suggestions. Now I believe $X_t$ is a Gaussian process.

How about for this integral $Y_t=\int_0^t{f(W_s)ds}$ if we assume that $f$ is some good function, say polynomial or exponential, i.e $Y_t=\int_0^t{e^{W_s}ds}$ $Y_t=\int_0^t{[a_n(W_s)^n+a_{n-1}(W_s)^{n-1}+...+a_0]ds}$

  • 0
    Did you get something out of one of the answers below?2019-05-20

3 Answers 3

5

As a linear functional of the Gaussian process $(W_s)_{0\le s\le t}$, $X_t$ is a Gaussian random variable. You indicate yourself that $X_t$ is centered and has variance $\sigma^2_t=\frac13t^3$, hence $X_t$ is distributed like $\sigma_tN$ with $N$ standard Gaussian.

Thus, for every $n$, $E((X_t)^{2n+1})=0$ and $E((X_t)^{2n})=(\sigma_t^2)^nE(N^{2n})$. If you know the moments of a standard Gaussian (and you should...), you are done.

  • 0
    @Jun Deng In about any textbook/lecture notes on Gaussian processes. Which one do you know/are you working on?2011-03-19
5

The random variable $X_t$ is Gaussian for the reasons in Didier's answer.

You can calculate the variance directly (without Ito's formula) as follows:

$\mathbb{E}\left[\left( \int^t_0 W_s ds \right)^2\right] = \int^t_0 \int^t_0 \mathbb{E}(W_r W_s) dr ds = \int^t_0 \int^t_0 (r\wedge s) dr ds ={t^3\over 3}.$

  • 0
    By the discussion before, we can suppose that t_1 put $\xi_1=W_{t_1}, \xi_2=W_{t_2}-W_{t_1},...,\xi_n=W_{t_n}-W_{t_{n-1}}$, which are independent Gaussian variables. Then, we have $E[W_{t_1}W_{t_2}...W_{t_n}]=E[\xi_1(\xi_1+\xi_2)...(\xi_1+\xi_2+...+\xi_n)],$ Which could be done. But I don't know the explicit formula.2011-03-19
1

Note that $X_T = \int_0^T W_t\,dt = T W_T - \int_0^T t\,dW_t$. Thus $X_T$ is a linear combination of normally distributed random variables. (The latter integral may be easier to show that it is normal.)