How to calculate $E[(\int_0^t{W_sds})^n], n \geq 2$

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 $$Y_t=\int_0^t{e^{W_s}ds}$$ $$Y_t=\int_0^t{[a_n(W_s)^n+a_{n-1}(W_s)^{n-1}+...+a_0]ds}$$

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 Re your update: are you really asking whether $Y_t$ is Gaussian? Wow. – Did Mar 19 '11 at 0:39 Did you get something out of one of the answers below? – Did Apr 7 '11 at 8:11

The random variable $X_t$ is Gaussian for the reasons in Didier's answer.
$$\mathbb{E}\left[\left( \int^t_0 W_s ds \right)^2\right] = \int^t_0 \int^t_0 \mathbb{E}(W_r W_s) dr ds = \int^t_0 \int^t_0 (r\wedge s) dr ds ={t^3\over 3}.$$
 Thanks. Your approach is much easier and easy to extend to high dimension. – Jun Deng Mar 18 '11 at 21:25 @Jun Deng You're welcome. I'm not sure how easy it is to extend this approach to higher moments. In order to work out the fourth moment, I needed to first deduce the formula $$\mathbb{E}(W_{t_1}W_{t_2}W_{t_3}W_{t_4})=t_{(1)} (2t_{(2)}+t_{(3)}),$$ where the $t_{(i)}$s are the order statistics of $t_i$. It works, but it's not that easy, and I'm not sure what the general pattern is. – Byron Schmuland Mar 18 '11 at 22:12 Let say for triple case, $$1=\sum_{\sigma(t1,t2,t3)}I_{t_1\leq t_2\leq t_3},$$ where $\sigma(t1,t2,t3)$ means all the permutation of $t_1,t_2,t_3$. Hence, we could get the integral by symmetry. Am I right? – Jun Deng Mar 18 '11 at 22:29 Yes. You are right. It can be done; but it is not obvious for the general formula. I will try again for the general case. – Jun Deng Mar 18 '11 at 22:48 By the discussion before, we can suppose that $t_1 As a linear functional of the Gaussian process$(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})$. If you know the moments of a standard Gaussian (and you should...), you are done. -  @Didier Piau:$X_t$is the limit of a serial of Gaussian random variables. We need to show that the limit is Gaussian. – Jun Deng Mar 18 '11 at 20:43 @Jun Deng Each$X_t$is the mean square limit of some linear combinations of$(W_s)_{0\le s\le t}$. The Gaussian property is preserved under linear combinations and mean square limits, thus$X_t$is a Gaussian random variable. And in fact the process$(X_t)_t$itself is Gaussian. – Did Mar 18 '11 at 21:04 @Didier Piau: Thanks. I should study limit theory later.^_^ – Jun Deng Mar 18 '11 at 21:19 @Didier: Do you do stuff with stochastic analysis? – Jonas Teuwen Mar 18 '11 at 22:22 It is possible to check all this (moment + gaussian nature of$X_t$) by studying the properties of the Fourrier transform of$X_t$(or its Laplace transform if you like it better). Regards. – TheBridge Mar 18 '11 at 22:34 show 6 more comments Note that$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.)