what is $\mathbb E[W(t)^n]$ where $W(t)$ is a wiener process?

2.2k Views Asked by At

What is the expectation of Wiener process raised to the power of $n$ where $W(t)$ is a wiener process or a standard Brownian motion?

E ( W ( t ) ) n )
2

There are 2 best solutions below

5
On

There are a few different approaches to take for this. One way is to know that since $W_t$ is a standard Wiener process, $W_t$ ~ $N(0, t)$. Then, $W^n_t$ is just a normal random variable with standard deviation $\sqrt t$ raised to the $n$-th power. The expectation can then be found by the moment-generating function (MGF) $M_X(t) = E(e^{tX})$, or specifically for a normal r.v.: $$M_X(t) = e^{t \mu + \frac{1}{2} \sigma^2 t^2}$$

Then, taking the n-th derivative with respect at $t = 0$ will yield the corresponding moment. Notice there is a little bit of a change in variable terminology due to the naming convention of $t$ for MGFs and the $t$ present in the Wiener process $W_t$. Since $\mu = 0$ and $\sigma = t$, let's call $t'$ the $t$ of the MGF as such:

$$ M_{W_t} = e^{\frac{1}{2} t {t'}^2}$$

Then, taking the n-th derivative w.r.t. $t'$ and setting $t = 0$ will give the corresponding moment (i.e. the expectation of the Wiener process to the power of $n$).

Another method related to the MGF is to use the characteristic function, but I am not too familiar with that.

Using an important result from stochastic calculus would be to use Ito's Lemma, but this is probably not the best example to demonstrate the power of the Lemma.

0
On

In the understanding that $W(t)\sim\mathsf{Norm}(0,t)$ we may conclude that:$$\mathbb{E}W\left(t\right)^{n}=\begin{cases} 0 & \text{if }n\text{ is odd}\\ t^{n/2}\left(n-1\right)!! & \text{if }n\text{ is even} \end{cases}$$

If $n$ is even then: $$\left(n-1\right)!!:=\left(n-1\right)\times\left(n-3\right)\times\cdots\times3\times1$$ Also see here.