Find $\mathbb{E}[W_t^3]$ and $\mathbb{E}[W_t^4]$

Question

I am very stuck on this past paper question.

$W_t$ is a brownian motion and find $\mathbb{E}[W_t^3]$ and $\mathbb{E}[W_t^4]$

I thought, since $W_t$ is normally distributed with density function $f(x)=\frac{1}{\sqrt{2\pi t}}e^{\frac{-x^2}{2t}}$, I obey the formula of the expected value of a continuous random variable,

$$\mathbb{E}[W_t^3]=\int_{-\infty}^{\infty}x^3 f(x) dx$$

$$\mathbb{E}[W_t^4]=\int_{-\infty}^{\infty}x^4 f(x) dx$$

But to be honest, the computation gets overly too complicated, involving integration by parts once or more and taking limits to infinity for fractions etc etc.

Anyway, this seems to be wrong given that each is only $3$ marks given for answering. I don't know any fancy stats method to solve this and I am very lost.

The answers are $0$ and $3t^2$ respectively but what is the method? please tell me, thank you

Answer 1

When you need to find moments consider the moment generating function. By definition this is $$m(u):=\mathbb{E}(e^{u W_t})=\sum_{m=0}^\infty {\mathbb{E}(W_t^m)\over m!}u^m.\tag1$$

On the other hand, for normal distributions we have $$m(u)=\exp(t u^2/2)=1+{tu^2\over 2}+{1\over 2!}\left({tu^2\over 2}\right)^2+\cdots \tag2 $$ Now equate the coefficients of $u^m$ for $m=3,4$ in (1) and (2) to get the answers.

Answer 2

Well, for $W_t^4$, I think the following might work. It appears that $W_t$ is a standard Brownian motion. Then $B_t = W/\sqrt t \sim N(0,1)$ and \begin{align*} E[W_t^4]&= t^2 E[B_t^4]\\ &=t^2E[[B_t^2]^2]\\ &=t^2\{\text{Var}[B_t^2]+E^2[B_t^2]\}\\ &=t^2[2+\{\text{Var}[B_t]+E^2[B_t]\}^2\\ &=t^2[2+(1+0)^2]\\ &=3t^2. \end{align*} Also, you might be able to argue that for the $W_t^3$ case, the expectation integrates an odd function so the integral will evaluate to $0$.

Answer 3

Using Ito you have that

$$W^2_t = 2 \int_0^t W_s ~ dw_s +t $$

therefore by Ito's isometry and Fubini's theorem we have that

\begin{align*} \mathbb E [W^4 _t ] &= 4 ~\mathbb E [(\int_0^t W_s ~ dw_s)^2 ] + 2t~\mathbb E [\int_0^t W_s ~ dw_s ] + t^ 2 \\ & = 4 ~ \int_0^t s ~ ds + t^2 = 3 t^ 2 \end{align*}

For $W^3_t$ you can either apply the same trick or noting that this implies computing a integral (defined in the all domain, $\mathbb R$) of a impar symmetric function (in relation to 0).

Answer 4

You can do the calculation for $W_1$ and then recall that $W_t$ has the same distribution as $\sqrt{t} W_1$. The calculation for $W_1$ winds up not being that hard: for $m \geq 2$, integration by parts tells us that

$$\int_{-\infty}^\infty x^m e^{-x^2/2} dx = (m-1) \int_{-\infty}^\infty x^{m-2} e^{-x^2/2} dx$$

Now run this calculation for $m=3$ and $m=4$. For $m=4$ you'll need to do it twice.