Find all moments of a Brownian motion random variable

Question

Let $Y_k = n(B_{t_k}-B_{t_{k-1}})^2$ where $B$ be the Brownian motion. Here $t_k = kt/n$.

First, I need to show that $Y_k$ are independently identically distributed (iid) to $Y := n(B_{t/n}− B_0)^2$

I know that the increments of Brownian motion are independent. But I don't know how to show the independence of the squared increments.

Second, I need to find all moments of $Y$

The first moment is easy $$E(Y)=E[n(B_{t/n}− B_0)^2]=nE[(B_{t/n}− B_0)^2]=Var(B_{t/n}− B_0)= n\frac{t}{n}=t$$ The second moment is then $$E(Y^2)=E[n^2(B_{t/n}− B_0)^4]=n^2E[(B_{t/n}− B_0)^4]=n^2E[(B_{t/n})^4+4(B_{t/n})^3(B_0)+6(B_{t/n})^2(B_0)^2+4(B_{t/n})(B_0)^3+(B_0)^4]$$ Now I am stuck here. It seems stupid to calculate the moment like this. But I don't know the probability density function of $(B_{t/n}− B_0)^2$.

Is there any better approach to this question? Thanks

Answer 1

Regarding Question 1: Use the following well-known statement:

Let $X_1,\ldots,X_n$ be independent random variables and $g: \mathbb{R} \to \mathbb{R}$ a measurable function. Then $g(X_1),\ldots,g(X_n)$ are independent random variables.

Regarding Question 2: Recall that $B_t$ is Gaussian with mean $0$ and variance $t$. This implies that

$$\begin{align*} \mathbb{E}(B_t^{2k}) &= \frac{1}{\sqrt{2\pi t}} \int_{\mathbb{R}} y^{2k} \exp \left( - \frac{y^2}{2t} \right) \, dy \\ &= 2 \frac{1}{\sqrt{2\pi t}} \int_{(0,\infty)} y^{2k} \exp \left(- \frac{y^2}{2t} \right) \, dy. \end{align*}$$

Now perform a change of variables, according to $z := y/\sqrt{2t}$, in order to express $\mathbb{E}(B_t^{2k})$ in terms of the Gamma function

$$\Gamma(r) = \int_{(0,\infty)} z^{-1+r} e^{-z} \, dz, \qquad r>0.$$

As $\mathbb{E}(Y^k) = n^k \mathbb{E}(B_{t/n}^{2k})$, this allows you to compute arbitrary moments of $Y$,