Find $E(B_{1}^2B_{2}B_{3})$, where $B_{i}$ is Brownian motion

Question

There is an exercise in a textbook by Durret. Find $E(B_{1}^2B_{2}B_{3})$, where $B_{i}$ is Brownian motion. I try to use the definition $E(B_{s}B_{t})=s\wedge t$ and $B_{t}$ has independent increments. I have got that

$$E(B_{1}^2B_{2}B_{3})=E\big(B_{1}^2(B_{2}-B_{1})(B_{3}-B_{2})\big)+E((B_{1}^2(B_{2}-B_{1})^2)+E(B_{1}^3(B_{2}-B_{1}))+E(B_{1}^3(B_{3}-B_{1}))+E(B_{1}^4).$$

But I am not sure that $E(B_{1}^4)=1$ and I have no idea about $B_{t}^2$ still standard normal Gaussian distribution. And if it is true, the answer is 1, since

$$E\big(B_{1}^2(B_{2}-B_{1})(B_{3}-B_{2})\big)=0$$ $$E((B_{1}^2(B_{2}-B_{1})^2)=0$$ $$E(B_{1}^3(B_{2}-B_{1}))=0$$ $$E(B_{1}^3(B_{3}-B_{1}))=0$$

Answer 1

$\begin{aligned}\begin{aligned}B_{1}^{2}\end{aligned} B_{2}B_{3} & =B_{1}^{2}B_{2}\left(B_{3}-B_{2}\right)+B_{1}^{2}B_{2}^{2}\\ & =B_{1}^{2}\left(B_{2}-B_{1}+B_{1}\right)\left(B_{3}-B_{2}\right)+B_{1}^{2}\left(B_{2}-B_{1}+B_{1}\right)^{2}\\ & =B_{1}^{2}\left(B_{2}-B_{1}\right)\left(B_{3}-B_{2}\right)+B_{1}^{3}\left(B_{3}-B_{2}\right)+B_{1}^{2}\left(B_{2}-B_{1}\right)^{2}+2B_{1}^{3}\left(B_{2}-B_{1}\right)+B_{1}^{4} \end{aligned} $

Then by linearity of expectation, independence and the fact that every $B_{t}$ has mean $0$ we find:

$\mathsf{E}\begin{aligned}B_{1}^{2}\end{aligned} B_{2}B_{3}=\mathsf{E}B_{1}^{2}\mathsf{E}\left(B_{2}-B_{1}\right)^{2}+\mathsf{E}B_{1}^{4}=1+3=4$

To find the equality $\mathsf{E}B_{1}^{4}=3$ see the answer of harmonicuser or have a look at the moments of normal distribution on wikipedia.

Answer 2

$E(B_1^{2}(B_2-B_1)^{2})$ cannot be 0 since $B_1^{2}(B_2-B_1)^{2}$ is a positive random variable. The correct value is $EB_1^{2}E(B_2-B_1)^{2}=1$ since $EX^{2}=Var (X)=1$ if $X$ is a standard normal variable. ($B_2-B_1)$ is independent of $B_1$ and has the same distribution as $B_1$. Also $EB_1^{4} \neq1$. $EB_1^{4}=\frac 1 {\sqrt {2\pi}} \int_{-\infty}^{\infty} x^{4}e^{-x^{2}/2} dx$. You have to integrate by parts repeatedly to evaluate this integral.

Answer 3

$\sqrt{2\pi}t^{1/2} = \int_{\mathbb{R}}e^{-\frac{x^{2}}{2t}}\,dx$. Differentiating twice with respect to $t$ we get \begin{equation*} \sqrt{2\pi}\frac{1}{2}\frac{3}{2}t^{-3/2} = \frac{1}{4t^{2}}\int_{\mathbb{R}}x^{4}e^{-\frac{x^{2}}{2t}}\,dt \end{equation*} Substituting $t = 1$ we get $E(B_{1}^{4}) = 3$.