Why is $\mathbb{E}[B_t^4] = 3t^2$?

Question

Where $B_t$ is general Brownian motion. Also, what is the $\mathbb{E}[(B_s^2)(B_t^2)$] where $s < t$?

Thank you for your help. I'm trying to wrap my head around these concepts, but I'm having quite a bit of difficulty with these proofs.

Answer 1

$B_t$ is distributed as $\sqrt{t}Z$ where $Z$ is a standard Normal random variable, so your question is equivalent to asking why $\mathbb EZ^4=3$. This can be computed from the definition, since $$ \mathbb EZ^4=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}x^4e^{-x^2/2}\ dx=3. $$

To answer the second question, we can write $B_t=\sqrt{s}Z+\sqrt{t-s}Y$ where $Z$ and $Y$ are independent standard Normal random variables, and therefore $$ \mathbb E(B_s^2B_t^2)=\mathbb E\bigl[sZ^2(\sqrt{s} Z+\sqrt{t-s}Y)^2\bigr]=\mathbb E\Bigl[sZ^2\bigl(s Z^2+2\sqrt{s(t-s)}ZY+(t-s)Y^2\bigr)\Bigr]. $$ Since $\mathbb EZ=\mathbb EY=0$ and $Z,Y$ are independent, the middle term is zero, leaving $$ s^2\mathbb EZ^4+s(t-s)\mathbb EZ^2\mathbb EY^2. $$ From the previous calculations $\mathbb EZ^4=3$ and since $Z,Y$ are standard Normal they have variance $1$, so that $\mathbb EZ^2=\mathbb EY^2=1$ and therefore $$ \mathbb E(B_s^2B_t^2)=3s^2+s(t-s). $$