What is the expected value of the cube of the increment of a Brownian Motion?

Question

If $(B_t)_{t\ge0}$ is a Standard Brownian Motion and $s < t$, what is the expectation $E[(B_t - B_s)^3]$?

Moreover, what is the trick to finding these expectations? Say I also need to find $E[(B_t - B_s)^4]$ or $E[B_t(B_t - B_s)^3]$ or the like -- what is the general mindset or approach that I should employ when attacking these problems? My professor doesn't shy away from giving lemmas for the values of the expectations of these, but he never actually tells us how they're proven. I'd like to learn.

Thank you so much in advance.

Answer 1

Since $B_t-B_s\sim\mathcal{N}(0,t-s)$, the expectation of its cube is zero (as is seen if you check that all moments exist and that the density function is symmetric around the origin). More generally, there are closed forms for the moments of a normal distribution, and a proof can be found here.

In particular

$$E[(B_t - B_s)^4]=3(t-s)^2. $$

Using independence of the increments can also come in handy, e.g.

$$E[B_s(B_t - B_s)^3]=E[B_s]E[(B_t - B_s)^3]=0\cdot 0= 0, $$

assuming that $B_0=0$.

Answer 2

If $X$ has standard normal distribution then $B_t-B _s$ has the same distribution as $\sqrt {t-s} X$. Hence $E(B_t-B_s)^{n}=(s-t)^{n/2}EX^{n}$ for any positive integer $n$.