brownian motion - expected value

Question

Let $B$ be an brownian motion and let $s \leq t$. Compute $\mathrm { E } \left[ B _ { s } B _ { t } ^ { 2 } \right]$.

I know, that the answer is $0$, but I can't see how this ends being $0$.

My attempt: $\mathrm { E } \left[ B _ { s } B _ { t } ^ { 2 } \right] = \mathrm { E } \left[ B _ { s } \left( B _ { t } - B _ { s } + B _ { s } \right) ^ { 2 } \right] = \mathrm { E } \left[ B _ { s } \left( B _ { t } - B _ { s } \right) ^ { 2 } \right] + 2 \mathrm { E } \left[ B _ { s } ^ { 2 } \left( B _ { t } - B _ { s } \right) \right] + \mathrm { E } \left[ B _ { s } ^ { 3 } \right].$

So, my question is: Because of independence, can I split this up?:

$$\mathrm { E } \left[ B _ { s } \left( B _ { t } - B _ { s } \right) ^ { 2 } \right] = \mathrm { E } \left[ B _ { s } \right] \cdot \mathrm { E } \left[ \left( B _ { t } - B _ { s } \right) ^ { 2 } \right].$$ Then I can use some of the basic Brownian motion proberties. If $\mathrm { E } \left[ B _ { s } \right] = 0$, then the whole first term is zero. My first thought was that $\mathrm { E } \left[ B _ { s } \right] = 0$, but now I'm not sure why this is true.

I can do the similar things with second term.

The third term is zero because of the rule: $\mathrm { E } \left[ X_t^{2k+1} \right] = 0$

Any help is appreciated.

Answer 1

What you did is correct! The expected value of both $B_s$ and $B_t-B_s$ are zero, because by the rules of brownian motion they are centered Gaussian random variables with variance respectively $s$ and $t-s$. Due to the fact that they are centered Gaussians, the expected values are null.

So as you said the third term vanishes because it is the third moment of a centered gaussian, which is zero.

The first and the second are splittable because of the fact that the Brownian motion has independent increments, so $B_s$ and $B_t-B_s$ are independent.

In the first case you have $\mathbb{E}[B_s(B_t-B_s)^2]=\mathbb{E}[B_s]\mathbb{E}[(B_t-B_s)^2]=0$ because the first factor is null.

In the second case you have $\mathbb{E}[B_s^2(B_t-B_s)]=\mathbb{E}[B_s^2]\mathbb{E}[(B_t-B_s)]=0$ because the second factor is null, as said before.

So the expected value is zero overall

Answer 2

The shortest proof is that the law of $B$ is preserved by flipping to $-B$, but that changes your product to its negative. So your variable has the same mean as its negative, so this mean must be zero. This does assume you know the product is integrable.