# Reality check: $\mathbb E \{ B_s B_t ^2\}=0$

I desire to calculate $\mathbb E \{ B_s B_t ^2\}$, where $B$ is a standard brownian motion starting from zero. I want to be sure I am not making any mistake on both reasoning and result, even if I calculated it by three differents methods.

1° solution:

If $s\leq t$, then we have \begin{align}\mathbb E \{ B_s B_t ^2\} &= \mathbb E \{ B_s (B_t ^2-t)\} \\&= \mathbb E \{ B_s \mathbb E \{(B_t ^2-t) | \mathcal F_s\}\} \\&= \mathbb E \{ B_s (B_s ^2-s)\} \\&= \mathbb E \{ B_s ^3\} \\&=s^{3/2}\mathbb E \{ B_1 ^3\} =0\end{align}

If $t\leq s$, then we have \begin{align}\mathbb E \{ B_s B_t ^2\} &= \mathbb E \{ B_t ^2\mathbb E \{ B_s | \mathcal F_t\}\} \\&= \mathbb E \{ B_t ^3\} =0\end{align}

2° solution:

We know that $B^2_t = t +2 \int_0^t B_u ~dB_u$ and $B_s = \int_0^s ~dB_u$ then

\begin{align}\mathbb E \{ B_s B_t ^2\} &= 2 E \{\int_0^s ~dB_u\int_0^t B_u ~dB_u\} \\&=2 E \{\int_0^{t\wedge s} B_u ~du\} \\&=\int_0^{t\wedge s} 2 E \{B_u\} ~du =0\end{align}

3° solution:

It's trivial that $\mathbb E \{ B_s B_t ^2\} =0$ since $B_t^2 \geq 0$ and the brownian motion is gaussien process so it has symetric law.

I will be thankful for any feedback.

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Both approaches seem good to me. – Bunder Apr 15 '13 at 10:05

The third solution uses a good idea but it needs to be slightly reformulated: the distributions of the processes $(B_t)_{t\geqslant0}$ and $(-B_t)_{t\geqslant0}$ are the same hence, for every $(s,t)$, the distributions of $(B_s,B_t)$ and $(-B_s,-B_t)$ are the same hence the distributions of $B_sB_t^2$ and $(-B_s)(-B_t)^2=-B_sB_t^2$ are the same, that is, the distribution of $B_sB_t^2$ is symmetric. In particular, it has mean zero.