Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
Both approaches seem good to me. – Bunder Apr 15 '13 at 10:05
up vote 2 down vote accepted

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.

share|cite|improve this answer
Thank you for the formalization. – Paul Apr 16 '13 at 17:03

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.