I need to compute the following:

$E\left[ B_t \int_0^tB_s^2 \, ds \right]$ for $t≥0$

Where $B_t$ is a standard Brownian motion.

I'm thinking this is really obvious, But I cannot get my head round it. I thought about using Fubinis theorem to move the expectation inside of the integral to get an $s$, but I don't think that works seeing as there's a product.

  • $\begingroup$ @ Lubo : Think about Itô's isometry i.e. $E[Y_t.X_t]=E[\int_0^t d<Y_s.X_s>]$ , where $<Y_s.X_s>$ is the co-quadratic variation of Y and X. best regards $\endgroup$
    – TheBridge
    Jan 21, 2016 at 9:38
  • 1
    $\begingroup$ This is $$\int_0^tE(B_tB_s^2)ds$$ hence the goal is to compute $E(B_tB_s^2)$ for $s<t$. Long road: $B_t=B_s+W$ with $W$ independent of $B_s$ and centered hence $E(B_tB_s^2)=E(B_s^3)+E(W)E(B_s^2)=$ $____$. Short road: $(-B_s,-B_t)$ is distributed like $(B_s,B_t)$ and $(-B_t)(-B_s)^2=-(B_tB_s^2)$ hence $E(B_tB_s^2)=E((-B_t)(-B_s)^2)=$ $____$. $\endgroup$
    – Did
    Jan 21, 2016 at 15:26

1 Answer 1


Since $B_t$ is symmetrical, $B_t$ and $-B_t$ have the same distribution, $B_t \int_0^tB_s^2$ and $-B_t \int_0^tB_s^2$ also have the same distribution. Therefore

$E\left[ B_t \int_0^tB_s^2 \, ds \right] = -E\left[ B_t \int_0^tB_s^2 \, ds \right]$

It is $0$.


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