Computing expectation of brownian motion

Question

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.

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 $0$.