# Expected value of a brownian motion times the deterministic integral of a brownian motion

Given a Standard Brownian motion $(B_t)_{t\in\mathbf{R}_{+}}$,

$E (B_t \int_0^tB_s^3ds)$ = ?

I try to turn the expected value into a double integral by rewriting the $B_t$ term as

1) $E(\int_0^t dB_s\int_0^tB_t^3ds) =$

2) $= E(\int_0^t\int_0^tB_t^3dB_sds)$

Then by linearity of expectation I could bring it inside the double summation

3) $= (\int_0^t\int_0^tE(B_t^3)dB_sds) = 0$

since $B_t^3=(\sqrt[]tN(0,1))^3=0$

On another attempt I try to consider it as a Covariance:

1)$Cov (B_t; \int_0^tB_t^3ds) = E (B_t \int_0^tB_t^3ds)- [E (B_t) E (\int_0^tB_t^3ds)]$

2)$= E (B_t \int_0^tB_t^3ds)- [E (B_t) (\int_0^tE(B_t^3)ds)]$

3)$= E (B_t \int_0^tB_t^3ds)- 0$ again, by linearity of expectations

At this point I end up in the same problem, but I do not know how to prove (and if) the covariance between a stoch.process and the deterministic integral of itself is zero.

• Hint: Compute $E(B_tB_s^3)$ for every $s$ in $(0,t)$ and integrate the result. – Did Jan 18 '15 at 14:07
• Correct. And please also replace $t$ by $s$ in all integrals, lot of typo's. – ujsgeyrr1f0d0d0r0h1h0j0j_juj Jan 18 '15 at 14:11