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.