# $\mathbb{E} \int_a^b W^3(t)\,dW(t)=?$

Is it true that

$\mathbb{E} \int_a^b W^3(t)\,dW(t)=0$, for $a < b \in \mathbb{R}$

I know that for an adapted process $\Delta(t), t\geq 0$, the integral

$\int_0^t \Delta(u)dW(u)$ is a martingale in $t$, and therefore had expectation $0$, if

$\mathbb{E} \int_0^T \Delta^2(t) dt < \infty$

How can I show the square-integrability condition, and is there an other way to show the expectation?

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If $X \sim \mathcal{N}(0,t)$, then $\Bbb E(|X|^6) = |t|^3\Bbb E(|\mathcal{N}(0,1)|^6)$. No? –  Siméon Nov 28 '13 at 18:00
Interchanging integral and mean (tonelli) gives us $\int_0^T E[W(t)^6]dt=\int_0^T 15 t^3 dt =15/4 \cdot T^4< \infty$ where we have used this scheme to find the 6'th moment.
$P$ and the lebesque measure are $\sigma-$finite, $W^6\geq 0$ and jointly measurable. Am I wrong? –  Henrik Nov 28 '13 at 19:16
Ahh.. Okay. There are different tricks from above when $f \geq 0$ we use tonelli. If say f is continuous an useful trick is f(W) is continuous so bounded on compact sets. Often times it is also possible to use ito or ito's product rule to express the integral as something else e.g. $W(t)^2=2\int_0^t W(s) dW(s)+t$ so $\int_0^t W(s) dW(s)= \frac{1}{2} (W(t)^2-t)$ so a martingale. That's the 3 tricks I use the most I think. –  Henrik Nov 29 '13 at 8:04