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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
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up vote 1 down vote accepted

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.

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How is Tonelli/Fubini justified here? –  Cherufe Nov 28 '13 at 19:12
    
$P$ and the lebesque measure are $\sigma-$finite, $W^6\geq 0$ and jointly measurable. Am I wrong? –  Henrik Nov 28 '13 at 19:16
    
No, you are not wrong. I was just looking for something more general for when one has f(W(t)) instead of W^3(t) inside. –  Cherufe Nov 28 '13 at 21:35
1  
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
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