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What is the right approach to take and find the moments of the following: $$\mathcal{Z}_t=\int\mathcal{W}_t^k\,d\mathcal{W}_t=?$$ $$\mathcal{W}_t \sim \mathcal{N}(0, t),\ k=2,3...$$ $$\operatorname{E}(\mathcal{Z}_t)=?$$ $$\operatorname{Var}(\mathcal{Z}_t)=?$$

I know that $\mathcal{Y}_t=\int\mathcal{W}_t\,d\mathcal{W}_t=\frac{\mathcal{W}_t^2-t}{2}$ and $\operatorname{E}(\mathcal{Y}_t)=0$. One can derive it from Ito's lemma and the fact that $\operatorname{Var}(\mathcal{W}_t)=t=\operatorname{E}(\mathcal{W}_t^2)$. Is there another way to prove it?

Is there also some generic rules to deal with $\int t\,d\mathcal{W}_t$ and $\int \mathcal{W}_t\,dt$?

Thanks!

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Hint: A stochastic integral is a martingale that starts at zero. What does that say about its mean? –  Byron Schmuland Apr 1 '11 at 23:21
    
@byron-schmuland Thanks! Yes, it is always zero. –  Viktor Apr 4 '11 at 15:53

1 Answer 1

You might try working formally with $$\mathbb{E}[\int e^{u \mathcal{W}_t}\,d\mathcal{W}_t] = \sum_{n=0}^\infty \frac{u^n}{n!} \int \mathcal{W}^n_t\,d\mathcal{W}_t$$ which should give you the right result.

To compute variances, recall the Ito isometry: $$\mathbb{E}[(\int X_t\,d\mathcal{W}_t)^2] = \mathbb{E}[\int X_t^2\,dt]$$ which gives you a simpler integral to deal with.

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Thanks! –  Viktor Apr 4 '11 at 15:56

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