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I am trying to find the distribution of the stochastic integral $X_t= \int^1_0\sqrt2 W_t dW_t $ where $W_t$ is a brownian motion.

I know this will be normal, so I have started to compute the moments for this: $E(X_t)= 0$ since $W_t $~$ N(0,t)$ But I am stuck on the variance. $$Var(X_t)=E(X_t^2) \\ =2\iint^1_0E(W_tW_s)dW_sdW_t$$ I know $E(W_tW_s)=min(t,s)$, but how do I compute the integral?

Am I on the right lines?

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    $\begingroup$ Why would that be normal? It's not. $\endgroup$ – Shalop Nov 22 '16 at 20:52
  • $\begingroup$ because $W_t$ is normal with mean 0 and variance t by definition, so surely $X_t$ is also normal? $\endgroup$ – Julie Nov 22 '16 at 20:55
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    $\begingroup$ No. Absolutely not. In fact that expression is equal to $\frac{\sqrt{2}}{2}(W_1^2-1)$ and the square of a normally distributed variable is definitely not normal. $\endgroup$ – Shalop Nov 22 '16 at 20:56
  • $\begingroup$ so you are saying it follows a chi-squared distribution? apologies, I am new to stochastic calculus and it has not been explained to me very well. $\endgroup$ – Julie Nov 22 '16 at 20:58
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    $\begingroup$ Yeah. Minus 1 and multiplied by $\sqrt{2}/2$. $\endgroup$ – Shalop Nov 22 '16 at 20:59
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Using the Ito isometry and Fubini's theorem, we have $$ \mathbb{E}\Big[\Big(\int_0^1\sqrt{2}W_t\;dW_t\Big)^2\Big]=\mathbb{E}\Big[\int_0^1(\sqrt{2}W_t)^2\;dt\Big]=\int_0^12\mathbb{E}[W_t^2]\;dt=\int_0^12t\;dt=1 $$

Another approach is to use Ito's lemma to obtain $$ \int_0^1\sqrt{2}W_t\;dW_t=\frac{\sqrt{2}}{2}(W_1^2-1)$$ and then compute $\frac{1}{2}\mathbb{E}[(W_1^2-1)^2]$. By the way, this calculation shows that $X_t$ is not normally distributed.

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  • $\begingroup$ Ah, thankyou. Using your first method, why does $dW_t$ change to just $dt$ when it is squared? $\endgroup$ – Julie Nov 22 '16 at 20:56
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    $\begingroup$ Have you seen the Ito isometry? $\endgroup$ – carmichael561 Nov 22 '16 at 20:58

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