Find $E[\int_{0}^{t}{W(s)dW(s)}]$ and $Var[\int_{0}^{t}{W(s)dW(s)}]$.
I think that, applying Ito's Lemma leads to $$\int_{0}^{t}W(s)dW(s)=\frac{W^{2}(t)}{2}-\frac{t^2}{2}$$ but I am not entirely sure.
How can someone approach these type of problems?
How to find the mean and variance of Ito's integral
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real-analysis
analysis
measure-theory
stochastic-processes
stochastic-integrals
1 Answers
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Yes, (almost, anyway). You use Ito's lemma to derive $$d(W_t^2) = 2W_tdW_t+dt$$ and integrate to get $$ W_t^2 = 2\int_0^t W_tdW_t + t$$ so $$ \int_0^t W_tdW_t = \frac{1}{2}W_t^2-\frac{1}{2}t.$$
The reason you know to look at $d(W_t^2)$ is because in regular calculus, integrating $\int fdf $ gives $\frac{1}{2}f^2$, so you know your answer will be that plus a correction from Ito's lemma.
Then, you can take the expected value and variance using knowledge of the moments of $W_t.$ $$\begin{align}E\left[ \int_0^t W_tdW_t\right] &= \frac{1}{2}E(W_t^2)-\frac{1}{2}t\\&=\frac{1}{2}t-\frac{1}{2}t\\&=0\end{align}$$ and $$\begin{align}\mathrm{Var}\left[ \int_0^t W_tdW_t\right] &= \frac{1}{4}\mathrm{Var}(W_t^2) \\&= \frac{1}{4}E(W_t^4)-\frac{1}{4}E(W_t^2)^2 \\&= \frac{1}{4}3t^2-\frac{1}{4} t^2\\&=\frac{1}{2}t^2 \end{align}$$