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How to evaluate this integral $$\int_0^T(W(s))^2 \, dW(s)$$ where $W(s)$ is random variable associated with brownian motion.

I am new to this .Thanks in advance.

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Hint: $\int_0^T W^2 dW = W(T)^3-W(0)^3$ – Fabian Oct 5 '12 at 16:30
how is this hint? and even if you consider making analogies to real $W$ then $\int_0^TW^2dW=(W(T)^3-W(0)^3)/3$ – Aang Oct 5 '12 at 16:33
sorry, you are right of course. Let me correct my hint (and also make it more accurate): $\int_0^T W^2 dW = \frac{1}{3}(W(T)^3 - W(0)^3) - \int_0^T W dt$. – Fabian Oct 5 '12 at 16:38
You get there by noting that $d[W(t)]^n =[W(t)+dW(t)]^n -W(t)^n =\sum_{j=1}^n \binom{n}{j} W(t)^{n-j} dW(t)^j$ and the fact that $dW(t)^r \to 0$ for $r>2$ and $dW(t)^2 = dt$. – Fabian Oct 5 '12 at 16:41
Avatar: What do you call evaluate? – Did Oct 5 '12 at 17:12


In comments under the question, you say your main concern is how to evaluate $\int_0^T W\,dt$.

By symmetry, $\mathbb E\left(\int_0^T W\,dt\right)=0$, so the substantial part is finding $\var\left(\int_0^T W\,dt\right)$. If linearity of covariance in each argument separately applies to integrals and not merely to the finite case (so you may have to either think about how this can be shown, or cite a theorem in some book), then we have $$ \cov\left( \int_0^T W(t)\,dt , \int_0^T W(s)\,ds \right) = \int_0^T\int_0^T \cov(W(t),W(s))\,dt\,ds $$ $$ = \int_0^T\int_0^T \min\{s,t\}\,dt\,ds. $$

Now do two integrals: one for the part of the square below the line $t=s$, and the other for the part above it. (But they're the same, by symmetry.)

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