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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$2012-10-05
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    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$2012-10-05
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    sorry, you are right of course. Let me correct my hint (and also make it more accurate): $\int_0^T W^2 dW = \frac{1}{3}(W(T)^3 - W(0)^3) - \int_0^T W dt$.2012-10-05
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    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 $dW(t)^2 = dt$.2012-10-05
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    Avatar: What do you call *evaluate*?2012-10-05
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    1. if you can do $\int e^{\lambda W} dW$, which is important, you can pick off the $\lambda^2$ terms. Similary, you are looking for a mnartingale starting $\frac {W^3} 3$ and the series expansion of the martingale $e^{\lambda W - \frac {\lambda^2} 2} $ (hermite polynoimial) is a good place to look.2012-10-05

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