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The value of a stochastic integral, in this case integrating a Wiener process with respect to itself $$\int_0^T W(t)\;dW(t)$$ is dependent on the chosen position of the endpoint of the subintervals.

For example the Ito integral (leftpoint chosen) gives: $$\int_0^T W(t)\;dW(t)=\frac{W(T)^2}{2}-\frac{T}{2}$$ The extra term $-T/2$ is the so called Ito correction term.

My question concerns this very term: If $W(t)$ wasn't stochastic but deterministic the classical calculus would be used and this extra term wouldn't be there. You can e.g. show via simulation that it disappears when you set $\sigma=0$ and thereby render Brownian motion deterministic.

My question
When the deterministic integral is interpreted as the limiting case ($\sigma \rightarrow 0$) of the stochastic integral where the extra term $-T/2$ vanishes when $\sigma$ reaches $0$ (and thereby becomes deterministic), how can it be that the extra term $-T/2$ is only dependent on $T$ and not on $\sigma$?

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The variance of $W(T)$ is $\sigma T$. – Michael Hardy Jan 24 '12 at 21:11
@Michael: Could you please elaborate on that and put it into an answer. That would be great - thank you! – vonjd Jan 24 '12 at 21:14
TYPO: It's $\sigma^2 T$. – Michael Hardy Jan 24 '12 at 21:18
up vote 1 down vote accepted

Let $B(t)$ be the standard Wiener process, so that $\operatorname{var}(B(t))=t$. Let $W(t) = \sigma B(t)$, so that $\operatorname{var}(W(t))=\sigma^2 t$. If $$ \int_0^T B(t)\;dB(t) = \frac{B(T)^2}{2} - \frac T2, $$ then $$ \begin{align} \int_0^T W(t) \; dW(t) & = \sigma^2\int_0^T B(t)\;dB(t) \\ \\ & = \sigma^2\left( \frac{B(T)^2}{2} - \frac T2 \right) = \frac{\sigma^2 B(T)^2}{2} - \frac{\sigma^2 T}{2} \\ \\ & = \frac{W(T)^2}{2} - \frac{\sigma^2 T}{2}. \end{align} $$ Now let $\sigma^2\to0$.

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Thank you, the result squares very well with my simulations! – vonjd Jan 25 '12 at 7:12

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