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On a lecture notes, there is a following arguement: To make $\int_0^T \pi_t dW_t$ well-defined, (maybe it means to make $\int_0^T \pi_t dW_t<\infty \ \ a.s.$) we only need $\int_0^T \pi_t^2 dt<\infty$. Where $W_t$ is a standard one dimention Brownian Motion, $\pi_t$ is a previsible process.

Question: I don't understand how we can use condition $\int_0^T \pi_t^2 dt<\infty$ to make the stochastic integral well-defined. Do we need to add some other conditions??


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The construction of the Ito integral starts with simple (=piecewise constant) functions for which the formula is just $$ –  Ilya Jan 31 '13 at 17:13

2 Answers 2

I think that you might have missed the fact that you cannot define a path-by-path Stochastic Integral with respect to Brownian Motion. That's because the paths of BM are of infinite variations on compacts almost surely.

On the other hand the quadratic variations of the paths of BM tends are equal to $t$ over the interval $[0,t]$ which is why this kind of conditions (among others like predictability which is also mandatory) makes it possible to define a stochastic integral with respect to a BM.

Best regards

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Thank you for your answer. I think we need condition like $E \int (...)^2dt<\infty$ to make its integrable well-defined. I don't understand why we don't need to take expectation in the condition. –  XXX11235 Jan 26 '13 at 16:37
Any comments??? –  XXX11235 Jan 29 '13 at 19:15

I think that the expectation-less condition (and others) is needed to define correctly the integral (and get its basic properties like linearity), just like TheBridge said. To get more, that is, to get the zero mean property or isometry property, then, you are right, an additional (or rather stronger) condition $$\int_0^T\mathbb{E}\left[\pi_t^2\right]dt < \infty $$ is needed. Hope this helps.

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