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What is $\int_{0}^{t}\sqrt{\frac{1}{T-s}}dW(s)$ where $0\le t <T$ and $T \in \mathbb{R}$ and W is a one dimensional brownian motion.

I tried applying Ito's rule with $f(x)= \sqrt{T-x}$. but I did not get the answer.

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Hint

Yes, Ito's rule won't help to my knowledge. That's useful for when you have $W$'s in the integrand. You're actually in the somewhat simpler territory where you don't. Think about it: the $dW_s's$ are independent normals with mean zero and variance $ds$ and they are multiplied by $\frac{1}{\sqrt{T-s}}.$ So you are adding up a bunch of independent normals with mean zero and variance $\frac{ds}{T-s}.$ What do you get?

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    $\begingroup$ Normal random variable with mean zero and variance 1/(T-t)??? $\endgroup$ – manhattan Mar 8 '17 at 3:41
  • $\begingroup$ @manhattan Not quite (wrong variance). $\endgroup$ – spaceisdarkgreen Mar 8 '17 at 3:43
  • $\begingroup$ variance is logT/(T-t)? $\endgroup$ – manhattan Mar 8 '17 at 3:44
  • $\begingroup$ @manhattan Hey you're not OP! (but yes) $\endgroup$ – spaceisdarkgreen Mar 8 '17 at 3:46
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    $\begingroup$ I should add that I don't believe there's simplified expression for the answer as a random variable (say, in terms of $W_t$ or whatever) so in a sense the question is a bit misleading but this gives its distribution, which is useful. $\endgroup$ – spaceisdarkgreen Mar 8 '17 at 3:50

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