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This is the integration $$\int_0^t {\tau\ dW(\tau)}$$ where $W(t)$ is Wiener Process.

I've check using wolfram mathematica that the solution is $$\frac{t}{\sqrt{3}} W(t)$$ But, I completely don't know why. I'm new in SDE.

The clue from the problem is you have to use Ito's lemma by define $G(t,X)=tX$ when $X(t)=W(t)$. But, when I try it, the solution is: $$(t-1)W(t)$$

Could you help me? How to solve it?

fahadh4ilyas
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  • Can you write out your calculation? My attempt was to construct a function $f(t,x)$ so that if for $g(t)=f(t,W(t))$ we have $dg=t dW(t)$. But then I get an inconsistent system of PDEs, namely $\frac{\partial f}{\partial x}=t$ and $\frac{\partial f}{\partial t} + \frac{1}{2} \frac{\partial^2 f}{\partial x^2} = 0$. The first gives $f(t,x)=tx+g(t)$, but then the second gives $x+g'(t)=0$ which is impossible. – Ian Mar 05 '17 at 01:21
  • Maybe I just made a silly mistake or maybe your final solution is incorrect. (I also tried computing it using the Paley-Wiener technique and I don't think the result I got agrees with yours, either.) – Ian Mar 05 '17 at 01:21
  • @Ian I think my calculation is incorrect because I blindly follow the example without even know the way the example use. – fahadh4ilyas Mar 05 '17 at 01:45
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    I actually don't even think the solution you were given is correct. Consider that $\int_0^t s dW(s) = t W(t) - \int_0^t W(s) ds$ (this amounts to "formal integration by parts", but it can be proven to be the same as the result of the Ito integral; search "Paley-Wiener integral" for details). Then I don't think it is true that $\int_0^t W(s) ds = \left ( 1-3^{-1/2} \right ) t W(t)$. – Ian Mar 05 '17 at 02:14
  • @Ian Ah that's it. My calculation stop at calculating $\int_0^t W(s) ds$. I thought the answer is just $W(t)$. That's why I feel like my calculation wrong. Oh, and $t/\sqrt{3} W(t)$ this solution I got from wolfram. So, I also don't know this is correct or not – fahadh4ilyas Mar 05 '17 at 02:43
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    No, $\int_0^t W(s) ds$ is not $W(t)$. I'm actually not sure if there is a good way to simplify that expression at all. – Ian Mar 05 '17 at 06:47

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