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Let $\{W_t\}_{t\ge0}$ be standard Wiener process. Show that process $\{Y_t\}_{t\ge0}$ defined below

$$Y_t:=\int_0^t W_s ds$$

satisfies the condition $\mathbb{E}Y_t^3=0$.


Firstly, I calculate

$$d(W_t^3)=3W_t^2dW_t+3W_tdt$$

So:

$$Y_t=\frac{1}{3}W_t^3-\int_0^tW_s^2dW_s$$

And:

$$\mathbb{E}Y_t^3=\mathbb{E}\left(\frac{1}{27}W_t^9\right)-\mathbb{E}\left(3\cdot \frac{1}{9}W_t^6 \int_0^tW_s^2dW_s\right)+\mathbb{E}\left(\frac{1}{3}W_t^3 \left(\int_0^tW_s^2dW_s\right)^2\right)-\mathbb{E}\left(\int_0^tW_s^2dW_s\right)^3$$

I know $\mathbb{E}\left(\frac{1}{27}W_t^9\right)$ is equal to $0$, but how to deal with the rest?

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  • $\begingroup$ Isn't there a square missing in the 3rd term of the RHS (last equation), when you expand the cube? (and the last sign is flipped, should be a minus) $\endgroup$ – Clement C. Oct 9 '15 at 14:11
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    $\begingroup$ $Y_t$ is a centered Gaussian variable, so all its odd-order moments are zero. $\endgroup$ – zhoraster Oct 9 '15 at 15:36
  • $\begingroup$ zhoraster, can you put it as an answer? and maybe explain a bit more? $\endgroup$ – luka5z Oct 9 '15 at 15:57
  • $\begingroup$ By this question, we have that $Y_t$ is normal with mean $0$ and therefore $\mathbb{E}(Y_t^3)=0$. $\endgroup$ – saz Oct 9 '15 at 16:51

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