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I wish to evaluate the following:

  • $E[W(t-1)W(t)^2]$

  • $E[W(t)^3]$

where $t > 1$, $W$ is a standard Brownian motion and we are at $\mathscr{F}_0$ now.

I know that $E[W(t-1)W(t)] = \min{(t-1,t)} = (t-1)$ when $W$ is a standard Brownian motion, but I'm not sure how to solve the above expectations. I could do it easily if I could go $E[W(t-1)W(t)^2]=E[W(t-1)]E[W(t)^2]=0$ (duh); however I can't do this beacuse they're dependent random variables.

Thanks.

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1 Answer 1

For the first one, let $X = W(t) - W(t-1)$ and write $W(t-1)W(t)^2 = W(t-1)(X + W(t-1))^2$. After expanding the square you have three terms, each of which is a product of independent random variables.

For the second one: all you need to know is that $W(t)$ is normally distributed with mean 0 and variance $t$, so you know its density function, and can compute the expectation directly from that (integration by parts will help).

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I beg to differ for the second one: the distribution of W(t) is symmetric hence E[u(W(t))]=0 for every odd function u, for example u(x)=x^3. –  Did Dec 13 '12 at 6:34

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