I have two jointly normal variables X and Y with mean both zeros and variances $\sigma^2_{X}$ and $\sigma^2_{Y}$ separately, the covariance is $\sigma_{XY}$. Now I want to calculate the expected value of $Z=X*Y^{2}$, $E(Z)$. Any ideas?
Thanks.
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1$\begingroup$ Using conditional expectation may help. $\endgroup$– vinuxCommented Apr 12, 2013 at 13:04
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2$\begingroup$ Assuming that $X$ and $Y$ are jointly normal (as I included in my edit), vinux's hint is the way to go. Given the value of $X$, $Y$ is a normal random variable whose mean and variance are known, and so $$E[Z\mid X] = E[XY^2\mid X] = XE[Y^2\mid X] = X(\sigma_{Y\mid X}^2 + \mu_{Y\mid X}^2 = g(X)$$ where $g(X)$ is a cubic in $X$ and so only the quadratic and constant term matter: $E[X] = E[X^3] = 0$ in this case. $\endgroup$– Dilip SarwateCommented Apr 12, 2013 at 13:46
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2 Answers
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$E[XY^2] = E[ E[XY^2/ Y] ] = E[Y^2 E[X\mid Y]]=\alpha E[Y^3 ] =0$
$\alpha = \frac{\sigma_X}{\sigma_Y} \rho$
$\rho = cor(X,Y)$
The thing is : your expectation is an integral of an odd $0$-symetric function on $[-\infty, +\infty] $ this is why it's equal to zero
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1$\begingroup$ You've made an assumption that $E\[X|Y\] = 0$, which I don't think is justified. $\endgroup$ Commented Apr 12, 2013 at 15:39
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$\begingroup$ Yes, I should have written $E[X/Y]= \rho Y$ and then $E[Y^3] =0$ where $\rho$ is the correlation between $X$ and $Y$ $\endgroup$– dfhgfhCommented Apr 12, 2013 at 16:37
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2$\begingroup$ +1 More simply, $E[XY^2]=E[(-X)(-Y)^2]=-E[XY^2]$ follows from the symmetry $(X,Y)\to (-X,-Y)$ and immediately implies the expectation is zero. $\endgroup$– whuber ♦Commented Apr 12, 2013 at 17:04
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You could compute the joint moment generating function of $X$ and $Y$, which has the form $M_{X,Y}(\vec{t})=\operatorname{exp}(\frac{1}{2} \vec{t}^{\operatorname{T}} \Sigma \vec{t})$ since in this case the mean is zero. Here $\Sigma$ is the variance-covariance matrix.
Then $E(X Y^2)=\left . \frac{\partial}{\partial t_1} \frac{\partial^2}{\partial t_2^2} M_{X, Y}(\vec{t}) \right |_{\vec{t}=\vec{0}}$.
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$\begingroup$ And when you do this computation, what answer do you get? :-) $\endgroup$– whuber ♦Commented Apr 12, 2013 at 17:04