1
$\begingroup$

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.

$\endgroup$
2
  • 1
    $\begingroup$ Using conditional expectation may help. $\endgroup$
    – vinux
    Apr 12, 2013 at 13:04
  • 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. $\endgroup$ Apr 12, 2013 at 13:46

2 Answers 2

3
$\begingroup$

$E[XY^2] = E[ E[XY^2/ Y] ] = E[Y^2 E[X\mid Y]]=\alpha E[Y^3 ] =0$

$\alpha = \frac{\sigma_X}{\sigma_Y} \rho$

$\rho = cor(X,Y)$

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

$\endgroup$
3
  • 1
    $\begingroup$ You've made an assumption that $E\[X|Y\] = 0$, which I don't think is justified. $\endgroup$ Apr 12, 2013 at 15:39
  • $\begingroup$ Yes, I should have written $E[X/Y]= \rho Y$ and then $E[Y^3] =0$ where $\rho$ is the correlation between $X$ and $Y$ $\endgroup$
    – dfhgfh
    Apr 12, 2013 at 16:37
  • 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
    Apr 12, 2013 at 17:04
0
$\begingroup$

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 $E(X Y^2)=\left . \frac{\partial}{\partial t_1} \frac{\partial^2}{\partial t_2^2} M_{X, Y}(\vec{t}) \right |_{\vec{t}=\vec{0}}$.

$\endgroup$
1
  • $\begingroup$ And when you do this computation, what answer do you get? :-) $\endgroup$
    – whuber
    Apr 12, 2013 at 17:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.