Can someone help me figure out how to take the expected value of the quantity $[(x-\mu_x)^2 (y-\mu_y)^2]$?

I know that $E[(x-\mu_x)^2]$ is of course $\sigma^2_x$, $E[(y-\mu_y)^2]$ is of course $\sigma^2_y$, and $E[(x-\mu_x)(y-\mu_y)]$ is $\sigma^2_{xy}$, but it's tripping me up to have four terms inside the brackets instead of two, and I'm not sure how to deal with it.

I tried expanding it to a sum of terms and then taking the expected value of each term, but I think I must be doing it wrong. Expanding yields:

$E[(x^2 -2x\mu_x + \mu^2_x)(y^2 -2y\mu_y + \mu^2_y)]$

$E[x^2 y^2] - 2\mu_y E[x^2 y] + \mu^2_y E[x^2] - 2\mu_x E[xy^2] + 4\mu_x \mu_y E[xy] - 2\mu_x \mu^2_y E[x] + \mu^2_x E[y^2] - 2\mu^2_x \mu_y E[y] + \mu^2_x \mu^2_y E[1]$

Then, because x and y are independent, this simplifies to:

$E[x^2]E[y^2] - 2\mu^2_y E[x^2] + \mu^2_y E[x^2] - 2\mu^2_x E[y^2] + 4\mu^2_x \mu^2_y - 2\mu^2_x \mu^2_y + \mu^2_x E[y^2] - 2\mu^2_x \mu^2_y + \mu^2_x \mu^2_y$

And then to:

$E[x^2]E[y^2] - \mu^2_y E[x^2] - \mu^2_x E[y^2] + \mu^2_x \mu^2_y$

$(E[x^2] - \mu^2_x)(E[y^2] - \mu^2_y)$

Which equals $\sigma^2_x \sigma^2_y$.

However, I feel like I must be doing something wrong. $x$ and $y$ are assumed to be jointly Gaussian, so what about $\sigma^2_{xy}$? Surely that must show up somewhere? Can anyone point out my mistake?


  • $\begingroup$ Welcome to math.stackexchange! What have you tried so far? It'd help to see your working for people to point out where your calculations can be simplified etc. $\endgroup$ – owen88 Mar 7 '18 at 21:41
  • $\begingroup$ $Ex^{2}y^{2}=Ex^{2}Ey^{2}$ assuming that the variables are independent. (Without the assumption of independence there is no answer to this question). Can you complete the computation now? $\endgroup$ – Kavi Rama Murthy Mar 8 '18 at 6:28
  • $\begingroup$ That helps, thank you! However, I think my computations are wrong, because my answer is only in terms of $\sigma^2_x$ and $\sigma^2_y$, when I would expect it to also be in terms of $\sigma^2_{xy}$ because $x$ and $y$ are jointly Gaussian. I updated my post with the expansion and simplification I got, in the hope that someone can explain where I'm going wrong. $\endgroup$ – Erika Lynn Mar 8 '18 at 7:52
  • $\begingroup$ If $x$ and $y$ are independent, then so are $(x-\mu_x)^2$ and $(y - \mu_y)^2$. So immediately you have $E[(x-\mu_x)^2(y - \mu_y)^2] = E[(x-\mu_x)^2]E[(y - \mu_y)^2] = \sigma_x^2 \sigma_y^2$, which agrees with the result obtained by your longer argument. $\endgroup$ – Bungo Mar 8 '18 at 8:13
  • $\begingroup$ Thanks for the response! Then why is $E[(x-\mu_x)(y-\mu_y)] = \sigma^2_{xy}$? $\endgroup$ – Erika Lynn Mar 8 '18 at 9:03

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