# Expectation of Univariate Quadratic Form under Multivariate Gaussian

Is there an obvious trick I am missing for solving the following integral:

$$\int_x P(y|x) W(x) (-x^TMx+2x^Tm -c)dx$$

Distributions are Gaussians and $M$ is symmetric.

I know how to do the expectation for a univariate Gaussian but I'm not sure of this multivariate case since $$P(y|x) W(x) = V(y,x)$$

$W$ is the PDF of $x$, $P$ is a conditional PDF of $y$ given $x$ and $V$ is the joint PDF of $x$ and $y$.

• Im sure this is answered before on this site, but on my mobile Phone now so too burdensone to search – kjetil b halvorsen Oct 4 '15 at 11:11
• Is $P$ th density? Waht is $W, V$? Please define! – kjetil b halvorsen Oct 4 '15 at 15:41
• $W$ is the PDF of $x$, $P$ is a conditional PDF of $y$ given $x$ and $V$ is the joint distribution density. All Gaussians. – user136228 Oct 4 '15 at 15:44
• Did you look at math.stackexchange.com/questions/442472/… ? And, what is the point of using $y$ in th expression above, when the quadratic form you want the expectation of do not depend on $y$? It shoul b irrlevant! – kjetil b halvorsen Oct 4 '15 at 15:52
• But the function $P(y|x)$ depends on $x$. We shouldn't be able to discard it and apply the expectation under $W(x)$ only. I mean now just from a calculus stand point, $P$ is inside the integral and it depends on $x$. – user136228 Oct 4 '15 at 16:04

So it seems there is a very easy trick to solve this. We only need to use the Bayes rule to get the marginal distribution $P(y)$: $$P(y)W(x|y) = P(y|x)W(x)$$
$P(y)$ can then be taken outside the integral and we end up with an expectation over a univariate conditional Gaussian $W(x|y)$ which can be calculated for example as stated in the matrix cookbook.
So, you have a multinormañ vector $(y,x)$ (where both components $y, x$ can be vectors), anf want the expectation of the quadratic form $-x^T M X +2x^T m - c$, $M$ an symmtric constant matrix, $m$ a constant vector and $c$ a constant scalar. Since this funtion (random variable) you want the expectation of, do not deend on $y$, the distribution of $y$ is irrlevant for the expectation. For the first part, you can just apply my answr to sum of squares of dependent gaussian random variables Then $\text{E} x^T m$ is simply $\text{E}(x)^T m$.