Integration In multidimensional domain

**pollo** · October 17th 2008, 01:50 AM

How to resolve?:

$\int_{X} \exp{ \left( -\frac{X^{T}QX}{2} \right)}\left( -\frac{X^{T}QX}{2} \right)dX$

where $X$ is a vector $[x_{1},...,x_{n}]^T$ and $Q$ a n*n matrix. The integral should be calculated over $\mathbb{R}^n$ ..

In analogy with the 1dimensional case, I supposed to consider the argument as a sort of $Qx^2$ , and resolve the integral by parts , dividing the last term and trying to obtain the derivate of the term $\left( -\frac{X^{T}QX}{2} \right)$ .
But I am not shure how to proceed in the multidimensional domain. Is my thought valid? Someone can help me?
Thanks
Pollo

**Laurent** · October 17th 2008, 05:54 AM

To simplify the question (and reduce to a more classical one), you could consider diagonalizing $Q$ .

This is not always possible, but for every $X$ , $X^TQ X=X^T\frac{Q+Q^T}{2}X$ , hence you can replace $Q$ by $A=\frac{Q+Q^T}{2}$ , which is real and symmetric. As a consequence, $A$ diagonalizes in an orthonormal basis. Let $\lambda_1,\ldots,\lambda_n$ be its eigenvalues. Changing from one orthonormal basis to another doesn't change the volume (this is a rotation, perhaps plus a symmetry, anyway it is an orthogonal transformation), hence your integral becomes, using this change of basis:
$\int \frac{1}{2}\left(\sum_{i=1}^n \lambda_i x_i^2\right) \exp\left(-\frac{1}{2}\sum_{i=1}^n \lambda_i x_i^2\right) dx_1\cdots dx_n$ .
(I swapped the factors to avoid confusion) You can expand the first sum so that you get the sum of $n$ integrals, and then integrate each of them rather simply. I think the integral converges iff the $\lambda_i$ 's are positive.

One question remains: what is the connection between the $\lambda_i$ 's and $Q$ ? Don't you in fact assume that $Q$ itself is symmetric (and positive)?

I think what you finally get is: $\frac{1}{2}(2\pi)^{n/2}{\rm Tr}(A)$ , which is also $\frac{1}{2}(2\pi)^{n/2}{\rm Tr}(Q)$ . The previous question becomes: for which $Q$ does the integral converge? (or: for which $Q$ is $A$ positive?) I don't know the answer. Does anyone?

**pollo** · October 17th 2008, 06:17 AM

Thanks for the response. The fact is: this integral came out from a more general problem of Information Theory: I need to prove that a gaussian vector $X$ having a covariance matrix $Q$ , and then having a probability function
$f(X)=\frac{1}{\sqrt{(2\pi)^n\det{(Q)}}}\exp\left(-\frac{X^{T}Q^{-1}X}{2}\right)$
has entropy
$h(X)=\frac{1}{2}\log\left((2\pi e)^n\det(Q)\right)$

where the general definition of the entropy is
$h(X)=\int_{X}-f(X)\log(f(X))dX$

then, developing the formula, and exploiting the property of the probability function $\int_{X}f(X)dX=1$ , I stop when I have to calculate the purposed integral.

**Laurent** · October 17th 2008, 06:31 AM

Originally Posted by pollo

covariance matrix $Q$

Hence $Q$ is symmetric (and probably positive, if the distribution is not degenerate), and you can reread what I wrote with $Q$ instead of $A$ .

I made a mistake in my (unwritten) computation: in fact I get $\frac{1}{2}(2\pi)^{n/2}n$ . I think this coincides with your result (this term gives the " $e$ ").

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