# Difficult Gaussian integral

1. Jun 12, 2013

### unchained1978

I'm dealing with multivariate normal distributions, and I've run up against an integral I really don't know how to do.

Given a random vector $\vec x$, and a covariance matrix $\Sigma$, how would you go about evaluating an expectation value of the form
$G=\int d^{n} x \left(\prod_{i=1}^{n} f_{i}(x_{i})\right) e^{-\frac{1}{2} \vec x \cdot \Sigma^{-1}\cdot \vec x}$
I've tried expanding the $f_{i}'s$ into a power series, and then using the moment generating function to obtain the powers of $x_{i}$, but this really doesn't simplify the problem much. The function I'm currently considering is $f_{i}(x_{i})=\cosh(x_{i})$.
I would extremely appreciate anyone's input on this problem.

2. Jun 12, 2013

### mathman

Transform the random vector x into a random vector y consisting of independent random variables. Each xi is a linear combination of yj's. cosh(xi) will then expand into a product of cosh(yj)'s and the integrals can then be evaluated.

As you see it may be quite messy, bur doable.

3. Jun 12, 2013

### unchained1978

I thought about that, but the argument of the cosh becomes a sum, which really doesn't simplify things at all. Thanks though. I've thought about using Isserlis's theorem, (also known as Wick's theorem) but I can't find a good summary/statement of the theorem that's more precise than Wikipedia's.

4. Jun 12, 2013

### Mute

Another option would be to write $\cosh(x_i) = (e^{x_i}+e^{-x_i})/2$; expanding out the products, the overall integral will look something like

$$G \propto \sum_{\sigma}\int d^{n} x \exp\left[-\frac{1}{2} \vec x \cdot \Sigma^{-1}\cdot \vec x + \vec J_\sigma \cdot \vec x\right],$$
where the $J_\sigma$ is a vector with entrees of $\pm 1$, where $\sigma$ denotes all of the permutations of the $\pm 1$ signs. You can now complete the square and evaluate each of the integrals in the sum. I don't know if you will be able to write the overall result in a nice, closed form, though.

5. Jun 13, 2013

### mathman

Mute's point is well taken. Once you write cosh(xi) = (exp(xi) + exp(-xi))/2, either approach works.