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Difficult Gaussian integral

  1. Jun 12, 2013 #1
    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 [itex]\vec x[/itex], and a covariance matrix [itex]\Sigma[/itex], how would you go about evaluating an expectation value of the form
    [itex]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}[/itex]
    I've tried expanding the [itex]f_{i}'s[/itex] into a power series, and then using the moment generating function to obtain the powers of [itex]x_{i}[/itex], but this really doesn't simplify the problem much. The function I'm currently considering is [itex]f_{i}(x_{i})=\cosh(x_{i})[/itex].
    I would extremely appreciate anyone's input on this problem.
     
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  3. Jun 12, 2013 #2

    mathman

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    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.
     
  4. Jun 12, 2013 #3
    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.
     
  5. Jun 12, 2013 #4

    Mute

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    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.
     
  6. Jun 13, 2013 #5

    mathman

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