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Multivariate Gaussian , prove normalization factor Watch

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    Hi,

    I am trying to follow my book's hint that to find the normalization factor one should

    "Diagnoalize \Sigma^{-1} to get n Gaussian which will have variance given by the eigenvalues of \Sigma . Then integrate gives \sqrt{2\pi}\Lambda_i, then use that the product of eigenvalues is the determinant

    . 2. Relevant equations

    What I know: \Sigma is symmetric and so it can be diagnolized \Sigma=PDP^{T} where P is the orthogonal matrix of eigenvectors and D is the matrix of eigenvalues

    3. The attempt at a solution I'm Stuck: I am blank as to where start explicitly to be honest, having not done any examples in this, if I have, at least for four years or so.

    Many thanks in advance
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    (Original post by xfootiecrazeesarax)
    Hi,

    I am trying to follow my book's hint that to find the normalization factor one should

    "Diagnoalize \Sigma^{-1} to get n Gaussian which will have variance given by the eigenvalues of \Sigma . Then integrate gives \sqrt{2\pi}\Lambda_i, then use that the product of eigenvalues is the determinant

    . 2. Relevant equations

    What I know: \Sigma is symmetric and so it can be diagnolized \Sigma=PDP^{T} where P is the orthogonal matrix of eigenvectors and D is the matrix of eigenvalues

    3. The attempt at a solution I'm Stuck: I am blank as to where start explicitly to be honest, having not done any examples in this, if I have, at least for four years or so.

    Many thanks in advance
    I believe this is the same question you have posted already, only slightly restated, is this correct?

    I think a problem is that no-one here is terribly familiar with the material of your course. I am very confident from the form of what you're trying to answer that the answer involves finding the square root S of Sigma, and then using a transform of the form y = Sx.

    For example, if f(x) is a "nice" function that vanishes as |x|->infty, then

    \int_{\mathbb{R}^n} f({\bf x}^T {\bf x})\, d{\bf x}= \sqrt{\det ( \Sigma )} \int_{\mahbb{R}^n} f({\bf x}^T\Sigma {\bf x}) \, d{\bf x}

    which can be shown by taking the square root of the diagonal matrix to find S with S^2 = Sigma, and then applying the linear change of variables y = Sx.

    If this result gives you what you want, I can explain why it holds. If it doesn't give you what you want you will need to supply a lot more detail. (And be prepared for me to go "OK, well, you need to do that bit, I don't know about Gaussian Fourier transforms in this context" or whatever).
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    (Original post by xfootiecrazeesarax)
    Hi,

    I am trying to follow my book's hint that to find the normalization factor one should

    "Diagnoalize \Sigma^{-1} to get n Gaussian which will have variance given by the eigenvalues of \Sigma . Then integrate gives \sqrt{2\pi}\Lambda_i, then use that the product of eigenvalues is the determinant

    . 2. Relevant equations

    What I know: \Sigma is symmetric and so it can be diagnolized \Sigma=PDP^{T} where P is the orthogonal matrix of eigenvectors and D is the matrix of eigenvalues

    3. The attempt at a solution I'm Stuck: I am blank as to where start explicitly to be honest, having not done any examples in this, if I have, at least for four years or so.

    Many thanks in advance
    Just to follow up DFranklin's excellent post (for which PRSOM), when you quote from a book, it would be very helpful if you could reference which book. My office is just over the bridge from the university library, so it's nice and easy to pop over there and look up the context of your question!

    Just to add to the substance, here's the heuristic of what's going on. When you are asked to normalize a probability distribution like the one you given, it amounts to performing a multi-dimensional integral - in this case of the function inside your red box. As \Sigma is symmetric, it diagonalizes nice and easily - and the point is that in the diagonal form the n-dimensional integral just becomes the product of n one-dimensional integrals each one of which is a standard Gaussian integral. You're going to get a \sqrt{2 \pi} \sigma_i from each of these where \sigma_i is an eigenvalue of \Sigma. The product of these eigenvalues is the determinant of \Sigma, hence the result follows.
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    (Original post by Gregorius)
    ..
    Thanks (also PRSOM!).
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    So I got the correct normalization constant by doing

    \tilde{y}=P^T \tilde{x},
    where \tilde{x}=x-\mu

    And \Sigma^{-1}=(PDP^T)^{-1} , D the eigenvalue matrix of \Sigma and P the eigenvector matrix

    I am looking at Kardar, which then says that 'similar manipulations can be used to find the characteristic function' , my lecture notes also say this, and mention we should convert back too !

    \tilde{p(\vec{k})}=e^{-i k.\lambda + \frac{1}{2} k.C.k}

    So the definition of \tilde{p(\vec{k})} is \tilde{p(\vec{k})} = <e^{-ik.x}>

    So using the same transformation as above I have  \frac{1}{\sqrt{det(2\pi C)}}\int e^{-1/2 y^T D^{-1} y} e^{-i k.(P.y)} dy

    Which doesn't look like it's simplified anything, since P isn't of an easy form?

    I'm guessing I need a different substitution? A

    ny help much appreciated anyone, thank you
 
 
 
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