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Math Help - Gaussian vector, again

  1. #1
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    Gaussian vector, again

    Let (\xi_1,\xi_2,...,\xi_n) be a Gaussian vector with zero mean and covariance matrix B. What is the distribution of the random variable defined by the conditional expectation \mathbb{E}(\xi_1|\xi_2,...,\xi_n)?

    The density of the random vector is of the form:
    p_{\xi}(x_1,x_2,...,x_n)=(\det B)^{-\frac{1}{2}}(2\pi)^{-\frac{n}{2}}\exp(-\frac{1}{2}(B^{-1}x,x))

    One possible approach is to use conditional densities (as described for example in Grimmett & Stirzaker), in order to get the density of the conditional expectation. Because of the correlation, explicit formula for the conditional density cannot be easily written. Using only the joint density, you could get to an expression of the form:

    \mathbb{E}(\xi_1|\xi_2\leq x_2,...,\xi_n\leq x_n)=\int_{-\infty}^{+\infty}\int_{-\infty}^{x_2}...\int_{-\infty}^{x_n}u_1p_{\xi}(u_1,u_2,...,u_n)du_1...du_  n

    A change of variable seems natural at this point, in order to get to the independence case, but difficulty arises in defining the image of the domain of integration. So this is not really helpful.

    But there surely is a better way of attacking the problem.
    Thanks for your help.
    Last edited by akbar; January 4th 2010 at 03:35 PM.
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  2. #2
    MHF Contributor

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    Hi,
    you may like to have a look at this book (thm 2 p303, available online!), for a slightly more general result (distribution of E[X|Y] when (X,Y) is a (d_X+d_Y)-dimensional Gaussian random vector) with a proof (using specific properties of Gaussian vectors, no density function).
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  3. #3
    Junior Member
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    Glad to see you're still awake at this time. Thanks for answering.
    Thanks for the link. It happens I have the reference.
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