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Math Help - Multivariate Normal Distribution

  1. #1
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    Multivariate Normal Distribution



    I know that f(x_1, x_2, x_3) = \frac{1}{(2 \pi)^{3/2}|\Sigma|^{1/2}}exp(-\frac{1}{2}x \Sigma^{-1} x) since n = 3 and mu = 0.
    I've never used the multivariate normal distribution. My prof just derived it, but never taught us how to use it.

    Any help would be appreciated.
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  2. #2
    MHF Contributor matheagle's Avatar
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    You may think this is a multivariate problem, but it isn't
    Just find the distribution of W=X_1-2X_2+X_3\sim N(0,\sigma^2)

    where \sigma^2=(1,-2,1)\Sigma(1,-2,1)^t
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  3. #3
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    W is a chi-squared distribution with 3 degrees of freedom.

    If that's true, what do I do with?
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  4. #4
    MHF Contributor matheagle's Avatar
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    why don't you get the square root of 15.36 and do it directly?
    REMEMBER x^2>4 is x>2 OR x<-2.
    I'm not sure about this chi-square 3 thingy
    It's not X_1^2 +X_2^2....
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  5. #5
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    So I did what you said:

    P(- \sqrt{15.36}<X_1 -2X_2 + X_3<\sqrt{15.36}

    Can I do the following:

    P(- \sqrt{15.36}<X_1<\sqrt{15.36}) + P(- \sqrt{15.36}<-2X_2<\sqrt{15.36}) + P(- \sqrt{15.36}<X_3<\sqrt{15.36})

    and <br />
\sigma^2=(1,-2,1)\Sigma(1,-2,1)^t = 5<br />
    Last edited by statmajor; October 7th 2009 at 06:42 PM.
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  6. #6
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by statmajor View Post
    So I did what you said:

    P(- \sqrt{15.36}<X_1 -2X_2 + X_3<\sqrt{15.36}

    Can I do the following:

    P(- \sqrt{15.36}<X_1<\sqrt{15.36}) + P(- \sqrt{15.36}<-2X_2<\sqrt{15.36}) + P(- \sqrt{15.36}<X_3<\sqrt{15.36})

    and <br />
\sigma^2=(1,-2,1)\Sigma(1,-2,1)^t = 5<br />

    no, those events are not mutually exclusive.
    JUST let

    W=X_1 -2X_2 + X_3

    Then figure out what the mean (0) and the variance of W is.
    THEN standardize

    Z={W-E(W)\over \sqrt{V(W)}}

    and use the st normal table
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  7. #7
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    Sorry to bother you again (and sound like an idiot), but I still can't figure out what what W is.

    It's probably something simple, but just can't get it.

    Sorry.
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