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Math Help - Joint pdf of standard normal and chi sq.

  1. #1
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    Joint pdf of standard normal and chi sq.

    X_{1},...,X_{N} is a random sample from a N(\mu,\sigma^2) distribution, where \mu and \sigma are unknown.

    I'm given:

      Z = \frac{\sqrt{n}(\overline{X} - \mu)}{\sigma}

    U = \frac{(n - 1)S^2}{\sigma^2}

    where

    \overline{X} = \sum\frac{X_{i}}{n}

    S^2 = \frac{\sum(X_{i} - \overline{X})^2}{n-1}

    Now, I think that Z is the standard normal dist, and U is dist. Chi Sq with n-1 degrees of freedom.

    I'm now asked to find the joint pdf of Z and U; how would I do this?

    I'm also given:

     T = \frac{\sqrt{n}(\overline{X} - \mu)}{S}

    and asked to deduce the joint pdf of T and U; again, I think T has the t dist. with n-1 degrees of freedom, but I'm not sure how to go about doing this question.
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  2. #2
    Moo
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    Hello,

    It may be easier to deal with standard normal distribution... : Z_1,\dots,Z_n \sim \mathcal{N}(0,1)

    By knowing that if A follows a normal distribution (mu,sigmaČ), aA+b follows a normal distribution (a*mu+b , aČ sigmaČ), you have :

    Z=\sqrt{n}\cdot \bar Z

    and U=\sum_{i=1}^n (Z_i-\bar Z)^2

    where \bar Z=\frac 1n \sum_{i=1}^n Z_i

    And it can be proved that U and \bar Z are independent. But the only proof I know uses Gaussian vectors...
    Last edited by Moo; August 8th 2009 at 05:01 AM.
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  3. #3
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    I think I've found a proof for it, just trying to get my head round it

    And... for the joint pdf of T and U, I've said:

    R = U

    T = \sqrt{n-1}U^{-\frac{1}{2}}Z

    Therefore Z = TR^{\frac{1}{2}}(n-1)^{-\frac{1}{2}}

    So I can find the joint pdf of T and R, using Jacobian transformation...

    However, my Jacobian keeps coming up as 0.

    I've got:

    \frac{dz}{dt} = r^{\frac{1}{2}}(n-1)^{-\frac{1}{2}}

    \frac{dz}{dr} =\frac{1}{2}tr^{-\frac{1}{2}}(n-1)^{-\frac{1}{2}}

    \frac{du}{dt} = -2z^2t^{-3}(n-1)

    \frac{du}{dr} = 1

    And then I calculate J and I get 0 :S... so I'm assuming one or more of my differentials is wrong; are they? Or am I just going about this the wrong way?
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