Here X and Y follow bivariate normal distribution and Q is a new variable including X and Y. Prove that the new variable Q follows chi square distribution with 2 degree of freedom.
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1The inevitable questions: What have you tried? Where did you stall in your solution? – MasB Jul 22 '23 at 15:04
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I was trying to use the uniqueness property of mgf to prove it follows chi square distribution. But I'm not able to simply it to the mgf formula of chi square distribution – Soham Raut Jul 22 '23 at 15:17
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If (x y) is a k-dimensional Gaussian random vector with mean vector $\mu$ and rank k covariance matrix C, then q=((x y) -$\mu^T$)$C^{-1}$((x y)$^T$-$\mu$) is chi squared distributed with k degrees of freedom. (https://en.wikipedia.org/wiki/Chi-squared_distribution)
Here is another interesting link explaining the relationship between multivariate normal distribution and chi squared distribution: https://statproofbook.github.io/P/mvn-chi2.html
Laura Keller
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This is not a proof, so much -- I think --, as a loud restatement of what the original poster wanted proving. – kimchi lover Jul 22 '23 at 16:24
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