Discussion
If \(X \sim N_{p}(\mu, \Sigma)\), then the distribution of \(N(\bar{X} - \mu)'\Sigma^{-1}(\bar{X} - \mu)\) would be :
Answer: C
The distribution is \(\chi^{2}\) with \(p\) degrees of freedom.
The mean \(\bar{X}\) of N independent \(N_{p}(\mu, \Sigma)\) vectors is \(N_{p}(\mu, \Sigma/N)\), so \(Z = \sqrt{N}\,\Sigma^{-1/2}(\bar{X} - \mu)\) is \(N_{p}(0, I)\). Then
\[ N(\bar{X} - \mu)'\Sigma^{-1}(\bar{X} - \mu) = Z'Z \]
is a sum of \(p\) independent squared standard normals, a \(\chi^{2}_{p}\). It is a scalar, not a multivariate normal vector. \(N - p\) degrees of freedom would arise only with \(\Sigma\) estimated (and then the result is an F, not a chi-square), and N degrees of freedom confuses the sample size with the dimension.
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