Discussion
Suppose a random vector \(\underline{X}\) has normal distribution with mean vector \(\underline{\mu}\) and dispersion matrix V. Then which of the following has a chi-square distribution ?
Answer: B
The chi-square variable is \((X - \mu)'V^{-1}(X - \mu)\).
If \(X \sim N_{p}(\mu, V)\) with V non-singular, write \(V^{-1} = (V^{-1/2})^{2}\) and set \(Z = V^{-1/2}(X - \mu)\). Then \(Z \sim N_{p}(0, I)\) and
\[ \begin{aligned} (X - \mu)'V^{-1}(X - \mu) &= Z'Z \\ &= \sum_{i=1}^{p} Z_{i}^{2} \sim \chi_{p}^{2} \end{aligned} \]
Using V instead of \(V^{-1}\) gives \(Z'V^{2}Z\), a weighted sum of chi-squares, which is chi-square only if \(V^{2} = I\). The exponentials are positive transforms of these quadratic forms and are not chi-square at all.
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