Discussion
If \((X, Y)'\) follows bivariate normal distribution with mean vector \((0, 0)'\) and variance-covariance matrix \(\Sigma = \begin{pmatrix} 1 & 0.5 \\ 0.5 & 1 \end{pmatrix}\), which of the following does not have a \(\chi^{2}\)-distribution ?
Answer: A
\(X^{2} + Y^{2}\) does not have a chi-square distribution.
X and Y are each standard normal, so \(X^{2}\) and \(Y^{2}\) are \(\chi^{2}_{1}\). But they are correlated (\(\rho = 0.5\)), so \(X^{2} + Y^{2}\) is not a sum of independent squares. Its distribution is that of \(\lambda_{1}Z_{1}^{2} + \lambda_{2}Z_{2}^{2}\), where \(\lambda_{1} = 1.5\) and \(\lambda_{2} = 0.5\) are the eigenvalues of \(\Sigma\), a weighted sum that is not chi-square.
The quadratic form \(\mathbf{x}'\Sigma^{-1}\mathbf{x}\) is always \(\chi^{2}_{2}\) for a bivariate normal with mean 0. Here
\[ \Sigma^{-1} = \frac{1}{0.75}\begin{pmatrix} 1 & -0.5 \\ -0.5 & 1 \end{pmatrix},\quad \mathbf{x}'\Sigma^{-1}\mathbf{x} = \frac{4}{3}(X^{2} - XY + Y^{2}) \]
so the last option is \(\chi^{2}_{2}\).
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