Discussion
Let \(X \sim N(0, \Sigma)\) where \(\Sigma\) is a +ve definite matrix, and A is an idempotent matrix of rank \(k \lt P\). Let \(Z = X^{1}\Sigma^{-1}AX\). Then mean and variance of \(z\) respectively equal :
Answer: B
The mean and variance are \(k\) and \(2k\).
Reading the printed \(X^{1}\) as the transpose \(X'\), \(Z = X'BX\) with \(B = \Sigma^{-1}A\). For \(X \sim N(0, \Sigma)\),
\[ \begin{aligned} E(Z) &= \mathrm{tr}(B\Sigma) = \mathrm{tr}(\Sigma^{-1}A\Sigma) \\ &= \mathrm{tr}(A) = \mathrm{rank}(A) = k \end{aligned} \]
since an idempotent matrix has trace equal to its rank. Also \((B\Sigma)^{2} = \Sigma^{-1}A^{2}\Sigma = B\Sigma\), so \(B\Sigma\) is idempotent of rank \(k\), and (with B symmetric, as the standard result assumes) Z is distributed as chi-square with \(k\) degrees of freedom, whose variance is \(2\,\mathrm{tr}(B\Sigma)^{2} = 2k\). Options in \(n\) use the dimension instead of the rank, and 'k, k' gives the right mean but a chi-square variable's variance is twice its degrees of freedom.
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