Discussion
If \(x_{1}\), \(x_{2}\), ........, \(x_{n}\) \((n \ge p + 1)\) are distributed independently each according to \(N_{p}(\mu, \Sigma)\), then the distribution of
\[ \begin{aligned} S = \frac{1}{n - 1}\sum_{\alpha=1}^{n} &(x_{\alpha} - \bar{x}) \\ &(x_{\alpha} - \bar{x})' \end{aligned} \]
is :
Answer: A
S has the distribution \(W_{p}(n - 1, \Sigma/(n - 1))\).
The matrix of sums of squares and products \(A = \sum(x_{\alpha} - \bar{x})(x_{\alpha} - \bar{x})'\) has the Wishart distribution \(W_{p}(n - 1, \Sigma)\): one degree of freedom is used in estimating the mean. Scaling a Wishart matrix by a constant scales its parameter matrix, so
\[ \begin{aligned} S &= \frac{A}{n - 1} \\ &\sim W_{p}\!\left(n - 1, \frac{\Sigma}{n - 1}\right) \end{aligned} \]
The condition \(n \ge p + 1\) makes S non-singular. Option B uses \(n\) degrees of freedom, which would be right only if \(\mu\) were known and used instead of \(\bar{x}\). S is a random matrix, so a normal distribution for it (C and D) makes no sense.
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