 Solution

Use the unbiased sample covariance matrixand assume $\Sigma$ is positive-definite and $n>p$, so $S$ is invertible almost surely. For a nonzero fixed direction $a$, the observations $Y_j=a^T{X}_j$ have mean $a^T\mu$ and variance $a^T\Sigma a$. Their squared Student t-test statistic for the projected null mean isPut $b=\bar{X}-{\mu }_0$. The Cauchy-Schwarz inequality giveswith equality when $a$ is proportional to $S^{-1}b$, if $b\ne 0$. For $b=0$ every projected statistic is zero. Maximizing the squared discrepancy over directions therefore gives Hotelling's T-squared statistic:This uses the squared generalized Rayleigh quotient. The preliminary formula on the printed cover has an unsquared numerator and is not valid as written: setting $a=t{C}^{-1}b$ makes that ratio $1/t$ for $b\ne 0$, so it is unbounded. The displayed Cauchy-Schwarz argument supplies the needed corrected identity independently.

Under the null, let $Z=\sqrt{n}(\bar{X}-{\mu }_0)$ and $C=(n-1)S$. Then $Z\sim N_p(0,\Sigma )$, $C\sim W_p(n-1,\Sigma )$, and $Z,C$ are independent. To see these sample properties, apply an orthogonal transformation to the $n$ observation indices with first row $n^{-1/2}(1,\dots ,1)$. Its first centered row is $Z$; its other $n-1$ rows are independent $N_p(0,\Sigma )$ residual contrasts, independent of the first. Their outer products sum to $C$, establishing the Wishart distribution and independence.

Since $T^2=(n-1)Z^T{C}^{-1}Z$, use the supplied Gaussian-Wishart quadratic-form law with $k=n-1$ to obtainThus a level-$\eta$ test rejects when this scaled statistic exceeds the $(1-\eta )$ quantile of the F-distribution. The Gaussian-Wishart law requires independence; it holds here because of the orthogonal sample decomposition. Its printed use of $X$ instead of the introduced $Z$ is a notational slip. A fixed-direction Student law alone would not calibrate the maximized statistic, because its maximizing direction depends on the sample.