Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 47 / 1

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.

Transform each observation to $Y_j=A^T{X}_j$. If $A^T$ has full row rank $m$, the transformed sample is multivariate normal with mean $A^T\mu$ and positive-definite covariance matrix $A^T\Sigma A$. Its sample mean is $A^T\bar{X}$ and its unbiased sample covariance matrix is $A^{T}SA$. The Hotelling test of linear hypotheses therefore usesHere $n>m$ is required. Reject for a large scaled statistic using the relevant upper F-distribution quantile. If the rows of $A^T$ are dependent, choose a basis for their row space, use that basis as the contrast matrix, and replace $m$ by its rank $r$. The null hypothesis is unchanged, but redundant contrasts must not be treated as independent dimensions. Rank zero gives no restrictions and hence no nontrivial test.

At equal spacing $h$, the adjacent increments of an affine mean profile are all $\beta h$. Their differences are therefore zero. Define the $(p-2)\times p$ matrix $B$ whose row associated with the interior time $i$ has coefficients $1,-2,1$ in columns $i-1,i,i+1$. ThenConversely these equations make all adjacent first differences equal, so ${\mu }_i$ is affine in $i$, equivalently affine in the equally spaced times. Thus the restriction is exactly $B\mu =0$, not just a necessary condition. Its null space has basis $1$ and $t=(t_1,\dots ,t_p{)}^T$, so $B$ has rank $r=p-2$.

For $p\ge 3$ and $n>r$, the second-difference test of a linear mean profile is the Hotelling test of linear hypotheses withReject the affine-profile hypothesis when the scaled statistic exceeds its level-$\eta$ upper F-distribution critical value. This removes the nuisance intercept and slope without estimating them separately.

For unequal ordered distinct times, put $h_i=t_{i+1}-t_i>0$ and replace each second-difference row by the adjacent-slope contrastIts three coefficients are $1/h_{i-1}$, $-(1/h_{i-1}+1/h_i)$ and $1/h_i$. Vanishing means all adjacent slopes agree, again giving null space $\mathrm{span}(1,t)$ and rank $p-2$. Use this new $B$ in the same statistic and null distribution. For $p\le 2$, every mean profile at distinct times is affine, so there is no lack-of-linearity restriction to test.