Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 32 / 1 / Solution

Put $G=X^{T}X$ and $H=X{G}^{-1}{X}^T$. The rank of a matrix assumption makes $G$ invertible, and $H$ is the orthogonal projection matrix onto the column space of the design matrix. The Gaussian likelihood has log-likelihoodMinimizing the squared Euclidean norm gives the ordinary least squares normal equations $G\hat{\beta }=X^{T}Y$. Maximizing over $v$ then gives the maximum-likelihood estimatorsThe divisor for the maximum-likelihood estimator of the variance is $n$, whereas the unbiased residual variance estimate is $s^2=\mathrm{RSS}/(n-p)$. The exceptional event $\mathrm{RSS}=0$ has probability zero under the Gaussian linear model with ${\sigma }^2>0$; on that event the likelihood is unbounded as $v\downarrow 0$.

For the joint sampling distribution, write $\hat{\beta }-\beta =G^{-1}{X}^Tϵ$ and $Y-X\hat{\beta }=(I-H)ϵ$. Their cross-covariance matrix vanishes because $X^T(I-H)=0$. These two vectors are jointly multivariate normal, so they are independent random variables. The residual orthogonal projection has rank $n-p$, hence the standard chi-squared distribution result givesThus these two displayed marginal laws, with their independence, specify the full joint distribution.

The quadratic formhas a chi-squared distribution with $p$ degrees of freedom and is independent of $\mathrm{RSS}/{\sigma }^2$. Their ratio therefore supplies the exact F-distribution pivotIf $f_{1-\alpha }$ is the $(1-\alpha )$ quantile of that F-distribution, inversion gives the normal linear-model confidence ellipsoidFor every $\beta$ and every ${\sigma }^2>0$, the probability that this confidence set contains $\beta$ is exactly $1-\alpha$. To test the null hypothesis, reject precisely when ${\beta }_0$ lies outside this confidence ellipsoid, equivalently when $T({\beta }_0)>f_{1-\alpha }$. Under the null hypothesis the rejection probability is exactly $\alpha$, independent of the unknown variance.

For the numerical calculation, the second column of the design matrix sums to zero and its sum of squares is $60$. Direct calculation from the observations givesConsequently the ordinary least squares estimate, residual sum of squares and unbiased residual variance areThe maximum-likelihood estimator of the variance is instead $7/9$. At the proposed null value the F-distribution statistic isTherefore reject the null hypothesis at the 5% level. The relevant critical value has $2$ numerator and $7$ residual degrees of freedom.