Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 4 / 27K (Principles of Statistics)

The part of the log-likelihood depending on $\theta$ isCompleting the square separates the sum into a term independent of $\theta$ and $n(\overline{X}-\theta {)}^T{\Sigma }^{-1}(\overline{X}-\theta )$. Positive definiteness makes its unique minimum occur at $\overline{X}$. Since linear combinations of independent multivariate normal vectors are normal,The Hessian matrix $-n{\Sigma }^{-1}$ is negative definite, confirming the global likelihood maximum.

The pivot $Z=\sqrt{n}{\Sigma }^{-1/2}(\overline{X}-\theta )$ has independent standard normal coordinates, so $Z^{T}Z$ has the chi-squared distribution with $d$ degrees of freedom. Let $q_{d,1-\alpha }$ be its $(1-\alpha )$ quantile. The confidence ellipsoidtherefore has exactly $1-\alpha$ coverage for every $\theta$. No large-sample approximation is required when the covariance is known.

With $\Sigma =I_d$, constrained likelihood maximization minimizes $\Vert \theta -\overline{X}{\Vert }^2$. On the unit sphere this is $1+\Vert \overline{X}{\Vert }^2-2{\theta }^T\overline{X}$. By the Cauchy-Schwarz inequality its minimizer isIf $\overline{X}=0$, every unit vector is a maximum-likelihood estimator; a specified measurable tie-breaking choice makes an estimator defined everywhere. This event has probability zero in the nonsingular normal model, but the optimization still has that exceptional case.

Decompose $\overline{X}$ into its components parallel and orthogonal to the unit vector $v$. Its nearest point in the hyperplane $v^T\theta =0$ is the orthogonal component, since the two components' squared distances add. ThusThis is a unique orthogonal projection; for $d=1$ the hyperplane is the singleton zero and the formula still applies.

Use the convention that the likelihood ratio is maximized alternative likelihood divided by null likelihood. The alternative excludes zero, but its supremum agrees with the unrestricted maximum, even on the exceptional sample $\overline{X}=0$. Completing the square givesUnder the null, $\sqrt{n}\overline{X}\sim {\mathcal{N}}_d(0,I_d)$, so $2\log \Lambda \sim {\chi }_d^2$ exactly for every $n$. Equivalently $\Lambda$ is distributed as $e^{Y/2}$ for $Y\sim {\chi }_d^2$. With the reciprocal likelihood-ratio convention the statistic is $-2\log \Lambda$ instead.

Wilks theorem predicts a limiting chi-squared distribution with degrees of freedom equal to the difference of parameter dimensions, here $d-0=d$, under regularity and an interior null point of the unrestricted model. This example strengthens the asymptotic conclusion to an exact finite-sample identity.