Past exam of the statistics course of the University of Oxford / 2023 / 1 / c

Put $S={\sum }_i(x_i-\bar{x}{)}^2$ and $q=S/{\sigma }_0^2$. Under the null hypothesis, maximizing over the nuisance mean gives ${\hat{\mu }}_0=\bar{x}$. The unrestricted normal mean and variance maximum-likelihood estimators are $\hat{\mu }=\bar{x}$, $\widehat{{\sigma }^2}=S/n$. Thus, for $S>0$,The likelihood-ratio test statistic is consequentlyThe usual model assumes $n\ge 2$. The event $S=0$ has probability zero under a nondegenerate normal distribution; at that observation the unrestricted likelihood is unbounded as the variance tends to zero, and the ratio has limiting value zero. This is the likelihood-ratio test for a normal variance with unknown mean.

The full parameter space has dimension two and the null space dimension one. The regular Wilks theorem therefore gives $W\stackrel{d}{\to }{\chi }_1^2$ under $H_0$ as $n\to \infty$. If $F_1$ denotes the cumulative distribution function of the chi-squared distribution with one degree of freedom, the approximate size-$\alpha$ rejection region isThe corresponding approximate p-value isThe one degree of freedom here describes the large-sample likelihood-ratio test statistic; it is different from the exact residual degrees of freedom $n-1$ in the next part.

The derivative of the log ratio is $\frac{1}{2}(n/q-1)$, so $\lambda (q)$ increases from zero to one on $(0,n)$ and decreases to zero on $(n,\infty )$. Rejecting for a sufficiently small ratio therefore gives two tails with $0<q_1<n<q_2$.

Under $H_0$, Cochran's theorem gives the exact sampling distribution $Q=S/{\sigma }_0^2\sim {\chi }_{n-1}^2$, independently of the unknown mean. Writing $F_{n-1}$ for its cumulative distribution function, the two cutoffs are determined byThe first equation calibrates the exact size of a statistical test. The second is $\lambda (q_1)=\lambda (q_2)$, so both tails belong to one likelihood ratio cutoff. Hence the exact critical region is $\boxed{\{q\le q_1\}\cup \{q\ge q_2\}}$. Its two tail probabilities are not generally equal: equal-tailed chi-squared cutoffs give a different test.