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

Sort the observations into order statistics $x_{(1)}\le \cdots \le x_{(n)}$. Choose plotting positions $p_i=(i-1/2)/n$ and calculate the standard normal quantiles $z_i={\Phi }^{-1}(p_i)$. The normal Q-Q plot consists of the pairs $(z_i,x_{(i)})$.

If the normal distribution model is appropriate, the points should lie approximately on a straight line with positive slope, because its population quantile function isThe intercept is approximately $\mu$ and the slope approximately $\sigma$; neither needs to be known in advance. The reference need not be the identity line. One justification is the probability integral transform: for an independent normal statistical sample, $\Phi ((X_{(i)}-\mu )/\sigma )$ are uniform order statistics, whose typical values are near their plotting positions. The equality is approximate for a finite sample, particularly at the extremes. Systematic curvature, unusually heavy tails, or isolated extreme departures argue against the proposed distributional shape. A straight marginal Q-Q plot does not itself establish independence of observations.

Expanding the exponent, the log-likelihood function iswhere $C$ is independent of $\theta$. Its score function and second derivative areThe first is positive below $\bar{y}$ and negative above it, so the unique maximum-likelihood estimator isUsing $\mathbb{E}{\sum }_i{Y}_i=n\theta$, the one-observation Fisher information is $I(\theta )=-n^{-1}\mathbb{E}{\ell }^{\prime \prime }(\theta )={\theta }^{-3}$. The support does not depend on the interior parameter $\theta >0$, and the regular asymptotic normality of a maximum likelihood estimator givesThis is Inverse Gaussian mean estimation with known shape one. Equivalently, the Inverse Gaussian distribution has variance ${\theta }^3$, so the same result follows directly from the central limit theorem applied to its sample mean.

The estimated standard error is ${\hat{\theta }}^{3/2}/\sqrt{n}$. Using the central 95% standard normal quantiles $\pm 1.96$, a Wald confidence interval isConsistency of $\hat{\theta }$ and Slutsky's theorem justify replacing the unknown scale ${\theta }^{3/2}$ by its estimate. The coverage probability is asymptotic, not exact at finite $n$. Intersecting with the positive parameter space removes impossible values without changing whether the true positive parameter is covered.

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.