Past exam of the statistics course of the University of Oxford / 2017 / 1 / b

Put $A={\sum }_i{x}_i$, $B={\sum }_i{x}_i^2$ and $M_2=B/n$. The normal distribution model givesand, up to a constant independent of $\mu$, its log-likelihood function isThe score function and its zero equation areFor $B>0$, this quadratic equation has exactly one positive root, giving the maximum-likelihood estimatorThe numerator $B-A\mu -n{\mu }^2$ of the score function is positive between its negative and positive roots and negative above its positive root. Thus the likelihood function increases up to $\hat{\mu }$ and decreases afterwards: this is its unique global maximum on $(0,\infty )$. Also, $\ell (\mu )\to -\infty$ at both ends. The all-zero observation vector has probability zero; on that exceptional vector $L(\mu )\propto {\mu }^{-n}$ and there is no attained maximum.

The second derivative and the relevant moments areTherefore the expected Fisher information isThe support of a probability distribution is the whole real line, independent of the positive statistical parameter. The law of large numbers gives $\overline{X}\to \mu$ and $M_2\to 2{\mu }^2$ in probability, so the displayed maximum-likelihood estimator is a consistent estimator. The regular asymptotic normality of a maximum likelihood estimator now yieldsThus the requested leading asymptotic variance on the unscaled estimator's scale is ${\mu }^2/(3n)$. The normal mean proportional to standard deviation model provides a reusable version of this calculation.

Let $z=z_{1-\alpha /2}$ be the $(1-\alpha /2)$ standard normal quantile. Since $\hat{\mu }$ is a consistent estimator, Slutsky's theorem givesIts estimated standard error is $\hat{\mu }/\sqrt{3n}$, giving the approximate $(1-\alpha )$ Wald confidence intervalFor each fixed $\mu >0$, this procedure's coverage probability tends to $1-\alpha$. Intersecting with the positive parameter space preserves coverage of the true positive value.

Write $S^2=Q/(n-1)$ and $S=\sqrt{S^2}$, with $n\ge 2$. Two exact facts about a normal sample areand independence of the normal sample mean and sample variance holds. A chi-squared distribution with $n-1$ statistical degrees of freedom is a sum of $n-1$ independent ${\chi }_1^2$ variables, each with mean one and variance two. The central limit theorem therefore givesFor $g(x)=\sqrt{x}$, the derivative at ${\mu }^2$ is $1/(2\mu )$ because $\mu >0$. The delta method gives the asymptotic normality of a normal sample standard deviation:The exact independence of $\overline{X}$ and $S$ makes their limiting normal distributions independent. Consequently,ThusThis has the same leading asymptotic variance as $\hat{\mu }$. The weights $1/3$ and $2/3$ are those in the optimal asymptotic combination of a normal sample mean and standard deviation. This does not require $S$ to be an exactly unbiased estimator of $\mu$.