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

For the independent statistical sample, write the likelihood function and log-likelihood function asThe expected Fisher information in the whole statistical sample isUnder the regularity conditions that permit differentiation under the integral sign, the score function has expected value zero andFor an interior statistical parameter, assuming identifiability and $0<I_1(\theta )<\infty$, the regular asymptotic normality of a maximum likelihood estimator statesThus the large-sample normal approximation is ${\hat{\theta }}_n\dot{\sim }N(\theta ,I_n(\theta {)}^{-1})$. The inverse information is the leading sampling variance approximation; the displayed convergence specifies its precise asymptotic meaning.

For $0<{\sigma }^2<\infty$, the central limit theorem givesIf $g$ is differentiable at $\mu$, the delta method applies to the first-order expansionwhere the remainder notation means that its product with $\sqrt{n}$ tends to zero in probability. HenceThe useful nondegenerate normal approximation requires $g^{\prime }(\mu )\ne 0$. When $g^{\prime }(\mu )=0$, the first-order limit is zero and a higher-order expansion may be needed.

For a Bernoulli distribution with fixed $0<p<1$, the mean and variance are $p$ and $p(1-p)$. The odds transformation $g(x)=x/(1-x)$ has derivative $g^{\prime }(x)=(1-x{)}^{-2}$. Applying the delta method from part (ii),More precisely, $\sqrt{n}\{\overline{X}/(1-\overline{X})-p/(1-p)\}$ converges to a normal distribution of mean zero and variance $p/(1-p{)}^3$, after defining the statistic arbitrarily on $\overline{X}=1$. That exceptional event has probability $p^n\to 0$, so this extension does not change the limit. The unextended ratio is undefined there and has no finite exact variance if interpreted as infinite. At $p=0$ it is zero almost surely; at $p=1$ the denominator is always zero. This is the delta method for Bernoulli odds.

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 $\mu$.