Past exam of the statistics course of the University of Oxford / 2021 / 1 / a

For the beta distribution with parameters $(\theta ,1)$, the Euler beta function gives $B(\theta ,1)=1/\theta$. Its probability density function is therefore $\theta x^{\theta -1}$ on $(0,1)$. Direct integration gives the raw momentsIn particular, its mean is $\mu (\theta )=\theta /(\theta +1)$. Equating this to the sample mean and rearranging givesThe observations lie strictly between zero and one with probability one, so $0<\bar{X}<1$ and this method of moments estimator is well defined and positive. This is an instance of estimation of a beta shape parameter.

Using the raw moments from part (i), the variance of one observation isThe central limit theorem for the independent statistical sample givesFor $g(x)=x/(1-x)$, its derivative at the mean is $g^{\prime }(\mu )=(1-\mu {)}^{-2}=(\theta +1{)}^2$. The delta method therefore givesEquivalently, the large-sample normal approximation isHere $v$ denotes the leading asymptotic variance, rather than an assertion about an exact finite-sample variance.

Write $A=-{\sum }_i\log X_i>0$, with $\log$ the natural logarithm. The likelihood function and log-likelihood function areThe score function is ${\ell }^{\prime }(\theta )=n/\theta -A$ and the second derivative is ${\ell }^{\prime \prime }(\theta )=-n/{\theta }^2<0$. Thus the log-likelihood function is strictly concave, its stationary point is the unique global maximum, and the maximum-likelihood estimator isThe one-observation Fisher information is $I(\theta )=1/{\theta }^2$. The support of a probability distribution here is independent of $\theta >0$, so the regular asymptotic normality of a maximum likelihood estimator givesThere is also a direct check: $U_i=-\log X_i$ has an exponential distribution with rate $\theta$, since $\mathbb{P}(U_i>u)=\mathbb{P}(X_i<e^{-u})=e^{-\theta u}$ for $u>0$. Its mean and variance are $1/\theta$ and $1/{\theta }^2$. Apply the central limit theorem to $A/n$ and the delta method to $h(u)=1/u$ to obtain the same result.

The specified asymptotic relative efficiency isFor every fixed $\theta >0$ this lies strictly between zero and one. Prefer $T_2$ on grounds of asymptotic precision: it has the smaller leading asymptotic variance and attains the inverse Fisher information. Both are consistent estimators, by the law of large numbers and the continuous mapping theorem applied respectively to $\bar{X}$ and $A/n$. The efficiency approaches zero as $\theta \downarrow 0$ and one as $\theta \to \infty$.

This preference is an asymptotic comparison. In particular, it does not assert that $T_2$ is an unbiased estimator: since $A$ has a gamma distribution of shape $n$ and rate $\theta$, $\mathbb{E}{T}_2=n\theta /(n-1)$ for $n>1$.

The standard error suggested by part (iii) is $\theta /\sqrt{n}$. Substituting the consistent estimator $T_2$ gives $T_2/\sqrt{n}$; Slutsky's theorem givesUsing the central standard normal quantiles $\pm 1.96$, an approximate 95% Wald confidence interval isIts coverage probability approaches 0.95 for each fixed $\theta >0$ as $n\to \infty$. The intersection enforces positivity and does not change whether the true positive statistical parameter is covered. The value 1.64 would give the wrong central two-sided coverage probability.