Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics III / 7

For the exponential distribution with rate $\theta$, the change of variables $u=\theta x$ givesThe integrals are one and two by integration by parts. Hence $\boxed{\mathbb{E}[X]=1/\theta ,\mathrm{Var}(X)=1/{\theta }^2}$.

For an estimator $T$ of $\theta$, its bias and mean squared error areWhen the second moment exists, expanding around ${\mathbb{E}}_{\theta }[T]$ gives the bias-variance decomposition ${\mathrm{MSE}}_{\theta }(T)={\mathrm{Var}}_{\theta }(T)+b_{\theta }(T{)}^2$.

For the independent random sample, the likelihood function on its support is $L(\theta )={\theta }^n{e}^{-\theta Y}$, where $Y=\sum X_i$. Its log-likelihood is $\ell (\theta )=n\log \theta -\theta Y$, withAlmost surely $Y>0$, and the unique stationary point is consequently the unique global maximum:The exceptional all-zero sample has probability zero; if $Y=0$ were nevertheless supplied, there would be no finite maximizing rate.

Use the gamma density and the normalization integral for shape $n-1$:The same argument, valid for $n\ge 3$, gives $\mathbb{E}[Y^{-2}]={\theta }^2/[(n-1)(n-2)]$.

The inverse moments in part i giveThus the bias is positive, and direct expansion of the mean squared error yieldsFor comparison, the variance is $n^2{\theta }^2/[(n-1{)}^2(n-2)]$; adding the squared bias gives the same answer by the bias-variance decomposition.

The central limit theorem givesWrite $z=z_{1-\alpha /2}$ for the indicated standard normal quantile. Inverting the approximate pivot inequality and substituting $Y/n=1/\hat{\theta }$ gives the approximate confidence intervalIts coverage is approximately $1-\alpha$ for large $n$. The intersection respects the positive rate parameter; asymptotically the lower endpoint is positive. This is an asymptotic interval, not an exact finite-sample confidence interval.