Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 3 / 20H (Statistics)

Write $S={\sum }_i{X}_i$. The likelihood function for the Poisson distribution is proportional to $e^{-n\theta }{\theta }^S$. For $S>0$, differentiation of the log-likelihood gives $-n+S/\theta =0$, and its second derivative is negative. For $S=0$, the likelihood decreases with $\theta$ and has its maximum at the boundary zero. HenceThe estimator is unbiased, since $\mathbb{E}\hat{\theta }=n^{-1}{\sum }_i\mathbb{E}{X}_i=\theta$. Its variance is $\theta /n$. If the parameter space excludes zero, an all-zero sample gives a supremum as $\theta \downarrow 0$ rather than an attained maximum.

For fixed $\theta >0$, the central limit theorem givesThe weak law of large numbers gives $\overline{X}\to \theta$, so Slutsky theorem permits replacing $\theta$ in the standard error by $\overline{X}$. With $z=z_{0.025}\approx 1.96$, an approximate 95% confidence interval isThe lower endpoint may be truncated at zero because the parameter is nonnegative. This is a large-sample normal approximation for fixed positive $\theta$; an all-zero or very sparse sample does not justify the plug-in normal approximation. At the degenerate parameter value zero, every observation is zero almost surely.

The specified gamma distribution uses shape $k$ and rate $\lambda$. Multiplying its density by the Poisson likelihood gives a posterior density proportional to ${\theta }^{k+S-1}{e}^{-(\lambda +n)\theta }$. Thus the conjugate prior calculation givesFor squared-error loss, the posterior expected loss isIt is minimized at the posterior mean, so the Bayes estimator isFor a fresh conditionally independent Poisson observation, the posterior predictive probability of zero is the posterior expectation of $e^{-\theta }$. Evaluating the gamma integral givesThe samples are independent conditional on $\theta$; integrating out the uncertain common parameter produces the predictive probability above.