Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 4 / 17H / c / Solution

The two variances are
$$
\operatorname{var}(\widehat\theta_{LS})
=\frac{\theta S_3}{S_2^2},
\qquad
\operatorname{var}(\widehat\theta_{MLE})
=\frac\theta{S_1}.
$$
Since $S_2^2\leq S_1S_3$ by Cauchy-Schwarz, maximum likelihood has no larger variance, with equality when all positive exposures are equal.


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 Incoming links: Variance comparison for Poisson exposure estimators