Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 1 / 7H / Solution

Let $S={\sum }_{i=1}^n{X}_i$. The likelihood ratio of rate two to rate one is $2^n{e}^{-S}$, which decreases strictly with $S$. By the Neyman-Pearson lemma, the most powerful test rejects for small values of the sum:Under the null, $S$ has the gamma distribution with shape $n$ and rate one. Thus $c_{\alpha }$ is its $\alpha$-quantile, equivalently half the $\alpha$-quantile of the chi-squared distribution with $2n$ degrees of freedom. The distribution is continuous, so boundary randomization is unnecessary for $0<\alpha <1$.

For every fixed alternative $\lambda >1$, the likelihood ratio against rate one is ${\lambda }^n{e}^{-(\lambda -1)S}$, again strictly decreasing. Therefore the same region is most powerful against every such alternative. Moreover, if $G$ has the shape-$n$, rate-one gamma distribution, then $S$ at rate $\lambda$ has the same distribution as $G/\lambda$. It follows that ${\mathbb{P}}_{\lambda }(S<c_{\alpha })=\mathbb{P}(G<\lambda c_{\alpha })$ increases with $\lambda$, so the maximum rejection probability over $\lambda \le 1$ is $\alpha$ at one. Any competing size-$\alpha$ test for the composite null is also of level at most $\alpha$ at rate one, and is bounded in power by the Neyman-Pearson lemma for each alternative. The test is uniformly most powerful for the composite hypotheses as well. This is the uniformly most powerful test for an exponential rate based on the sample sum.