Past exam of the mathematics course of the University of Cambridge / 2002 / ib / Paper 2 / 3H (Statistics)

Represent a possibly randomized statistical test by a function $\phi (Y)$ taking values in $[0,1]$, its conditional probability of rejection. Its power function is $\pi (\rho )={\mathbb{E}}_{\rho }\phi (Y)$, and its size of a statistical test is the supremum of this probability over the null hypothesis. At a prescribed significance level $\alpha$, a uniformly most powerful test has at least as much power as every other level-$\alpha$ test at every parameter in the alternative.

Put $S={\sum }_{j=1}^n{Y}_j$. For any fixed alternative $0<{\rho }_1<{\rho }_0$, the likelihood ratio iswhich is strictly increasing in $S$. The Neyman-Pearson lemma therefore selects the upper tail $S>c_{\alpha }$. Its cutoff is determined solely by the null distribution, not by ${\rho }_1$, so the same statistical test is most powerful against every allowed alternative and hence uniformly most powerful.

Under rate $\rho$, $S$ has a gamma distribution with shape $n$ and rate $\rho$. For completeness, convolution starts with $\rho e^{-\rho s}$ and, if the $n$-term density is ${\rho }^n{s}^{n-1}{e}^{-\rho s}/(n-1)!$, convolving once more gives ${\rho }^{n+1}{s}^n{e}^{-\rho s}/n!$. Integrating this density by parts repeatedly gives its upper-tail probability. Consequently reject precisely when $S>c_{\alpha }$, whereand the power function isFor $0<\alpha <1$ the continuous strictly decreasing null tail gives a unique positive cutoff, with exact size $\alpha$ and no boundary randomization.