Past exam of the Probability, statistics and options II course of the University of Oxford / 2010 / O / O5

For a rejection function $\phi (x)\in [0,1]$, its power is $\beta (\theta )={\mathbb{E}}_{\theta }\phi (X)$. A test has level at most $\alpha$ if ${\sup }_{\theta \in H_0}\beta (\theta )\le \alpha$; its size is that supremum. In a simple null the significance probability is just the rejection probability under that null.

For two simple hypotheses with joint sample densities $f_0,f_1$, the Neyman–Pearson lemma states that a test rejecting where $f_1>k{f}_0$, accepting where $f_1<k{f}_0$, and randomizing if necessary on equality to have size $\alpha$, is most powerful among level-$\alpha$ tests. Let ${\phi }_{\ast }$ be that test and $\phi$ any competitor. Pointwise,Integration givessince $k\ge 0$, the first test has size $\alpha$, and the other has size at most $\alpha$. This proves the claimed power bound, allowing null-density-zero regions as well. Equality implies agreement away from the likelihood-ratio boundary, up to null sets, when the inequalities are strict.

For a fixed $\lambda <{\lambda }_0$, the likelihood ratio against the null isstrictly increasing in $S={\sum }_i{x}_i$. The Neyman–Pearson critical region is therefore an upper tail in $S$, and its null threshold depends only on the desired size, not on which $\lambda <{\lambda }_0$ was chosen. Thus the same test is uniformly most powerful for the entire one-sided alternative.

Under the null $S\sim \mathrm{Gamma}(n,{\lambda }_0)$ with the second parameter interpreted as rate. Its continuous distribution permits the exact thresholdUnder any fixed alternative its rejection probability is $\boxed{Q(\lambda )=1-F(c_{\alpha };n,\lambda )}$. Applying the lemma separately to each such $\lambda$ shows that every level-$\alpha$ test has power at most this value there. The constructed test attains the bound for all $\lambda <{\lambda }_0$ simultaneously.