Past exam of the mathematics course of the University of Cambridge / 2019 / ib / Paper 1 / 19H (Statistics)

The Neyman-Pearson lemma says that for simple hypotheses with densities $f_0,f_1$, a size-$\alpha$ test that rejects where $f_1/f_0>k$ and accepts where $f_1/f_0<k$, with possible randomization on equality to attain size $\alpha$, is most powerful among all tests of size at most $\alpha$. To prove it, let ${\phi }^{\ast }$ be this test and $\phi$ any competing test. Pointwise,After integration,which proves maximal power.

For the exponential sample, with $S={\sum }_i{X}_i=n\overline{X}$,Since ${\lambda }_1<{\lambda }_0$, this ratio is increasing in $S$. The likelihood-ratio test therefore rejects for large $\overline{X}$:where $G_{n,\lambda }$ is the distribution function of the rate-parameterized Gamma distribution $\Gamma (n,\lambda )$.

The power function of a test is the probability of rejection as a function of the true parameter. Since $S=n\overline{X}\sim \Gamma (n,\lambda )$,At the null value, $\beta ({\lambda }_0)=\alpha$.

A level-$\alpha$ test is uniformly most powerful for a composite alternative when its power is at least that of every other level-$\alpha$ test at every parameter value in that alternative.

For every fixed ${\lambda }_1<{\lambda }_0$, the likelihood ratio is increasing in the same statistic $S$, and the size-$\alpha$ threshold is always the null quantile $n{c}_{\alpha }$, independent of ${\lambda }_1$. The Neyman-Pearson lemma therefore makes the test from part (a) most powerful against every individual ${\lambda }_1<{\lambda }_0$. Consequently it is