Past exam of the statistics course of the University of Oxford / 2017 / 2 / a / Solution

The Neyman-Pearson lemma concerns a simple hypothesis $H_0:\theta ={\theta }_0$ against another simple hypothesis $H_1:\theta ={\theta }_1$. For a continuous statistical sample, put $f_j(x)={\prod }_{i}f(x_i;{\theta }_j)$ for the joint probability density functions, $j=0,1$. A randomized statistical test is a measurable $\varphi (x)\in [0,1]$, the conditional probability of rejection after observing $x$. Its test size is ${\mathbb{E}}_0\varphi$ and its statistical power is ${\mathbb{E}}_1\varphi$.

Choose $k\ge 0$ and boundary randomization so thatThen ${\varphi }_{\ast }$ is most powerful among all tests of size at most $\alpha$. Thus one rejects for sufficiently large alternative-to-null likelihood ratio. If the equality set has zero probability under $H_0$, the test needs no randomization. Continuous observations alone do not guarantee that their likelihood ratio has a continuous probability distribution.

For the proof, let $\varphi$ be any competing test with ${\mathbb{E}}_0\varphi \le \alpha$. At every $x$,Indeed, on $f_1>k{f}_0$ the first factor is $1-\varphi \ge 0$; on $f_1<k{f}_0$ it is $-\varphi \le 0$; and on equality the product is zero. Integration giveswhich proves the claimed optimality. Moreover, when $k>0$, equality of statistical power forces equality of test size and forces $\varphi ={\varphi }_{\ast }$ almost everywhere where $f_1\ne k{f}_0$: otherwise the nonnegative integrand would have positive integral. This proves the usual necessity assertion as well, up to equality sets and sets of zero probability under both hypotheses.