Past exam of the mathematics course of the University of Cambridge / 2017 / ib / Paper 4 / 19H (Statistics)

Let $P_0,P_1$ be simple hypotheses with densities $p_0,p_1$ (their Radon-Nikodym derivatives) relative to a common dominating measure. A possibly randomised statistical test is a measurable function $\phi \in [0,1]$; its size of a statistical test is $E_0\phi$ and its statistical power is $E_1\phi$. The Neyman-Pearson lemma says that a likelihood ratio threshold testchosen to have size $\alpha$, is a most powerful test among tests of size at most $\alpha$. A constant randomisation on the equality set suffices to reach the required size; sets where $p_0=0<p_1$ are always rejected. For $0<\alpha <1$, a quantile of the likelihood ratio under $P_0$ gives such a threshold. If the threshold is zero, randomisation where $p_1=0$ may be needed to use the remaining size. At $\alpha =0$, one can reject only a $P_0$-null set and optimally reject the part supporting $P_1$; at $\alpha =1$, always reject.

For the proof, every competing $\phi$ satisfiespointwise: the signs agree off the equality set, where the product is zero. Integrating givesThus the likelihood-ratio threshold test has maximum power at the stated level. If $c>0$, a competing most-powerful test must have size $\alpha$ and agree with ${\phi }_{\ast }$ off the equality set, apart from sets null for $P_0+P_1$; both inequalities must then be equalities. For $c=0$, equal size is not necessary for optimality, because changing decisions where $p_1=0$ does not affect power. These qualifications account for ties and singular supports rather than assuming all densities are strictly positive.

Under the null hypothesis $\theta =0$, $X$ has the uniform distribution on $[0,1]$. Against $\theta =1$, the likelihood ratio is $2x$, increasing on that interval. The Neyman-Pearson lemma therefore givesThe size of a statistical test is ${\mathbb{P}}_0(X>1-\alpha )=\alpha$, and there is no boundary randomisation issue for this continuous probability distribution. Its statistical power against $\theta =1$ is ${\int }_{1-\alpha }^{1}2xdx=2\alpha -{\alpha }^2$.

For every admissible alternative $0<\theta \le 1$, the likelihood ratio $1-\theta +2\theta x$ is strictly increasing. The same upper-tail rejection set has size $\alpha$ and is a most powerful test against each fixed such alternative, again by the Neyman-Pearson lemma. Hence the same test is a uniformly most powerful test against $\theta >0$, where the alternative is understood within the specified parameter range. Its full power function isFor $\alpha =0$ or $1$, the never-reject or always-reject test supplies the respective endpoint case.