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

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.

Let $S={\sum }_i{X}_i$. For the exponential distribution with mean $\theta$, the likelihood function is $L(\theta )={\theta }^{-n}{e}^{-S/\theta }$ on positive observations. Fix any $0<{\theta }_1<{\theta }_0$. The alternative-to-null likelihood ratio isSince $1/{\theta }_1-1/{\theta }_0>0$, this is strictly decreasing in $S$. The Neyman-Pearson lemma says to reject for small $S$. Under the null hypothesis, the given sum rule yields $S\sim \mathrm{Gamma}(n,1/{\theta }_0)$, where the second argument is a rate, and henceLet $q_{\alpha }$ be the lower $\alpha$ quantile of this chi-squared distribution. The required critical region isIts test size is exactly $\alpha$ because the null distribution is continuous. Its cutoff is the same for every ${\theta }_1<{\theta }_0$, so it is simultaneously most powerful against every such value: it is a uniformly most powerful test. Equivalently, reject if $S$ is below the lower $\alpha$ quantile of the specified gamma distribution.

Put $A={\sum }_i{X}_i$, $B={\sum }_j{Y}_j$ and $N=n+m$. Independence gives the likelihood functionThe unrestricted maximum-likelihood estimators are $\hat{\theta }=A/n$ and $\hat{\mu }=B/m$. Under the null hypothesis, the common mean $\eta =\theta =\mu$ maximizes ${\eta }^{-N}{e}^{-(A+B)/\eta }$ at $\hat{\eta }=(A+B)/N$. These are maxima because each one-parameter score function has the sign of its observed sum minus its sample size times its proposed mean.

Define the generalized likelihood ratio to be the constrained maximum divided by the unrestricted maximum. The exponential factors cancel, givingSince $T=A/(A+B)$ and $1-T=B/(A+B)$,It has maximum one at $T=n/N$, when the two fitted means agree, and is small toward either endpoint. A likelihood-ratio test rejects for small $\Lambda$, or equivalently large $W=-2\log \Lambda$. This is the likelihood-ratio test for equality of exponential means.

Under the null hypothesis, the full model has two free positive statistical parameters and the constrained model has one. The Wilks theorem therefore givesas both sample sizes tend to infinity, with $n/(n+m)$ tending to a number strictly between zero and one. For the observed $t$, calculateThe large-sample p-value is the upper-tail probabilitywhere $\Phi$ is the cumulative distribution function of the standard normal distribution. The last equality uses the fact that the square of a standard normal random variable has a chi-squared distribution with one statistical degree of freedom. Values of $T$ on either side of $n/(n+m)$ are already included by the transformation to $W$.