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Use the convention that the likelihood ratio is maximized alternative likelihood divided by null likelihood. The alternative excludes zero, but its supremum agrees with the unrestricted maximum, even on the exceptional sample $\overline{X}=0$. Completing the square givesUnder the null, $\sqrt{n}\overline{X}\sim {\mathcal{N}}_d(0,I_d)$, so $2\log \Lambda \sim {\chi }_d^2$ exactly for every $n$. Equivalently $\Lambda$ is distributed as $e^{Y/2}$ for $Y\sim {\chi }_d^2$. With the reciprocal likelihood-ratio convention the statistic is $-2\log \Lambda$ instead.

Wilks theorem predicts a limiting chi-squared distribution with degrees of freedom equal to the difference of parameter dimensions, here $d-0=d$, under regularity and an interior null point of the unrestricted model. This example strengthens the asymptotic conclusion to an exact finite-sample identity.

A simple hypothesis specifies a single probability distribution. For a test with rejection region $R$, its size of a statistical test is the rejection probability under the simple null hypothesis, while its statistical power against the simple alternative is the rejection probability under that alternative. The Neyman-Pearson lemma states that among tests of size at most $\alpha$, a test rejecting for the largest values of the likelihood ratio $f_1/f_0$ has greatest power, with boundary randomization if needed.

Herewhich decreases with $|x|$. The best test therefore rejects for $|X|\le c$. Under $H_0$, ${\mathbb{P}}_0(|X|\le c)=c$, so size $0.05$ requires $c=0.05=1/20$. Its power isThus reject $H_0$ exactly when $|X|\le 0.05$.

The Neyman-Pearson lemma states that, for testing one simple hypothesis with density $f_0$ against another with density $f_1$, a size-$\alpha$ test that rejects for the largest values of the likelihood ratio $f_1/f_0$ is most powerful among all tests of size at most $\alpha$, with boundary randomization if needed.

Let $\Delta ={\theta }_1-{\theta }_0>0$. For one observation from the unit-variance normal location family,The sample log-likelihood ratio is therefore $n{T}_n$. By the Neyman-Pearson lemma, the level-$\alpha$ most powerful test rejects for a sufficiently large likelihood ratio, equivalently when $T_n>C_{\alpha }$, with $C_{\alpha }$ chosen to give null rejection probability $\alpha$.

For ${\theta }_1>{\theta }_0$, the likelihood ratio iswhich is increasing in $x$. Thus the logistic location family has a monotone likelihood ratio in $X$.

The upper-tail rejection probability is increasing in the location parameter, since $X=\theta +Z$ with $Z$ having the standard logistic distribution. Hence for the critical value from part (i),The test has size $\alpha$ for the composite null.

Now fix any alternative ${\theta }_1>0$. The Neyman-Pearson lemma applied to the simple hypotheses $\theta =0$ and $\theta ={\theta }_1$ says that this same upper-tail test is most powerful among all tests whose rejection probability at $0$ is at most $\alpha$. Every test of size at most $\alpha$ for the composite null satisfies that restriction. The upper-tail test is therefore most powerful at every ${\theta }_1>0$, so it is the uniformly most powerful upper-tail test for a logistic location.

The likelihood ratio of the simple alternative to the simple null isThis is a strictly increasing function of $x$. By the Neyman-Pearson lemma, the most powerful test therefore rejects for $X>c$. The logistic distribution hasThe size condition ${\mathbb{P}}_0(X>c)=\alpha$ givesBecause the distribution is continuous, no boundary randomization is needed. The most powerful size-$\alpha$ test is therefore

For $0<\pi <1$, let ${\delta }_{\pi }$ be the Bayes classifier with class-zero prior $\pi$, and define its two class-conditional error probabilities byWriting the likelihood ratio as $L=f_1/f_0$, the rule chooses class one whenThe Gaussian likelihood-ratio level sets have probability zero, so dominated convergence shows that $e_0$ and $e_1$ depend continuously on $\pi$. As $\pi \downarrow 0$, the threshold tends to zero and the classifier chooses class one almost surely, givingAs $\pi \uparrow 1$, it chooses class zero almost surely, giving the opposite limit $(0,1)$. The intermediate value theorem therefore provides ${\pi }^{\ast }\in (0,1)$ such that

The rule ${\delta }_{{\pi }^{\ast }}$ is an equalizer rule with worst-case risk $r$. For any classifier $\delta$,because ${\delta }_{{\pi }^{\ast }}$ minimizes the integrated risk for its prior. Hence ${\delta }_{{\pi }^{\ast }}$ is a minimax decision rule, as summarized by the minimax Gaussian Bayes classifier from equal class errors.

The prior is indeed least favorable. Its Bayes risk is $r$, while for any other prior $q$ the Bayes risk is at most the integrated risk of the same equalizer rule ${\delta }_{{\pi }^{\ast }}$, namelyThus no prior has larger Bayes risk.

If $L_j=d{P}_j/dQ$ and the nonnegative mixture weights sum to one, the likelihood ratio of the mixture model is $L={\sum }_j{w}_j{L}_j$. Expanding ${\mathbb{E}}_Q{L}^2-1$ proves the displayed chi-squared divergence identity. For independent random variables sampled from a fixed component, one-observation cross moments multiply across observations. The component is drawn once for the whole sample; redrawing it per observation gives a different mixture model.

A statistical test is a measurable rule for deciding whether to reject a null hypothesis. Allowing randomisation, it is represented by a function $\phi (x)\in [0,1]$ giving the conditional rejection probability after observing $x$. A deterministic test takes only values zero and one. Under a simple null, $E_0\phi$ is its size of a statistical test; under an alternative, $E_1\phi$ is its statistical power. Randomisation on a likelihood ratio equality set allows a Neyman-Pearson lemma threshold test to attain a prescribed size even for discrete data.

For independent samples from an exponential distribution, a test of rate at most ${\lambda }_0$ against rate greater than ${\lambda }_0$ rejects for small sample sums. All simple-alternative likelihood ratios against ${\lambda }_0$ decrease with that sum; its gamma distribution calibrates the size at ${\lambda }_0$. Monotonicity of the rejection probability in the rate proves that this is a uniformly most powerful test for the composite hypotheses.