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At level $\alpha$, a most powerful test against a specified simple alternative maximises statistical power among statistical tests whose size of a statistical test is at most $\alpha$. The Neyman-Pearson lemma constructs it by a likelihood ratio threshold. A uniformly most powerful test is most powerful simultaneously against every member of the specified alternative family; maximising power at one alternative alone need not give uniform optimality.

For two unit-scale Laplace distributions with locations ${\theta }_0<{\theta }_1$, the likelihood ratio is nondecreasing in $X$, with constant tails and an increasing middle interval. Therefore an upper-tail statistical test of exact size of a statistical test $\alpha$ is a most powerful test. Its cutoff is $c_{\alpha }={\theta }_0-\log (2\alpha )$ for $\alpha \le 1/2$, and $c_{\alpha }={\theta }_0+\log (2(1-\alpha ))$ otherwise. Under location ${\theta }_1$, its statistical power is $\frac{1}{2}{e}^{{\theta }_1-c_{\alpha }}$ if $c_{\alpha }\ge {\theta }_1$, and $1-\frac{1}{2}{e}^{c_{\alpha }-{\theta }_1}$ otherwise. Constant likelihood ratio tails can make the most powerful test nonunique.

Represent a possibly randomized statistical test by a function $\phi (Y)$ taking values in $[0,1]$, its conditional probability of rejection. Its power function is $\pi (\rho )={\mathbb{E}}_{\rho }\phi (Y)$, and its size of a statistical test is the supremum of this probability over the null hypothesis. At a prescribed significance level $\alpha$, a uniformly most powerful test has at least as much power as every other level-$\alpha$ test at every parameter in the alternative.

Put $S={\sum }_{j=1}^n{Y}_j$. For any fixed alternative $0<{\rho }_1<{\rho }_0$, the likelihood ratio iswhich is strictly increasing in $S$. The Neyman-Pearson lemma therefore selects the upper tail $S>c_{\alpha }$. Its cutoff is determined solely by the null distribution, not by ${\rho }_1$, so the same statistical test is most powerful against every allowed alternative and hence uniformly most powerful.

Under rate $\rho$, $S$ has a gamma distribution with shape $n$ and rate $\rho$. For completeness, convolution starts with $\rho e^{-\rho s}$ and, if the $n$-term density is ${\rho }^n{s}^{n-1}{e}^{-\rho s}/(n-1)!$, convolving once more gives ${\rho }^{n+1}{s}^n{e}^{-\rho s}/n!$. Integrating this density by parts repeatedly gives its upper-tail probability. Consequently reject precisely when $S>c_{\alpha }$, whereand the power function isFor $0<\alpha <1$ the continuous strictly decreasing null tail gives a unique positive cutoff, with exact size $\alpha$ and no boundary randomization.

A simple hypothesis specifies one probability distribution. Let the null hypothesis and alternative have probability density functions $f_0,f_1$ with respect to a common measure. A possibly randomized statistical test is a measurable function $\phi (x)\in [0,1]$, its conditional probability of rejecting the null hypothesis. Its size of a statistical test is ${\mathbb{E}}_0\phi$, and its statistical power at the alternative is ${\mathbb{E}}_1\phi$. A most powerful test at level $\alpha$ maximizes power among tests with size at most $\alpha$.

The Neyman-Pearson lemma says to choose $k\ge 0$ and a test ${\phi }_{\ast }$ of size $\alpha$ with ${\phi }_{\ast }=1$ where $f_1>k{f}_0$ and ${\phi }_{\ast }=0$ where $f_1<k{f}_0$, choosing values on the equality set to reach the specified size. Equivalently, threshold the likelihood ratio, treating positive alternative density with zero null density as an infinite ratio. Randomization on a threshold atom ensures exact size when needed. If $k=0$, all remaining null-only mass can be used to reach size without changing power.

For any test $\phi$ of size at most $\alpha$, the signs of the two factors givepointwise. Integrating provesThis is the required optimality proof. If $k>0$ and another test is equally powerful, both inequalities must be equalities. It must have size $\alpha$ and agree with ${\phi }_{\ast }$ wherever $f_1\ne k{f}_0$, apart from null sets for the relevant densities. Thus freedom is confined to the equality set and sets null under both hypotheses; at $k=0$, additional size need not improve power. This also explains the nonuniqueness below.

For the unit-scale Laplace distribution, put $\delta ={\theta }_1-{\theta }_0>0$. Its likelihood ratio isIt is nondecreasing, so an upper-tail critical region has the Neyman-Pearson lemma form, even if the cutoff lies inside a constant-ratio tail. Solving ${\mathbb{P}}_{{\theta }_0}(X>c_{\alpha })=\alpha$ gives the most powerful test for a Laplace location shift,Continuity of the distribution makes this nonrandomized test exactly size $\alpha$. It need not be the unique most powerful test: within a flat extreme likelihood ratio region, other selections having the same null probability are equally good.

Its statistical power isHere $\beta$ denotes power, not the probability of a type-II error. For $\alpha =0.05$, $c_{\alpha }={\theta }_0+\log 10$. Thus the power is $0.05e^{\delta }$ for $\delta \le \log 10$, and $1-5e^{-\delta }$ otherwise. The first branch is at most $1/2$. On the second branch, power at least $0.95$ is equivalent to $5e^{-\delta }\le 0.05$. Therefore the required pairs are exactly ${\theta }_1-{\theta }_0\ge \log 100$, with no restriction on the common location.

A simple hypothesis specifies the entire probability distribution. A critical region $C$ is the set of observations for which the null hypothesis is rejected. The size of a statistical test is ${\sup }_{\theta \in H_0}{\mathrm{Pr}}_{\theta }(X\in C)$, which reduces to ${\mathrm{Pr}}_0(C)$ for a simple hypothesis. The power function is ${\mathrm{Pr}}_{\theta }(C)$ as a function of the true parameter or distribution. At a specified alternative, the Type II error probability is ${\mathrm{Pr}}_{\theta }(C^c)=1-{\mathrm{Pr}}_{\theta }(C)$, the chance of failing to reject a false null hypothesis. The Type I error probability is the corresponding rejection probability under the null hypothesis.

The Neyman-Pearson lemma says that, for two simple hypotheses, a test rejecting where $f_1>k{f}_0$, accepting where $f_1<k{f}_0$, and if necessary randomizing on equality to achieve prescribed size $\alpha$, has maximal statistical power among tests of size at most $\alpha$. Conversely a most powerful test can be taken in this likelihood ratio form, up to equality sets and null sets.

Here the likelihood ratio for $x\ne 0$ isIts value is infinite at $0$, which has probability zero under both continuous distributions. Large likelihood ratios correspond to small $|x|$. Thus the best critical region is $|X|<c_{\alpha }$, whereThere is no boundary randomization to worry about. The complete answer is

For the minimum sum of errors in a simple hypothesis test, write the error sum as ${\int }_C{f}_0+{\int }_{C^c}{f}_1$. At each observation its smaller contribution is obtained by rejecting exactly where $f_1>f_0$. This proves optimality over all tests, including randomized ones, rather than just optimizing within an assumed family. Here it gives $c_{\ast }=\sqrt{2/\pi }$, and thereforeThe minimum error sum is $1-e^{-1/\pi }+2[1-\Phi (\sqrt{2/\pi })]$. This criterion weights the two error probabilities equally; it does not impose a conventional $5\%$ significance level.

A critical region $C$ is the set of sample outcomes on which a test rejects its null hypothesis. Its power function is $\beta (\theta )={\mathbb{P}}_{\theta }(X\in C)$; its size of a statistical test is ${\sup }_{\theta \in {\Theta }_0}\beta (\theta )$. Randomized tests replace the indicator of $C$ by a rejection probability $\varphi (X)\in [0,1]$.

The Neyman-Pearson lemma states that for two simple hypotheses with densities $p_0,p_1$, the test ${\varphi }^{\ast }$ that is one where $p_1>k{p}_0$, zero where $p_1<k{p}_0$, and randomized on equality to have null rejection probability $\alpha$, maximizes power among tests of size at most $\alpha$. Here $k\ge 0$. To prove it, let $\varphi$ be any competing test. Pointwise, $({\varphi }^{\ast }-\varphi )(p_1-k{p}_0)\ge 0$, since both factors have the same sign off the equality set. Integrating givesThis proves the lemma, including equality-set randomization and points where $p_0=0$.

Take the exponential parameter to be its rate: $p_{\lambda }(x)=\lambda e^{-\lambda x}$ for $x\ge 0$. For $T={\sum }_i{X}_i$, the joint likelihood ratio is $({\lambda }_1/{\lambda }_0{)}^n{e}^{-({\lambda }_1-{\lambda }_0)T}$, decreasing in $T$. Hence for $0<\alpha <1$ the most powerful test isThere is no boundary randomization because $T$ has a continuous gamma distribution. Its density follows by repeated convolution: $p_T(t)={\lambda }^n{t}^{n-1}{e}^{-\lambda t}/(n-1)!$. Integrating by parts gives the exponential-rate likelihood-ratio test powerThe threshold is the unique positive solution of this formula at $\lambda ={\lambda }_0$ with value $\alpha$. Differentiating the finite sum yieldsThus the power function is strictly increasing, and its supremum over $0<\lambda \le {\lambda }_0$ is $\beta ({\lambda }_0)=\alpha$. Therefore the same test has size $\alpha$ for the composite null $\lambda \le {\lambda }_0$. Since the threshold does not depend on the particular larger rate, it is also uniformly most powerful against all $\lambda >{\lambda }_0$. The endpoint sizes zero and one have their trivial constant tests.

Put $S=X_1+X_2+X_3$. Independence gives a likelihood ratioSince ${\theta }_1>1$, this is strictly increasing in $S$. A threshold chosen between its values at $S=5$ and $S=6$ gives precisely the stated upper-tail critical region, so it is a likelihood-ratio test.

The sum has a Poisson distribution with mean $3\theta$. Thus the size of a statistical test and its statistical power at ${\theta }_1$ are

For the grouped-area problem, let $S_1={\sum }_{i=1}^m{Y}_i$, $S_2={\sum }_{i=m+1}^{2m}{Y}_i$, $T=S_1+S_2$. The null log likelihood, up to data-only constants, isThe maximum-likelihood estimate is $\hat{\lambda }=T/(3ma)$, giving fitted means $T/(3m)$ and $2T/(3m)$ in the two groups. Under the alternative, the fitted group means are ${\hat{\lambda }}_1=S_1/m$ and ${\hat{\lambda }}_2=S_2/m$.

Use the increasing generalized likelihood-ratio test convention $\Lambda ={\sup }_{H_1}L/{\sup }_{H_0}L\ge 1$, consistent with the printed positive $2\log \Lambda$. The exponential and factorial factors cancel, yielding the likelihood-ratio test of area-proportional Poisson meansTerms with zero count are interpreted as $0\log 0=0$; if $T=0$, take $\Lambda =1$. If the inverse ratio is used, the reported test statistic is instead $-2\log \Lambda$.

Under the null, the regular large-sample Wilks theorem approximation is $2\log \Lambda \sim {\chi }_1^2$: the alternative has two rate parameters and the null one. The value $15.67$ far exceeds the usual $1\%$ critical value, about $6.63$, and has approximate tail probability $7.5\times 10^{-5}$. Thus reject proportionality of the expected counts to area: the two groups show evidence of different territory densities. The statistic alone does not say which group has greater density; that requires $S_1/m$ versus $S_2/(2m)$. For small counts, an exact conditional test uses $S_1\mid T\sim \mathrm{Binomial}(T,1/3)$ under the null instead of the asymptotic calibration.

A unit-root autoregressive process retains shocks permanently, whereas a causal time series with $|\varphi |<1$ reverts towards its mean. Testing the unit root determines whether stationary autoregressive analysis is appropriate or differencing is needed. For the model without an intercept or trend, use the Dickey–Fuller test against the lower-sided alternative $\varphi <1$ near the null. PutThis is the ordinary regression statistic for a zero coefficient when $\Delta X_t$ is regressed on $X_{t-1}$, but its null probability distribution is not the usual Student law. Under the standard unit-root initialization $X_1=o_P(\sqrt{n})$ and innovations independent of the starting value,where $W$ is standard Brownian motion. If $d_{\alpha }$ is the lower $\alpha$-quantile of $D$, the asymptotic level-$\alpha$ critical region isThe deterministic terms and null initialization must match the critical-value table. For exact finite-sample size of a statistical test, calibrate the statistic from its Gaussian random walk null with the specified initial condition and noise scale; for a zero starting value its distribution is scale-free. The printed two-sided recurrence alone specifies neither an initial law nor a universal finite-sample critical value. Ordinary normal quantiles do not give the intended size of a statistical test.

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.

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 derivative of the log ratio is $\frac{1}{2}(n/q-1)$, so $\lambda (q)$ increases from zero to one on $(0,n)$ and decreases to zero on $(n,\infty )$. Rejecting for a sufficiently small ratio therefore gives two tails with $0<q_1<n<q_2$.

Under $H_0$, Cochran's theorem gives the exact sampling distribution $Q=S/{\sigma }_0^2\sim {\chi }_{n-1}^2$, independently of the unknown mean. Writing $F_{n-1}$ for its cumulative distribution function, the two cutoffs are determined byThe first equation calibrates the exact size of a statistical test. The second is $\lambda (q_1)=\lambda (q_2)$, so both tails belong to one likelihood ratio cutoff. Hence the exact critical region is $\boxed{\{q\le q_1\}\cup \{q\ge q_2\}}$. Its two tail probabilities are not generally equal: equal-tailed chi-squared cutoffs give a different test.

For a critical region $C$, the size of a statistical test of the simple null hypothesis $P_0$ is $\alpha =P_0(\mathbf{X}\in C)$, the Type I error probability. Its statistical power against the simple alternative hypothesis $P_1$ is $P_1(\mathbf{X}\in C)$, the probability of correctly rejecting that null. For a randomized test function $\phi (\mathbf{x})\in [0,1]$, replace these by ${\mathbb{E}}_0\phi$ and ${\mathbb{E}}_1\phi$.

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.