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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.

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.

The Neyman-Pearson lemma concerns a simple hypothesis with density $f_0$ against a simple alternative with density $f_1$. Allow a randomized test $0\le \phi \le 1$. Choose a threshold $c\ge 0$ and, if necessary, randomize on the equality set so thatThen ${\phi }_{\ast }$ is a most powerful test among all tests of size at most $\alpha$. To prove it, let $E_0\phi \le \alpha$. Pointwise, $({\phi }_{\ast }-\phi )(f_1-c{f}_0)\ge 0$: above threshold both factors are nonnegative; below threshold both are nonpositive; on the threshold their product vanishes. ThereforeThis proves the lemma, including regions with $f_0=0$, where a positive alternative density must be rejected. It also shows that ties in the likelihood ratio are the only discretionary region when the inequality is strict off the threshold.

For the stated family, $R=|X|$ has the Gamma distribution with shape $\theta$ and unit rate. At $\theta =1$ it has the exponential distribution, and the likelihood ratio for $\theta =2$ against $\theta =1$ isbecause $\Gamma (1)=\Gamma (2)=1$. The most powerful test therefore rejects for large $|X|$. Since ${\mathrm{Pr}}_1(|X|>c)=e^{-c}$, its exact size-$\alpha$ critical value and rejection rule areThe distribution is continuous, so boundary randomization is unnecessary. Under $\theta =2$, integration by parts gives the power function of a statistical testFor every fixed $\theta >1$, the likelihood ratio against the same null is $|x{|}^{\theta -1}/\Gamma (\theta )$, strictly increasing in $|x|$. The size constraint still selects exactly the same critical value $c=-\log \alpha$, independent of the alternative. The Neyman-Pearson lemma therefore makes this single test most powerful against each $\theta >1$ separately. Hence it is uniformly most powerful for the composite alternative $\theta >1$.

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.

The Neyman-Pearson lemma concerns testing a simple null density $f_0$ against a simple alternative $f_1$. A test that rejects where $f_1/f_0>c$, accepts where the ratio is below $c$, and if necessary randomizes on the boundary to have null rejection probability $\alpha$, is a most powerful test among tests of size at most $\alpha$. Thus a likelihood-ratio threshold maximizes power at the specified simple alternative. The boundary randomization is relevant for discrete distributions.

The Neyman-Pearson lemma concerns testing a simple null density $f_0$ against a simple alternative $f_1$. A test that rejects where $f_1/f_0>c$, accepts where the ratio is below $c$, and if necessary randomizes on the boundary to have null rejection probability $\alpha$, is a most powerful test among tests of size at most $\alpha$. Thus a likelihood-ratio threshold maximizes power at the specified simple alternative. The boundary randomization is relevant for discrete distributions.

Put $T={\sum }_i{X}_i$. For interior parameters $0<{\theta }_0<{\theta }_1<1$, the alternative-to-null likelihood ratio isThe second base is greater than one, so this is strictly increasing in $T$. The Neyman-Pearson lemma therefore gives an upper-tail rejection rule. Under the null hypothesis, $T\sim \mathrm{Bin}(n,{\theta }_0)$, and an ordinary conservative threshold can be chosen asThere is an important discrete qualification: an integer cutoff need not have test size exactly $\alpha$. To obtain exact size, choose an integer $b$ with ${\mathrm{Pr}}_0(T>b)\le \alpha \le {\mathrm{Pr}}_0(T\ge b)$, reject certainly when $T>b$, and at $T=b$ reject with probabilityThis randomized statistical test is a most powerful test at exact size $\alpha$; when a binomial upper tail equals $\alpha$, no randomization is needed and the stated region is literal. Otherwise the unrandomized upper-tail region has size at most $\alpha$, and the exact-size result includes the boundary randomization. Endpoint parameter cases retain the same upper-tail ordering by taking limits, with ties resolved on the null-supported outcomes.

For two simple statistical hypotheses with joint probability density functions $g_0$ and $g_1$, fix a significance level $\alpha$. A randomized statistical test $\phi (x)$ is the conditional probability of rejection given the data. Choose $k\ge 0$ and, if necessary, randomization on the equality set so thatThe Neyman-Pearson lemma says that this is a most powerful test among all tests with test size at most $\alpha$:Thus the optimal ordering rejects where the alternative-to-null likelihood ratio is largest. On null-impossible observations with $g_1>0$, the rejection probability is one. Randomization is dispensable whenever an ordinary threshold already attains the required test size.

Let $S={\sum }_i{X}_i$, which has a binomial distribution with parameters $(n,p)$. For $0<p_1<1/2$, the alternative-to-null likelihood ratio of the independent Bernoulli trials isSince $p_1/(1-p_1)<1$, this is strictly decreasing in $S$. A threshold on large $R$ is therefore a lower-tail likelihood-ratio test, and a threshold between $R(2)$ and $R(1)$ gives the stated critical region $S<2$. The Neyman-Pearson lemma also proves it is a most powerful test at its attained test size.

Using the binomial probability mass, its test size and statistical power are respectivelyThe associated power function is $\pi (p)=(1-p{)}^n+np(1-p{)}^{n-1}$. These formulas hold for positive integer $n$; when $n=1$ the test always rejects and has size one. If the boundary value $p_1=0$ is admitted, all alternative probability is at $S=0$; the same test still has statistical power one and can be represented using a threshold with ties in the likelihood ratio.

The Neyman-Pearson lemma states that a test which rejects where the joint likelihood ratio $L_1/L_0$ is largest is a most powerful test of its size: its power is at least that of any test with no larger size. More precisely, choose a threshold $k$ and reject if $L_1/L_0>k$, accept if it is smaller, and if necessary randomize on equality to attain the specified size.

For an independent and identically distributed statistical sample, takeThe $Y_i$ are independent and identically distributed random variables under either simple hypothesis, and the log joint ratio is their sum. Thus the nonrandomized threshold form can be writtenwith $c=-\log k$, using either strict or non-strict inequality when the boundary has probability zero. The usual finite-log expression assumes positive common support; zero densities can instead be handled by the extended likelihood ratio on the relevant supports. Continuous observations alone do not ensure a continuous likelihood-ratio statistic: ties may require the boundary randomization in the lemma. This qualification preserves the exact-size formulation.