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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.

Assume equal allocation of $n$ independent offenders to each arm and a two-sided statistical test of significance level $\alpha$. Write $p_0=0.60$ for P30 and $p_1=0.56$ for initial community-service assignment, with $\Delta =p_0-p_1=0.04$ and $\bar{p}=(p_0+p_1)/2=0.58$. The two event counts have independent binomial distributions. Under the stated alternative, the difference of sample proportions is approximately a normal distribution with mean $\Delta$ and varianceThe pooled two-proportion test has approximate upper rejection threshold $z_{1-\alpha /2}\sqrt{2\bar{p}(1-\bar{p})/n}$ at the planning rates, where $z_q$ is the standard normal quantile. To obtain statistical power $0.80$ against a positive difference, place the alternative mean $z_{0.80}$ alternative-standard-errors above that threshold:Solving yields the usual sample size for comparing two proportions:per arm, rounded upwards, with total sample size $2n$. This targets the dominant rejection tail under the planning alternative; an exact power calculation can include the very small opposite-tail probability or use the binomial distributions directly.

For illustration, at $\alpha =0.05$, $z_{0.975}\simeq 1.960$ and $z_{0.80}\simeq 0.842$, giving $n\simeq 2388.8$. ThusFor a prespecified one-sided test, replace $z_{1-\alpha /2}$ by $z_{1-\alpha }$. The calculation assumes complete endpoint ascertainment, independent individual allocation and that the two initial-policy rates are appropriate; losses or clustered allocation would require corresponding design adjustments.

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.

Total variation distance asks how much two probability distributions can disagree about the same event. For probability measures $P,Q$ on one measurable space $(S,\mathcal{F})$, its definition isThe supremum ranges over all measurable sets, so $d_{\mathrm{TV}}(P,Q)\le \epsilon$ guarantees that every yes-or-no prediction has probability error at most $\epsilon$. This metric takes values between zero and one; it is zero exactly when the two probability measures agree. This makes it useful for comparing a sampling algorithm's output with its target probability distribution, and for defining the mixing time of a Markov chain.

On a finite set, add the excess probability mass wherever $P$ exceeds $Q$. That excess is also half the sum of all absolute differences:The last expression reveals the common probability mass that the two probability distributions can share.^[ref] The coupling characterization of total variation distance turns this overlap into an actual joint construction; minimum sum of errors in a simple hypothesis test gives it a statistical interpretation. One quantity connects probability bookkeeping, comparison of sampling methods, and the limits of a statistical test.

Suppose one observation comes from either $P$ or $Q$, each with prior probability $1/2$. A decision rule that chooses $P$ on an event $E$ and $Q$ otherwise has success probabilityOn a finite set, choosing $E=\{x:P(x)>Q(x)\}$ therefore gives $\frac{1}{2}(1+d_{\mathrm{TV}}(P,Q))$. This is the equal-prior version of minimum sum of errors in a simple hypothesis test; allowing a statistical test to use an auxiliary random variable cannot improve it, because for each outcome the better of the two decisions already maximizes its contribution.

For the four-outcome example, choose $P$ after $A$ and $Q$ after $B$ or $C$, with either decision allowed after $D$. Choosing $Q$ after $D$ yields $\frac{1}{2}(0.4+0.9)=0.65$. The best success probability is 65%, with a 35% error probability. A smaller total variation distance means the two sources are harder to distinguish from one observation. Unequal prior probabilities, unequal costs, or multiple observations change this particular decision problem; the $65\%$ figure is not a universal bound for every experiment.