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For independent exponential distribution observations with rate $\lambda$, the likelihood ratio for a larger simple rate is decreasing in their sum. The Neyman-Pearson lemma therefore selects a lower-tail sum threshold $c$. The sum has a gamma distribution, giving the displayed power; its derivative is $c{e}^{-\lambda c}(\lambda c{)}^{n-1}/(n-1)!>0$. Calibrating at ${\lambda }_0$ gives size $\alpha$ for the composite null $\lambda \le {\lambda }_0$ and a uniformly most powerful test against larger rates.

For independent $N(0,{\sigma }^2)$ observations, testing a fixed variance against larger variances uses the sum of squares $S={\sum }_i{X}_i^2$. The likelihood ratio for any simple larger variance is strictly increasing in $S$. The Neyman-Pearson lemma therefore gives the displayed rejection rule, where $F_n$ is the chi-squared distribution function with $n$ degrees of freedom. Its threshold is independent of the selected larger variance, so it is a uniformly most powerful test for the one-sided composite alternative. Its power function is $1-F_n(({\sigma }_0^2/{\sigma }^2)F_n^{-1}(1-\alpha ))$, increasing continuously from zero to one and equaling $\alpha$ at the null. A known nonzero mean is handled by replacing each observation by its difference from that mean.

A statistician associated with the Neyman-Pearson lemma, confidence intervals and the Neyman Type A distribution.

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.

The Neyman-Pearson lemma says that an optimal region accepting $P$ has the likelihood-ratio formwith possible randomization on the boundary.

A simple hypothesis specifies a single probability law, with no unknown parameter remaining. For a possibly randomized hypothesis test, let $\phi (x)\in [0,1]$ be its conditional rejection probability. Its test size is $E_0\phi (X)$ and its statistical power against the specified alternative is $E_1\phi (X)$.

The Neyman-Pearson lemma states that for two densities $f_0,f_1$ with respect to a common measure, a likelihood-ratio test rejecting when $f_1/f_0>k$, accepting when it is below $k$, and randomizing on equality to obtain size $\alpha$, maximizes power among all tests of size at most $\alpha$.

Here the likelihood ratio isWriting $r=|x|$, its logarithm apart from the constant is $(r^2-r)/2$. This decreases until $r=1/2$ and then increases. In particular it is nonpositive for $0\le r\le 1$ and positive for $r>1$. The supplied normal-tail value and $\alpha <1/4$ imply that the quantile $t$ defined by $2[1-\Phi (t)]=\alpha$ satisfies $t>1$. Hence the threshold $\Lambda (t)$ is above every central value on $r\le 1$; outside that interval the ratio is strictly increasing. The Neyman-Pearson lemma therefore selects precisely the two tails:The equality boundary has probability zero under both continuous laws, so no randomization is needed. Under the alternative density, the statistical power isThe central dip in the likelihood ratio is why checking the threshold against the entire central region, rather than assuming monotonicity in $|x|$ everywhere, is necessary.

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.

Put $S={\sum }_i{X}_i$, and let $z_{.95}$ be the exact standard-normal 95th percentile. Under parameter $\theta$, $S=n\theta +\sqrt{n}Z$ with $Z$ standard normal. This distribution can be justified directly: the joint density of the centered observations is proportional to $\exp [-{\sum }_i(X_i-\theta {)}^2/2]$. An orthogonal change of coordinates with first row $(1,\dots ,1)/\sqrt{n}$ has unit absolute Jacobian and preserves the sum of squares. Its density factors into $n$ one-dimensional standard-normal densities, making the first coordinate $Z=(S-n\theta )/\sqrt{n}$ standard normal.

Consider the test ${\varphi }_{\ast }=1_{\{S>\sqrt{n}{z}_{.95}\}}$. Its rejection probability isincreasing in $\theta$. Therefore its maximum rejection probability over the null $\theta \le 0$ is achieved at zero and equals $\alpha =1/20$. This proves it has the required test size.

To prove the uniformly most powerful test property, rather than quote the Neyman-Pearson lemma, fix any ${\theta }_1>0$. The likelihood ratio of the full sample against $\theta =0$ isstrictly increasing in $S$. Set $K=\exp ({\theta }_1\sqrt{n}{z}_{.95}-n{\theta }_1^2/2)$. For any competing possibly randomized test $0\le \varphi \le 1$ with composite-null size at most $\alpha$, we have ${\mathbb{E}}_0\varphi \le \alpha ={\mathbb{E}}_0{\varphi }_{\ast }$. Pointwise,because both factors are nonnegative above the threshold and both are nonpositive below it. Integrating givesThis comparison proves the likelihood-ratio optimality result needed here in full. Since the same critical region works for every ${\theta }_1>0$, it proves the uniformly most powerful test property over the entire alternative.

For $n=9$, the exact critical region and its quoted-table approximation areUsing the supplied rounded percentile $1.65$ gives $\sum X_i>4.95$. The exact cutoff uses $z_{.95}\approx 1.644854$ and hence $\overline{X}>0.548285$; the rounded cutoff is the intended numerical approximation, not a literally exact size-$0.05$ threshold. The Gaussian distribution is continuous, so equality at the cutoff has probability zero and no randomization there is needed.

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 $\theta >1$.

Interpret $\varphi (x)$ as the conditional rejection probability. The Type I error is $\alpha =E_0\varphi (X)$, the Type II error is $\beta =E_1[1-\varphi (X)]$, the size is $\alpha$, and the power is $1-\beta$. Their weighted sum isPointwise minimization over $0\le \varphi \le 1$ forces rejection where $p_1>c{p}_0$ and acceptance where $p_1<c{p}_0$, with arbitrary randomization on equality. Conversely those choices minimize every integrand. The equivalence holds almost everywhere, since changing a test on a common null set cannot change its errors. With both weights strictly positive, any simultaneous improvement of the two errors, strict in at least one, would lower this weighted sum. Every such positive-threshold likelihood ratio test is admissible for the two simple hypotheses.

For the specified densities the likelihood ratio isIt increases strictly on $(\theta ,1)$ from zero to $1/(1-\theta )$. For $0<c<1/(1-\theta )$, the likelihood-ratio test rejects when $x>\theta /[1-c(1-\theta {)}^2]$. Larger thresholds accept almost surely. Threshold zero rejects everywhere above $\theta$ and permits arbitrary rejection below $\theta$, where both weighted choices tie; values at individual endpoints do not affect errors. These describe all nonnegative-threshold likelihood ratio tests, including their limiting accept-all and reject-all cases.

The upper-tail test ${\varphi }_k$ has size $1-k^2$. If $\theta <k$, it is a positive-threshold likelihood ratio test and is admissible. If $\theta =k$, it rejects exactly the alternative's support, with power one; no test can retain power one with smaller size, so it remains admissible. If $\theta >k$, it still has power one but wastes rejection probability on $(k,\theta )$, where the alternative density vanishes. The test ${\varphi }_{\theta }$ has the same power and smaller size, soNevertheless ${\varphi }_k$ is a uniformly most powerful test at size $1-k^2$ against all $0<\theta <1$: for $\theta <k$, apply the Neyman-Pearson lemma; for $\theta \ge k$, its power is already the maximum one. It is also admissible against the composite alternative. Indeed choose any $0<\theta <k$. Strict monotonicity of its likelihood ratio makes ${\varphi }_k$ the unique positive-threshold minimizer up to null sets. A test dominating it for the composite problem would either improve the weighted simple risk at this $\theta$, impossible, or tie that risk and coincide almost everywhere. Since $p_0>0$ throughout $(0,1)$, such coincidence also precludes improvement at any other $\theta$.

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.

Write $H_0,H_1$ for the two simple hypotheses, with prior probabilities ${\pi }_0,{\pi }_1$ and densities $f_0,f_1$. Under zero-one loss, the posterior expected losses of choosing $H_0$ and $H_1$ are respectively the posterior probabilities of $H_1$ and $H_0$. A Bayes decision rule therefore chooses $H_1$ ifchooses $H_0$ for the reverse inequality, and may randomize at equality. With positive priors this is a likelihood-ratio test with cutoff $k={\pi }_0/{\pi }_1$.

For a test $\varphi$, its prior risk is ${\pi }_0{E}_0\varphi +{\pi }_1(1-E_1\varphi )$. Choose $k$ and boundary randomization so that the likelihood-ratio test ${\varphi }_k$ has size $\alpha$, and choose priors with ratio $k$. Equivalence of normal and extensive analyses makes the pointwise posterior minimizer minimize this prior risk over all randomized tests. Thus for any $E_0{\varphi }^{\prime }\le \alpha$,This proves the Neyman-Pearson lemma: a likelihood-ratio test of size $\alpha$ has greatest power among tests of size at most $\alpha$. Such a cutoff and randomization come from a quantile of the likelihood ratio under $H_0$; always reject where $f_0=0<f_1$. Endpoint cutoffs are handled directly or by their limiting rules.

For the remaining assertions use the usual common-support formulation with positive densities. A family has a monotone likelihood ratio in $T$ if for every ${\theta }_1>{\theta }_0$, $f_{{\theta }_1}(x)/f_{{\theta }_0}(x)$ is a nondecreasing function $r(T(x))$. A monotone test here is an upper-cutoff test $\varphi =1_{T>c}+\eta 1_{T=c}$, $0\le \eta \le 1$, including constant tests. It does not mean an arbitrary smoothly increasing randomized rejection function.

Put $D(\theta )=\gamma (\theta )-{\gamma }^{\prime }(\theta )$. For a finite cutoff, choose $k>0$ between the values of $r$ below and above $c$, with $k=r(c)$ when the boundary has positive probability. On $T>c$, $\varphi -{\varphi }^{\prime }\ge 0$ and $r-k\ge 0$; on $T<c$, both are nonpositive. At a boundary with positive mass $r-k=0$. ThereforeWhen there is probability on both sides, finite positive monotone ratios below the cutoff bound those above it away from zero, so $k$ can be positive and finite. A boundary with positive mass instead supplies the positive finite value $r(c)$. If the cutoff yields a constant test on the support, positivity on the common support proves strict propagation directly: a test identically zero cannot have greater power, and for a test identically one any positive-probability nonrejection by the competitor remains positive-probability at the larger parameter. Hence $D({\theta }_0)>0$ implies $D({\theta }_1)>0$.

Let ${\theta }^{\ast }=\inf \{\theta :D(\theta )>0\}$, taking ${\theta }^{\ast }=+\infty$ when this set is empty. Below ${\theta }^{\ast }$ there is no positive difference. Above a finite ${\theta }^{\ast }$ there is a smaller parameter with positive difference, which propagates upward; the same reasoning applies when ${\theta }^{\ast }=-\infty$. ThusNo assertion at ${\theta }^{\ast }$ itself is needed.

For the final request fix any cutoff $a$ and prior density $\pi$. Set $A(x)={\int }_{\theta >a}\pi (\theta )f_{\theta }(x)d\theta$ and $B(x)={\int }_{\theta \le a}\pi (\theta )f_{\theta }(x)d\theta$. The posterior odds are $A(x)/B(x)$. If $T(x_2)\ge T(x_1)$, thenEvery bracket is nonnegative by monotone likelihood ratio. Equal statistics make the bracket zero, so the ratio depends only on $T$. Dividing where the posterior is defined proves the posterior odds are nondecreasing in $T$, with extended values allowed when one posterior mass is zero. This proves that posterior odds are monotone under a monotone likelihood ratio without restricting the prior's shape.

The common positive-support condition is material to the printed strict inequality. Under the broader definition permitting zero densities, take a two-parameter family on $T\in \{0,1,2\}$ with $f_0=(1/3,1/3,1/3)$ and $f_1=(0,0,1)$. Its likelihood ratio $(0,0,3)$ is nondecreasing. The cutoff test rejecting $T\ge 1$ has greater power than the test rejecting only $T=2$ at parameter zero, but both have power one at parameter one. Thus the strict assertion needs the usual positivity convention; it is false without it.

Let $S={\sum }_{i=1}^n{X}_i$. The likelihood ratio of rate two to rate one is $2^n{e}^{-S}$, which decreases strictly with $S$. By the Neyman-Pearson lemma, the most powerful test rejects for small values of the sum:Under the null, $S$ has the gamma distribution with shape $n$ and rate one. Thus $c_{\alpha }$ is its $\alpha$-quantile, equivalently half the $\alpha$-quantile of the chi-squared distribution with $2n$ degrees of freedom. The distribution is continuous, so boundary randomization is unnecessary for $0<\alpha <1$.

For every fixed alternative $\lambda >1$, the likelihood ratio against rate one is ${\lambda }^n{e}^{-(\lambda -1)S}$, again strictly decreasing. Therefore the same region is most powerful against every such alternative. Moreover, if $G$ has the shape-$n$, rate-one gamma distribution, then $S$ at rate $\lambda$ has the same distribution as $G/\lambda$. It follows that ${\mathbb{P}}_{\lambda }(S<c_{\alpha })=\mathbb{P}(G<\lambda c_{\alpha })$ increases with $\lambda$, so the maximum rejection probability over $\lambda \le 1$ is $\alpha$ at one. Any competing size-$\alpha$ test for the composite null is also of level at most $\alpha$ at rate one, and is bounded in power by the Neyman-Pearson lemma for each alternative. The test is uniformly most powerful for the composite hypotheses as well. This is the uniformly most powerful test for an exponential rate based on the sample sum.

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 Neyman-Pearson lemma says that for simple hypotheses with densities $f_0,f_1$, a size-$\alpha$ test that rejects where $f_1/f_0>k$ and accepts where $f_1/f_0<k$, with possible randomization on equality to attain size $\alpha$, is most powerful among all tests of size at most $\alpha$. To prove it, let ${\phi }^{\ast }$ be this test and $\phi$ any competing test. Pointwise,After integration,which proves maximal power.

For the exponential sample, with $S={\sum }_i{X}_i=n\overline{X}$,Since ${\lambda }_1<{\lambda }_0$, this ratio is increasing in $S$. The likelihood-ratio test therefore rejects for large $\overline{X}$:where $G_{n,\lambda }$ is the distribution function of the rate-parameterized Gamma distribution $\Gamma (n,\lambda )$.

A level-$\alpha$ test is uniformly most powerful for a composite alternative when its power is at least that of every other level-$\alpha$ test at every parameter value in that alternative.

For every fixed ${\lambda }_1<{\lambda }_0$, the likelihood ratio is increasing in the same statistic $S$, and the size-$\alpha$ threshold is always the null quantile $n{c}_{\alpha }$, independent of ${\lambda }_1$. The Neyman-Pearson lemma therefore makes the test from part (a) most powerful against every individual ${\lambda }_1<{\lambda }_0$. Consequently it is