Why the Likelihood-Ratio Rule Matters
When two fully specified models are being compared, the evidence is not equally informative for every possible outcome. The Neyman–Pearson lemma formalizes a useful principle: for a fixed probability of rejecting the null model when it is true, rank outcomes by how much more likely they are under the alternative than under the null. This gives a test with the greatest power among tests of that size.
The result applies to simple hypotheses: each hypothesis specifies one complete probability distribution. It does not, by itself, give a most powerful test for a composite hypothesis, which allows multiple parameter values within a hypothesis.
The Lemma and Its Conditions
Suppose the observation \(X\) has density or probability mass function \(f_0\) under \(H_0\) and \(f_1\) under \(H_1\), with both defined relative to the same dominating measure. For discrete outcomes, these are probability masses; for continuous observations, they are densities. The likelihood ratio is the alternative likelihood divided by the null likelihood.
A large value of \(\Lambda(x)\) means that the observed outcome is relatively more supportive of \(H_1\) than of \(H_0\). When \(f_0(x)=0\) but \(f_1(x)>0\), the outcome can be treated as having an infinite likelihood ratio: it cannot occur under the null model, but it can occur under the alternative. When \(f_1(x)=0\) and \(f_0(x)>0\), its ratio is zero.
\(\phi(x)=1\) when \(\Lambda(x)>c\), \(\phi(x)=\gamma\) when \(\Lambda(x)=c\), and \(\phi(x)=0\) when \(\Lambda(x)<c\), where \(0\leq\gamma\leq1\). Choose \(c\) and \(\gamma\) so that \(E_0[\phi(X)]=\alpha\), when exact size \(\alpha\) is attainable. The resulting test has greatest power among all tests whose size is at most \(\alpha\).
The rule includes the infinite- and zero-ratio cases just described. It also explains why a critical region need not be a single upper tail of the observed measurement: the region is an upper set in likelihood-ratio values. The shape of that set depends on the two specified distributions.
Boundary Randomization and the Size Constraint
In a discrete model, the null probability of outcomes can be too coarse to hit a requested size using only reject-or-do-not-reject decisions. A test may reject for every outcome above the boundary ratio and for a fraction of the outcomes tied at the boundary. If the boundary outcome occurs with null probability \(P_0(\Lambda(X)=c)\), the size is
Solving this expression for \(\gamma\) gives the randomization probability. In particular, if \(P_0(\Lambda(X)=c)>0\), then
This value is valid when it lies between 0 and 1. Equivalently, select a boundary \(c\) for which
Here is the short derivation: outcomes with ratio above \(c\) are always rejected, contributing \(P_0(\Lambda(X)>c)\) to the null rejection probability. Outcomes tied at \(c\) are rejected with probability \(\gamma\), contributing \(\gamma P_0(\Lambda(X)=c)\). Adding these contributions and setting their sum equal to \(\alpha\) yields the displayed formula. If the likelihood ratio has no probability mass at the boundary, no boundary randomization is needed; the critical value can be chosen to give the desired size when such a value exists.
Write the null and alternative distributions completely. The lemma compares one specified distribution against another.
Calculate \(f_1(x)/f_0(x)\) where \(f_0(x)>0\), and account separately for outcomes with zero probability under one hypothesis.
Reject first for the largest ratios, choosing the boundary so that the null rejection probability does not exceed the target size.
If the boundary outcomes have positive null probability, solve for \(\gamma\) to reach the target size exactly.
Average the test function using the alternative distribution, not the null distribution.
Worked Applications
Worked Example: Randomizing at a Discrete Boundary
A hypothetical inspection system records one of three labels, \(A\), \(B\), or \(C\). Under \(H_0\), the probabilities are \(0.50\), \(0.30\), and \(0.20\); under \(H_1\), they are \(0.20\), \(0.30\), and \(0.50\), respectively. Construct a Neyman–Pearson test of size \(0.30\) and calculate its power. The distributions are the stated categorical distributions; interest and mortality assumptions are not applicable.
Method. Calculate the alternative-to-null probability ratio for each label, then take the labels with the largest ratios first.
The ordering is \(C\), then \(B\), then \(A\). Rejecting whenever \(C\) occurs uses size \(P_0(C)=0.20\). The remaining size allowance is \(0.30-0.20=0.10\). Since \(P_0(B)=0.30\), the required boundary randomization probability is
Thus the test rejects for \(C\), rejects with probability \(1/3\) for \(B\), and does not reject for \(A\). Its size is \(0.20+(1/3)(0.30)=0.30\). Its power is
Check. The null rejection probability is exactly \(0.30\), and the alternative rejection probability is \(0.60\). The common mistake is to randomize over \(A\), the lowest-ratio label, rather than over \(B\), the boundary label. That would spend size on an outcome less favorable to \(H_1\) while leaving a more favorable outcome unrejected.
Worked Example: A Continuous Likelihood-Ratio Threshold
A hypothetical measurement \(X\) lies in \([0,1]\). Under \(H_0\), \(X\) has a Uniform\((0,1)\) distribution, with density \(f_0(x)=1\). Under \(H_1\), its density is \(f_1(x)=2x\) on \([0,1]\). Find the Neyman–Pearson test of size \(0.25\) and its power. These are the complete model specifications; interest and mortality assumptions are not applicable.
Method. Divide the alternative density by the null density, then find the upper likelihood-ratio region with null probability \(0.25\).
Because \(2x\) increases with \(x\), the test rejects for large \(x\). Under the uniform null, \(P_0(X>c)=1-c\). Set this equal to \(0.25\), obtaining \(c=0.75\). The test rejects when \(x>0.75\); the boundary itself has probability zero under these continuous distributions, so boundary randomization is unnecessary.
Under the alternative, the power is the probability that \(X>0.75\):
Check. The size is \(1-0.75=0.25\), and the power is \(0.4375\). The common mistake is to use the null probability for power. Under \(H_0\), the rejection probability is \(0.25\); power must instead be calculated by integrating the alternative density over the same rejection region.
Worked Example: An Outcome Impossible Under the Null
A hypothetical monitoring device reports \(A\), \(B\), or \(C\). Under \(H_0\), the probabilities are \(0.60\), \(0.40\), and \(0\); under \(H_1\), they are \(0.20\), \(0.50\), and \(0.30\). Construct a size-\(0.20\) test and calculate its power. The model is categorical with the stated probabilities; interest and mortality assumptions are not applicable.
Method. Include outcomes that are possible under the alternative but impossible under the null, then order the remaining outcomes by their ratios.
Outcome \(C\) has \(f_0(C)=0\) and \(f_1(C)=0.30\), so it can be rejected without increasing the size. For the other outcomes,
Rejecting \(B\) with probability \(0.50\), in addition to always rejecting \(C\), gives size \(0.50(0.40)+0(0.60)+1(0)=0.20\). The corresponding power is
Check. Under \(H_0\), \(C\) contributes no rejection probability, and the randomized \(B\) contributes \(0.20\). Under \(H_1\), \(C\) contributes \(0.30\) and randomized \(B\) contributes \(0.25\). The common mistake is to divide by the zero null probability for \(C\) and call the ratio an ordinary finite number; it is clearer to handle this outcome separately as an infinite-ratio outcome.
What the Guarantee Does—and Does Not—Say
The lemma’s optimality statement compares tests for the same two simple hypotheses and the same size constraint. It says that no competing test with size at most \(\alpha\) has greater power than the likelihood-ratio test constructed for that size. It does not say that the test has high power for every alternative distribution, or that the null hypothesis is the more plausible one whenever the test does not reject.
For a discrete distribution, a nonrandomized test may be unable to attain size exactly \(\alpha\). If the decision must be deterministic, one can often select the largest likelihood ratios whose total null probability is no greater than \(\alpha\); the resulting size may be below the allowance. The randomized version reaches the requested size when the boundary probability permits it. In an application, whether randomized decisions are operationally meaningful is a separate practical consideration.
Check Your Understanding
For each problem, identify the likelihood-ratio ordering, apply the size constraint, and calculate power where requested.
- A categorical observation has probabilities \((0.40,0.40,0.20)\) under \(H_0\) and \((0.10,0.40,0.50)\) under \(H_1\), in the order \(A,B,C\). Find the size-\(0.30\) likelihood-ratio test, including any boundary randomization, and its power. (Answer: Reject \(C\), reject \(B\) with probability \(0.25\), and do not reject \(A\); power \(0.60\).)
- Under \(H_0\), \(X\) is Uniform\((0,2)\), with density \(1/2\) on \([0,2]\); under \(H_1\), its density is \(x/2\) on \([0,2]\). Find the size-\(0.20\) rejection region and power. (Answer: Reject for \(X>1.6\); power \(0.36\).)
- For outcomes \(A,B,C\), let the probabilities under \(H_0\) be \((0.50,0.25,0.25)\) and under \(H_1\) be \((0.20,0.50,0.30)\). Find the size-\(0.40\) test and power. (Answer: Reject \(B\), reject \(C\) with probability \(0.60\), and do not reject \(A\); power \(0.68\).)
- Under \(H_0\), the probabilities of \(A,B,C\) are \((0.70,0.30,0)\); under \(H_1\), they are \((0.10,0.60,0.30)\). Find a size-\(0.15\) test and its power. (Answer: Always reject \(C\), reject \(B\) with probability \(0.50\), and do not reject \(A\); power \(0.60\).)
- Under \(H_0\), probabilities for \(A,B,C\) are \((0.50,0.30,0.20)\); under \(H_1\), they are \((0.40,0.20,0.40)\). Find a size-\(0.10\) test and power. (Answer: Reject \(C\) with probability \(0.50\), and do not reject \(A\) or \(B\); power \(0.20\).)