Uniqueness, Ties, and Equivalent Most Powerful Tests

Why Ties Matter in a Most Powerful Test

For two simple hypotheses, the Neyman–Pearson method ranks possible observations by how strongly they support the alternative relative to the null. A most powerful test rejects observations with the largest likelihood ratios first. Sometimes several outcomes have exactly the same ratio. If the target size requires rejecting only part of this tied group, there may be more than one way to construct a most powerful test.

The distinction matters in discrete problems: a test can be optimal without being the only optimal test. Moreover, two tests can differ in which tied outcomes they reject and still have exactly the same size and power. The key is that the likelihood ratio is constant across the tied boundary.

Definition: For simple hypotheses \(H_0\) and \(H_1\), let \(f_0\) and \(f_1\) be the probability mass functions or density functions under the two hypotheses, defined relative to the same underlying measure. A test function \(\phi(x)\) gives the probability of rejecting \(H_0\) after observing \(x\), so \(0\leq \phi(x)\leq 1\). A nonrandomized test has \(\phi(x)\in\{0,1\}\). The size is \(E_0[\phi(X)]\), the probability of rejection under \(H_0\); the power is \(E_1[\phi(X)]\), the probability of rejection under \(H_1\). The likelihood ratio is \(\Lambda(x)=f_1(x)/f_0(x)\) wherever \(f_0(x)>0\).

A randomized test can be implemented by an auxiliary random draw: after observing a boundary outcome \(x\), reject with probability \(\phi(x)\). In a discrete model, randomization lets the test attain a target size that a nonrandomized rejection region may not attain.

Theorem: For testing two simple hypotheses at target size \(\alpha\), a Neyman–Pearson test rejects when \(\Lambda(x)>c\), accepts when \(\Lambda(x)<c\), and may choose any rejection probabilities between zero and one when \(\Lambda(x)=c\), provided its total size is \(\alpha\). Outcomes with \(f_0(x)=0\) and \(f_1(x)>0\) have infinite likelihood ratio and are rejected; outcomes with both probabilities or densities zero do not affect size or power.

Here, the set \(\{x:\Lambda(x)=c\}\) is the tied boundary. On the strict-rejection region, the test must reject; on the strict-acceptance region, it must accept. The possible choices that matter for uniqueness are on the tied boundary, including how much randomization to use there.

Why Boundary Choices Preserve Power

Suppose two tests, \(\phi_1\) and \(\phi_2\), agree outside the tied boundary and both have size \(\alpha\). Write \(B=\{x:\Lambda(x)=c\}\). On \(B\), \(f_1(x)=c f_0(x)\). Therefore, their power difference is determined directly by their size difference on that set.

\begin{aligned} E_1[\phi_1(X)]-E_1[\phi_2(X)] &=\int_B(\phi_1(x)-\phi_2(x))f_1(x)\,dx\\ &=c\int_B(\phi_1(x)-\phi_2(x))f_0(x)\,dx. \end{aligned}

Because the tests agree off \(B\), the integral of \((\phi_1-\phi_2)f_0\) over \(B\) is their total size difference. If their sizes are equal, that integral is zero, and their powers are equal as well. For a discrete sample space, the integrals in this argument are sums. This justifies the central fact: among outcomes with the same likelihood ratio, equal null probability assigned to rejection produces equal alternative probability of rejection.

The general Neyman–Pearson comparison also explains why an optimal test must follow the likelihood-ratio ordering. If \(\phi^*\) is a threshold test of size \(\alpha\), and \(\phi\) is any other test of size at most \(\alpha\), then

E_1[\phi^*]-E_1[\phi] =\int(\phi^*-\phi)(f_1-cf_0) +c\big(E_0[\phi^*]-E_0[\phi]\big)\geq 0.

The first term is nonnegative: above the threshold, \(\phi^*=1\) and \(f_1-cf_0>0\); below it, \(\phi^*=0\) and \(f_1-cf_0<0\); on the boundary, \(f_1-cf_0=0\). The second term is nonnegative because \(c\geq0\), \(E_0[\phi^*]=\alpha\), and \(E_0[\phi]\leq\alpha\). Thus no test of size at most \(\alpha\) has greater power. For equal-size tests, differences on the boundary make no contribution to the first term, which is why different boundary allocations can remain equally powerful.

Worked Examples: Recognizing Nonuniqueness

Worked Example: A Randomized Choice on a Tied Boundary

A hypothetical diagnostic has outcomes \(U,V,W,Z\). Under \(H_0\), their probabilities are \(0.10,0.15,0.25,0.50\); under \(H_1\), they are \(0.20,0.15,0.25,0.40\), respectively. Construct two most powerful tests of size \(0.35\), and check their powers.

Method. Calculate the likelihood ratios and take the outcomes with the largest ratios first.

For each outcome, divide its probability under \(H_1\) by its probability under \(H_0\):

\Lambda(U)=\frac{0.20}{0.10}=2,\quad \Lambda(V)=\frac{0.15}{0.15}=1,\quad \Lambda(W)=\frac{0.25}{0.25}=1,\quad \Lambda(Z)=\frac{0.40}{0.50}=0.8.

Outcome \(U\) has the largest ratio, so reject it first. Its null probability is \(0.10\), leaving \(0.35-0.10=0.25\) of size to allocate. Outcomes \(V\) and \(W\) are tied at ratio \(1\), and \(W\) has null probability \(0.25\).

Test 1: reject \(U\) and \(W\). Its size is \(0.10+0.25=0.35\). Its power is \(0.20+0.25=0.45\).

Test 2: reject \(U\) and \(V\), and reject \(W\) with probability \(0.40\). Its size is \(0.10+0.15+0.40(0.25)=0.10+0.15+0.10=0.35\). Its power is \(0.20+0.15+0.40(0.25)=0.20+0.15+0.10=0.45\). Both tests are most powerful: they reject \(U\), and allocate the remaining size only among outcomes tied at the next-highest likelihood ratio.

Check. The second test has the required size \(0.35\) and the same power \(0.45\), as does the first.

Common mistake and correction. Rejecting \(U\) and \(V\) alone does not give size \(0.35\): its size is \(0.10+0.15=0.25\), and its power is \(0.20+0.15=0.35\). To reach size \(0.35\), include \(W\) with probability \(0.40\). This correction also shows why simply swapping tied outcomes does not always produce another deterministic test of the required size.

Worked Example: Two Different Deterministic Most Powerful Tests

For a hypothetical inspection system, outcomes \(A,B,C\) have probabilities \((0.20,0.20,0.60)\) under \(H_0\) and \((0.30,0.30,0.40)\) under \(H_1\). Find two different nonrandomized most powerful tests of size \(0.20\), and verify their powers.

Method. Compare likelihood ratios, then select one full outcome from the tied highest-ratio group.

The ratios are \(\Lambda(A)=0.30/0.20=1.5\), \(\Lambda(B)=0.30/0.20=1.5\), and \(\Lambda(C)=0.40/0.60=2/3\). Thus \(A\) and \(B\) are tied for the highest ratio. Each has null probability \(0.20\), exactly the target size.

One test rejects only \(A\). Its size is \(P_0(A)=0.20\), and its power is \(P_1(A)=0.30\). A second test rejects only \(B\). Its size is \(P_0(B)=0.20\), and its power is \(P_1(B)=0.30\). Both are most powerful because each spends its entire rejection probability on a highest-ratio outcome.

Check. The tests have the same size and power, even though their rejection regions differ. The equality follows from the tied ratios and equal null masses. A test rejecting \(A\) with probability \(r_A\) and \(B\) with probability \(r_B\) also has size \(0.20(r_A+r_B)\) and power \(0.30(r_A+r_B)\); setting \(r_A+r_B=1\) gives size \(0.20\) and power \(0.30\).

Common mistake. A tie does not mean both outcomes must be rejected. Rejecting both \(A\) and \(B\) would have size \(0.40\), not \(0.20\). The target size determines how much of the tied group can be included.

Worked Example: A Unique Most Powerful Test When There Is No Boundary Tie

A hypothetical screening procedure produces outcomes \(R,S,T\). Their probabilities under \(H_0\) are \((0.20,0.30,0.50)\), and under \(H_1\) they are \((0.50,0.30,0.20)\). At size \(0.20\), identify the most powerful test and explain whether it is unique.

Method. Rank the likelihood ratios and check whether the target size ends at a tied boundary.

The likelihood ratios are \(\Lambda(R)=0.50/0.20=2.5\), \(\Lambda(S)=0.30/0.30=1\), and \(\Lambda(T)=0.20/0.50=0.4\). The largest ratio is for \(R\), whose null probability is exactly \(0.20\). The test rejects \(R\) and accepts \(S\) and \(T\).

Its size is \(P_0(R)=0.20\), and its power is \(P_1(R)=0.50\). The cutoff can be chosen between \(1\) and \(2.5\), so no outcome lies on the cutoff boundary. The test is unique as a test function, apart from changes on events with null probability zero: any different allocation of the same size would replace some rejection probability at ratio \(2.5\) with probability at a lower ratio, reducing power.

Check. The stated test meets the size constraint and attains power \(0.50\). Unlike the preceding examples, there is no tied boundary on which to exchange rejection probability.

Common mistake. Uniqueness does not follow merely because the target size is met exactly. It follows here from the strict likelihood-ratio ordering at the cutoff.

When Is the Most Powerful Test Unique?

For tests treated as functions \(\phi(x)\), a useful sufficient condition for uniqueness is that the likelihood ratio equals the cutoff only on a set of null probability zero. Then the threshold rule determines the test, apart from changes on outcomes that have probability zero under \(H_0\). Such changes also have no effect on power when the alternative is absolutely continuous with respect to the null on those outcomes.

A positive-probability tied boundary does not automatically imply nonuniqueness. The amount of rejection probability required on the boundary matters, as does the structure of the outcomes and the class of tests being considered. If the target size requires rejecting none or all of the boundary, there may be no alternative boundary allocation. If a single boundary outcome must be rejected with a fixed fractional probability, the test function at that outcome is fixed, even though its random implementation can differ. Conversely, when rejection probability can be distributed in more than one way across the boundary, distinct test functions can have the same size and power.

For example, in the first worked example the total null mass needed from the tied outcomes was \(0.25\). That mass could be obtained by taking all of \(W\), or by taking all of \(V\) plus \(40\%\) of \(W\). The resulting rejection functions differ, but each assigns the same total null probability to outcomes whose likelihood ratio is \(1\). The boundary identity \(f_1=cf_0\) then guarantees equal alternative probability as well.

Key takeaway: A most powerful test is determined by strict likelihood-ratio inequalities away from the cutoff. On the tied boundary, equal-size allocations have equal power. Uniqueness depends on whether the boundary permits genuinely different test functions that meet the size constraint, not merely on whether a tie exists.

Common Traps in Tie Problems

Assuming every tied outcome can be exchanged deterministically. Their null probabilities may differ, so swapping one for another may change the size. Check the actual null mass before claiming a second test has the same size.
Reporting a size without summing all rejection probabilities. For a randomized test, multiply each outcome's null probability by its rejection probability, then add.
Checking size but not power. Calculate power separately using probabilities under \(H_1\). On a tied boundary, equal null rejection mass implies equal alternative rejection mass because the ratio is constant; it does not mean that the power equals the size.
Calling any tie proof of nonuniqueness. Check whether the boundary allocation can actually vary while preserving the required size, and clarify whether uniqueness refers to the test function or to the random mechanism used to implement it.

Check Your Understanding

For each problem, identify the likelihood-ratio boundary where relevant, and verify both size and power when a test is requested.

A device has outcomes \(M,N,R\), with probabilities \((0.25,0.25,0.50)\) under \(H_0\) and \((0.40,0.40,0.20)\) under \(H_1\). At size \(0.25\), give two distinct most powerful nonrandomized tests and their common power. (Answer: Reject \(M\) or reject \(N\); each has size \(0.2500\) and power \(0.4000\).)
Outcomes \(A,B,C,D\) have probabilities \((0.10,0.20,0.30,0.40)\) under \(H_0\) and \((0.20,0.40,0.15,0.25)\) under \(H_1\). At size \(0.20\), give two different most powerful randomized tests using \(A\) and \(B\), and state their common power. (Answer: Reject \(B\) and accept \(A\); or reject \(A\) and reject \(B\) with probability \(0.50\). Each has size \(0.2000\) and power \(0.4000\).)
Outcomes \(J,K,L\) have probabilities \((0.50,0.30,0.20)\) under \(H_0\) and \((0.20,0.60,0.20)\) under \(H_1\). At size \(0.40\), find a most powerful test and its power. (Answer: Reject \(K\) and reject \(L\) with probability \(0.50\); size \(0.3000+0.50(0.2000)=0.4000\), power \(0.6000+0.50(0.2000)=0.7000\).)
For outcomes \(G,H,I\), the null probabilities are \((0.20,0.30,0.50)\) and the alternative probabilities are \((0.50,0.30,0.20)\). At size \(0.20\), is the most powerful test unique as a test function, up to null-probability events, and what is its power? (Answer: Yes; reject \(G\) only. The likelihood ratios are \(2.5,1,0.4\), so there is no boundary tie at the cutoff; power is \(0.5000\).)
Suppose two tests agree off a boundary on which \(\Lambda(x)=1.7\), and both have size \(0.12\). What is their power difference if their only differences are on that boundary? (Answer: Zero; their power difference is \(1.7\) times their size difference on the boundary, which is zero.)

What You'll Learn