Why a Most Powerful Region Can Be Disconnected
For two simple hypotheses, the Neyman–Pearson principle ranks outcomes by how strongly they favor the alternative over the null. It does not say that outcomes should be ranked by their numerical size. If the likelihood ratio is nonmonotone in the observed value, the most favorable outcomes may lie in several separated ranges. A rejection region can therefore consist of multiple intervals, tails, or isolated outcomes.
The practical task is to calculate the likelihood ratio, order the possible outcomes by that ratio, and choose the cutoff that gives the required size. The resulting region may look unusual if one expects a single upper or lower tail. Its shape follows from the likelihood ratio, not from a default rule about the direction of the data.
A test is represented by a function \(\phi\) that gives the probability of rejecting \(H_0\) after observing the data. A nonrandomized test has \(\phi(x)\) equal to 0 or 1; a randomized test can have a value between 0 and 1. Its size is \(E_0[\phi]\), the probability of rejection under \(H_0\), and its power is \(E_1[\phi]\), the probability of rejection under \(H_1\). A test with size at most \(\alpha\) controls the null rejection probability at the chosen level \(\alpha\).
The Likelihood-Ratio Cutoff and the Shape of the Region
Under the absolute-continuity condition in the definition, the Neyman–Pearson test rejects for the largest likelihood ratios. In particular, a cutoff test has the form \(\phi(x)=1\) when \(\Lambda(x)>k\), \(\phi(x)=0\) when \(\Lambda(x)<k\), and may randomize when \(\Lambda(x)=k\). The constant \(k\) is chosen to attain the desired size when possible. In a discrete problem, randomization can be necessary because outcome probabilities do not usually permit every target size exactly.
When \(\Lambda(x)\) rises and then falls, a set of the form \(\{x:\Lambda(x)>k\}\) need not be one interval. It may include outcomes from both ends of the sample space while excluding values in between. In a continuous model, it can be two tails or several separated intervals. In a discrete model, it can be a scattered set of outcomes.
Why the Cutoff Test Is Most Powerful
Here is a short justification of the optimality claim. Let \(\phi\) be a cutoff test of size \(\alpha\), and let \(\psi\) be any other test with size at most \(\alpha\). Since \(P_1\) is absolutely continuous with respect to \(P_0\), expectations under the alternative can be written using \(\Lambda\):
Where \(\Lambda>k\), the cutoff test has \(\phi=1\), so \(\phi-\psi\geq 0\). Where \(\Lambda<k\), it has \(\phi=0\), so \(\phi-\psi\leq 0\). Thus \((\Lambda-k)(\phi-\psi)\geq 0\) in either case; the product is zero at the boundary. Also, \(k\geq 0\) and \(E_0[\phi-\psi]\geq 0\), because the cutoff test has size \(\alpha\) and \(\psi\) has size at most \(\alpha\). Both terms are therefore nonnegative, giving \(E_1[\phi]\geq E_1[\psi]\). This proves that the cutoff test has at least as much power as any competing test of size at most \(\alpha\).
The change from \(E_1\) to \(E_0\) in this argument depends on absolute continuity: \(E_1[g]=E_0[\Lambda g]\). If an outcome has positive probability under \(H_1\) and zero probability under \(H_0\), the ratio is not an ordinary finite ratio there, and this identity cannot simply be applied to that outcome. Such outcomes require separate treatment, illustrated below.
Worked Example: Selecting Two Separated Discrete Outcomes
A hypothetical inspection records an integer score \(X\in\{0,1,2,3,4\}\). Under \(H_0\), the probabilities are \(0.20\) for each score. Under \(H_1\), the probabilities for scores \(0,1,2,3,4\) are \(0.40, 0.05, 0.10, 0.40, 0.05\), respectively. Construct a most powerful test of size \(0.40\). Give its power. The method is likelihood-ratio ordering; it avoids assuming that larger scores must favor the alternative.
Step 1: Calculate the ratios. Divide each alternative probability by its null probability:
Step 2: Order outcomes by ratio. Scores 0 and 3 tie for the largest ratio, 2. Their combined null probability is \(0.20+0.20=0.40\), exactly the target size. Reject for \(X=0\) or \(X=3\); no boundary randomization is needed.
Step 3: Compute power and check size. The alternative probability of the rejection region is \(0.40+0.40=0.80\). Under the null, the probability of that same region is \(0.20+0.20=0.40\). The test therefore has size \(0.40\) and power \(0.80\), as required.
Common mistake avoided: Rejecting the two largest numerical scores would select \(X=3,4\), with size \(0.40\) but power \(0.40+0.05=0.45\). The numerical tail is not the likelihood-ratio ordering; selecting scores 0 and 3 gives greater power.
Worked Example: A Continuous Test with Two Tails
Suppose one measurement \(X\) follows \(N(0,1)\) under \(H_0\) and \(N(0,4)\) under \(H_1\), where the second argument is the variance. Construct the likelihood-ratio test that rejects when \(|X|>2\), and calculate its size and power. The method is to express both probabilities using standard normal variables. The common mistake avoided is treating this likelihood ratio as increasing with \(X\) rather than with \(|X|\).
Step 1: Find the likelihood ratio. The standard deviations are 1 and 2. The densities are \(f_0(x)=(2\pi)^{-1/2}e^{-x^2/2}\) and \(f_1(x)=(2\sqrt{2\pi})^{-1}e^{-x^2/8}\). Hence
This ratio is a function of \(x^2\): it is smallest at 0 and increases as \(|x|\) increases. At the boundary \(|x|=2\), the ratio is \(\frac12 e^{3(4)/8}=\frac12e^{1.5}\approx 2.2408\). Thus the rejection set is \(\{x:\Lambda(x)>2.2408\}=\{x:|x|>2\}\), apart from the boundary, which has probability zero under these continuous models.
Step 2: Calculate size under the null. Under \(H_0\), \(X\sim N(0,1)\), so
Step 3: Calculate power under the alternative. Under \(H_1\), \(X/2\sim N(0,1)\). Therefore
The test has size \(0.0455\) and power \(0.3173\), with probabilities rounded to four decimals. Its two-tail shape follows directly from the ratio: large positive and large negative measurements both favor the larger-variance alternative. A one-sided upper-tail rule would omit evidence in the negative tail.
Boundary Randomization and Null-Impossible Outcomes
For a discrete distribution, an exact target size may fall between the sizes of two nonrandomized cutoff tests. If the next outcome in likelihood-ratio order has null probability \(p\), reject on it with probability \(\gamma\), where \(0\leq\gamma\leq1\). Choose \(\gamma\) so that the added size is the remaining amount. Apply the same rejection probability at that outcome when calculating power, using its alternative probability.
There is an additional case when the alternative assigns positive probability to an outcome that the null makes impossible. Such an outcome should be rejected with probability 1 before using the finite likelihood ratios on outcomes with positive null probability. This adds power but no size. On the common support, likelihood-ratio ordering then determines how to spend any remaining size. The next example carries out this construction without using the ratio identity on the null-impossible outcome.
Worked Example: Randomization After a Null-Impossible Outcome
A hypothetical categorical observation has outcomes \(A,B,C,D\). Under \(H_0\), the probabilities are \(0.40,0.30,0.30,0\), respectively; under \(H_1\), they are \(0.10,0.20,0.40,0.30\). Construct a most powerful test of size \(0.45\) and calculate its power. The method first includes an alternative-positive, null-zero outcome, then orders the common-support outcomes. The common mistake avoided is treating the ratio at \(D\) as an ordinary finite quotient.
Step 1: Include \(D\). Since \(P_0(D)=0\) and \(P_1(D)=0.30\), rejecting whenever \(D\) occurs uses no size and contributes \(0.30\) power. The ordinary ratio \(P_1(D)/P_0(D)\) is undefined, so it is not used as a finite ratio.
Step 2: Order the remaining outcomes. On the outcomes with positive null probabilities, the ratios are
Reject on \(C\) first. This uses size \(0.30\) and contributes power \(0.40\). The remaining size allowance is \(0.45-0.30=0.15\). Outcome \(B\) has null probability \(0.30\), so randomize there with probability \(\gamma=0.15/0.30=0.50\). Do not reject on \(A\).
Step 3: Check size and power. The size is \(0+0.30+0.50(0.30)+0.40(0)=0.45\). The power is \(0.30+0.40+0.50(0.20)+0(0.10)=0.80\). Thus the test attains the target size and has power \(0.80\). The contribution from \(D\) is calculated directly under \(H_1\), not through a likelihood-ratio identity requiring absolute continuity.
Common Traps and a Construction Checklist
A disconnected region is not a defect or an exception to the Neyman–Pearson rule. It is the natural result when separated observations have large likelihood ratios. Before stating a final test, check both the ordering and the probability calculations.
Write the null and alternative probabilities or densities, including their supports. Identify outcomes that have zero probability under the null.
Use alternative likelihood divided by null likelihood. Do not reverse the ratio: large values must favor the alternative.
List all outcomes or solve the inequality \(\Lambda(x)>k\). Keep every separated component of the resulting set.
Add null probabilities in likelihood-ratio order. If the target falls inside a boundary outcome, randomize by the fraction of its null probability needed.
Use null probabilities for size and alternative probabilities for power. Include any null-impossible outcomes directly in the alternative power calculation.
Two errors are especially costly. First, assuming a simple alternative automatically implies a one-tail test confuses the simplicity of the hypotheses with the shape of the likelihood ratio. Second, using \(P_1(A)=E_0[\Lambda\,1_A]\) without checking absolute continuity can omit probability on outcomes that are possible only under \(H_1\). State the support condition before using that identity, or handle null-impossible outcomes separately.
Check Your Understanding
For each problem, identify the likelihood-ratio ordering before deciding on the rejection set. Answers use probabilities rounded to four decimals where needed.
- For \(X\in\{-2,-1,0,1,2\}\), let the null probabilities in that order be \(0.10,0.20,0.40,0.20,0.10\), and the alternative probabilities be \(0.30,0.05,0.30,0.05,0.30\). Find a most powerful test of size \(0.20\) and its power. (Answer: Reject for \(X=-2\) or \(X=2\); size \(0.20\), power \(0.60\).)
- Suppose \(X\sim N(1,1)\) under \(H_0\) and \(X\sim N(1,9)\) under \(H_1\). For the region \(|X-1|>2\), give the likelihood-ratio shape, size, and power. (Answer: The ratio increases with \(|X-1|\), so this is a two-tail region; size \(2[1-\Phi(2)]\approx0.0455\), power \(2[1-\Phi(2/3)]\approx0.5050\).)
- For outcomes \(1,2,3,4\), let \(P_0=(0.25,0.25,0.25,0.25)\) and \(P_1=(0.10,0.50,0.10,0.30)\). Construct a size \(0.40\) most powerful test, including any needed randomization, and find its power. (Answer: Reject at 2 and reject at 4 with probability \(0.60\); size \(0.25+0.60(0.25)=0.40\), power \(0.50+0.60(0.30)=0.68\).)
- For outcomes \(A,B,C\), let \(P_0=(0.50,0.50,0)\) and \(P_1=(0.20,0.30,0.50)\). Find a most powerful test of size \(0.25\) and its power. (Answer: Always reject at \(C\), reject at \(B\) with probability \(0.50\), and do not reject at \(A\); size \(0.25\), power \(0.50+0.50(0.30)=0.65\).)