Why Monotonicity Determines the Rejection Tail
For two simple hypotheses, the Neyman–Pearson lemma orders sample outcomes by their likelihood ratio. The practical question is often how to interpret that ordering: does stronger evidence for the alternative correspond to a large or a small value of a statistic? A monotone likelihood ratio answers this directly. It converts a likelihood-ratio rule into a one-sided threshold rule, while its direction identifies which tail favors the alternative.
This conclusion depends on the direction of monotonicity, not on whether the alternative parameter is numerically larger or smaller. A larger alternative parameter often makes large observations favor the alternative, but that is not a definition and should not be assumed without checking the ratio.
The usual threshold conclusion assumes the likelihood ratio can be represented by a monotone function of one real-valued statistic on the relevant common support. When supports depend on the hypothesis, or when the ratio is not a function of a single statistic, the argument must be reconsidered rather than applied automatically. Zeros in one likelihood can also produce ratios of zero or infinity; those outcomes can be ordered directly, but they should not be silently treated as ordinary finite ratios.
From the Neyman–Pearson Rule to a Threshold
For testing two simple hypotheses at a specified size, the Neyman–Pearson rule rejects \(H_0\) when the likelihood ratio exceeds a critical constant \(k\). If the ratio equals \(k\) at an outcome boundary, a nonrandomized test may include or exclude that boundary, while a randomized test may reject there with a probability between zero and one to attain the target size.
For evidence in favor of \(H_1\), use \(L_1/L_0\), not its reciprocal.
Separate factors that depend on the data through \(T\) from factors that do not vary with the observed sample.
An increasing ratio favors \(H_1\) at larger values; a decreasing ratio favors \(H_1\) at smaller values.
Solve \(g(T)>k\) for \(T\), respecting whether \(T\) is continuous or takes only discrete values.
The threshold statement follows from order alone. If \(g\) is nondecreasing and \(g(t_0)>k\), then for every \(t>t_0\), \(g(t)\geq g(t_0)>k\). Therefore all larger statistic values are also in the rejection region. For a nonincreasing \(g\), if \(g(t_0)>k\), then every \(t<t_0\) has \(g(t)\geq g(t_0)>k\), so the rejection region extends toward smaller values. This proves the one-sided structure; it does not by itself determine the numerical cutoff \(k\) needed for a specified size.
Worked Examples
Worked Example: A Binomial Sample and a Randomized Upper Tail
Suppose \(X_1,\ldots,X_5\) are independent Bernoulli observations with success probability \(p\). Test \(H_0:p=0.3\) against \(H_1:p=0.6\), using the statistic \(T=\sum_{i=1}^5 X_i\). Find the form of the most powerful size-\(0.10\) test, including any boundary randomization, and calculate its power.
Method and direction. The joint likelihood ratio is \( \Lambda(x)= (0.6/0.3)^T(0.4/0.7)^{5-T}=2^T(4/7)^{5-T} \). For adjacent values, \( \Lambda(T+1)/\Lambda(T)=2/(4/7)=3.5>1 \). Thus the ratio increases with \(T\), so the most powerful rejection region is an upper tail, with possible randomization at its smallest included value.
Under \(H_0\), \(T\sim\operatorname{Binomial}(5,0.3)\). The two relevant upper-tail probabilities are \(P_0(T\geq4)=5(0.3)^4(0.7)+(0.3)^5=0.02835+0.00243=0.03078\), and \(P_0(T=3)=\binom{5}{3}(0.3)^3(0.7)^2=10(0.027)(0.49)=0.1323\). The nonrandomized region \(T\geq3\) has size \(0.03078+0.1323=0.16308\), which is too large. The region \(T\geq4\) has size \(0.03078\), so randomize at \(T=3\).
Let \(\gamma\) be the probability of rejecting when \(T=3\). Set the size equal to \(0.10\): \(0.03078+\gamma(0.1323)=0.10\). Hence \( \gamma=(0.10-0.03078)/0.1323=0.06922/0.1323=0.5232048\ldots \). The test rejects for \(T\geq4\), rejects with probability \(0.5232\) when \(T=3\), and otherwise does not reject. The randomization probability is rounded to four decimals.
Under \(H_1\), \(T\sim\operatorname{Binomial}(5,0.6)\). Therefore \(P_1(T\geq4)=5(0.6)^4(0.4)+(0.6)^5=0.2592+0.07776=0.33696\), and \(P_1(T=3)=10(0.6)^3(0.4)^2=0.3456\). The power is \(0.33696+(0.5232048\ldots)(0.3456)=0.5177796\ldots\), or \(0.5178\) to four decimals.
Check. The size is \(0.03078+(0.5232048\ldots)(0.1323)=0.1000\) to four decimals. The calculation avoids the mistake of including all of \(T=3\), which would exceed the target size.
Worked Example: A Normal Mean Alternative That Favors Small Sums
Let \(X_1,\ldots,X_8\) be independent with \(X_i\sim N(\mu,1)\). Test \(H_0:\mu=0.75\) against \(H_1:\mu=0.25\). Show that the likelihood ratio is decreasing in \(T=\sum_{i=1}^8 X_i\), determine the direction of the threshold region for a general critical ratio \(k\), and check the ratio at \(T=0\) and \(T=8\).
Method and direction. Using the normal density under each hypothesis, \( \Lambda(x)=\prod_{i=1}^8 \exp\{-\tfrac12[(x_i-0.25)^2-(x_i-0.75)^2]\} \). For one observation, \((x_i-0.25)^2-(x_i-0.75)^2=x_i-0.5\). Thus \( \Lambda(x)=\exp\{-\tfrac12\sum_{i=1}^8(x_i-0.5)\} =\exp(2-0.5T) \). Its exponent decreases as \(T\) increases, so the ratio is decreasing. The test rejects for sufficiently small \(T\).
For a general \(k>0\), solve the likelihood-ratio inequality: \( \exp(2-0.5T)>k \) if and only if \( 2-0.5T>\ln k \), which is equivalent to \( T<4-2\ln k \). At the illustrative cutoff \(k=1\), this becomes \(T<4\). The test's actual cutoff for a prescribed size must be chosen to meet that size; \(k=1\) is used here only to show the direction and a simple boundary.
At \(T=0\), the ratio is \(e^2=7.389056\ldots\), or \(7.389\) to three decimals. At \(T=8\), it is \(e^{-2}=0.135335\ldots\), or \(0.1353\) to four decimals. Thus the smaller sum has the larger likelihood ratio, as the decreasing formula predicts.
Check. Substituting \(T=4\) gives \(\exp(2-0.5(4))=\exp(0)=1\), consistent with the cutoff \(k=1\). The common mistake is to say \(e^2=2\); in fact \(e^2\approx7.389\). Confusing the alternative's smaller mean with an upper-tail test would also reverse the correct direction.
Worked Example: A Poisson Sample and an Integer Threshold
Suppose \(X_1,X_2,X_3\) are independent Poisson counts, each with rate \(\lambda\). Test \(H_0:\lambda=4\) against \(H_1:\lambda=2\). Express the likelihood ratio using \(T=X_1+X_2+X_3\), identify which tail favors \(H_1\), and find the integer values of \(T\) for which the ratio exceeds 1.
Method and direction. For one count \(x_i\), the likelihood ratio is \( \{e^{-2}2^{x_i}/x_i!\}/\{e^{-4}4^{x_i}/x_i!\}=e^2(1/2)^{x_i} \). Multiplying the three ratios gives \( \Lambda(x)=e^6\,2^{-T} \). Each increase of \(T\) by one multiplies the ratio by \(1/2\), so it is decreasing. Small total counts favor \(H_1\), and the rejection region for any likelihood-ratio cutoff is a lower tail.
For the requested cutoff \(k=1\), \(e^6 2^{-T}>1\) is equivalent to \(6-T\ln2>0\), or \(T<6/\ln2=8.656\ldots\). Because \(T\) is an integer, the ratio exceeds 1 exactly when \(T\leq8\). At \(T=8\), \( \Lambda=e^6/256=403.4288\ldots/256=1.5759\ldots \). At \(T=9\), \( \Lambda=e^6/512=403.4288\ldots/512=0.7879\ldots \).
Check. The ratio crosses 1 between the consecutive attainable values 8 and 9, matching the derived integer condition. The cutoff 1 identifies outcomes with larger likelihood under \(H_1\); it is not automatically the cutoff for every requested test size.
Interpreting Ties, Thresholds, and the Alternative
A monotone likelihood ratio establishes the shape of the Neyman–Pearson region, but several distinctions matter in applications:
- Increasing versus decreasing is the key decision. An increasing ratio means large values of \(T\) favor \(H_1\); a decreasing ratio means small values favor \(H_1\). The numerical direction of a parameter change alone is not enough.
- A threshold may not be an attainable value. For a discrete statistic, solving an inequality can produce a noninteger boundary. Translate it into the appropriate set of attainable values, as in the Poisson example.
- Size may require randomization. If the probability under \(H_0\) jumps past the target when an entire boundary value is added, randomization at that value can attain the target size. The randomization probability is chosen using the boundary's probability under \(H_0\).
- Weak monotonicity can create ties. If \(g\) is constant over a range of statistic values, those values have the same likelihood ratio. The threshold region may include or exclude tied outcomes together, or use boundary randomization as allowed by the Neyman–Pearson rule.
- The ratio's orientation matters. Replacing \(L_1/L_0\) with \(L_0/L_1\) reverses its ordering. Always state which hypothesis the ratio favors before describing a tail.
Check Your Understanding
For each problem, identify the likelihood-ratio direction and translate it into a threshold or tail statement. All hypotheses below are simple.
- For four independent Bernoulli observations, test \(H_0:p=0.25\) against \(H_1:p=0.5\). Let \(T\) be the number of successes. Write the likelihood ratio as a function of \(T\) and state the favored tail. (Answer: \( \Lambda(T)=2^T(2/3)^{4-T} \), with adjacent ratio \(3\); it increases, so large \(T\) favors \(H_1\).)
- For one observation \(X\sim\operatorname{Poisson}(\lambda)\), test \(H_0:\lambda=3\) against \(H_1:\lambda=6\). Express the likelihood ratio and identify the tail for an NP rejection region. (Answer: \( \Lambda(x)=e^{-3}2^x \), which increases in \(x\); reject for sufficiently large counts.)
- Let \(X_1,\ldots,X_5\) be independent \(N(\mu,4)\) observations. Test \(H_0:\mu=2\) against \(H_1:\mu=3\), with \(T=\sum X_i\). Find \(\Lambda\) and the condition for \(\Lambda>1\). (Answer: \( \Lambda=\exp(T/4-25/8) \); it increases in \(T\), and \(\Lambda>1\) exactly when \(T>25/2=12.5\).)
- For six independent Bernoulli observations, test \(H_0:p=0.7\) against \(H_1:p=0.4\). Let \(T\) count successes. Find the factor by which the likelihood ratio changes when \(T\) increases by one, then state the favored tail. (Answer: the factor is \(\{0.4(0.3)\}/\{0.7(0.6)\}=2/7\); the ratio decreases, so small \(T\) favors \(H_1\) and the rejection region is a lower tail.)