Why Different Supports Require Care
For two simple hypotheses, the Neyman–Pearson principle orders observations by how much more likely they are under the alternative than under the null. Uniform models make the densities especially simple, but different support endpoints create an important complication: some observations may be impossible under one hypothesis and possible under the other. Such outcomes cannot be handled safely by dividing two densities without first checking where each density is positive.
This tutorial considers one observation \(X\) from one of two specified continuous uniform distributions. The same support logic applies to joint densities for a sample, but the examples here use a single observation so the support regions and probability calculations are explicit. The tests are aligned with the SOA VEE Mathematical Statistics syllabus.
Likelihood Ratios Across the Supports
A likelihood ratio compares the density under \(H_1\) with the density under \(H_0\). On the part of the real line where both densities are positive, it is their ordinary ratio. Where only one density is positive, the interpretation follows directly from the hypotheses: an alternative-only outcome can occur under \(H_1\) but has null probability zero; a null-only outcome can occur under \(H_0\) but has alternative probability zero.
If \(f_0(x)>0\) and \(f_1(x)=0\), then \(\Lambda(x)=0\). If both densities are zero, the observation is impossible under both hypotheses, so the test can be defined arbitrarily there. For uniform models, the ratio is constant on the overlap of the two supports because each density is constant. Thus the ordering usually has just three levels: alternative-only outcomes have infinite ratio, overlap outcomes have a positive constant ratio, and null-only outcomes have ratio zero.
| Location of \(x\) | Null density | Alternative density | Likelihood-ratio role |
|---|---|---|---|
| Alternative support only | Zero | Positive | Infinite ratio; reject \(H_0\) |
| Both supports | Positive | Positive | Compare the constant density ratio |
| Null support only | Positive | Zero | Zero ratio; do not spend size here unless needed |
| Neither support | Zero | Zero | Impossible under both; irrelevant to size and power |
How the Neyman–Pearson Rule Applies
A test can be represented by a function \(\phi(x)\) between zero and one: after observing \(x\), reject \(H_0\) with probability \(\phi(x)\). A nonrandomized test has \(\phi(x)\) equal to zero or one. Its size is \(E_0[\phi(X)]\), the probability of rejection under \(H_0\), and its power is \(E_1[\phi(X)]\), the probability of rejection under \(H_1\). The goal for a size \(\alpha\) test is to maximize power among tests whose size is no greater than \(\alpha\).
The Neyman–Pearson construction rejects for the largest likelihood ratios first. In particular, alternative-only outcomes are rejected because they add power without adding null probability. On an overlap where the likelihood ratio is constant, all points tie: one may reject on a suitable part of that region or randomize there to use the desired null probability. But a subset of the tied region can attain the target size only when the required null probability is no greater than the null probability of the whole tied region. If the target is larger, rejection must also be assigned to lower-ratio outcomes, such as the null-only region, or randomized there.
The basic justification also explains why the support cases matter. Let \(\phi^*\) be a likelihood-ratio test with cutoff \(c\geq 0\), and let \(\phi\) be any competing test with no greater size. On points above the cutoff, \(\phi^*=1\); below it, \(\phi^*=0\); and at a tie the selected value is between zero and one. Therefore, pointwise, \((\phi^*-\phi)(f_1-cf_0)\geq 0\). Integrating gives a nonnegative quantity. Because \(E_0[\phi^*]\geq E_0[\phi]\), we also have \(c(E_0[\phi^*]-E_0[\phi])\geq 0\). Adding these two quantities yields the difference in power:
Where \(f_0=0<f_1\), rejecting contributes positively to power and nothing to size, consistent with the ordering. Where \(f_1=0<f_0\), rejection contributes nothing to power, so such outcomes are used only if the target size cannot be reached with higher-ratio outcomes. The argument assumes the tests are being compared at the same maximum allowed size and that the cutoff and any tie handling produce a test of that size.
Constructing and Checking a Test
Worked Example: The Target Size Exceeds the Overlap Probability
Suppose \(H_0:X\sim U(0,4)\) and \(H_1:X\sim U(0,2)\). Construct a most powerful test of size \(\alpha=0.75\), and calculate its power. Use the likelihood-ratio ordering, including the part of the null support that is absent under \(H_1\).
Method and supports. The null density is \(f_0(x)=1/4\) on \((0,4)\); the alternative density is \(f_1(x)=1/2\) on \((0,2)\). On the overlap \((0,2)\), \( \Lambda(x)=(1/2)/(1/4)=2\). On \((2,4)\), \(f_1(x)=0\), so \(\Lambda(x)=0\). There is no alternative-only region.
Use the highest ratio first. Reject throughout \((0,2)\). Its null probability is \(2(1/4)=0.50\), which is less than the target \(0.75\). The tied region is the whole overlap, with null probability only \(0.50\); therefore a subset of that region cannot supply the remaining \(0.25\) null probability. The test must also reject on a part of the null-only region.
Under \(H_0\), an interval of length \(L\) in \((2,4)\) has probability \(L/4\). To add size \(0.25\), solve \(L/4=0.25\), giving \(L=1\). One valid rejection region is \(C=(0,2)\cup(2,3)\). Its size is \(2/4+1/4=0.75\).
Under \(H_1\), all of the support \((0,2)\) is rejected, and the additional interval \((2,3)\) has probability zero. Thus the power is \(1\). The size check is \(0.75\), and the power cannot exceed \(1\), so the result is consistent. Common mistake avoided: treating the overlap as if it could supply a size of \(0.75\), despite its null probability being only \(0.50\).
Worked Example: Alternative-Only Outcomes and an Overlap Tie
Suppose \(H_0:X\sim U(0,2)\) and \(H_1:X\sim U(1,4)\). Construct a most powerful test of size \(\alpha=0.25\) and find its power. State what happens in each support region.
Method and densities. The null density is \(1/2\) on \((0,2)\), and the alternative density is \(1/3\) on \((1,4)\). The null-only interval is \((0,1)\), the overlap is \((1,2)\), and the alternative-only interval is \((2,4)\).
On the overlap, \(\Lambda(x)=(1/3)/(1/2)=2/3\). On \((0,1)\), the ratio is zero. On \((2,4)\), the null density is zero and the alternative density is positive, so the ratio is infinite. Reject on all of \((2,4)\) first; this adds no size under \(H_0\).
The alternative-only interval contributes null probability zero, so the remaining size to allocate is \(0.25\). The null density on the overlap is \(1/2\), so an interval of length \(L\) there has null probability \(L/2\). Setting \(L/2=0.25\) gives \(L=0.50\). Choose \((1,1.5)\) as that part of the overlap. The rejection region is \(C=(1,1.5)\cup(2,4)\), and its null probability is \((0.50)(1/2)=0.25\).
Under \(H_1\), the alternative-only region has probability \(2/3\), since its length is \(2\) and the density is \(1/3\). The selected overlap interval has probability \((0.50)(1/3)=1/6\). Total power is \(2/3+1/6=4/6+1/6=5/6\), or \(0.8333\) to four decimal places. The null-only interval is not rejected. Common mistake avoided: omitting the alternative-only interval because a finite density ratio cannot be calculated there; its likelihood-ratio role is to be rejected without using size.
Worked Example: Both Distributions Have Exclusive Support Regions
Suppose \(H_0:X\sim U(2,5)\) and \(H_1:X\sim U(0,4)\). Find a most powerful test of size \(\alpha=0.20\) and calculate its power. Verify that neither exclusive support region is confused with the overlap.
Method and densities. Under \(H_0\), \(f_0(x)=1/3\) on \((2,5)\). Under \(H_1\), \(f_1(x)=1/4\) on \((0,4)\). The alternative-only region is \((0,2)\), the overlap is \((2,4)\), and the null-only region is \((4,5)\).
On the overlap, \(\Lambda(x)=(1/4)/(1/3)=3/4\). The ratio is infinite on \((0,2)\) and zero on \((4,5)\). Reject on \((0,2)\) at once. Its null probability is zero, while its alternative probability is \(2/4=0.50\).
The target size \(0.20\) must be taken from the overlap, where the null density is \(1/3\). A selected interval of length \(L\) has null probability \(L/3\), so \(L/3=0.20\) gives \(L=0.60\). Choose \((2,2.6)\). The rejection region is \(C=(0,2)\cup(2,2.6)\), and its size is \(0+0.60/3=0.20\).
The selected overlap contributes alternative probability \(0.60/4=0.15\). Adding the alternative-only probability gives power \(0.50+0.15=0.65\). The size is exactly \(0.20\), and the rejected region contains no null-only outcomes. Common mistake avoided: using the density \(1/3\) to calculate the alternative-only probability; that region has zero null density and must instead be evaluated under the alternative density \(1/4\).
Practical Checks and Common Traps
When constructing a test for uniform models with different endpoints, draw or list the intervals before computing probabilities. The overlap determines where an ordinary density ratio exists; the exclusive pieces determine where the ratio is zero or infinite. Then allocate null probability in likelihood-ratio order until the requested size is reached. Since the ratio is constant across the overlap, a selected interval or randomized fraction of that region is adequate when the required null probability fits within its available null mass.
A frequent error is to assume that every exact size can be achieved using only the overlap. If the required null probability is larger than the null probability of the entire tied region, this cannot work: no subset of a region has more probability than the whole region. Continue into the null-only region, which has lower likelihood ratio, or randomize there as needed. Such additional rejection does not add power when the alternative density is zero there, but it can make the size exact.
- Check supports before dividing. A ratio calculated only on the overlap does not describe alternative-only or null-only outcomes.
- Use the correct density for each probability. Size is calculated under \(H_0\); power is calculated under \(H_1\).
- Distinguish size from power. A rejection region can have positive alternative probability but zero null probability, as happens on an alternative-only region.
- Check the available mass before selecting a subset. A tied region can supply only up to its full null probability.
- Remember the role of null-only outcomes. They have likelihood ratio zero and are not useful for power, but may be needed to attain a large target size.
Check Your Understanding
For each problem, identify the relevant support regions, construct a likelihood-ratio rejection region at the stated size, and check the final size and power.
- Let \(H_0:X\sim U(0,5)\) and \(H_1:X\sim U(0,2)\). Find a most powerful test of size \(0.60\) and its power. (Answer: Reject on \((0,2)\) and a length-1 interval in \((2,5)\), for example \((2,3)\); size \(0.60\), power \(1.0000\).)
- Let \(H_0:X\sim U(0,3)\) and \(H_1:X\sim U(1,5)\). At size \(0.30\), reject all alternative-only outcomes and choose an interval in the overlap to complete the test. Find the interval length and power. (Answer: Alternative-only probability \(0.50\); overlap interval length \(0.90\); size \(0.30\), power \(0.7250\).)
- Let \(H_0:X\sim U(0,4)\) and \(H_1:X\sim U(0,8)\). At size \(0.25\), construct a most powerful test and calculate its power. (Answer: Reject on \((4,8)\) and an overlap interval of length \(1\); size \(0.25\), power \(0.6250\).)
- Let \(H_0:X\sim U(0,2)\) and \(H_1:X\sim U(0,4)\). At size \(0.80\), how much of the overlap's null probability must be rejected, and what is the resulting power? (Answer: Reject all of \((2,4)\) and a subset of \((0,2)\) with null probability \(0.80\), such as an interval of length \(1.6\); power \(0.9000\).)
- Let \(H_0:X\sim U(0,3)\) and \(H_1:X\sim U(2,6)\). At size \(0.10\), find the length of a selected overlap interval and the power. (Answer: Reject \((3,6)\) and an overlap interval of length \(0.30\); size \(0.10\), power \(0.8250\).)