Why a One-Sided Poisson Test Has a Boundary Case
A Poisson rate test asks whether an event rate is consistent with a specified range. In a one-sided test, the alternative points in only one direction: for example, whether a hypothetical service has more incidents per month than a benchmark rate. The direction matters because an observation far below the benchmark is not evidence for an upper-sided alternative.
The likelihood-ratio method compares the best likelihood allowed by the null hypothesis with the best likelihood when the rate is unrestricted. For a one-sided composite null, the unrestricted maximum-likelihood estimate may already satisfy the null. In that case, the two best likelihoods are equal and the likelihood-ratio statistic is zero. If the unrestricted estimate falls outside the null range, the constrained estimate is at the boundary, and the statistic is positive.
The model assumes that the counts are independent, the exposures are known and measured in consistent units, and the counts share the same rate. The Poisson mean for count \(K_j\) is \(t_j\lambda\), not \(\lambda\) unless \(t_j=1\). If rates differ across observations, or if events are dependent in a way that changes the count distribution, the formulas below need not apply.
Likelihood and the Two One-Sided Hypotheses
Ignoring factors that do not depend on \(\lambda\), the likelihood for the observed counts is proportional to \(e^{-T\lambda}\lambda^K\). Its log-likelihood, also omitting terms independent of \(\lambda\), is \(K\log\lambda-T\lambda\). For \(K>0\), differentiating gives \(K/\lambda-T=0\), so the unrestricted estimate is \(K/T\). When \(K=0\), the likelihood is largest at \(\lambda=0\), which is also \(K/T\).
For an upper-sided alternative, the hypotheses are \(H_0:\lambda\leq\lambda_0\) against \(H_1:\lambda>\lambda_0\), where \(\lambda_0>0\) is specified. The null parameter space includes its boundary \(\lambda_0\). If \(K/T\leq\lambda_0\), the unrestricted estimate is allowed under the null, so the constrained estimate equals \(K/T\). If \(K/T>\lambda_0\), the likelihood increases up to the unrestricted estimate, which is outside the null; the best null value is then \(\lambda_0\).
For a lower-sided alternative, the hypotheses are \(H_0:\lambda\geq\lambda_0\) against \(H_1:\lambda<\lambda_0\). Here the constrained estimate is \(K/T\) when \(K/T\geq\lambda_0\), and is \(\lambda_0\) when \(K/T<\lambda_0\).
| Test direction | Unrestricted estimate inside null | Constrained null estimate | When the statistic is positive |
|---|---|---|---|
| Upper-sided: \(H_0:\lambda\leq\lambda_0\) | \(K/T\leq\lambda_0\) | \(K/T\) | \(K/T>\lambda_0\) |
| Lower-sided: \(H_0:\lambda\geq\lambda_0\) | \(K/T\geq\lambda_0\) | \(K/T\) | \(K/T<\lambda_0\) |
Let \(\Lambda\) be the ratio of the maximum likelihood under the null to the unrestricted maximum likelihood, so \(0<\Lambda\leq1\). The likelihood-ratio statistic is \(G=-2\log\Lambda\). For either test direction, it is zero when the unrestricted estimate is inside the null. When it is outside, the constrained estimate is \(\lambda_0\), giving the following result.
Why the Statistic Has This Form
When the unrestricted estimate is outside the null, the numerator likelihood is evaluated at \(\lambda_0\), while the denominator is evaluated at \(\hat{\lambda}=K/T\). Taking the log-likelihood difference explains both the formula and its nonnegative value.
After canceling terms that do not involve \(\lambda\), \(\log\Lambda=(K\log\lambda_0-T\lambda_0)-(K\log\hat{\lambda}-T\hat{\lambda})\).
Substitute \(\hat{\lambda}=K/T\), so \(T\hat{\lambda}=K\), and rearrange to obtain \(\log\Lambda=-K\log(K/(T\lambda_0))-T\lambda_0+K\).
Multiplying the negative log ratio by two gives \(G=2[K\log(K/(T\lambda_0))-(K-T\lambda_0)]\).
This derivation applies only when the unrestricted estimate falls outside the null range. Otherwise the null and unrestricted maxima coincide, \(\Lambda=1\), and \(G=-2\log(1)=0\). The statistic describes the likelihood comparison; its calibration into a decision rule is a separate step.
Worked Example: Testing Whether a Hypothetical Incident Rate Is Higher
A hypothetical monitoring system records \(K=20\) incidents over \(T=12\) system-years. Assume a Poisson count with common rate \(\lambda\) per system-year, and test \(H_0:\lambda\leq1\) against \(H_1:\lambda>1\). Find the likelihood-ratio statistic. No interest or mortality assumption is involved.
Method. Use the upper-sided Poisson likelihood-ratio statistic. The observed count has distribution \(K\sim\operatorname{Pois}(12\lambda)\), and the null boundary is \(\lambda_0=1\) incident per system-year.
First calculate the unrestricted estimate: $$ \hat{\lambda}=\frac{K}{T}=\frac{20}{12}=\frac{5}{3}=1.6667 $$ incidents per system-year. Since \(1.6667>1\), the unrestricted estimate is outside the upper-sided null, so the constrained null estimate is \(\lambda_0=1\).
The expected count at the null boundary is \(T\lambda_0=12(1)=12\). Therefore $$ G=2\left[20\log\left(\frac{20}{12}\right)-(20-12)\right] =2\left[20\log\left(\frac{5}{3}\right)-8\right] =4.433024950639628\ldots $$ Thus, to four decimal places, \(G=4.4330\).
Check. The estimated rate exceeds the null boundary, so the statistic must be positive; the computed value is positive. The common mistake is to use \(K\log(20/12)\) but forget to subtract \(K-T\lambda_0=8\). That omission would not be the likelihood-ratio statistic.
Worked Example: An Upper-Sided Test With No Evidence in the Tested Direction
A hypothetical call center observes \(K=9\) calls in \(T=10\) hours. Assume a Poisson count with rate \(\lambda\) calls per hour, and test \(H_0:\lambda\leq1.2\) against \(H_1:\lambda>1.2\). Calculate the statistic. No interest or mortality assumption is involved.
Method. Compare the unrestricted estimate with the null boundary before evaluating a likelihood ratio. Here \(K\sim\operatorname{Pois}(10\lambda)\) and \(\lambda_0=1.2\) calls per hour.
The unrestricted estimate and null expected count are $$ \hat{\lambda}=\frac{9}{10}=0.9 \qquad\text{and}\qquad T\lambda_0=10(1.2)=12. $$ Because \(0.9\leq1.2\), the unrestricted estimate is allowed by the null. Thus the constrained and unrestricted estimates are both \(0.9\), their maximized likelihoods are equal, and $$ \Lambda=1,\qquad G=-2\log(1)=0. $$
Check. The observed rate is below the boundary, so it cannot create evidence in favor of the stated higher-rate alternative through this likelihood-ratio statistic. The common mistake is to plug the data into the positive-deviance expression regardless of direction; the constrained-estimate check must come first.
Worked Example: Testing Whether a Hypothetical Defect Rate Is Lower
A hypothetical production line records \(K=12\) defects during \(T=8\) operating shifts. Assume a Poisson count with rate \(\lambda\) defects per shift, and test \(H_0:\lambda\geq2\) against \(H_1:\lambda<2\). Find the statistic. No interest or mortality assumption is involved.
Method. This is a lower-sided test, so evidence in the alternative direction occurs when the observed rate is below the null boundary. The count distribution is \(K\sim\operatorname{Pois}(8\lambda)\), with \(\lambda_0=2\) defects per shift.
The unrestricted estimate is $$ \hat{\lambda}=\frac{12}{8}=1.5 $$ defects per shift. Since \(1.5<2\), it is outside the lower-sided null; the constrained estimate is \(2\). The expected count at the boundary is \(8(2)=16\). Hence $$ \begin{aligned} G &=2\left[12\log\left(\frac{12}{16}\right)-(12-16)\right]\\ &=2\left[12\log(0.75)+4\right]\\ &=2[-3.4521848694+4]\\ &=1.0956302612\ldots \end{aligned} $$ Therefore, to four decimal places, \(G=1.0956\).
Check. Because \(12/8<2\), the estimate is outside the null in the lower direction, so the statistic is positive. A common mistake is to reverse the inequality used to decide whether the estimate is on the null side; the lower-sided test has a positive statistic for \(\hat{\lambda}<\lambda_0\), not for \(\hat{\lambda}>\lambda_0\).
Using Exposure Correctly and Avoiding Common Traps
For \(n\) independent observations each with unit exposure, \(T=n\) and \(K\) is the sum of the observed counts. More generally, differing known exposures can be combined by summing both the counts and exposures. The rate estimate is total count divided by total exposure, not the unweighted average of separate rates when exposures differ.
- Choose the direction before calculating. For \(H_0:\lambda\leq\lambda_0\), only an estimate above \(\lambda_0\) gives a positive statistic. For \(H_0:\lambda\geq\lambda_0\), only an estimate below \(\lambda_0\) does.
- Keep rate and count units distinct. The comparison is \(K/T\) versus \(\lambda_0\), or equivalently \(K\) versus \(T\lambda_0\). Comparing \(K\) directly with \(\lambda_0\) is generally dimensionally incorrect.
- Do not force a positive value. A one-sided likelihood-ratio statistic can be exactly zero because the unrestricted estimate satisfies the composite null.
- Handle zero counts by a limit. When \(K=0\), the term \(K\log(K/(T\lambda_0))\) is defined as zero. This avoids an undefined-looking \(0\log 0\) calculation.
- Keep the likelihood ratio convention consistent. Here \(\Lambda\) is null maximum likelihood divided by unrestricted maximum likelihood, so \(G=-2\log\Lambda\) is nonnegative.
Check Your Understanding
For each problem, identify whether the unrestricted estimate is inside the null before calculating the statistic. Give numerical statistics to four decimal places where needed.
- A hypothetical sensor records \(K=7\) events over \(T=10\) days. Assume \(K\sim\operatorname{Pois}(10\lambda)\), where \(\lambda\) is events per day. For \(H_0:\lambda\leq0.8\) against \(H_1:\lambda>0.8\), find \(G\). (Answer: \(\hat{\lambda}=0.7\); \(G=0\).)
- A hypothetical dispatch unit records \(K=18\) calls over \(T=12\) hours. Assume a Poisson rate \(\lambda\) calls per hour. For \(H_0:\lambda\leq1.25\) against \(H_1:\lambda>1.25\), calculate \(G\). (Answer: \(\hat{\lambda}=1.5\); \(G=0.5636\).)
- A hypothetical process records \(K=6\) flaws over \(T=5\) batches. Assume \(K\sim\operatorname{Pois}(5\lambda)\), with \(\lambda\) flaws per batch. For \(H_0:\lambda\geq1.5\) against \(H_1:\lambda<1.5\), calculate \(G\). (Answer: \(\hat{\lambda}=1.2\); \(G=0.3223\).)
- Two independent hypothetical sites share a Poisson event rate \(\lambda\) per operating hour. Site A observes \(5\) events over \(4\) hours, and site B observes \(11\) over \(6\) hours. For \(H_0:\lambda\leq1.2\) against \(H_1:\lambda>1.2\), aggregate the data and calculate \(G\). (Answer: \(K=16\), \(T=10\), \(\hat{\lambda}=1.6\); \(G=1.2058\).)
- A hypothetical monitoring network records zero events over \(T=4\) years. Assume \(K\sim\operatorname{Pois}(4\lambda)\), with rate \(\lambda\) per year. For \(H_0:\lambda\geq0.5\) against \(H_1:\lambda<0.5\), calculate the statistic, using the limiting convention for the zero-count term. (Answer: \(\hat{\lambda}=0\); \(G=2[0-(0-2)]=4\).)