Why Test a Specified Poisson Rate?
A Poisson model is often used for counts of events in a fixed unit of exposure, such as incidents per policy-year or equipment failures per month. A rate test asks whether the data are compatible with a particular proposed rate, while allowing the likelihood under the alternative to use whichever rate best fits the observations. The likelihood ratio measures how much fit is lost by imposing the specified rate.
For this tutorial, assume \(X_1,\ldots,X_n\) are independent and identically distributed with \(X_i\sim\operatorname{Poisson}(\lambda)\), where \(\lambda\) is the mean count per observation and \(\lambda>0\). The hypotheses are \(H_0:\lambda=\lambda_0\), for a specified \(\lambda_0>0\), against the unrestricted alternative \(H_1:\lambda\ne\lambda_0\), with \(\lambda\ge 0\). The observations must use the same exposure unit; for example, counts observed over months should not be mixed with counts observed over years without adjusting the model.
Deriving the Statistic From the Sample Total
Let \(T=\sum_{i=1}^n X_i\). Under the stated model, the likelihood for the observed counts is
When \(T>0\), maximizing the log-likelihood \(\ell(\lambda)=-n\lambda+T\log\lambda-\sum_i\log(x_i!)\) gives \(\widehat{\lambda}=T/n\). If \(T=0\), the likelihood is maximized at the boundary \(\widehat{\lambda}=0\). Thus the unrestricted fit is determined by the sample total, not by the arrangement of the individual counts.
Substitute \(\lambda_0\) into the likelihood. Its likelihood is \(e^{-n\lambda_0}\lambda_0^T/\prod_i x_i!\).
For \(T>0\), substitute \(\widehat{\lambda}=T/n\). For \(T=0\), use the maximizing boundary value \(\widehat{\lambda}=0\).
Take \(-2\) times the logarithm of the null likelihood divided by the unrestricted maximum likelihood.
For \(T>0\), the data-only factorial terms cancel in the ratio. Substitution and simplification give
and therefore
This derivation shows why the statistic depends only on \(T\): after the likelihoods are divided, every factor involving the individual factorials disappears. It also shows why the statistic compares the observed total \(T\) with its null expectation \(n\lambda_0\). The logarithmic term measures the relative discrepancy, while the linear term completes the likelihood comparison.
When \(T=0\), the unrestricted maximum likelihood is at \(\lambda=0\), and the null-to-maximum likelihood ratio is \(e^{-n\lambda_0}\). The corresponding statistic is \(D=2n\lambda_0\). This agrees with the formula above if the term \(T\log(T/(n\lambda_0))\) is defined to be zero at \(T=0\), using the limit \(t\log t\to0\) as \(t\) decreases to zero.
Using the Chi-Square Approximation
Under standard regularity conditions, Wilks' theorem says that under the null hypothesis, \(D\) is approximately chi-square with degrees of freedom equal to the difference in the number of free parameters in the unrestricted and null models. Here the unrestricted model estimates one rate, while the null model fixes that rate, so the difference is one. For a sufficiently informative sample with \(\lambda_0>0\), the reference distribution is approximately \(\chi^2_1\).
At significance level \(\alpha\), the approximate test rejects \(H_0\) when \(D\) exceeds the upper-tail critical value \(\chi^2_{1,1-\alpha}\). At the 5% level this value is approximately 3.841. An approximate p-value is \(P(\chi^2_1\ge D_{\mathrm{obs}})\). The test is two-sided in its alternative: a total much smaller than the null expectation or much larger than it can produce a large deviance.
The approximation is not an exact finite-sample result. Under the null, \(T\) takes integer values, so the exact distribution of \(D\) is discrete. If \(n\lambda_0\) is small, attainable values of the statistic may be far apart and the chi-square tail can give a misleading calibration. In that situation, an exact likelihood ratio p-value can be calculated by summing null probabilities for all totals whose deviance is at least the observed deviance:
This exact calculation uses the null distribution \(T\sim\operatorname{Poisson}(n\lambda_0)\) and the same likelihood ratio ordering as the test statistic. It can differ from the chi-square approximation, especially when the expected total is low.
Worked Applications
Worked Example: A Higher-Than-Expected Event Total
A hypothetical insurer models the number of specified equipment incidents per policy-year as independent Poisson counts with common rate \(\lambda\). For \(n=20\) policy-years, the observed total is \(T=80\). Test \(H_0:\lambda=3\) against the unrestricted alternative at the 5% level using the chi-square approximation.
Method. Use the Poisson likelihood ratio statistic based on the total. The null expected total is \(n\lambda_0=20(3)=60\), and the unrestricted estimate is \(\widehat{\lambda}=T/n=80/20=4\) incidents per policy-year.
The statistic is \(D=2[80\log(80/60)-(80-60)]\). Since \(80/60=4/3\) and \(\log(4/3)\approx0.287682\), \(80\log(4/3)\approx80(0.287682)=23.0146\). Thus \(D=2(23.0146-20)=6.0292\).
The 5% critical value is 3.841, and \(6.0292>3.841\). The approximate chi-square p-value is 0.0141, so the test rejects the specified rate at the 5% level. The data favor a rate above 3 in this sample, although the test itself compares the specified rate with an unrestricted alternative.
Check. The statistic is positive, as required because the unrestricted fit cannot have lower likelihood than the null fit. The calculation also uses 60, not 3, as the null expected total. This avoids the common mistake of comparing the observed count with the per-policy rate without accounting for all 20 policy-years.
Worked Example: A Lower-Than-Expected Total
In a hypothetical monitoring study, \(n=15\) independent devices each have a Poisson number of alarms with common rate \(\lambda\) per observation period. The specified rate is \(\lambda_0=2\), and the observed total is \(T=18\). Test the specified rate at the 5% level using the chi-square approximation.
Method. First calculate the null expected total and the unrestricted estimate. They are \(n\lambda_0=15(2)=30\) and \(\widehat{\lambda}=18/15=1.2\) alarms per period, respectively.
Then \(D=2[18\log(18/30)-(18-30)]\). The ratio is \(18/30=0.6\), and \(\log(0.6)\approx-0.510826\). Therefore \(18\log(0.6)\approx18(-0.510826)=-9.1949\), while \(18-30=-12\). It follows that \(D=2[-9.1949-(-12)]=2(2.8051)=5.6102\).
Because \(5.6102>3.841\), reject \(H_0\) at the 5% level. The approximate p-value is 0.0179. Here the observed total is below its null expectation, but the two-sided likelihood ratio test still rejects because the null rate fits substantially worse than the estimated rate.
Check. The contribution from the linear term is \(-(T-n\lambda_0)=-(-12)=12\); it must be added, not subtracted as a positive 12 from the negative logarithmic contribution. The resulting deviance is positive.
Worked Example: Small Expected Total and Exact Calibration
A hypothetical laboratory observes counts from \(n=3\) independent test periods. Under the specified model, each period has a Poisson count with \(\lambda_0=0.5\), and the total observed count is \(T=0\). Calculate the statistic, then compare the chi-square approximation with the exact likelihood ratio p-value.
Method. The null expected total is \(n\lambda_0=3(0.5)=1.5\). Since \(T=0\), use the boundary case \(D=2n\lambda_0=2(1.5)=3\). The unrestricted estimate is \(\widehat{\lambda}=0/3=0\).
The chi-square approximation gives \(P(\chi^2_1\ge3)\approx0.0833\). It would not reject at the 5% level. To find the exact likelihood ratio p-value, include all totals \(k\) for which \(D(k)\ge3\), under \(T\sim\operatorname{Poisson}(1.5)\). At \(k=0\), \(D(0)=3\). At \(k=4\), \(D(4)=2[4\log(4/1.5)-(4-1.5)]\approx2.8466\), which is below 3. At \(k=5\), \(D(5)=2[5\log(5/1.5)-(5-1.5)]\approx5.0397\). For larger totals the deviance increases, so the included totals are \(k=0\) and \(k\ge5\).
Under the null, \(P(T=0)=e^{-1.5}\approx0.2231\). Also, \(P(T\ge5)=1-\sum_{k=0}^{4}e^{-1.5}1.5^k/k!\). The probabilities for \(k=0,1,2,3,4\) are approximately \(0.223130,\ 0.334695,\ 0.251021,\ 0.125511,\ 0.047067\), which sum to \(0.981424\). Thus \(P(T\ge5)\approx0.018576\), and the exact p-value is \(0.223130+0.018576=0.241706\), or 0.2417 to four decimals.
The exact p-value does not reject at 5%, unlike what a careless use of the approximate cutoff might suggest if the statistic were compared incorrectly. In this case both calibrations do not reject, but their p-values differ substantially: 0.0833 versus 0.2417. The small null expected total warns that the chi-square approximation is not dependable.
Check. The observed zero total is included in the exact tail because its deviance equals the observed value. Totals 1 through 4 are not included; in particular, \(D(4)\) is below 3. This avoids treating the exact two-sided likelihood ratio tail as merely the probability of a zero count.
Assumptions and Common Exam Traps
The model requires independent observations with a common Poisson rate in the same exposure unit. If observations are dependent, or if rates differ across observations, the likelihood and the distribution of the total need not be those used here. The formula also tests a positive specified rate against a model that allows the rate to vary over nonnegative values. If the null rate is zero, the null is on the boundary and the standard chi-square conclusion does not follow from the regular interior-parameter argument given above.
- Using the rate instead of the expected total. Compare \(T\) with \(n\lambda_0\), not with \(\lambda_0\) alone.
- Reversing the likelihood ratio. With \(\Lambda=L(\lambda_0)/L(\widehat{\lambda})\), the statistic is \(-2\log\Lambda\), so a worse null fit produces a larger nonnegative value.
- Forgetting the zero-total case. At \(T=0\), use \(D=2n\lambda_0\), or apply the stated zero-term convention. Do not substitute \(T=0\) into a logarithm without addressing the limit.
- Using the wrong degrees of freedom. There is one free rate parameter in the unrestricted model and none in the specified-rate null model, so the asymptotic reference is \(\chi^2_1\).
- Treating an approximation as exact. When the null expected total is small, discrete exact calibration can be materially different from a chi-square tail probability.
Check Your Understanding
For each problem, use the stated model and show the sample total, null expected total, and likelihood ratio statistic before interpreting the result.
- Eight independent periods have Poisson counts with common rate \(\lambda\). Under \(H_0\), \(\lambda_0=1.5\), and the observed total is 20. Find \(\widehat{\lambda}\) and \(D\), then decide whether to reject at 5% using the chi-square approximation. (Answer: \(\widehat{\lambda}=2.5\), \(D\approx4.4330\), reject at 5%.)
- Twenty-five independent observations have a common Poisson rate, with \(\lambda_0=0.8\) and observed total \(T=14\). Calculate \(D\) and give the approximate 5% decision. (Answer: \(n\lambda_0=20\), \(D\approx2.0131\), do not reject at 5%.)
- Five independent periods have Poisson counts, \(\lambda_0=0.2\), and \(T=0\). Find the statistic and explain whether the chi-square approximation should be trusted. (Answer: \(n\lambda_0=1\), \(D=2\), the approximate p-value is 0.1573, but the expected total is small so exact calibration is preferable.)
- Two independent periods have Poisson counts with \(\lambda_0=0.5\), and \(T=0\). Find the observed deviance and the exact likelihood ratio p-value by identifying the null totals with at least that deviance. (Answer: \(D=2\); the qualifying totals are 0 and totals at least 3, giving exact p-value \(P(T=0)+P(T\ge3)\approx0.4482\).)
- For the setting in question 4, compare the exact p-value with the \(\chi^2_1\) approximation and state the 5% conclusion from each. (Answer: the approximate p-value is \(P(\chi^2_1\ge2)\approx0.1573\), while the exact p-value is approximately 0.4482; neither rejects at 5%, and the substantial difference cautions against relying on the approximation for this small expected total.)