Likelihood Ratio Tests For Equality Of Two Poisson Rates

Why Compare Poisson Rates?

When two groups are observed for different amounts of time or over different numbers of opportunities, comparing their raw counts can be misleading. A group with more events may simply have had greater exposure. A likelihood ratio test compares the underlying event rates after accounting for exposure, and it measures how much fit is lost when the two rates are required to be equal.

For example, an analyst might compare hypothetical claim counts from two portfolios observed over different policy-years. The question is not whether the counts match; it is whether one common rate can plausibly describe both portfolios. This tutorial develops that test and explains its pooled estimate, statistic, and large-sample reference distribution. It is aligned with the SOA VEE Mathematical Statistics syllabus.

Model: Let $X_1$ and $X_2$ be independent counts, with $X_i\sim\operatorname{Poisson}(t_i\lambda_i)$, where $t_i>0$ is the known exposure for group $i$ and $\lambda_i\geq 0$ is its event rate per unit of exposure. The equality test is $H_0:\lambda_1=\lambda_2$ against $H_1:\lambda_1\ne\lambda_2$. The exposures must be measured in compatible units, and the Poisson counts must be independent under the stated model.

Separate Estimates and the Pooled Null Estimate

Under the alternative, each group has its own rate. The likelihood for the observations $x_1,x_2$ is

L(\lambda_1,\lambda_2) =\prod_{i=1}^2\frac{(t_i\lambda_i)^{x_i}e^{-t_i\lambda_i}}{x_i!}.

Maximizing the two factors separately gives $\widehat{\lambda}_i=x_i/t_i$. Under the null, both rates equal a single rate $\lambda$, so the likelihood depends on the total count and total exposure. Maximizing that constrained likelihood gives the pooled estimate.

Result: Under the alternative, $\widehat{\lambda}_1=x_1/t_1$ and $\widehat{\lambda}_2=x_2/t_2$. Under $H_0$, the maximum likelihood estimate is $$ \widehat{\lambda}_0=\frac{x_1+x_2}{t_1+t_2}. $$ The null-fitted expected counts are $\widehat{\mu}_{i,0}=t_i\widehat{\lambda}_0$. These are exposure-weighted fitted counts, not generally the arithmetic average of the observed counts.

The pooled rate is a weighted average of the two separate rate estimates:

\widehat{\lambda}_0 =\frac{t_1}{t_1+t_2}\widehat{\lambda}_1 +\frac{t_2}{t_1+t_2}\widehat{\lambda}_2.

Thus a group with greater exposure receives greater weight. The two null-fitted counts are equal when the exposures are equal, or when the pooled estimated rate is zero. If the pooled estimated rate is positive, the fitted counts are equal only when the exposures are equal.

Likelihood Ratio Statistic and Its Derivation

The likelihood ratio is the maximized likelihood under the null divided by the maximized likelihood under the alternative. Its deviance form is $G^2=-2\log\Lambda$, where $\Lambda$ denotes that ratio. A large value means the common-rate restriction reduces the maximized likelihood substantially.

Test statistic: If $x_1+x_2>0$, the likelihood ratio deviance is $$ G^2 =2\sum_{i=1}^2 x_i\log\left(\frac{x_i}{\widehat{\mu}_{i,0}}\right) =2\left[ x_1\log\left(\frac{x_1}{t_1\widehat{\lambda}_0}\right) +x_2\log\left(\frac{x_2}{t_2\widehat{\lambda}_0}\right) \right]. $$ For a zero observed count, its term is defined as zero when the corresponding fitted count is positive. If $x_1=x_2=0$, both maximized likelihoods are attained at zero rates and $G^2=0$; do not evaluate a $0/0$ ratio.

To see why the deviance has this form, write the log likelihood for a Poisson count with mean $\mu_i=t_i\lambda_i$ as $x_i\log\mu_i-\mu_i-\log(x_i!)$. Under the alternative, the fitted mean is $x_i$, and under the null it is $\widehat{\mu}_{i,0}$. Subtract the null maximized log likelihood from the alternative maximized log likelihood. The factorial terms cancel, and the linear mean terms cancel because $\widehat{\mu}_{1,0}+\widehat{\mu}_{2,0}=x_1+x_2$. The difference is therefore $\sum_i x_i\log(x_i/\widehat{\mu}_{i,0})$, giving $G^2$ after multiplication by 2. When $x_i=0$ and the fitted mean is positive, the contribution is the continuous limiting value $\lim_{x\to0^+}x\log(x/\widehat{\mu}_{i,0})=0$. If both counts are zero, the fitted means are also zero, so this limiting convention is not a definition of a $0/0$ term; the likelihoods must instead be evaluated directly.

Worked Example: Comparing Rates Across Unequal Exposures

Two hypothetical inspection programs record 13 events over 32 thousand inspections and 19 events over 64 thousand inspections. Assume independent Poisson counts with exposure measured in thousands of inspections. Test equality of the event rates using the likelihood ratio deviance, and report the statistic to four decimal places.

Method and estimates. Under the alternative, the rate estimates are $13/32=0.40625$ and $19/64=0.296875$ events per thousand inspections. Under the null, the pooled rate is $\widehat{\lambda}_0=(13+19)/(32+64)=32/96=1/3$ event per thousand inspections.

The null-fitted counts are $32(1/3)=32/3$ and $64(1/3)=64/3$. Their sum is $32$, which checks against the total observed count $13+19=32$.

Deviance calculation. The observed-to-fitted ratios are $13/(32/3)=39/32$ and $19/(64/3)=57/64$. Hence

\begin{aligned} G^2 &=2\left[13\log(39/32)+19\log(57/64)\right]\\ &\approx 2\left[13(0.197826)+19(-0.115832)\right]\\ &=2(2.571738-2.200808)\\ &\approx 0.7419. \end{aligned}

The computation uses the observed counts in the logarithm weights and the pooled fitted counts in the denominators. A common mistake is to compare 13 and 19 directly, ignoring that the second program had twice the exposure. The separate estimated rates, rather than the raw counts, are the relevant comparison. The fitted counts sum to the observed total, and the deviance is nonnegative, as required.

Worked Example: One Observed Count Is Zero

Two hypothetical service centers have independent Poisson event counts. Center 1 records 6 events over 3 thousand service-hours, and Center 2 records 12 events over 9 thousand service-hours. Assume the rates are constant within each center. Calculate the equality-test deviance to four decimal places.

Method and pooled fit. The total count is $6+12=18$, and the total exposure is $3+9=12$ thousand service-hours. Thus $\widehat{\lambda}_0=18/12=1.5$ events per thousand service-hours. The fitted counts are $3(1.5)=4.5$ and $9(1.5)=13.5$, which sum to 18.

The observed-to-fitted ratios are $6/4.5=4/3$ and $12/13.5=8/9$. Therefore,

\begin{aligned} G^2 &=2\left[6\log(4/3)+12\log(8/9)\right]\\ &\approx 2\left[6(0.2876821)+12(-0.1177830)\right]\\ &=2(1.7260926-1.4133964)\\ &\approx 0.6254. \end{aligned}

This example also checks why the pooled fit is exposure-weighted: the second center has three times the exposure, so its null-fitted count is three times the first center's fitted count. A common mistake is to use the average observed count, 9, as the fitted count in each group. That would ignore the unequal exposures and would not be the constrained maximum likelihood fit.

Worked Example: A Zero Count and the Zero-Count Convention

In a hypothetical trial, Group 1 has zero events in 2 thousand observation-days and Group 2 has 10 events in 3 thousand observation-days. Under independent Poisson sampling, calculate the equality-test deviance to four decimal places.

Method and pooled fit. The pooled estimate is $\widehat{\lambda}_0=(0+10)/(2+3)=2$ events per thousand observation-days. Thus the null-fitted counts are $2(2)=4$ and $3(2)=6$. Their sum is 10, matching the observed total.

The first count contributes zero because its observed count is zero and its null-fitted count is positive. The second contribution gives

\begin{aligned} G^2 &=2\left[0+10\log(10/6)\right]\\ &=20\log(5/3)\\ &\approx 20(0.5108256)\\ &\approx 10.2165. \end{aligned}

The statistic is large compared with the previous examples, reflecting the difference between the observed rates $0$ and $10/3$ events per thousand observation-days. The zero contribution is valid here because its fitted count is $4>0$. If both observed counts had been zero, both null-fitted counts would also be zero and the correct statistic would instead be $G^2=0$.

Asymptotic Reference Distribution

Under $H_0$, if the common true rate is positive and the model is regular, Wilks' theorem gives an asymptotic chi-square reference distribution. The alternative has two free rates, while the null imposes one equality constraint and has one free rate. The difference in parameter dimensions is one.

Asymptotic result: Under a common positive rate, with independent Poisson observations and known positive exposures, $G^2$ is approximately distributed as $\chi^2_1$ for sufficiently large expected counts. The degrees of freedom are $2-1=1$, the difference between the alternative and null parameter dimensions. This approximation is not exact in small or sparse samples.

At a chosen significance level $\alpha$, the approximate upper-tail test rejects equality when $G^2$ exceeds the $1-\alpha$ quantile of $\chi^2_1$. For example, the 5% upper-tail cutoff is approximately 3.841. The asymptotic result concerns the distribution under a positive true common rate; if the true rate is zero, the parameter is on the boundary and the usual regularity argument does not apply. An observed all-zero sample gives $G^2=0$, but that data outcome does not establish that the regular chi-square approximation is suitable for every sparse-data problem. A poor approximation should not be repaired by treating the asymptotic reference distribution as exact.

Exam Traps and Interpretation

Comparing counts instead of rates. Counts must be interpreted relative to their exposures. Equality of rates does not mean equality of expected counts unless the exposures are equal.
Using an unweighted average. The pooled rate is total count divided by total exposure, not the average of the two separate rate estimates unless the exposures happen to be equal.
Reversing the likelihood ratio. With $\Lambda=L_0/L_1$, the deviance is $-2\log\Lambda$, so evidence against the null produces a larger, not a negative, statistic.
mishandling zero counts. A zero observed count contributes zero when its null-fitted mean is positive. The all-zero case must be treated separately because it would otherwise produce a ratio of zero to zero.
Overusing the chi-square approximation. The one-degree-of-freedom chi-square calibration is asymptotic and relies on regularity, including a positive common rate. Sparse expected counts can make its finite-sample accuracy poor.

Key takeaway: Fit one rate per group under the alternative and one exposure-weighted pooled rate under equality. Compare the observed counts with the pooled fitted counts using the Poisson deviance. Under a positive common rate and adequate sample size, its asymptotic reference distribution has one degree of freedom.

Check Your Understanding

For each problem, use independent Poisson counts and known exposures. Unless stated otherwise, calculate the likelihood ratio deviance under the equality-of-rates null. Give statistics to four decimal places.

A hypothetical monitoring network records 8 events over 4 thousand sensor-days at Site A and 12 events over 8 thousand sensor-days at Site B. Find the pooled rate, the two null-fitted counts, and $G^2$. (Answer: pooled rate $=20/12=1.6667$ events per thousand sensor-days; fitted counts $=6.6667,13.3333$; $G^2=0.3885$.)
Two hypothetical laboratories record 0 and 9 instrument faults over exposures of 2 and 4 thousand operating-hours. Find the pooled rate and the deviance, applying the zero-count convention correctly. (Answer: pooled rate $=1.5$ faults per thousand operating-hours; fitted counts $=3,6$; $G^2=18\log(1.5)=7.2984$.)
Two independent hypothetical districts record 15 and 5 incidents, each over 5 thousand resident-months. Calculate the pooled fit and deviance. (Answer: pooled rate $=2$ incidents per thousand resident-months; fitted counts $=10,10$; $G^2=2[15\log(1.5)+5\log(0.5)]=5.2325$.)
A hypothetical pair of systems has zero recorded events in each system, with exposures 7 and 11 thousand operating-days. What are the maximum likelihood rates under the alternative and null, and what is $G^2$? Is the usual positive-rate Wilks argument applicable at a true common rate of zero? (Answer: all fitted rates are $0$, $G^2=0$, and the usual regular chi-square argument is not applicable when the true common rate is zero.)
A hypothetical insurer observes 4 claims over 2 thousand policy-years in one block and 16 claims over 6 thousand policy-years in another. Find the pooled rate, null-fitted counts, and deviance. (Answer: pooled rate $=20/8=2.5$ claims per thousand policy-years; fitted counts $=5,15$; $G^2=2[4\log(4/5)+16\log(16/15)]=0.2801$.)

What You'll Learn