A Score Test for a Poisson Rate with an Offset

Why Exposure Changes the Test

Poisson counts are often collected over unequal amounts of time, person-time, area, or another known exposure. A count of 12 events over 3,000 person-years does not represent the same rate as 12 events over 300 person-years. The model must account for exposure before testing a rate. Here we derive the score test for a common Poisson rate when each observation has its own known exposure, then identify exactly which score and information belong in the statistic.

Model and hypotheses: Let $Y_1,\ldots,Y_n$ be independent counts with $Y_i\sim\operatorname{Poisson}(\lambda t_i)$, where each exposure $t_i\geq 0$ is known and the common rate $\lambda>0$ is unknown. Write $T=\sum_{i=1}^n t_i$ and $Y=\sum_{i=1}^n Y_i$. We test $H_0:\lambda=\lambda_0$, for a specified $\lambda_0>0$, against either a two-sided or a directional alternative.

The exposure $t_i$ is an offset: it is known and fixes the scale of the mean, rather than being a parameter to estimate. We assume $T>0$. If $t_i=0$, the model assigns $Y_i=0$ with probability one; such an observation contributes nothing to the likelihood. The results below also apply if those zero-exposure observations are omitted.

Likelihood, Score, and Null Information

For positive exposures, the joint log-likelihood, up to no omitted terms that depend on $\lambda$, is

\ell(\lambda) =\sum_{i=1}^n\left\{y_i\log(\lambda t_i)-\lambda t_i-\log(y_i!)\right\} =Y\log\lambda-\lambda T+C,

where $C=\sum_i\{y_i\log t_i-\log(y_i!)\}$ is constant in $\lambda$. Differentiating gives the score and the observed information:

U(\lambda)=\frac{\partial\ell(\lambda)}{\partial\lambda} =\frac{Y}{\lambda}-T, \qquad J(\lambda)=-\frac{\partial^2\ell(\lambda)}{\partial\lambda^2} =\frac{Y}{\lambda^2}.

The score test evaluates the score at the null value. Its denominator is the Fisher information under the null, not the information at the unrestricted estimate. Since $E_\lambda[Y]=\lambda T$, the Fisher information is

I(\lambda)=E_\lambda[J(\lambda)] =\frac{E_\lambda[Y]}{\lambda^2} =\frac{T}{\lambda}.

Score-test statistic: For $H_0:\lambda=\lambda_0$, the signed standardized score and its square are $$ Z_S=\frac{U(\lambda_0)}{\sqrt{I(\lambda_0)}} =\frac{Y-\lambda_0T}{\sqrt{\lambda_0T}}, \qquad Q_S=Z_S^2=\frac{(Y-\lambda_0T)^2}{\lambda_0T}. $$ The numerator and the information in the denominator are both evaluated at the null rate. Under regular large-sample conditions, $Q_S$ is compared with a chi-square distribution with one degree of freedom.

Under the null, $E_{\lambda_0}[Y]=\lambda_0T$ and $\operatorname{Var}_{\lambda_0}(Y)=\lambda_0T$. Thus $Z_S$ is the null-centered total count divided by its null standard deviation. For a two-sided test at level $\alpha$, reject when $Q_S>\chi^2_{1,1-\alpha}$, where $\chi^2_{1,1-\alpha}$ denotes the $1-\alpha$ quantile of the chi-square distribution with one degree of freedom. Equivalently, reject when $|Z_S|>z_{1-\alpha/2}$. For the upper-tailed alternative $\lambda>\lambda_0$, reject for a sufficiently large positive $Z_S$; for the lower-tailed alternative, reject for a sufficiently negative one.

Why the Null Evaluation Is Essential

The unrestricted maximum likelihood estimate, when $Y>0$, is $\hat{\lambda}=Y/T$, obtained by solving $U(\lambda)=0$. It follows immediately that $U(\hat{\lambda})=0$. A score test cannot use the unrestricted score: it would always produce a zero statistic. Instead, it asks whether the score points away from the null when evaluated at the null, and scales that score by the variability predicted under the null.

The distinction applies to information as well. The observed information at the null is $J(\lambda_0)=Y/\lambda_0^2$, while the null Fisher information is $I(\lambda_0)=T/\lambda_0$. The score statistic defined here uses the latter. Replacing it with observed information gives a different statistic; replacing it with information at $\hat{\lambda}$ while keeping the null score creates a hybrid that is not this score test.

Worked Example: Unequal Exposures and a Two-Sided Test

Suppose four independent Poisson counts are observed with exposures $1,2,3,4$, respectively, and the counts are $2,5,7,6$. Test $H_0:\lambda=1.5$ against a two-sided alternative.

The total exposure is $T=1+2+3+4=10$, and the total count is $Y=2+5+7+6=20$. Under the null, the expected total count is $\lambda_0T=1.5(10)=15$, and the null Fisher information is

I(\lambda_0)=\frac{10}{1.5}=\frac{20}{3}.

The null score is $U(\lambda_0)=20/1.5-10=10/3$. Therefore

Z_S=\frac{10/3}{\sqrt{20/3}}=\frac{5}{\sqrt{15}}\approx1.291, \qquad Q_S=\frac{25}{15}\approx1.667.

The two-sided chi-square reference gives a p-value of approximately $0.197$ (equivalently, the two-sided normal p-value from $Z_S$ is approximately $0.197$). At the 5% level, the test does not reject $H_0$. Unequal exposures are handled through their sum $T$; it would be incorrect to treat the four counts as if they had equal exposure.

Worked Example: An Upper-Tailed Test

Suppose counts $9,19,29,39$ are observed over exposures $10,20,30,40$. Test $H_0:\lambda=0.8$ against $H_1:\lambda>0.8$.

Here $Y=9+19+29+39=96$ and $T=10+20+30+40=100$. The null expected count is $0.8(100)=80$, and the null information is $I(\lambda_0)=100/0.8=125$. The score and standardized score are

U(\lambda_0)=\frac{96}{0.8}-100=20, \qquad Z_S=\frac{20}{\sqrt{125}}\approx1.789.

The upper-tail normal p-value is $P(Z\geq1.789)\approx0.037$, so this test rejects at the 5% level. The direction matters: for a lower-tailed alternative the observed positive score would not count as evidence in the rejection direction. For a two-sided test, the corresponding statistic is $Q_S=1.789^2=3.2$, with a p-value of approximately $0.074$.

Derivation and Large-Sample Calibration

The following derivation explains both the form of the statistic and its asymptotic reference distribution. It also shows why, in this common-rate model, the individual exposure-count pairs enter the score test only through their totals.

Result: Under $H_0:\lambda=\lambda_0$, the total $Y=\sum_iY_i$ has a Poisson distribution with mean $\lambda_0T$. If a sequence of such experiments has $\lambda_0T\to\infty$, then $$ Z_S=\frac{Y-\lambda_0T}{\sqrt{\lambda_0T}}\xrightarrow{d}N(0,1), \qquad Q_S\xrightarrow{d}\chi^2_1. $$

Proof. Independence gives the probability generating function of $Y$, for $0\leq s\leq1$:

E_{\lambda_0}[s^Y] =\prod_{i=1}^n E_{\lambda_0}[s^{Y_i}] =\prod_{i=1}^n \exp\{\lambda_0t_i(s-1)\} =\exp\{\lambda_0T(s-1)\}.

This is the probability generating function of a Poisson random variable with mean $\lambda_0T$, proving the stated distribution. Put $\mu=\lambda_0T$. The characteristic function of the standardized total is

E_{\lambda_0}\left[\exp\left\{iu\frac{Y-\mu}{\sqrt{\mu}}\right\}\right] =\exp\left\{\mu\left(e^{iu/\sqrt{\mu}}-1-\frac{iu}{\sqrt{\mu}}\right)\right\}.

For fixed real $u$, Taylor expansion gives $e^{iu/\sqrt{\mu}}=1+iu/\sqrt{\mu}-u^2/(2\mu)+O(\mu^{-3/2})$ as $\mu\to\infty$. Substitution makes the exponent $-u^2/2+O(\mu^{-1/2})$, so the characteristic function converges to $e^{-u^2/2}$, that of $N(0,1)$. The continuity theorem for characteristic functions yields $Z_S\xrightarrow{d}N(0,1)$. By the continuous mapping theorem, squaring gives $Q_S\xrightarrow{d}\chi^2_1$. $\square$

This argument requires the null expected total count $\lambda_0T$ to grow. It does not say that the chi-square approximation is exact for small expected counts. In fact, for finite $T$, the null distribution is determined exactly by $Y\sim\operatorname{Poisson}(\lambda_0T)$, and discreteness can make a nominal large-sample test’s actual size differ from its target.

Worked Example: Null and Unrestricted Information Are Different

Suppose the total observed count is $Y=30$, the total exposure is $T=20$, and the null rate is $\lambda_0=1$. Compute the score test and compare the information values.

At the null, $U(1)=30/1-20=10$, and the null Fisher information is $I(1)=20/1=20$. Thus the score statistic is $Q_S=10^2/20=5$, with $Z_S=10/\sqrt{20}\approx2.236$. Its two-sided large-sample p-value is approximately $0.025$.

The unrestricted estimate is $\hat{\lambda}=Y/T=30/20=1.5$. As required by the likelihood equation, its score is $U(1.5)=30/1.5-20=0$. The unrestricted Fisher information is $I(1.5)=20/1.5\approx13.333$. Meanwhile, the observed information at the null is $J(1)=30$, not the null Fisher information $I(1)=20$. Using the null score but the unrestricted information would give $10^2/13.333\approx7.5$, not the score statistic $5$. The test is defined by evaluating both the score and its Fisher-information scaling at the null.

Common Mistake and Practical Interpretation

Wrong reasoning: “The unrestricted estimate is $Y/T$, so I should evaluate the score and information there.” This makes the score zero because the unrestricted estimate maximizes the likelihood. It therefore cannot measure departure from the null. Correction: evaluate $U(\lambda_0)$ and $I(\lambda_0)$; the score test uses the slope and its null variability at the restricted value.

For this model, the likelihood depends on the data through $Y$ and $T$, so the score statistic also depends on them through those totals. This does not make the offset optional: $T$ must be the sum of the actual known exposures, not the number of observations or an unweighted average exposure. The construction also assumes one common rate multiplied by known exposures; if rates vary by group or covariates enter the model, the score and information must reflect that larger model.

Key takeaway: For a Poisson rate with known exposure offsets, the score test compares the observed total count with its null expected total, $\lambda_0T$, and scales by the null variance, also $\lambda_0T$. The score is evaluated at $\lambda_0$, not at the unrestricted estimate.

Check Your Understanding

Work through these questions to check the construction and interpretation of the test.

For independent $Y_i\sim\operatorname{Poisson}(\lambda t_i)$, derive the score $U(\lambda)$ and Fisher information $I(\lambda)$ in terms of $Y$ and $T$.
If $Y=42$, $T=50$, and $\lambda_0=0.75$, calculate $Z_S$ and $Q_S$. Which direction of evidence does the signed score indicate?
Why does evaluating the score at the unrestricted estimate fail to produce a score test?
Distinguish the null Fisher information $I(\lambda_0)$ from the observed information $J(\lambda_0)$.
What condition on the null expected total count supports the large-sample chi-square calibration, and why can that approximation be unreliable when the expected count is small?