Checking The Proportional Exposure Assumption

Why Check Proportional Exposure?

A Poisson rate model with exposure assumes that expected counts increase in direct proportion to the amount of exposure, such as person-time or population at risk. If exposure doubles while other predictors stay fixed, the expected count doubles. This is the assumption encoded by using log exposure as an offset. It is useful to check whether the data are compatible with that assumption: a systematically changing rate as exposure grows can make an offset model's predictions misleading.

The check compares the offset model with a model that estimates the exposure coefficient. It asks whether that coefficient is compatible with one, conditional on the other predictors in the model. It does not, by itself, establish that the model is otherwise well specified or that exposure has a causal effect.

Models and the Proportionality Hypothesis

Suppose \(Y_i\) is a count for observation \(i\), \(t_i>0\) is its known exposure, and \(z_i\) is a vector of predictors other than exposure. Assume the counts are conditionally independent with \(Y_i\sim\operatorname{Poisson}(\mu_i)\). The estimated-exposure-coefficient model is

\log \mu_i=\beta_0+z_i^\mathsf{T}\beta+\gamma\log t_i.

Here \(\gamma\) is estimated from the data. The offset model fixes its value at one:

\log \mu_i=\beta_0+z_i^\mathsf{T}\beta+\log t_i, \qquad \mu_i=t_i\exp(\beta_0+z_i^\mathsf{T}\beta).

The exposure must be positive to use \(\log t_i\). The same observations and the same other predictors must be used in both models. The model matrix, including the column \(\log t_i\), must have full column rank in the estimated-coefficient model; otherwise \(\gamma\) is not separately identifiable.

Definition: The proportional exposure assumption is the hypothesis that, conditional on the other predictors, the expected count is proportional to exposure. In the model above it is \(H_0:\gamma=1\). The alternative is \(H_1:\gamma\ne 1\), unless a one-sided alternative is specified in advance.

The rate is the expected count per exposure unit. Dividing the mean by exposure gives

\frac{\mu_i}{t_i} =\exp(\beta_0+z_i^\mathsf{T}\beta)t_i^{\gamma-1}.

Thus, if \(\gamma=1\), the rate does not vary with exposure when the other predictors are held fixed. If \(\gamma\ne1\), the model implies a rate that varies with exposure. More precisely, multiplying exposure by a factor \(c>0\) multiplies the expected count by \(c^\gamma\), and the rate by \(c^{\gamma-1}\).

How to Test the Assumption

Fit the estimated-coefficient model and test \(H_0:\gamma=1\). A Wald test uses the estimated coefficient and its standard error:

z=\frac{\hat{\gamma}-1}{\operatorname{SE}(\hat{\gamma})}.

Under the null and standard large-sample regularity conditions, this statistic is approximately standard normal. The corresponding two-sided test rejects at level \(\alpha\) when \(|z|>z_{1-\alpha/2}\). The estimated-coefficient model must have a finite, interior maximum-likelihood estimate, and the standard error must be based on a valid covariance estimate. The usual Poisson model also presumes conditionally independent counts with the stated Poisson distribution; material departures from that distribution can invalidate the conventional standard error and test calibration.

A likelihood-ratio test compares the maximized log-likelihoods of the same estimated-coefficient and offset models:

G^2=2\{\ell(\hat{\beta}_0,\hat{\beta},\hat{\gamma})- \ell(\tilde{\beta}_0,\tilde{\beta},1)\}.

The tildes denote estimates under the restriction \(\gamma=1\). Under the null, independent Poisson sampling, identifiability, an interior true parameter, and the usual large-sample conditions for likelihood inference, \(G^2\) is approximately distributed as \(\chi^2_1\). The models differ by one freely estimated parameter, which accounts for the one degree of freedom. This is the likelihood-ratio test for a nested restriction.

Result: With identical observations and other predictors, the offset model is exactly the estimated-exposure-coefficient model restricted to \(\gamma=1\). To see this, substituting \(\gamma=1\) into \(\log\mu_i=\beta_0+z_i^\mathsf{T}\beta+\gamma\log t_i\) gives \(\log\mu_i=\beta_0+z_i^\mathsf{T}\beta+\log t_i\), the offset specification. Therefore a test of \(\gamma=1\) directly compares the proportional-exposure model with this one-parameter extension.

For Poisson observations, the individual log-likelihood contribution is \(y_i\log\mu_i-\mu_i-\log(y_i!)\). The two models use the same observations, so the factorial terms cancel in their likelihood difference. Consequently,

G^2 =2\sum_i\left[ y_i\log\left(\frac{\hat{\mu}_{i,\mathrm{full}}} {\hat{\mu}_{i,\mathrm{offset}}}\right) -\hat{\mu}_{i,\mathrm{full}}+\hat{\mu}_{i,\mathrm{offset}} \right].

This follows by subtracting the restricted log-likelihood from the full log-likelihood and multiplying by two. The formula also makes clear that the test measures the improvement in fit from allowing the exposure coefficient to differ from one; it is not a general test of every feature of the model.

Worked Interpretations

Worked Example: A Rate That Rises With Exposure

Suppose two independent Poisson counts have exposures \(t_1=10\) and \(t_2=20\), with observed counts \(y_1=20\) and \(y_2=80\). Use the model \(\log\mu_i=\beta_0+\gamma\log t_i\), with no other predictors. The estimated-coefficient model has two parameters and fits the two observed means exactly: \(\hat{\mu}_1=20\) and \(\hat{\mu}_2=80\).

The estimated exposure coefficient is the change in log mean divided by the change in log exposure:

\hat{\gamma} =\frac{\log(80)-\log(20)}{\log(20)-\log(10)} =\frac{\log 4}{\log 2}=2.

The fitted rate is \(\hat{\mu}_i/t_i\). It is \(20/10=2\) at the first exposure and \(80/20=4\) at the second. Doubling exposure is associated with a fourfold expected count, or a twofold rate, in this fitted model.

Under the offset null, the common rate estimate is total count divided by total exposure: \((20+80)/(10+20)=10/3\). The restricted fitted means are \(100/3\) and \(200/3\). Hence

G^2=2\left[ 20\log\left(\frac{20}{100/3}\right) +80\log\left(\frac{80}{200/3}\right) \right] =2\{20\log(0.6)+80\log(1.2)\} \approx 8.74.

Against a \(\chi^2_1\) reference distribution, this gives a two-sided likelihood-ratio \(p\)-value of about \(0.003\). It is evidence against proportional exposure under the model assumptions. The fitted coefficient is \(\hat{\gamma}=2\), not one. Because this example has only two exposure groups and the full model fits both means exactly, the result is best read as an illustration of the comparison, not as a broad diagnostic of model adequacy.

Worked Example: A Test That Does Not Reject

Now suppose the counts are \(20\) and \(40\) at exposures \(10\) and \(20\), respectively, in the same two-parameter model. The fitted rates are both \(2\), and

\hat{\gamma} =\frac{\log(40/20)}{\log(20/10)} =\frac{\log 2}{\log 2}=1.

The restricted and unrestricted fits coincide, so \(G^2=0\). For these two fitted means, the approximate standard error of the slope is

\operatorname{SE}(\hat{\gamma}) =\frac{\sqrt{1/20+1/40}}{\log 2} =\frac{\sqrt{0.075}}{0.6931} \approx 0.395.

The Wald statistic is \((1-1)/0.395=0\), so the test does not reject. A 95% Wald interval is \(1\pm1.96(0.395)\), or approximately \((0.226,1.774)\). This interval includes one, but it also includes substantial departures from one. Not rejecting proportionality is not proof that the rate is constant; the data may simply be too imprecise to distinguish plausible alternatives.

Worked Example: Translating an Adjusted Coefficient

Suppose an estimated-coefficient Poisson model, adjusted for the same set of predictors in both fits, gives \(\hat{\gamma}=0.80\) with standard error \(0.10\). The Wald statistic for proportional exposure is

z=\frac{0.80-1}{0.10}=-2.00.

The two-sided normal-reference \(p\)-value is approximately \(0.0455\), just below \(0.05\). The fitted model says that doubling exposure multiplies the expected count by \(2^{0.80}\), but multiplies the rate by

2^{0.80-1}=2^{-0.20} =\exp(-0.20\log 2) \approx \exp(-0.1386) \approx 0.871.

Thus, conditional on the other predictors, the fitted rate at twice the exposure is about 87% of the rate at the original exposure. A 95% Wald interval for \(\gamma\) is \(0.80\pm1.96(0.10)=(0.604,0.996)\). Transforming its endpoints through \(2^{\gamma-1}\) gives an approximate interval of \(2^{-0.396}\) to \(2^{-0.004}\), or \(0.760\) to \(0.997\), for the rate ratio on doubling exposure. The interpretation is conditional and model-based, not a causal conclusion.

What a Departure Can—and Cannot—Tell You

A coefficient below one means the fitted rate decreases as exposure grows; a coefficient above one means it increases. These patterns can signal that a constant-rate description is inadequate. They can also reflect omitted predictors, an incorrect functional form for another covariate, dependence between observations, or a problem with the exposure measurement. Inspect exposure ranges and fitted-versus-observed patterns alongside the coefficient test. A single coefficient represents one power-law relationship across the modeled range; it may conceal curvature or different patterns in subgroups.

The test is conditional on the included predictors. If exposure is associated with an omitted risk factor, the estimated \(\gamma\) can differ from one even when a simpler scientific description would posit a constant rate within correctly defined groups. Conversely, a non-significant test does not show that every exposure-specific rate is identical. The result should be considered together with uncertainty, the scientific design, and the plausibility of the Poisson assumptions.

Common Mistakes

Testing whether the coefficient is zero. Zero would mean the expected count does not change with exposure. Proportional exposure requires \(\gamma=1\), so the null value is one.
Interpreting \(\gamma\) as the rate multiplier. It is the exposure power for the expected count. The rate multiplier when exposure is multiplied by \(c\) is \(c^{\gamma-1}\).
Comparing models with different predictors or observations. Then the likelihood difference does not isolate the exposure restriction. Keep the other terms and the analysis sample the same.
Treating a large \(p\)-value as confirmation. Failure to reject can reflect limited information. Report the estimate and its interval as well as the test result.
Calling every rejection evidence of a changing biological rate. The coefficient test can respond to misspecification, omitted variables, dependence, or measurement problems, not only to a true exposure-dependent rate.

Key takeaway: Compare an offset Poisson model with the otherwise identical model that estimates the coefficient of log exposure. Test \(H_0:\gamma=1\), then interpret the estimate and uncertainty through the implied rate change \(c^{\gamma-1}\). The test is a focused check of proportionality, not a substitute for broader model assessment.

Check Your Understanding

Answer these questions by relating the exposure coefficient to the count and rate scales.

In a Poisson model with \(\log\mu_i=\beta_0+z_i^\mathsf{T}\beta+\gamma\log t_i\), what value of \(\gamma\) represents proportional exposure?
If \(\gamma=1.3\), by what factor does the fitted rate change when exposure is doubled?
Why must the offset and estimated-coefficient models use the same predictors and observations for the likelihood-ratio comparison?
A test of \(\gamma=1\) has \(p=0.42\). What can you conclude, and what can you not conclude?
Name two reasons, other than a genuinely exposure-dependent rate, that an estimated coefficient might depart from one.