Deriving The Exposure Offset

From Event Rate to Expected Count

A count is not interpretable without knowing how much opportunity there was for events to occur. Two patients may have the same underlying event rate but different expected counts if one is observed for longer. The exposure offset incorporates that difference while keeping the regression effects focused on the event rate.

The key modeling assumption is that, for a given covariate profile, the event rate is constant per unit of exposure. The expected count is then proportional to exposure. If counts are modeled as Poisson, taking the logarithm of that mean produces the offset form.

Definition: For observation $i$, let $Y_i$ be a count, $E_i$ a known, strictly positive exposure, and $x_i$ a covariate vector. A Poisson rate model specifies $$ Y_i\mid x_i,E_i \sim \operatorname{Poisson}(E_i\lambda_i), \qquad \log(\lambda_i)=x_i^T\beta. $$ Here $\lambda_i$ is the event rate per exposure unit and $\beta$ is the regression coefficient vector. The equivalent log-mean model is $$ \log E[Y_i\mid x_i,E_i]=\log(E_i)+x_i^T\beta. $$ The known quantity $\log(E_i)$, whose coefficient is fixed at one, is the exposure offset.

The exposure and rate must be expressed in compatible units: multiplying the exposure by the rate gives an expected count. The covariates describe how the rate varies across observations; exposure accounts for how much opportunity each observation has to accumulate events. The exposure is treated as known in this model, and $E_i>0$ is needed to take its logarithm.

Deriving the Offset

First consider the mean. If an observation is exposed for $E_i$ units and the event rate per unit is $\lambda_i$, the constant-rate assumption gives the expected count $\mu_i=E_i\lambda_i$. Model the rate with a log-linear predictor, $\lambda_i=\exp(x_i^T\beta)$. Substitution gives a Poisson mean of $E_i\exp(x_i^T\beta)$. The logarithm separates this product into a known exposure term and the modeled log rate.

Result: Suppose $E_i>0$ is known and the conditional count model is Poisson with mean $E_i\lambda_i$, where $\log(\lambda_i)=x_i^T\beta$. Then the conditional log mean has the offset form $$ \log E[Y_i\mid x_i,E_i]=\log(E_i)+x_i^T\beta. $$ The coefficient of $\log(E_i)$ is one because exposure multiplies the rate in the mean.

Derivation. The expectation of a Poisson random variable is its mean, so $E[Y_i\mid x_i,E_i]=E_i\lambda_i$. Since both factors are positive, apply the logarithm and use its product rule:

\begin{aligned} \log E[Y_i\mid x_i,E_i] &=\log(E_i\lambda_i)\\ &=\log(E_i)+\log(\lambda_i)\\ &=\log(E_i)+x_i^T\beta. \end{aligned}

This derivation explains the fixed coefficient. There is no unknown multiplier on exposure in the assumed mean: one exposure unit contributes one unit of opportunity at the modeled rate, two units contribute twice that expected count, and so on. The logarithm turns this multiplication into addition without changing the multiplier on $\log(E_i)$.

The constant-rate assumption about the mean does not, by itself, specify a Poisson distribution. It gives the proportional-mean relationship $E[Y_i\mid x_i,E_i]=E_i\lambda_i$. To specify the full Poisson count model as well, one may assume that events arise from a homogeneous Poisson process during the exposure, conditional on the covariates. In that case, the count over exposure $E_i$ has a Poisson distribution with mean $E_i\lambda_i$. The offset in the log-mean model follows from the mean relationship; the Poisson assumption supplies the count distribution.

Worked Rate-Model Calculations

Worked Example: One Constant Rate and One Exposure

Suppose events occur at a constant rate of $0.05$ per person-month, and a person is observed for $12$ person-months. Under a Poisson count model, find the expected count, write its log mean as an offset model, and calculate the probability of zero events.

The mean count is the rate multiplied by exposure: $12(0.05)=0.60$. Thus $Y\sim\operatorname{Poisson}(0.60)$. The rate has log value $\log(0.05)$, so the log mean can be written in either of the equivalent ways $\log(0.60)=\log(12)+\log(0.05)$.

For a Poisson random variable with mean $0.60$, the probability of zero events is $P(Y=0)=\exp(-0.60)0.60^0/0!=\exp(-0.60)\approx0.5488$. The exposure offset accounts for the $12$ person-months; the rate is still $0.05$ per person-month.

Worked Example: Covariates Model the Rate

Let $x_i$ be a binary covariate, with rate model $\log(\lambda_i)=\beta_0+\beta_1x_i$. Suppose $\beta_0=\log(0.03)$, $\beta_1=\log(2)$, and an observation has $x_i=1$ and exposure $E_i=8$. Find its rate and expected count, and verify the offset expression.

The rate for $x_i=1$ is $\lambda_i=\exp\{\log(0.03)+\log(2)\}=0.03(2)=0.06$ events per exposure unit. Multiplying by the exposure gives the expected count $\mu_i=8(0.06)=0.48$.

Using the log-mean form gives the same answer: $\log(\mu_i)=\log(8)+\log(0.03)+\log(2)=\log(0.48)$. The covariate changes the rate from $0.03$ to $0.06$; the exposure of $8$ then scales that rate to the expected count of $0.48$.

Worked Example: Exposure Changes the Count Mean

Suppose two observations share the same covariates and have rate $\lambda=0.25$ per exposure unit. Their exposures are $E_1=2$ and $E_2=6$. Under independent Poisson count models, calculate their means and the probabilities of zero events.

For the first observation, $\mu_1=2(0.25)=0.5$, so $Y_1\sim\operatorname{Poisson}(0.5)$. Therefore $P(Y_1=0)=\exp(-0.5)\approx0.6065$.

For the second observation, $\mu_2=6(0.25)=1.5$, so $Y_2\sim\operatorname{Poisson}(1.5)$, and $P(Y_2=0)=\exp(-1.5)\approx0.2231$. The second observation has three times the exposure and three times the expected count, while both observations have the same rate. Their probabilities of zero events differ because their Poisson means differ.

Why the Coefficient Is Fixed at One

In a log-link model, an offset is a known term added to the linear predictor with its coefficient constrained to one. Here the linear predictor for the mean is $\log(E_i)+x_i^T\beta$. The coefficient is not estimated from the count data because the rate model already says how exposure enters: mean count equals exposure times rate.

The same point can be checked directly through the rate implied by the mean. Since $E_i>0$, divide the conditional mean by exposure:

Result: Under the Poisson exposure-offset model, $$ E\left[\frac{Y_i}{E_i}\middle|x_i,E_i\right]=\lambda_i=\exp(x_i^T\beta), \qquad \operatorname{Var}\left(\frac{Y_i}{E_i}\middle|x_i,E_i\right)=\frac{\lambda_i}{E_i}. $$ Thus the model assigns the covariate-dependent mean to the rate, while the exposure determines the count mean.

Proof. By the Poisson mean and variance formulas, $E[Y_i\mid x_i,E_i]=E_i\lambda_i$ and $\operatorname{Var}(Y_i\mid x_i,E_i)=E_i\lambda_i$. Because $E_i$ is fixed and positive conditional on the model inputs, division by $E_i$ gives

E\left[\frac{Y_i}{E_i}\middle|x_i,E_i\right] =\frac{E_i\lambda_i}{E_i} =\lambda_i, \qquad \operatorname{Var}\left(\frac{Y_i}{E_i}\middle|x_i,E_i\right) =\frac{E_i\lambda_i}{E_i^2} =\frac{\lambda_i}{E_i}.

The observed rate $Y_i/E_i$ is random, but its conditional expectation is the modeled rate. Its conditional variance decreases as exposure increases under this Poisson model. $\square$

Likelihood Form and a Common Mistake

For independent observations with $E_i>0$, the conditional Poisson likelihood makes the offset’s role explicit. Write $\lambda_i=\exp(x_i^T\beta)$. The probability of observing $y_i$ is $e^{-E_i\lambda_i}(E_i\lambda_i)^{y_i}/y_i!$. Taking the product over observations and then its logarithm gives

\begin{aligned} \ell(\beta) &=\sum_i\left\{-E_i\exp(x_i^T\beta) +y_i\log\!\left(E_i\exp(x_i^T\beta)\right) -\log(y_i!)\right\}\\ &=\sum_i\left\{-E_i\exp(x_i^T\beta) +y_i\log(E_i)+y_i x_i^T\beta -\log(y_i!)\right\}. \end{aligned}

The known exposure appears both in the Poisson mean and in the log-probability term. It is not an additional covariate effect to be interpreted like a component of $\beta$; it is part of the specified relationship between exposure and expected count.

Common mistake: “The log-mean predictor should estimate a coefficient for $\log(E_i)$, just as it estimates coefficients for the other predictors.”
Correction: The rate model assumes $E[Y_i\mid x_i,E_i]=E_i\exp(x_i^T\beta)$. Taking logs gives $\log(E_i)+x_i^T\beta$, with a coefficient of exactly one on the known exposure term. Estimating that coefficient would no longer impose the stated proportional relationship.
Common mistake: “A constant rate alone means the counts must be Poisson.”
Correction: A constant rate specifies proportionality of the expected count to exposure. The Poisson distribution is an additional count-distribution assumption, such as the one supplied by a homogeneous Poisson process.

Key takeaway: Model the rate as $\lambda_i=\exp(x_i^T\beta)$, multiply by known positive exposure to obtain the expected count, and take logs. The result is a Poisson log-mean model with $\log(E_i)$ as an offset whose coefficient is fixed at one.

Check Your Understanding

Use the rate-to-count relationship and the Poisson mean formula to answer these questions.

A rate is $0.08$ events per exposure unit and exposure is $5$ units. What is the Poisson mean?
If $\log(\lambda_i)=x_i^T\beta$, derive the log-mean expression for a known exposure $E_i>0$.
Why is the coefficient of $\log(E_i)$ fixed at one in the exposure-offset model?
Does a constant expected event rate by itself imply a Poisson count distribution? Explain.
Under the offset model, what is the conditional expectation of $Y_i/E_i$?

What You'll Learn