Deriving the Log-Exposure Offset

From a Rate to an Expected Count

A count of claims, incidents or other events depends both on the underlying rate and on how much opportunity there was for an event to occur. A policy observed for six months contributes less exposure than one observed for three years, even if their modeled rates are the same. A Poisson model accounts for this by multiplying the rate by exposure. With a log link, that multiplication becomes addition of the log exposure.

The result is useful only when the exposure and rate refer to compatible units and the assumed rate structure is reasonable. The derivation below makes those conditions explicit. It is aligned with the SOA Exam PA (Predictive Analytics) syllabus.

Definition: Let \(N_i\) be a nonnegative integer event count for observation \(i\), and let \(e_i>0\) be its known exposure, measured in a stated unit such as policy-years. Under a Poisson rate model, conditional on the explanatory variables and exposure, \(N_i\) has a Poisson distribution with mean \(e_i\lambda_i\), where \(\lambda_i>0\) is the event rate per unit of exposure. With a log-rate model \(\log(\lambda_i)=\eta_i=x_i^\mathsf{T}\beta\), the expected count is \(e_i\exp(\eta_i)\).

The standard assumptions are that the count is Poisson conditional on the modeled information, that its conditional mean equals its conditional variance, and that the rate is constant over the exposure represented by the observation after accounting for the explanatory variables. Counts from different observations are commonly assumed conditionally independent. In a constant-rate Poisson process, these assumptions correspond to events occurring at a stable rate per exposure unit, with independent increments.

The rate model is positive because it is an exponential: \(\lambda_i=\exp(\eta_i)\). Exposure is a multiplier, not another quantity to be added to the rate. Therefore, the expected count is

\mu_i=E[N_i\mid x_i,e_i]=e_i\lambda_i=e_i\exp(x_i^\mathsf{T}\beta).

This formula applies when \(e_i\) measures the amount of exposure at the modeled rate. If the rate changes materially during the observation period, a single rate multiplied by total exposure may not represent the expected count. A model may instead need to account for the changing rate over the period.

Why Log Exposure Is an Offset

A log link models the logarithm of the expected count as a linear predictor. Taking the logarithm of the expected-count formula gives

\log(\mu_i)=\log(e_i)+\eta_i =\log(e_i)+x_i^\mathsf{T}\beta.

The term \(\log(e_i)\) is known from the exposure data. It is included in the linear predictor with its coefficient fixed at one; it is not an estimated coefficient. This known term is called the log-exposure offset. The model is therefore a regression for the log rate plus a known adjustment that converts the rate to an expected count.

The offset's coefficient must be one because the original relationship is multiplication by exposure. If exposure doubles while the modeled rate stays fixed, the expected count doubles. An estimated coefficient on log exposure would allow the data to replace this specified proportional relationship with a different one; that would be a different model.

Derivation: For a Poisson count \(N_i\) with mean \(\mu_i=e_i\exp(\eta_i)\), its probability mass function is \(P(N_i=n_i)=\exp(-\mu_i)\mu_i^{n_i}/n_i!\). Taking logarithms and substituting the mean gives

\begin{aligned} \ell_i &=-\mu_i+n_i\log(\mu_i)-\log(n_i!)\\ &=-e_i\exp(\eta_i)+n_i\{\log(e_i)+\eta_i\}-\log(n_i!). \end{aligned}

The exposure appears in the model through the known quantity \(\log(e_i)\) in the log mean, while \(e_i\) also determines the mean itself. Thus, placing \(\log(e_i)\) in the linear predictor with coefficient one reproduces the Poisson likelihood for a count whose mean is exposure times rate.

The derivation assumes \(e_i>0\), so that \(\log(e_i)\) is defined. It also assumes the Poisson distribution is appropriate. If event counts have more variation than the Poisson model allows, or observations are dependent in ways not represented by the model, the log-exposure term alone does not resolve those issues.

Worked Examples

Worked Example: Calculate Expected Counts From Modeled Rates

A hypothetical insurer models annual claim rates for two groups using \(\eta_i=\log(0.08)+x_i\log(2)\), where \(x_i=0\) for Group A and \(x_i=1\) for Group B. Group A has 2.5 policy-years of exposure and Group B has 4 policy-years. Find each group’s modeled rate and expected claim count.

Method. First calculate each log rate, then exponentiate to obtain the rate per policy-year. Multiply each rate by the exposure in policy-years.

For Group A, \(x_A=0\), so \(\eta_A=\log(0.08)+0\log(2)=\log(0.08)\). Thus \(\lambda_A=\exp(\eta_A)=0.08\) claims per policy-year, and \[ \mu_A=2.5(0.08)=0.20 \] expected claims.

For Group B, \(x_B=1\), so \(\eta_B=\log(0.08)+\log(2)=\log(0.16)\). Thus \(\lambda_B=0.16\) claims per policy-year, and \[ \mu_B=4(0.16)=0.64 \] expected claims.

Check. Group B has twice Group A’s modeled rate and \(4/2.5=1.6\) times its exposure. Its expected count is therefore \(2(1.6)=3.2\) times Group A’s: \(0.20(3.2)=0.64\). The expected counts need not be integers. This method avoids treating the rate itself as the expected count.

Worked Example: Convert Exposure Units Without Changing the Mean

A hypothetical coverage has 18 months of exposure. Its modeled event rate is 0.12 events per year. Find the expected count using years and then using months, and verify that the result is unchanged.

Method. Pair the exposure with a rate expressed in the same time unit.

In years, the exposure is \(18/12=1.5\) years. The expected count is \[ \mu=1.5(0.12)=0.18. \]

In months, the rate is \(0.12/12=0.01\) events per month. The expected count is \[ \mu=18(0.01)=0.18. \]

The log-link calculation gives the same check. Using years, \[ \log(1.5)+\log(0.12)=0.405465-2.120264=-1.714799. \] Using months, \[ \log(18)+\log(0.01)=2.890372-4.605170=-1.714798. \] The small difference is due to rounding of the displayed logarithms; both expressions equal \(\log(0.18)\) before rounding. The expected count is 0.18 in either unit system. This avoids combining, for example, exposure in months with a rate per year without converting one of them.

Worked Example: Use the Offset in Poisson Probabilities

A hypothetical study has two conditionally independent groups. Group 1 has exposure \(e_1=5\) policy-years and modeled rate \(\lambda_1=0.20\) claims per policy-year. Group 2 has exposure \(e_2=10\) policy-years and modeled rate \(\lambda_2=0.10\). The observed counts are \(n_1=0\) and \(n_2=2\). Calculate the expected count in each group and the probability of observing these two counts under the model.

Method. Calculate each Poisson mean from exposure times rate, then multiply the group probabilities using conditional independence.

The means are \[ \mu_1=5(0.20)=1,\qquad \mu_2=10(0.10)=1. \] The first count has probability \[ P(N_1=0)=\frac{\exp(-1)1^0}{0!}=\exp(-1)\approx 0.3679. \] The second count has probability \[ P(N_2=2)=\frac{\exp(-1)1^2}{2!} =\frac{\exp(-1)}{2}\approx 0.1839. \] Therefore, the joint probability is \[ P(N_1=0,N_2=2)=0.3679(0.1839)\approx 0.0677, \] with the product evaluated using unrounded probabilities.

Check. Both groups have expected count 1 despite different exposures because their rates differ in the opposite direction. The displayed probabilities multiply to approximately 0.0677. This method avoids assuming that equal expected counts require equal rates.

Common Mistake: Adding Exposure to the Rate

A frequent error is to write \(\mu_i=\exp(\eta_i+e_i)\). This incorrectly puts exposure inside the exponential as an additive input. For a rate of 0.10 per exposure unit and exposure \(e=4\), the correct calculation is \[ \mu=4(0.10)=0.40. \] The incorrect calculation is \[ \exp\{\log(0.10)+4\}=0.10\exp(4)\approx 5.4598, \] which is not exposure times rate. The correct log-mean expression is \(\log(\mu)=\log(e)+\eta\), not \(\eta+e\).

Another related error is estimating a coefficient for \(\log(e_i)\) when the modeling goal specifies that exposure is a known proportional multiplier. The offset has coefficient one precisely because the expected count is defined as exposure times rate. Changing that coefficient changes the assumed relationship between exposure and expected count.

Key takeaway: Specify a positive rate per exposure unit, model its logarithm as \(\eta_i\), and multiply the rate by exposure to obtain the expected count. Taking logs yields \(\log(\mu_i)=\log(e_i)+\eta_i\), where \(\log(e_i)\) is a known offset with coefficient one.

Check Your Understanding

For each problem, identify the rate and exposure units before calculating.

A hypothetical portfolio has a modeled rate of 0.06 claims per policy-year and 7 policy-years of exposure. Find its expected claim count. (Answer: 0.42 claims)
A group has a modeled rate of 0.15 events per month and 8 months of exposure. Write its log-mean expression and calculate its expected count. (Answer: \(\log(\mu)=\log(8)+\log(0.15)\); \(\mu=1.20\))
A log-rate model gives \(\eta=\log(0.04)+\log(3)\) for an observation with 5 units of exposure. Find the modeled rate and expected count. (Answer: rate \(=0.12\) per exposure unit; expected count \(=0.60\))
A hypothetical observation has 24 months of exposure and a rate of 0.09 events per year. Convert to compatible units and find the expected count. (Answer: 2 years at 0.09 per year, so \(\mu=0.18\))
For a Poisson count with exposure \(e=3\) and modeled log rate \(\eta=\log(0.2)\), an analyst writes \(\log(\mu)=\eta+3\). State the correct log mean and expected count. (Answer: \(\log(\mu)=\log(3)+\log(0.2)=\log(0.6)\); \(\mu=0.60\))

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