Why Frequency Models Need an Exposure Term
A policy observed for more time, or over more insured units, has more opportunity to generate claims. A count model should distinguish that opportunity from the underlying claim frequency. For example, a policy with half a vehicle-year of exposure should not generally be expected to produce as many claims as an otherwise identical policy with one vehicle-year of exposure.
The exposure term makes this proportional relationship explicit. In a log-link model, it appears as the logarithm of exposure added to the predictor. Its coefficient is fixed at one because the expected count is proportional to exposure, not because the data estimate that relationship as an ordinary rating effect.
Here, \(e\) denotes exposure, such as earned vehicle-years. It is not the base of the exponential function; expressions such as \(\exp(\eta)\) denote exponentiation. The logarithm is the natural logarithm. Exposure must be positive for \(\log(e)\) to be defined.
From Expected Count to Offset
The key starting point is that the expected claim count is exposure multiplied by the expected claim count per exposure unit. Taking the logarithm of both sides separates those two components. This is the whole reason the offset has the form \(\log(e)\).
For rating characteristics \(x\), define \(\lambda(x)\) as the expected number of claims per exposure unit. For exposure \(e\), the expected total count is \(E[N\mid x,e]=e\lambda(x)\).
Because \(e>0\) and \(\lambda(x)>0\), take natural logarithms: \(\log E[N\mid x,e]=\log(e\lambda(x))\).
Use \(\log(ab)=\log(a)+\log(b)\) for positive \(a\) and \(b\): \(\log E[N\mid x,e]=\log(e)+\log(\lambda(x))\).
With a log link, write \(\log(\lambda(x))=x^\mathsf{T}\beta\), where \(x^\mathsf{T}\beta\) is the linear predictor for frequency per exposure unit. Then the count predictor is \(\log(e)+x^\mathsf{T}\beta\).
The derivation uses the proportional-mean assumption and the logarithm identity. It does not depend on a Poisson distribution specifically. A Poisson assumption is commonly added for a count GLM: conditional on rating characteristics and exposure, \(N\mid x,e\sim\operatorname{Poisson}(e\lambda(x))\). This specifies a count distribution with mean and variance both equal to \(e\lambda(x)\). The offset derivation itself concerns the mean and still holds if a different count distribution is used with the same mean structure.
The exposure offset is therefore a known term in the linear predictor, not an additional rating factor to estimate. The coefficients in \(x^\mathsf{T}\beta\) describe how the frequency per exposure unit varies with the modeled characteristics. The offset accounts for how much exposure was observed.
Worked Applications
Worked Example: Converting a Per-Unit Frequency to an Expected Count
A hypothetical policy has \(0.75\) vehicle-years of exposure. Its modeled frequency is \(0.12\) claims per vehicle-year. Assume that, conditional on this frequency and exposure, its claim count is Poisson with mean \(e\lambda\). Find the expected claim count and its log-link predictor. State the offset.
Method. Multiply the per-unit frequency by exposure to obtain the Poisson mean, then take its logarithm. This keeps exposure separate from the frequency rate.
The exposure is \(e=0.75\), and the frequency is \(\lambda=0.12\). Thus,
The assumed count distribution is \(N\sim\operatorname{Poisson}(0.09)\). The exposure offset is \[ \log(e)=\log(0.75)=-0.287682. \] The log frequency is \[ \log(\lambda)=\log(0.12)=-2.120264. \] Therefore, the count-scale linear predictor is \[ \log(e)+\log(\lambda) =-0.287682-2.120264 =-2.407946. \] Checking directly, \(\log(0.09)=-2.407946\), so exponentiating the predictor returns the expected count, \(\exp(-2.407946)=0.09\). The expected count is \(0.09\) claims.
Common mistake avoided: Using \(0.12\) as the expected count would treat the rate per vehicle-year as the count for a policy with \(0.75\) vehicle-years. The correct mean is \(0.75(0.12)=0.09\).
Worked Example: Adding a Rating Relativity to the Offset
A hypothetical class has a base frequency of \(0.04\) claims per exposure unit. A rating characteristic multiplies that frequency by \(1.5\). A policy in the class has \(2.5\) exposure units. Assume a Poisson count conditional on the modeled mean. Find the class frequency, the expected count, and the full log-link predictor.
Method. First apply the multiplicative relativity to the base frequency. Then multiply by exposure. On the log scale, the relativity contributes the logarithm of its factor, while exposure contributes the offset.
The class frequency per exposure unit is \[ \lambda=0.04(1.5)=0.06. \] The expected count is \[ E[N]=2.5(0.06)=0.15, \] so \(N\sim\operatorname{Poisson}(0.15)\). The log frequency can be expressed as the base log frequency plus the log relativity: \[ \log(\lambda)=\log(0.04)+\log(1.5) =-3.218876+0.405465 =-2.813411. \] The exposure offset is \[ \log(2.5)=0.916291. \] Thus, the full predictor is \[ \log(2.5)+\log(0.04)+\log(1.5) =0.916291-3.218876+0.405465 =-1.897120. \] Checking on the mean scale, \(\log(0.15)=-1.897120\), and \(\exp(-1.897120)=0.15\). The expected count is \(0.15\) claims.
Common mistake avoided: Applying the relativity to the exposure rather than to the frequency would not represent the stated rating structure. The factor \(1.5\) changes claims per exposure unit; the exposure \(2.5\) then scales the resulting expected count.
Worked Example: Comparing Counts at Different Exposures
Two hypothetical risks have the same modeled frequency, \(0.20\) claims per exposure unit. Risk A has \(0.25\) exposure units; Risk B has \(1.75\). Assume independent Poisson counts conditional on their means. Calculate each mean and predictor, then check the aggregate expected count.
Method. Use the same log frequency for both risks and add each risk’s own log-exposure offset. This illustrates that the rate is held constant while the expected count changes with exposure.
For Risk A, \[ E[N_A]=0.25(0.20)=0.05. \] Its offset is \(\log(0.25)=-1.386294\), and its log frequency is \(\log(0.20)=-1.609438\). The predictor is \[ -1.386294-1.609438=-2.995732=\log(0.05). \] Thus \(N_A\sim\operatorname{Poisson}(0.05)\).
For Risk B, \[ E[N_B]=1.75(0.20)=0.35. \] Its offset is \(\log(1.75)=0.559616\). Its predictor is \[ 0.559616-1.609438=-1.049822=\log(0.35). \] Thus \(N_B\sim\operatorname{Poisson}(0.35)\).
The combined exposure is \(0.25+1.75=2.00\) units, and the total expected count is \(0.05+0.35=0.40\). Applying the common frequency to the combined exposure gives \(2.00(0.20)=0.40\), the same result. The aggregate log mean also checks: \[ \log(2.00)+\log(0.20) =0.693147-1.609438 =-0.916291 =\log(0.40). \]
Common mistake avoided: Giving both risks an expected count of \(0.20\) confuses a per-unit frequency with a count. Their rates match, but their expected counts differ because their exposures differ.
Model Specification and Interpretation
If the frequency per exposure unit is modeled by a log link, a standard specification for the expected count is
for observation \(j\), with positive exposure \(e_j\). Equivalently, the expected count is
The intercept and rating coefficients determine the modeled frequency per exposure unit. The offset \(\log(e_j)\) converts that frequency into an expected count for the observed exposure. In particular, if two risks have identical rating characteristics and their exposures are in the ratio \(c\), their expected counts are also in the ratio \(c\), because \[ \frac{e_2\lambda(x)}{e_1\lambda(x)}=\frac{e_2}{e_1}. \] This proportionality is the modeling assumption represented by fixing the offset coefficient at one.
The Poisson version of this specification assumes that, conditional on the modeled characteristics and exposure, the count has a Poisson distribution with mean \(e_j\lambda(x_j)\). The log-mean relationship alone does not establish that distributional assumption. A modeler should therefore distinguish the choice of mean structure from the choice of count distribution.
Common Exam Traps
A frequent error is to omit exposure and model the count as though every observation had the same amount of exposure. Another is to include exposure once as an offset and then multiply the predicted mean by exposure again. The latter double-counts exposure.
For example, suppose the modeled frequency is \(0.20\) and exposure is \(1.75\). The correct mean is \(1.75(0.20)=0.35\). If a predictor already includes the offset \(\log(1.75)\), exponentiating it already incorporates the factor \(1.75\). Multiplying that result by \(1.75\) again would give \(1.75(0.35)=0.6125\), which is not the specified mean.
A further check is dimensional: \(\lambda\) is a frequency per exposure unit, whereas \(e\lambda\) is an expected count. Keep those quantities distinct in calculations and interpretation. The offset derivation assumes the exposure is known and positive and that the expected count is proportional to exposure; it is not suitable without adjustment if that proportionality assumption does not describe the risk being modeled.
Check Your Understanding
For each problem, identify the exposure contribution and keep the per-unit frequency separate from the expected count.
- A hypothetical risk has \(0.6\) exposure units and frequency \(0.15\) claims per unit. Under a Poisson count model, find its expected count and log-mean predictor. (Answer: Mean \(=0.09\); predictor \(=\log(0.09)\approx-2.4079\).)
- A class has base frequency \(0.05\) per exposure unit and a rating factor of \(1.2\). For exposure \(3\), find the expected count. (Answer: Frequency \(=0.06\); expected count \(=0.18\).)
- A log-link model has frequency predictor \(\log(\lambda)=-2\), and a risk has exposure \(2\). Find the offset, full count predictor, and expected count. (Answer: Offset \(=\log(2)\approx0.6931\); predictor \(\approx-1.3069\); mean \(=2\exp(-2)\approx0.2707\).)
- Two risks share frequency \(0.08\) per exposure unit and have exposures \(0.5\) and \(1.5\). Find their expected counts and verify the aggregate mean using total exposure. (Answer: Means \(=0.04\) and \(0.12\); aggregate mean \(=0.16=2.0(0.08)\).)
- A risk has frequency \(0.10\) per unit and exposure \(4\). A calculation uses the offset in the predictor, obtains mean \(0.40\), and then multiplies by \(4\) again. Identify the error and the correct mean. (Answer: Exposure was counted twice; the correct expected count is \(4(0.10)=0.40\), not \(1.60\).)