Offset Variable

Study guides: Exam MAS-I, Exam PCPA

An Offset Variable is a predictor in a Generalized Linear Model whose coefficient is fixed at exactly $1$ rather than estimated from the data — used to force a known, exact relationship (most commonly exposure) into the linear predictor.

$g(\mu_i) = \beta_0 + \beta_1 x_{i1} + \cdots + \beta_p x_{ip} + \text{offset}_i$

$\text{log link, exposure offset:} \quad \ln\mu_i = \ln(\text{Exposure}_i) + \beta_0 + \beta_1 x_{i1} + \cdots + \beta_p x_{ip}$

The offset must be on the same scale as the linear predictor: with a log link it is the logged quantity — $\ln(\text{Exposure})$ , not exposure. Several offsets are added together after each is transformed
For claim counts $N_i \sim \text{Poi}(\mu_i)$ , since $E[N_i] = \text{Exposure}_i \times \text{Frequency}_i$ , including $\ln(\text{Exposure}_i)$ as an offset converts a model for expected claim counts into a model for the underlying claim Frequency rate per unit of exposure
An offset differs from an ordinary Control Variable: its coefficient is not estimated — it is forced to equal $1$ , consistent with the known proportional relationship between exposure and expected claims
If exposure were instead entered as an ordinary covariate, its estimated coefficient need not equal $1$ , which would mask the intended exactly-proportional relationship
Offsets are also used to carry forward previously-determined relativities (e.g., an established deductible or territory relativity) into a new model without re-estimating them — again on the linear-predictor scale: under a log link a deductible factor of $0.90$ enters as $\ln 0.90 = -0.1054$ , and a base loss cost of $265$ with that factor as $\ln 265 + \ln 0.90 = 5.4744$

Fitting a Poisson Frequency Model with an Exposure Offset A Poisson GLM for claim counts uses $\ln(\text{Exposure})$ as an offset and a single rating variable, driver age, with fitted equation $\ln\hat{\mu}_i = \ln(\text{Exposure}_i) - 2.5 + 0.01 \cdot \text{age}_i$ . For a driver age 40 with $2$ car-years of exposure, find the expected claim count and the fitted frequency.

Answer $\ln\hat{\mu} = \ln(2) - 2.5 + 0.01(40) = 0.693 - 2.5 + 0.4 = -1.407$ $\hat{\mu} = e^{-1.407} \approx 0.245 \text{ expected claims}$ Since the offset coefficient is fixed at $1$ , dividing back out the exposure gives the fitted frequency: $\hat{\mu}/\text{Exposure} = 0.245/2 \approx 0.122$ claims per car-year.