From a Link Function to a Ratemaking Calculation
A frequency model is useful for ratemaking only when its predictions can be translated into quantities such as expected claims per policy-year or expected claims over a particular policy period. A log link supplies a natural multiplicative structure for risk factors, while a known exposure offset scales the annualized frequency to the time or amount of risk actually covered.
This workshop carries one hypothetical frequency model from its specification through predictions for new risks. The emphasis is on keeping three quantities distinct: the linear predictor, the frequency rate per unit exposure, and the expected claim count over the observed exposure. No claim-count likelihood calculation is needed to make these mean predictions.
The modeling assumptions for this mean structure should be explicit. Exposure is positive, measured on a consistent basis, and known rather than estimated as a rating effect. Conditional on the included rating characteristics, expected claim count is proportional to exposure, and the risk characteristics are treated as representative of the risk during the exposure period. The log link constrains the fitted mean and rate to be positive. These assumptions concern the mean structure; they do not, by themselves, specify the full probability distribution of claim counts.
For categorical rating variables, one level is typically a reference level. Its coefficient is zero under reference coding, and each other level has a coefficient representing its change on the log scale relative to that reference. Exponentiating a coefficient gives the corresponding multiplicative relativity. A positive coefficient produces a relativity above one; a negative coefficient produces one below one.
Why the Offset Produces a Rate and a Count
The offset is not an extra rating factor to estimate. Its value is fixed by the known exposure. The mean structure follows directly from the assumption that, for a given risk profile, doubling the exposure doubles the expected number of claims.
Write the expected count as exposure times the annualized frequency: \(E[N_j\mid \mathbf{x}_j,e_j]=e_j\lambda_j\).
Because \(e_j\) and \(\lambda_j\) are positive, \(\log E[N_j\mid \mathbf{x}_j,e_j]=\log(e_j)+\log(\lambda_j)\).
Set \(\log(\lambda_j)=\beta_0+\mathbf{x}_j^\mathsf{T}\boldsymbol{\beta}\). Substitution gives \(\log E[N_j\mid \mathbf{x}_j,e_j]=\log(e_j)+\beta_0+\mathbf{x}_j^\mathsf{T}\boldsymbol{\beta}\), with \(\log(e_j)\) as the offset.
Exponentiating gives \(E[N_j\mid \mathbf{x}_j,e_j]=e_j\exp(\beta_0+\mathbf{x}_j^\mathsf{T}\boldsymbol{\beta})\). Thus the rate depends on risk characteristics, and exposure scales that rate to an expected count.
This derivation also explains why the offset must have coefficient one. If its coefficient were estimated freely, the model would no longer impose proportionality between expected count and exposure. That would be a different mean structure, not the standard known-exposure frequency specification.
Worked Predictions From a Fitted Model
Suppose a hypothetical insurer has fitted a log-link frequency model with the following annualized baseline rate and multiplicative factors. The baseline is for a reference risk with one unit of each reference rating characteristic.
Here each indicator equals one when the named characteristic applies and zero otherwise. The factors are equivalent to log-link coefficients \(\log(1.50)\), \(\log(1.20)\), and \(\log(0.80)\). The baseline rate is \(0.10\) expected claims per policy-year. These are hypothetical model values, not observed industry results.
Worked Example: Predicting a Count for a Part-Year Risk
A mature driver in an urban area insures a high-power vehicle for \(0.75\) policy-years. Using the model above, calculate the annualized frequency rate and fitted expected claim count.
Method. First multiply the relevant factors to obtain the risk relativity. Then multiply the baseline rate by that relativity. Finally apply exposure to obtain the expected count.
Risk relativity. All three non-reference characteristics apply, so \(1.50\times1.20\times0.80=1.80\times0.80=1.44\).
Annualized rate. \(0.10\times1.44=0.144\) expected claims per policy-year.
Fitted expected count. \(0.75\times0.144=0.108\) expected claims over the insured period. The result is an expectation, not a claim count that must be an integer.
Check. The exposure is less than one policy-year, so the fitted count \(0.108\) is less than the annualized rate \(0.144\), as it should be. The common mistake is to report \(0.144\) as the part-year expected count; that omits the \(0.75\) exposure multiplier.
Worked Example: Aggregating Fitted Counts Across Risks
Consider three hypothetical risks under the same model. Risk A is a reference risk with \(2.00\) policy-years of exposure. Risk B is mature and urban, but has a standard-power vehicle, with \(0.50\) policy-years of exposure. Risk C has a high-power vehicle only, with \(1.25\) policy-years of exposure. Find each fitted count, the portfolio total, and the exposure-weighted portfolio frequency rate.
Risk A. Its relativity is \(1.00\). Its rate is \(0.10\times1.00=0.10\) claims per policy-year, and its fitted count is \(2.00\times0.10=0.200\).
Risk B. Its relativity is \(1.50\times1.20=1.80\). Its rate is \(0.10\times1.80=0.18\) claims per policy-year, and its fitted count is \(0.50\times0.18=0.090\).
Risk C. Its relativity is \(0.80\). Its rate is \(0.10\times0.80=0.08\) claims per policy-year, and its fitted count is \(1.25\times0.08=0.100\).
Portfolio expected count. Add the fitted counts: \(0.200+0.090+0.100=0.390\).
Total exposure and portfolio rate. Total exposure is \(2.00+0.50+1.25=3.75\) policy-years. The exposure-weighted rate is \(0.390/3.75=0.104\) expected claims per policy-year.
Check. The portfolio rate lies between the smallest individual rate, \(0.08\), and the largest, \(0.18\). A common mistake is to take the unweighted average of the three rates, \((0.10+0.18+0.08)/3=0.12\). That is not the portfolio rate because the risks have unequal exposure. The correct aggregate divides the sum of expected counts by total exposure.
Worked Example: Comparing New Risks Without Confusing Counts and Rates
A hypothetical rating team compares a reference risk with a new risk that is mature and has a high-power vehicle, but is not urban. The reference risk has \(1.00\) policy-year of exposure; the new risk has \(0.40\) policy-years. Calculate the new risk's rate, its expected count, and the count ratio relative to the reference risk. Explain why the count ratio is not the risk relativity.
Reference risk. Its relativity is \(1.00\), so its annualized rate is \(0.10\) and its expected count is \(1.00\times0.10=0.100\).
New risk relativity. Only the mature and high-power factors apply: \(1.50\times0.80=1.20\). The new risk's frequency rate is therefore \(0.10\times1.20=0.12\) claims per policy-year.
New risk expected count. Apply its own exposure: \(0.40\times0.12=0.048\) expected claims.
Count ratio. The fitted count ratio is \(0.048/0.100=0.48\). This ratio is smaller than one because the new risk has much less exposure, even though its annualized rate is higher. The risk relativity is the rate ratio \(0.12/0.10=1.20\), not the count ratio.
Check. The count ratio also equals the product of the exposure ratio and the rate ratio: \((0.40/1.00)\times1.20=0.48\). The common mistake is to interpret \(0.48\) as the risk-factor relativity. It compares counts over unequal exposure periods; the appropriate relativity compares annualized rates.
Using the Model for New Risks
For a new risk, the practical calculation is to assign its rating variables to the model's reference levels and fitted coefficients, form the linear predictor, exponentiate to get the annualized rate, and then multiply by the requested exposure. Keeping the order explicit prevents a common units error: the exponentiated linear predictor is a rate only when the model has been defined as the log of the annualized rate.
The relative frequency of two risks \(a\) and \(b\), measured on the same exposure basis, is \(\widehat{\lambda}_a/\widehat{\lambda}_b\). Under a log link, the log of this ratio is the difference in their linear predictors. If the risks differ in one factor only, their rate ratio is the factor's exponentiated coefficient. If they differ in several factors and the model has no interactions, their relativity is the product of the applicable factor relativities.
Do not confuse a factor relativity with a percentage change. A relativity of \(1.20\) means a rate \(20\%\) above the reference rate; a relativity of \(0.80\) means a rate \(20\%\) below it. Nor should a rate be described as an expected count: a rate is expressed per exposure unit, while the count prediction also depends on the particular risk's exposure.
Common Specification and Calculation Traps
- Leaving exposure out of the prediction. The exponentiated risk portion gives the annualized rate. Multiply by exposure to obtain an expected count.
- Adding factors on the response scale. Log-link effects add on the linear-predictor scale and multiply on the rate scale. For example, factors \(1.50\) and \(1.20\) combine as \(1.50\times1.20=1.80\), not \(1.50+1.20\).
- Using the wrong exposure unit. If the fitted rate is per policy-year, exposure must be expressed in policy-years. A half-year should be entered as \(0.50\), not \(6\) months as though the unit were a year.
- Comparing expected counts to obtain a relativity when exposures differ. First convert predictions to rates, or adjust the count ratio for the exposure ratio.
- Treating an estimated exposure coefficient as a standard offset. A known exposure offset has a fixed coefficient of one on the log scale. Estimating that coefficient changes the assumed relationship between exposure and expected count.
A disciplined final check is to attach units to every result. The linear predictor is dimensionless; its exponential is a frequency per stated exposure unit; multiplying by exposure gives an expected number of claims. This unit check often reveals an omitted offset or a mistaken time conversion before the result is used in a rating indication.
Check Your Understanding
For each original problem, identify the relevant rate or relativity before applying exposure.
- A hypothetical model has a baseline frequency of \(0.12\) claims per policy-year. A new risk has relativity \(1.25\) and exposure \(0.40\) policy-years. What are its annualized rate and fitted expected count? (Answer: rate \(0.15\) claims per policy-year; expected count \(0.06\).)
- A risk's baseline rate is \(0.08\), and its two applicable factor relativities are \(1.40\) and \(0.75\). For exposure of \(1.50\) policy-years, find the combined relativity, annualized rate, and expected count. (Answer: relativity \(1.05\); rate \(0.084\) claims per policy-year; expected count \(0.126\).)
- Two risks differ only in rating class. Their factors relative to a common base are \(1.60\) and \(1.20\). What is the first risk's rate relativity compared with the second? (Answer: \(1.60/1.20=1.3333\), approximately 33.33% higher.)
- A hypothetical model has baseline rate \(0.09\). Risk A has relativity \(1.50\) and exposure \(0.60\); Risk B has relativity \(0.80\) and exposure \(1.20\). Find each fitted count and the aggregate rate. (Answer: Risk A count \(0.081\); Risk B count \(0.0864\); total count \(0.1674\), total exposure \(1.80\), aggregate rate \(0.093\) claims per policy-year.)
- A fitted log mean for a risk is \(\log(e)+(-2.20)+0.30\), where \(e=0.50\) policy-years and the last two terms form the risk portion of the predictor. Find the annualized rate and expected count. (Answer: rate \(\exp(-1.90)\approx0.1496\) claims per policy-year; expected count \(0.50\exp(-1.90)\approx0.0748\).)