Likelihood Contributions in Ratemaking GLMs

Why Likelihood Contributions Matter

A GLM likelihood gives a precise account of how well a set of model parameters describes the observed data. In ratemaking, frequency and severity data are recorded at different levels: a policy or rating cell supplies a claim count and exposure, while each reported claim can supply a separate severity observation. Writing the likelihood correctly requires keeping those units distinct.

This tutorial constructs the likelihood for a Poisson frequency model and a Gamma severity model. It emphasizes the exposure offset in the frequency component and shows how independent observations combine. The model assumptions and likelihood contributions are aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus.

Definition: A likelihood is the joint probability mass or density of the observed data, viewed as a function of unknown model parameters. For discrete observations it uses probability mass functions; for continuous observations it uses densities. Under independence, the joint likelihood is the product of the individual contributions.

Poisson Frequency With an Exposure Offset

For policy \(i\), let \(N_i\) be the observed claim count, \(e_i>0\) its exposure measured in policy-years (or another consistent exposure unit), and \(\lambda_i\) the expected claim frequency per exposure unit. Assume that, conditional on the rating variables, \(N_i\) has a Poisson distribution with mean \(e_i\lambda_i\). A log-link model for the rate is

\log(\lambda_i)=\eta_i=\mathbf{x}_i^\mathsf{T}\boldsymbol{\beta}, \qquad N_i\sim\operatorname{Poisson}(e_i\lambda_i).

The equivalent model for the expected count has linear predictor \(\log(e_i)+\eta_i\). Thus \(\log(e_i)\) is an offset: its coefficient is fixed at one, rather than estimated. Exposure scales the expected number of claims, not the rate per exposure unit.

The Poisson probability mass function gives the contribution from one policy:

L_{F,i} =P(N_i=n_i) =\frac{\exp(-e_i\lambda_i)(e_i\lambda_i)^{n_i}}{n_i!}.

The factorial \(n_i!\) is part of the probability. It does not depend on the regression coefficients, so it may be omitted when maximizing the likelihood over those coefficients alone; it must be retained when writing the full likelihood or calculating its numerical value.

Worked Example: Two Policies With Different Exposures

In a hypothetical portfolio, claim frequency is modeled as a constant \(0.08\) claims per policy-year. Policy A has \(2.5\) policy-years of exposure and no claims; Policy B has \(1.25\) policy-years and one claim. Assuming independent Poisson counts, calculate the joint frequency likelihood.

Method and assumptions. Use a Poisson count with mean \(e_i\lambda_i\) for each policy. The model rate is \(\lambda_i=0.08\) for both policies, and the two policy counts are independent.

For Policy A, the expected count is \(2.5(0.08)=0.20\). With observed count \(n_A=0\),

L_{F,A}=\frac{\exp(-0.20)(0.20)^0}{0!} =\exp(-0.20) \approx 0.818731.

For Policy B, the expected count is \(1.25(0.08)=0.10\). With observed count \(n_B=1\),

L_{F,B}=\frac{\exp(-0.10)(0.10)^1}{1!} =0.10\exp(-0.10) \approx 0.090484.

Independence gives the joint contribution:

L_F=L_{F,A}L_{F,B} =\exp(-0.20)\bigl(0.10\exp(-0.10)\bigr) =0.10\exp(-0.30) \approx 0.074082.

The expected counts sum to \(0.30\), and the product retains the \(0.10\) factor required by Policy B's one observed claim. Common mistake avoided: using the rate \(0.08\) as the Poisson mean for both policies would ignore their different exposures.

Gamma Severity Contributions

Let \(Y_{ij}\) be the amount of claim \(j\) for policy \(i\), measured in a stated monetary unit. Given the rating variables, assume claim severities are conditionally independent and have a Gamma distribution with mean \(\mu_{ij}>0\) and dispersion \(\phi>0\). Use the mean-dispersion parameterization

\operatorname{Var}(Y_{ij}\mid\mathbf{x}_{ij})=\phi\mu_{ij}^{2}, \qquad \text{shape}=\frac{1}{\phi}, \qquad \text{scale}=\phi\mu_{ij}.

With a log link, \(\log(\mu_{ij})=\mathbf{z}_{ij}^{\mathsf T}\boldsymbol{\gamma}\). The Gamma density for an observed positive amount \(y_{ij}\) is

f(y_{ij}\mid\mu_{ij},\phi) = \frac{y_{ij}^{1/\phi-1} \exp\!\left(-y_{ij}/(\phi\mu_{ij})\right)} {\Gamma(1/\phi)(\phi\mu_{ij})^{1/\phi}}, \qquad y_{ij}>0.

The severity likelihood uses one density contribution per observed claim. It is not one contribution per policy-year: a policy with no claims supplies no claim-level severity observation. In this model, exposure affects the expected claim count, while the severity mean is specified by its own rating variables. Adding an exposure offset to severity is not automatic; it would require a separate, justified model specification.

Worked Example: Two Gamma Claim Amounts

A hypothetical rating cell has two observed claim amounts, \(500\) and \(1{,}000\) dollars. Assume conditionally independent Gamma severities, each with mean \(\mu=1{,}000\) dollars and dispersion \(\phi=0.25\). Calculate the likelihood contribution from these two claims.

Method and assumptions. The shape is \(1/\phi=4\), and the scale is \(\phi\mu=250\) dollars. Thus each amount has a Gamma distribution with shape \(4\) and scale \(250\), and its density is \(y^3\exp(-y/250)/(\Gamma(4)250^4)\). Here \(\Gamma(4)=3!=6\).

For the \(500\)-dollar claim, \(y/250=2\) and \(y^3=125{,}000{,}000\). The denominator is \(6(250^4)=23{,}437{,}500{,}000\), so

f(500) = \frac{125{,}000{,}000\exp(-2)} {23{,}437{,}500{,}000} \approx 0.000721788 \quad\text{per dollar}.

For the \(1{,}000\)-dollar claim, \(y/250=4\) and \(y^3=1{,}000{,}000{,}000\). Therefore,

f(1{,}000) = \frac{1{,}000{,}000{,}000\exp(-4)} {23{,}437{,}500{,}000} \approx 0.000781467 \quad\text{per dollar}.

The joint severity contribution is the product of the two densities:

L_S=f(500)f(1{,}000) \approx (0.000721788)(0.000781467) \approx 5.6405\times 10^{-7} \quad\text{per dollar squared}.

The units are squared because this is a joint density for two continuous observations. Common mistake avoided: the Gamma scale is \(250\), not the mean \(1{,}000\); using the mean as the scale would change both density contributions.

Combining Policies, Rating Cells, and Claim Records

Suppose policies have independent frequency counts conditional on their rating variables. The frequency likelihood is the product of their Poisson contributions. Given the counts and relevant rating variables, suppose observed claim severities are conditionally independent and follow the specified Gamma model. The severity likelihood is then the product of the claim-level densities. If the frequency and severity components are also conditionally independent under the model, their product is the joint likelihood:

L(\boldsymbol{\beta},\boldsymbol{\gamma},\phi) = \prod_i \frac{\exp(-e_i\lambda_i)(e_i\lambda_i)^{n_i}}{n_i!} \; \prod_i\prod_{j=1}^{n_i} f(y_{ij}\mid\mu_{ij},\phi).

This factorization is a modeling assumption, not a consequence of using two GLMs. It is appropriate when the conditional frequency and severity distributions have the stated independence structure. If unobserved risk characteristics create dependence between a policy's frequency and claim severities, the joint data model may need to represent that dependence.

For a rating cell containing several policies, the cell frequency contribution can be written as a product over policies if their counts are conditionally independent and each policy's exposure and predictors are available. A cell total may instead be modeled as one Poisson count when the cell model specifies a Poisson total with an appropriate mean. For severity, a cell with \(m\) observed claims contributes \(m\) density terms when the individual amounts are observed. A total paid amount alone is not interchangeable with those individual observations under a Gamma claim-level likelihood.

Worked Example: A Joint Frequency-Severity Likelihood

Two independent hypothetical rating cells have Poisson expected claim counts \(0.4\) and \(0.8\). Their observed counts are respectively one and zero. The one observed claim has amount \(1\) thousand dollars. Assume its severity is exponential with mean \(1\) thousand dollars, equivalently Gamma with shape \(1\), scale \(1\), and dispersion \(\phi=1\). Calculate the joint likelihood.

Method and assumptions. Multiply the two independent Poisson probability contributions and the density of the one observed claim. Frequency and severity are conditionally independent, and the claim amount is measured in thousands of dollars.

For the first cell, with mean count \(0.4\) and observed count one,

L_{F,1} =\frac{\exp(-0.4)(0.4)^1}{1!} =0.4\exp(-0.4) \approx 0.268128.

For the second cell, with mean count \(0.8\) and observed count zero,

L_{F,2} =\frac{\exp(-0.8)(0.8)^0}{0!} =\exp(-0.8) \approx 0.449329.

The frequency likelihood is \(L_F=0.4\exp(-1.2)\). For an exponential severity with mean \(1\), the density at \(y=1\) is

f(1)=\exp(-1)\approx 0.367879 \quad\text{per thousand dollars}.

Consequently, the joint likelihood is

L=L_F f(1) =0.4\exp(-1.2)\exp(-1) =0.4\exp(-2.2) \approx 0.0443213.

As a check, multiplying the three observed-data factors gives \(0.268128(0.449329)(0.367879)\approx0.0443213\). Rounded to four decimal places, the likelihood is \(0.0443\). Common mistake avoided: multiplying the frequency and severity components is valid here because the example assumes their conditional independence; it is not a universal rule for every frequency-severity model.

Why the Product Form Holds—and When to Check It

The Poisson offset and the product across observations follow directly from the assumed distributions. For one policy, the Poisson mean is \(m_i=e_i\exp(\eta_i)\). Substituting this mean in the Poisson mass function gives the contribution below. For independent policies, the joint probability is the product of those masses. The same product rule applies to independent claim severities, using densities instead of probability masses.

P(N_i=n_i) =\frac{\exp(-m_i)m_i^{n_i}}{n_i!}, \qquad m_i=e_i\exp(\eta_i) \quad\Longrightarrow\quad L_{F,i} =\frac{\exp[-e_i\exp(\eta_i)]\,[e_i\exp(\eta_i)]^{n_i}}{n_i!}.

This also shows why the offset is \(\log(e_i)\): taking the logarithm of the expected count gives \(\log(m_i)=\log(e_i)+\eta_i\). The offset preserves the interpretation of \(\exp(\eta_i)\) as a rate per exposure unit. The product likelihood requires the stated conditional independence; if observations are clustered or otherwise dependent, multiplying marginal contributions alone does not give their joint likelihood.

A practical check is to ask what each row of data represents before writing a likelihood. The frequency product has one Poisson term per policy or modeled cell. The severity product has one density term per observed claim, unless the specified data and model use a different severity observation. Also check that monetary units are consistent: a Gamma density for amounts in dollars is numerically different from a density for amounts in thousands of dollars.

Common Likelihood Errors

Using exposure as an estimated coefficient. The frequency offset is \(\log(e_i)\) with fixed coefficient one. The modeled rate remains \(\lambda_i=\exp(\eta_i)\).
Applying exposure to severity without a model reason. In the stated frequency-severity setup, exposure scales expected claim count. Severity has a claim-level mean model.
Dropping normalizing terms in a numerical likelihood. Omitting \(n_i!\) or the Gamma normalizer may preserve a coefficient maximizer in some settings, but it does not give the full likelihood value.
Using one severity contribution per policy. A policy with \(n_i\) observed claims supplies \(n_i\) claim-level densities under the stated model, not one density merely because it is one policy.
Multiplying contributions without checking dependence. The product is the joint likelihood only when the observations are independent under the specified conditioning assumptions.

Key takeaway: Start with the observation unit and its assumed distribution. Use a Poisson mass with mean \(e_i\lambda_i\) for each frequency observation, a Gamma density for each claim severity, and multiply the contributions only under the model's independence assumptions.

Check Your Understanding

For each problem, identify the required likelihood contribution and keep frequency exposure distinct from claim severity.

A hypothetical policy has exposure \(4\) policy-years, frequency rate \(0.06\) per policy-year, and zero observed claims. What is its Poisson frequency contribution? (Answer: \(\exp(-0.24)\approx0.7866\).)
A rating cell has expected claim count \(0.30\) and observes two claims. Calculate its Poisson contribution, including the factorial. (Answer: \(\exp(-0.30)(0.30)^2/2\approx0.03334\).)
A hypothetical claim amount is \(400\) dollars. Its severity is Gamma with mean \(800\) dollars and dispersion \(\phi=0.5\). Find the density at \(400\), to seven significant figures. (Answer: shape \(2\), scale \(400\), so \(f(400)=\exp(-1)/400\approx0.000919698\) per dollar.)
Two independent policies have frequency rate \(0.05\) per policy-year. Policy A has exposure \(2\) years and one claim; Policy B has exposure \(1.5\) years and no claims. Find their joint frequency likelihood. (Answer: \((0.1\exp(-0.1))\exp(-0.075)=0.1\exp(-0.175)\approx0.0839457\).)
Two independent frequency cells have expected counts \(0.2\) and \(0.3\), with observed counts one and zero. The one claim has amount \(2\) in the stated monetary unit and has an exponential severity distribution with mean \(2\) units. Find the joint frequency-severity likelihood. (Answer: \((0.2\exp(-0.2))\exp(-0.3)(0.5\exp(-1))=0.1\exp(-1.5)\approx0.0223130\), with density measured per unit.)

What You'll Learn