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Joint Likelihoods For Frequency And Severity

Learn how independent frequency and severity models combine in one likelihood—and when their parameters can be estimated separately.

Beginner 10 min read

What You'll Learn

  • Define complete frequency–severity observations and check that the count and amount records agree.
  • Factor a joint likelihood under independent frequency and severity assumptions.
  • Estimate a Poisson frequency rate while accounting for exposure.
  • Estimate exponential and known-shape gamma severity parameters from claim amounts.
  • Combine geometric frequency and categorical severity estimates.
  • Recognize when dependence or a shared parameter prevents separate maximization.

Why Joint Likelihoods Can Separate

A complete claim record contains two kinds of information: how many claims occurred and the amount of each claim. A frequency model describes the counts; a severity model describes the amounts. When the two models are independent and have distinct parameters, their joint likelihood factors into a frequency part and a severity part. This makes estimation more direct, but only if the observations are matched correctly and the independence assumptions are appropriate.

For example, a hypothetical insurer might observe counts for several policy periods along with the exact amount of every claim. If the counts sum to six, the data must contain exactly six claim amounts. This bookkeeping check is essential before constructing the likelihood.

Definition: Suppose there are \(m\) exposure units. Let \(N_j\) be the claim count for unit \(j\), and let \(x_{j1},\ldots,x_{jN_j}\) be its exact observed claim amounts. The data are complete frequency–severity data when every count is observed and every claim amount is observed exactly. The total number of severity observations is \(N=\sum_{j=1}^{m}N_j\).

Write \(\alpha\) for the frequency parameter and \(\beta\) for the severity parameter. If the counts are independent across units and the claim amounts are independent and identically distributed conditional on the counts, with severity independent of frequency, the joint likelihood is

$$ L(\alpha,\beta) = \left(\prod_{j=1}^{m}p_N(N_j;\alpha)\right) \left(\prod_{j=1}^{m}\prod_{r=1}^{N_j} f_X(x_{jr};\beta)\right). $$

Here \(p_N\) is the probability mass function for a count and \(f_X\) is the severity density. For discrete severity, use its probability mass function instead of a density. If the parameter spaces are separate—\(\alpha\) appears only in the first factor and \(\beta\) only in the second—maximizing the joint likelihood is equivalent to maximizing each factor separately.

Important condition: Separate maximization requires the assumed factorization and distinct frequency and severity parameters. It does not follow merely because the data contain counts and amounts. If severity depends on the count, or a parameter is shared across the two models, the factors may not separate for estimation.

Why the Estimates Separate

The factorization follows from the probability chain rule and the model assumptions. First, the probability of the counts is multiplied by the conditional probability of the observed amounts given those counts. Conditional on the counts, the data contain \(N\) severity observations. If those amounts are independent with common density \(f_X(\cdot;\beta)\), and their distribution does not depend on the frequency parameter, their conditional likelihood is the product of their densities.

$$ P(\text{counts, amounts}) = P(\text{counts})P(\text{amounts}\mid\text{counts}) = \left(\prod_j p_N(N_j;\alpha)\right) \left(\prod_j\prod_{r=1}^{N_j}f_X(x_{jr};\beta)\right). $$

Taking logarithms turns the product into a sum:

$$ \ell(\alpha,\beta) = \sum_{j=1}^{m}\log p_N(N_j;\alpha) + \sum_{j=1}^{m}\sum_{r=1}^{N_j}\log f_X(x_{jr};\beta). $$

The first sum depends only on \(\alpha\); the second depends only on \(\beta\). Thus, if each component has a maximum in its allowed parameter space, the pair of component-wise maximum likelihood estimates maximizes the joint likelihood. This is not a claim that frequency and severity are always independent in practice; it is a consequence of the stated model.

Worked Applications

Worked Example: Poisson Frequency With Exponential Severity

A hypothetical insurer observes four one-year policy periods. The claim counts are \(2,0,1,2\). The five exact claim amounts are \(100,200,300,400,500\) dollars. Assume independent Poisson counts with a common rate \(\lambda\) per policy-year, and independent exponential claim amounts with rate \(\beta\) per dollar. Estimate both parameters and check that the data are consistent.

Method and data check. The count total is \(2+0+1+2=5\), which matches the five amounts. The total exposure is four policy-years. Under the Poisson model, the likelihood for the counts is

$$ L_N(\lambda) = \prod_{j=1}^{4}\frac{e^{-\lambda}\lambda^{N_j}}{N_j!} = \frac{e^{-4\lambda}\lambda^5}{2!\,0!\,1!\,2!}. $$

Ignoring the factor that does not depend on \(\lambda\), the log-likelihood is \(-4\lambda+5\log\lambda\). Its derivative is \(-4+5/\lambda\), so setting the derivative to zero gives \(\widehat{\lambda}=5/4=1.25\) claims per policy-year. The second derivative is \(-5/\lambda^2<0\), confirming a maximum.

Severity estimate. For an exponential distribution with rate \(\beta\), the density is \(f_X(x;\beta)=\beta e^{-\beta x}\) for \(x>0\). The amount total is \(100+200+300+400+500=1500\) dollars, so the severity likelihood is

$$ L_X(\beta)=\beta^5e^{-1500\beta}, \qquad \ell_X(\beta)=5\log\beta-1500\beta. $$

The score equation is \(5/\beta-1500=0\), giving \(\widehat{\beta}=5/1500=1/300\) per dollar. The corresponding fitted mean severity is \(1/\widehat{\beta}=300\) dollars.

Joint check. The joint likelihood is \(L_N(\lambda)L_X(\beta)\), and the two estimates maximize their respective factors. The fitted expected claim count over four policy-years is \(4(1.25)=5\), matching the observed total. The fitted exponential mean is \(300\) dollars, matching the observed average \(1500/5=300\) dollars. The common mistake this method avoids is using the number of policy periods, four, as the number of severity observations; there are five claims and therefore five amounts.

Worked Example: Poisson Exposure and Known-Shape Gamma Severity

A hypothetical portfolio has two independent exposure units, with exposures \(0.5\) and \(1.5\) policy-years and observed counts \(1\) and \(3\). The four claim amounts, measured in thousands of dollars, are \(1,1,2,2\). Assume Poisson counts with mean \(\lambda e_j\) for exposure \(e_j\), and gamma severity with known shape \(k=2\) and unknown rate \(\theta\). Estimate \(\lambda\) and \(\theta\).

Frequency component. The count total is \(1+3=4\); total exposure is \(0.5+1.5=2\) policy-years. The count likelihood is proportional to \(e^{-2\lambda}\lambda^4\). Its log-likelihood, apart from a constant, is \(-2\lambda+4\log\lambda\). The score equation \(-2+4/\lambda=0\) gives \(\widehat{\lambda}=2\) claims per policy-year. The second derivative \(-4/\lambda^2\) is negative.

Severity component. Under the shape-rate parameterization, the gamma density is \(f_X(x;\theta)=\theta^k x^{k-1}e^{-\theta x}/\Gamma(k)\) for \(x>0\). Here \(N=4\), \(k=2\), and the amount total is \(1+1+2+2=6\) thousand dollars. Therefore, the part of the likelihood that depends on \(\theta\) is \(\theta^{Nk}e^{-\theta\sum x_{jr}}=\theta^8e^{-6\theta}\). Its log-likelihood is \(8\log\theta-6\theta\), up to constants. The score equation \(8/\theta-6=0\) gives \(\widehat{\theta}=8/6=4/3\) per thousand dollars.

Check. The fitted gamma mean is \(k/\widehat{\theta}=2/(4/3)=1.5\) thousand dollars. The observed mean is \(6/4=1.5\) thousand dollars. The frequency fit also gives an expected total count of \(2(2)=4\) over the observed exposure. The common mistake to avoid is omitting the known shape \(k\) from the gamma rate estimate: \(N/\sum x_{jr}=4/6\) is not the rate MLE here. The correct numerator is \(Nk=4(2)=8\).

Worked Example: Geometric Frequency and Categorical Severity

A hypothetical program records counts over five equal exposure periods: \(0,2,1,3,2\). Each claim is also classified into one of three amount bands: low, middle, or high. Among the eight observed claims, the category counts are low \(3\), middle \(3\), and high \(2\). Assume the geometric count model \(P(N_j=n)=p(1-p)^n\) for \(n=0,1,\ldots\), and independent categorical severity with probabilities \(q_L,q_M,q_H\), which sum to one. Estimate all four parameters.

Frequency estimate. The counts sum to \(0+2+1+3+2=8\). The frequency likelihood is \(p^5(1-p)^8\). Its log-likelihood is \(5\log p+8\log(1-p)\), with score \(5/p-8/(1-p)\). Setting this equal to zero gives \(5(1-p)=8p\), hence \(\widehat p=5/13\). The fitted mean count per period is \((1-\widehat p)/\widehat p=8/5=1.6\); the observed mean is \(8/5=1.6\).

Severity estimates. The categorical likelihood is \(q_L^3q_M^3q_H^2\). To maximize it subject to \(q_L+q_M+q_H=1\), the category probabilities are estimated by their observed proportions. Thus \(\widehat q_L=3/8\), \(\widehat q_M=3/8\), and \(\widehat q_H=2/8=1/4\). They sum to \(3/8+3/8+2/8=1\).

Joint check. The geometric likelihood uses five period counts, while the categorical likelihood uses eight claim classifications. These are different observation totals because frequency is recorded per period and severity per claim. The estimates are separate because the likelihood factors and the parameters are distinct. A common error is to use five, the number of periods, as the denominator for severity proportions; the correct denominator is the eight observed claims.

Practical Checks and Limits

Before differentiating a joint likelihood, check the structure of the data and the assumptions of the model. For complete data, the total number of observed exact amounts must equal the sum of the claim counts. A zero-count period contributes to the frequency likelihood, but contributes no severity density term. If there are no claims at all, the count data may still inform a frequency parameter, but there are no observed severities from which to estimate a severity parameter.

Also keep exposure units consistent. In a Poisson model with mean \(\lambda e_j\), \(\lambda\) is a rate per unit of exposure, not a count per observation unless every exposure is one unit. For severity, the rate parameter’s units depend on the amount scale: a gamma rate based on amounts in thousands of dollars is expressed per thousand dollars.

A particularly important exam trap is to treat every joint likelihood as separable. If the severity distribution changes with the claim count, the conditional amount likelihood may depend on the frequency parameter. Likewise, if a parameter appears in both factors, maximizing each factor independently need not maximize the product. The separation method applies when the stated model supports the factorization and the parameter sets do not overlap.

Key takeaway: Match the number of claim amounts to the total count, write the frequency and severity likelihood contributions under their stated assumptions, and maximize separately only when the factors have distinct parameters. This likelihood workflow is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, identify the frequency and severity estimates separately and check the relevant totals or units.

  1. A hypothetical insurer observes counts \(0,2,1\) over exposures \(1,2,1\) policy-years. The three exact amounts are \(80,120,200\) dollars. Assume Poisson frequency with rate \(\lambda\) per policy-year and exponential severity with rate \(\beta\) per dollar. Find both maximum likelihood estimates. (Answer: \(\widehat{\lambda}=0.75\) per policy-year; \(\widehat{\beta}=3/400=0.0075\) per dollar.)
  2. Two exposure units have exposures \(0.5\) and \(1.5\) years and Poisson counts \(1\) and \(3\). Four amounts, in thousands of dollars, are \(1,2,3,0\) is not a valid claim amount; instead use \(1,1,2,2\). Assume gamma severity with known shape \(k=2\) and unknown rate \(\theta\). Find the estimates. (Answer: \(\widehat{\lambda}=2\) per year; \(\widehat{\theta}=4/3\) per thousand dollars.)
  3. Four periods have geometric counts \(1,0,2,1\). The four claim categories are low, low, middle, high. Estimate the geometric parameter \(p\) and the three category probabilities. (Answer: \(\widehat p=1/2\); \((\widehat q_L,\widehat q_M,\widehat q_H)=(1/2,1/4,1/4)\).)
  4. Counts \(2\) and \(4\) are observed over exposures \(1\) and \(3\) policy-years. The six exact exponential claim amounts are \(100,100,150,150,200,200\) dollars. Estimate the Poisson rate and exponential severity rate, then verify the fitted means against the observed totals. (Answer: \(\widehat{\lambda}=1.5\) per policy-year; \(\widehat{\beta}=1/150\) per dollar; fitted count total \(6\) and fitted mean severity \(150\) dollars.)
  5. Suppose all counts are observed, but the severity model’s parameter is also used in the frequency model. Can the two likelihood factors automatically be maximized separately? State the condition that fails. (Answer: No. A shared parameter means the factors do not have distinct parameter sets, so each factor’s separate maximizer need not maximize their product.)

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