Likelihood Construction for Frequency and Severity Samples

Why Likelihood Construction Starts With the Data

A likelihood records how well a specified probability model accounts for the observations. Constructing it correctly begins by identifying what was observed: a claim count, an exact loss amount, a grouped amount, or a censored amount. These observations do not all contribute the same kind of probability expression. A count contributes a probability mass; an exact observation from a continuous severity model contributes a density.

The likelihood is a function of the model parameter with the observed data held fixed. It is not, in general, the probability that the parameter has a particular value, and a density evaluated at an exact continuous observation is not the probability of that observation. The central task here is to translate the sampling assumptions and recorded data into the correct product of mass or density contributions.

Definition: For observed data \(D\) and a model indexed by parameter \(\theta\), the likelihood \(L(\theta;D)\) is the joint probability mass, density, or mixed mass-density of the observed data, viewed as a function of \(\theta\) with \(D\) fixed. For independent observations, the joint expression factors into the product of their individual contributions. Factors that do not depend on \(\theta\) may be omitted when comparing parameter values for estimation, but retaining them makes the likelihood expression complete.

Count Observations Contribute Probability Masses

Suppose \(N_1,\ldots,N_m\) are independent claim counts, and count \(N_j=n_j\) is observed. If the probability mass function is \(p_n(\theta)=P(N=n;\theta)\), then the likelihood is

L(\theta;n_1,\ldots,n_m)=\prod_{j=1}^m p_{n_j}(\theta).

For example, if each count follows a Poisson distribution with mean \(\lambda t_j\), where \(t_j\) is the exposure and \(\lambda\) is the claim rate per exposure unit, then

P(N_j=n_j)=\frac{e^{-\lambda t_j}(\lambda t_j)^{n_j}}{n_j!}, \qquad n_j=0,1,2,\ldots.

This expression requires that the counts are independent and that the count model is Poisson with the stated mean for each exposure. The exposure scales the mean; it is not itself a count. If the model is not Poisson, or if counts are dependent, this product of Poisson contributions does not apply.

Worked Example: Independent Poisson Counts With Unequal Exposure

A hypothetical portfolio has three independent observation periods with exposures of 1, 2, and 1 policy-years. The observed claim counts are 0, 2, and 1, respectively. Assume the count in a period with exposure \(t\) is Poisson with mean \(\lambda t\), where \(\lambda\) is measured in claims per policy-year. Construct the likelihood and evaluate it at \(\lambda=1\).

Method and assumptions. Use the Poisson probability mass for each period and multiply because the three period counts are assumed independent. For exposure 1 and count 0, the contribution is \(e^{-\lambda}\). For exposure 2 and count 2, it is \(e^{-2\lambda}(2\lambda)^2/2!\). For exposure 1 and count 1, it is \(e^{-\lambda}\lambda\).

Multiplying the three contributions gives

L(\lambda)=e^{-\lambda} \left(e^{-2\lambda}\frac{(2\lambda)^2}{2!}\right) \left(e^{-\lambda}\lambda\right) =2\lambda^3e^{-4\lambda}.

At \(\lambda=1\), the separate contributions are \(e^{-1}\), \(2e^{-2}\), and \(e^{-1}\). Thus \(L(1)=e^{-1}(2e^{-2})e^{-1}=2e^{-4}\). Using \(e^{-4}\approx0.0183156389\), the likelihood is \(2(0.0183156389)=0.0366312778\), or \(0.036631\) to six decimal places.

Check. The exponent is \(-4\lambda\), the sum of the three Poisson means, and the power \(\lambda^3\) comes from the total of three observed claims. The factor 2 comes from \((2\lambda)^2/2!=2\lambda^2\). A common mistake is to use \(e^{-\lambda}\lambda^2/2!\) for the period with exposure 2; its mean is \(2\lambda\), not \(\lambda\).

Exact Severity Observations Contribute Densities

Let \(X_1,\ldots,X_n\) be independent, identically distributed claim amounts with continuous density \(f_X(x;\theta)\). For exact observed amounts \(x_1,\ldots,x_n\), the severity likelihood is

L(\theta;x_1,\ldots,x_n)=\prod_{k=1}^n f_X(x_k;\theta).

This construction assumes that each recorded amount is an exact observation from the stated continuous model and that the amounts are independent with the same parameter. If the data are rounded into intervals, only known to exceed a threshold, or otherwise censored, an exact-value density contribution is not appropriate for those records. Their contributions must reflect the event actually observed.

Worked Example: Exact Exponential Severity Data

A hypothetical sample contains three independent, exact claim amounts of 1, 2, and 3 thousand dollars. Assume each amount has an exponential distribution with rate \(\beta=0.5\) per thousand dollars, whose density is \(f_X(x;\beta)=\beta e^{-\beta x}\) for \(x\geq0\). Construct the likelihood and evaluate it at the stated rate.

Method and assumptions. Since the three amounts are exact and independent, multiply their densities. The contributions at \(x=1,2,3\) are, respectively, \(0.5e^{-0.5}\), \(0.5e^{-1}\), and \(0.5e^{-1.5}\). Therefore

L(\beta)=\prod_{k=1}^{3}\beta e^{-\beta x_k} =\beta^3 e^{-\beta(1+2+3)} =\beta^3e^{-6\beta}.

At \(\beta=0.5\), the rate factor is \((0.5)^3=0.125\), and the exponential factor is \(e^{-3}\approx0.0497870684\). Their product is \(0.125(0.0497870684)=0.00622338355\), or \(0.006223\) to six decimal places. Because the amounts are expressed in thousands of dollars, the product of three densities has units of \((\text{thousand dollars})^{-3}\).

Check. The exponent uses the sum of observed amounts, \(1+2+3=6\), and there is one factor of \(\beta\) for each of three observations. A common mistake is to call this the probability of observing exactly these three amounts. Under a continuous model that probability is zero; the calculated quantity is a joint density value, not an event probability.

Combining Frequency and Severity Data

Suppose observations consist of counts for several periods and exact claim amounts for the claims in those periods. A mixed data set has a joint mass-density: it is discrete in the counts and continuous in the severity values. Under the assumption that the severity amounts are independent and identically distributed conditional on the observed counts, with density \(f_X(x;\theta)\), the appropriate factorization is

f(\text{counts},x_1,\ldots,x_N) = P(\text{counts})\,f(x_1,\ldots,x_N\mid\text{counts}).

If, in addition, the severities are conditionally independent of one another and their distribution does not depend on the counts, then

f(x_1,\ldots,x_N\mid\text{counts}) =\prod_{k=1}^{N} f_X(x_k;\theta).

The first factor is a probability mass for the observed count data; the second is a conditional density for the observed severities. The identity is a mixed mass-density identity, not an assertion that exact severity values have positive probability. If frequency and severity share parameters, the same factorization can still be used, but both factors must be expressed using the shared parameters. Independence between periods or claims must not be assumed unless the model specifies it.

Identify the observation units.
List the periods or exposures for counts, and determine which exact claim amounts are recorded.

Write each count contribution.
Use the probability mass for the observed count, with the correct exposure or other model inputs.

Write each severity contribution.
Use a density for each exact continuous amount, or the probability of the recorded value if severity is discrete.

Factor only under the stated assumptions.
Multiply independent contributions and use the conditional mass-density factorization for combined data.

Check what was observed.
Replace an exact-value density contribution when the record is grouped, rounded, truncated, or censored.

Worked Example: Joint Likelihood for Counts and Exact Losses

In a hypothetical two-year study, the annual claim counts are 1 and 2. The three recorded exact claim amounts are 2, 4, and 1 thousand dollars. Assume independent annual Poisson counts with rate \(\lambda\) claims per year, and claim amounts that are independent exponential variables with rate \(\beta\) per thousand dollars, independent of the counts. Construct the joint likelihood and evaluate it at \(\lambda=1\) and \(\beta=0.5\).

Method and assumptions. First multiply the two count masses. The count of 1 contributes \(e^{-\lambda}\lambda\), and the count of 2 contributes \(e^{-\lambda}\lambda^2/2!\). Thus

L_F(\lambda)=\frac{\lambda^3}{2}e^{-2\lambda}.

The three exact losses contribute exponential densities. Their sum is \(2+4+1=7\), so

L_S(\beta)=\beta^3e^{-7\beta}.

Under the stated independence assumptions, the joint mass-density is the product:

L(\lambda,\beta)=\frac{\lambda^3}{2}e^{-2\lambda}\,\beta^3e^{-7\beta}.

At \(\lambda=1\), \(L_F(1)=\frac12e^{-2}\approx0.0676676416\). At \(\beta=0.5\), \(L_S(0.5)=(0.5)^3e^{-3.5}=0.125(0.0301973834)\approx0.00377467292\). Multiplying gives \(L(1,0.5)\approx0.0002554\), or \(0.000255\) to six decimal places. Equivalently, the joint expression is \(\frac{1}{16}e^{-5.5}\), which gives the same result.

Check. The frequency factor has two exponential terms because there are two annual observations, while the severity factor has three powers of \(\beta\) because there are three claims. A common mistake is to treat the observed severities as part of the count probability. They instead contribute a conditional density; the count likelihood and severity likelihood describe different components of the data.

Common Traps and Scope of the Likelihood

Likelihood construction is sensitive to the observation process as well as to the distribution. If a claim amount is known only to exceed a deductible or reporting threshold, the data do not provide an exact amount below that threshold. If an amount is recorded only as lying in an interval, its contribution is the probability of that interval, not a density at an arbitrarily chosen point. Likewise, if counts are recorded by unequal exposures, using the same mean for every period can change the likelihood.

Using a density as a point probability. A continuous exact observation has probability zero. Its density contributes to the likelihood.
Using a mass for a continuous exact amount. A density is required for an exact observation from a continuous severity model.
Ignoring exposure. For a Poisson rate model, exposure \(t\) gives mean \(\lambda t\), affecting both the exponential and power terms.
Multiplying without independence. A product of marginal contributions requires the relevant independence assumptions. Dependence requires a joint model.
Misrepresenting incomplete records. A censored or grouped observation contributes the probability or density appropriate to what was actually observed.

Likelihoods provide a common framework for frequency and severity data, but they do not remove the need to state the sampling model. A correct expression makes the observations, distributional assumptions, exposure units, and independence conditions visible. This foundation is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, identify the appropriate mass or density contributions before simplifying.

Independent Poisson counts are 2 and 1 over exposures of 1 and 3 years, respectively. The rate is \(\lambda\) claims per year. Write the likelihood as a function of \(\lambda\). (Answer: \(L(\lambda)=\frac{3}{2}\lambda^3e^{-4\lambda}\).)
Two exact independent loss amounts are 0.5 and 1.5 thousand dollars from an exponential model with rate \(\beta=2\) per thousand dollars. Evaluate their joint density. (Answer: \(4e^{-4}\approx0.073263\) per \((\text{thousand dollars})^2\).)
Two independent binomial observations have 2 successes in 5 trials and 4 successes in 6 trials, with common success probability \(p\). Write their joint likelihood. (Answer: \(150p^6(1-p)^5\).)
One annual Poisson count is 0 and another is 1; their common rate is \(\lambda=0.5\) per year. The one observed claim has exact amount 2 thousand dollars and exponential severity rate \(\beta=0.25\) per thousand dollars. Find the joint mass-density. (Answer: \(0.125e^{-1.5}\approx0.027891\).)
Under an exponential severity model with rate \(\beta=0.2\), one claim amount is observed exactly as 2 and another is known only to exceed 3, with amounts in the same units. Assuming independent severities, write the likelihood. (Answer: \(\beta e^{-2\beta}e^{-3\beta}=\beta e^{-5\beta}\), which is approximately \(0.073576\) at \(\beta=0.2\).)

What You'll Learn