Likelihood Construction for Claim Counts and Rates

From Claim Data to a Likelihood

A claim-count observation is not itself a model. To construct a likelihood, first state what was observed, what quantity is unknown, and how much exposure or how many trials produced the observation. These distinctions matter in actuarial work: a count observed over a long exposure period should not be interpreted like the same count observed over a short period, and a count among a fixed set of policies is naturally described differently from a count per policy-year.

This tutorial focuses on likelihood construction for Poisson claim counts and binomial claim indicators. A likelihood expresses how compatible the observed data are with different values of an unknown risk parameter, under a specified probability model. It is formed from the probability mass function of the data, viewed as a function of the parameter.

Definition: A likelihood is the probability of the observed data, written as a function of an unknown model parameter. If the observed data are \(x\) and the parameter is \(\theta\), the likelihood is \(L(\theta;x)=P_\theta(X=x)\). The data \(x\) are held fixed while \(\theta\) varies. A likelihood is not a probability distribution over possible values of \(\theta\).

For each problem, write down four items before substituting into a formula: the observed data, the unknown parameter, the exposure or trial count, and the assumed distribution. Here, exposure means the amount of opportunity for claims to occur, such as policy-years or vehicle-years. A trial is a defined unit that results in one of two outcomes, such as whether an insured location has at least one claim during a specified period.

Model assumptions used here: Unless a problem states otherwise, Poisson counts have a specified mean and are conditionally independent across distinct observation periods or groups. Binomial trials are independent, each has the same claim probability \(p\), and the number of trials is fixed. Interest and mortality assumptions do not enter these claim-count models.

Poisson Likelihoods With Exposure

Suppose \(N\) is the number of claims observed over exposure \(e>0\), and \(\lambda\) is the unknown expected claim frequency per unit of exposure. Under a homogeneous Poisson frequency model, \(N\mid\lambda,e\sim\operatorname{Poisson}(e\lambda)\). Thus the expected count is exposure multiplied by the frequency rate. The probability mass function gives the likelihood directly.

L(\lambda;n,e) =P(N=n\mid\lambda,e) =\frac{(e\lambda)^n e^{-e\lambda}}{n!}, \qquad \lambda\geq 0.

The observed value \(n\) is fixed in this expression; \(\lambda\) is the variable. The rate estimate \(n/e\) is a data summary, not a replacement for the observed count in the likelihood. In particular, the probability of observing the count is not obtained by treating \(n/e\) as though it were a Poisson count.

Worked Example: A Claim Count Over Policy-Year Exposure

A hypothetical small commercial portfolio has 4.5 policy-years of exposure and records 3 claims. Construct the likelihood for the unknown claim frequency \(\lambda\), measured per policy-year, and evaluate it at \(\lambda=0.40\) and \(\lambda=0.20\). Assume the count follows a Poisson distribution with mean equal to exposure times frequency.

Identify the model inputs. The observed count is \(n=3\), the exposure is \(e=4.5\) policy-years, and the unknown parameter is \(\lambda\). The Poisson mean is \(e\lambda=4.5\lambda\). No interest or mortality assumption applies.

Construct the likelihood.

L(\lambda;3,4.5) =\frac{(4.5\lambda)^3 e^{-4.5\lambda}}{3!} =\frac{(4.5\lambda)^3 e^{-4.5\lambda}}{6}.

At \(\lambda=0.40\), the Poisson mean is \(4.5(0.40)=1.8\). The observed-count likelihood is

L(0.40;3,4.5) =\frac{1.8^3e^{-1.8}}{6} =\frac{5.832}{6}(0.165299) =0.160671.

At \(\lambda=0.20\), the mean is \(4.5(0.20)=0.9\), so

L(0.20;3,4.5) =\frac{0.9^3e^{-0.9}}{6} =\frac{0.729}{6}(0.406570) =0.049398.

The likelihood at \(0.40\) is larger for these observed data than the likelihood at \(0.20\). These are likelihood values for the specified observation, not posterior probabilities that either frequency is true. Method and trap avoided: the exposure is multiplied by the rate to obtain the Poisson mean. Using \(3\) as the mean, or inserting the observed rate \(3/4.5\) as the count, would describe a different calculation.

Why Exposure Multiplies the Rate

The exposure adjustment follows from the meaning of a frequency rate. If the expected number of claims per unit of exposure is \(\lambda\), then multiplying the exposure by \(e\) gives expected count \(e\lambda\). Under the Poisson assumption, the count probability is determined by that mean, not by \(\lambda\) alone. Substitution of the Poisson mass function therefore yields the likelihood above.

Specify the rate and exposure.
\(\lambda\) is claims per unit of exposure, and \(e\) is the observed amount of exposure.

Obtain the expected count.
The assumed mean count is \(e\lambda\), so the count model is \(N\mid\lambda,e\sim\operatorname{Poisson}(e\lambda)\).

Insert the observed count into the mass function.
For \(N=n\), the probability is \((e\lambda)^n e^{-e\lambda}/n!\), which is then viewed as a function of \(\lambda\).

This construction assumes the rate is constant over the exposure represented by the observation, conditional on the model parameter. If the rate varies by period or group, use the appropriate rate for each observation rather than applying one common rate without adjustment. A Poisson model may also be unsuitable if its assumptions do not reasonably describe the data; a likelihood formula is always conditional on the model chosen.

Independent Counts Across Periods or Groups

For independent Poisson observations with possibly different exposures, the likelihood for the complete observed data is the product of the period-specific probabilities. If the underlying frequency is a common \(\lambda\), and observation \(j\) has exposure \(e_j\) and count \(n_j\), then \(N_j\mid\lambda,e_j\sim\operatorname{Poisson}(e_j\lambda)\). More generally, a known multiplier \(c_j\) can adjust the rate in observation \(j\), giving mean \(e_jc_j\lambda\). The multiplier is treated as known, not as an additional unknown parameter.

L(\lambda;\boldsymbol{n},\boldsymbol{e},\boldsymbol{c}) =\prod_{j=1}^{J} \frac{(e_jc_j\lambda)^{n_j}e^{-e_jc_j\lambda}}{n_j!}.

If only the total count is observed, rather than the count in each period or group, the likelihood must be based on that total observation. When independent Poisson counts have means \(e_jc_j\lambda\), their sum is Poisson with mean \(\lambda\sum_j e_jc_j\). But observing the individual counts provides more detailed data than observing only the sum; the two data descriptions need not have identical likelihood values, even when their likelihoods are proportional as functions of \(\lambda\).

Worked Example: Separate Counts With a Known Rate Multiplier

A hypothetical insurer observes counts in two segments. Segment 1 has exposure \(e_1=2\), known multiplier \(c_1=1\), and count \(n_1=1\). Segment 2 has exposure \(e_2=3\), known multiplier \(c_2=2\), and count \(n_2=4\). Assume independent Poisson counts with mean \(e_jc_j\lambda\), where \(\lambda\) is unknown. Construct the likelihood from the separate observed counts and evaluate it at \(\lambda=0.50\).

Identify the inputs. The unknown is the base frequency \(\lambda\). The two Poisson means are \(2(1)\lambda=2\lambda\) and \(3(2)\lambda=6\lambda\). The data are the pair of counts \((1,4)\); the counts are assumed independent conditional on \(\lambda\). No interest or mortality assumption applies.

Multiply the two count probabilities.

L(\lambda;1,4) =\frac{(2\lambda)^1e^{-2\lambda}}{1!} \frac{(6\lambda)^4e^{-6\lambda}}{4!} =108\lambda^5e^{-8\lambda}.

For the coefficient, \(2\cdot6^4/4!=2\cdot1296/24=108\); the exponential terms combine as \(e^{-2\lambda}e^{-6\lambda}=e^{-8\lambda}\). At \(\lambda=0.50\), the means are \(1\) and \(3\), and

L(0.50;1,4) =\frac{1^1e^{-1}}{1!}\frac{3^4e^{-3}}{4!} =e^{-1}\frac{81e^{-3}}{24} =3.375e^{-4} =0.061815.

If the only recorded datum were the total count \(5\), its distribution would be Poisson with mean \(8\lambda\), and its likelihood at \(0.50\) would be \(e^{-4}4^5/5!=0.156293\). That is not the likelihood for the separately recorded pair \((1,4)\): the event “total equals 5” includes other allocations between the segments. Method and trap avoided: multiply period-specific probabilities when the individual counts are observed; do not silently replace them with a probability for the total.

Binomial Likelihoods From a Fixed Number of Trials

A binomial model is appropriate when the number of trials \(m\) is fixed, each trial has two possible outcomes, and each trial has the same probability \(p\) of a claim. Let \(K\) be the number of claim-positive trials. Under independent trials, \(K\mid p,m\sim\operatorname{Binomial}(m,p)\). The observed data for this count model are \(k\) claims among \(m\) trials.

L(p;k,m) =P(K=k\mid p,m) =\binom{m}{k}p^k(1-p)^{m-k}, \qquad 0\leq p\leq 1.

The factor \(\binom{m}{k}\) counts the ways that \(k\) claim-positive outcomes can occur among \(m\) trials. Include it when calculating the probability of the observed count. For likelihood comparisons that use the same data and model, it is constant with respect to \(p\), so it can be omitted when writing a likelihood proportional to the full likelihood. It must not be omitted if the requested quantity is the actual probability of the count.

The binomial probability \(p\) refers to one trial’s claim probability over the trial’s specified period. It is not automatically a frequency per policy-year. If one policy is observed over multiple periods, decide whether the data are claim counts per period or a single indicator of whether any claim occurred; those are different observations and generally call for different models.

Worked Example: Claim-Positive Locations Among a Fixed Set

A hypothetical insurer studies 12 independent storage locations for one specified year. Each location is classified as either having at least one claim or having none. Three locations are claim-positive. Construct the likelihood for the common annual claim probability \(p\), then calculate the probability of exactly three claim-positive locations at \(p=0.15\) and \(p=0.25\).

Identify the model inputs. The observed number of claim-positive locations is \(k=3\); the fixed number of trials is \(m=12\); and the unknown parameter is the per-location annual probability \(p\). Under the binomial assumptions, \(K\mid p\sim\operatorname{Binomial}(12,p)\). No interest or mortality assumption applies.

Construct the likelihood. There are \(\binom{12}{3}=220\) ways to choose the three claim-positive locations, so

L(p;3,12) =\binom{12}{3}p^3(1-p)^9 =220p^3(1-p)^9.

At \(p=0.15\), \(p^3=0.003375\) and \((1-p)^9=0.85^9=0.231616946\). Therefore,

$$ L(0.15;3,12) =220(0.003375)(0.231616946) =0.171976. $$

At \(p=0.25\), \(p^3=0.015625\) and \((1-p)^9=0.75^9=0.075084686\). Thus,

$$ L(0.25;3,12) =220(0.015625)(0.075084686) =0.258104. $$

The likelihood is larger at \(p=0.25\) for this observation. Method and trap avoided: the observed fraction \(3/12\) does not replace the binomial probability mass function. In particular, \(p^3(1-p)^9\) alone is the probability of one specific arrangement of which locations had claims, not the probability of exactly three claim-positive locations.

Likelihood Construction: Checks and Common Traps

A correct likelihood begins with a clearly stated data-generating model. Before finalizing an expression, check that the exponent corresponds to the observed count or number of successes, that exposure is in the same units used to define the frequency, and that the likelihood treats the unknown parameter as variable while holding the data fixed.

Using a rate as if it were a count. In a Poisson model, use the observed integer count \(n\) in the mass function and use exposure \(e\) in the mean \(e\lambda\).
Forgetting the exposure unit. If \(\lambda\) is claims per policy-year, \(e\) must be measured in policy-years. A rate per vehicle-year cannot be multiplied by policy-years without an appropriate conversion.
Confusing count and indicator data. A binomial trial records a binary outcome. If the data record the number of claims, including multiple claims for one risk, a binomial model of claim-positive risks may not match the observation.
Dropping the binomial coefficient when an exact probability is requested. It is constant in \(p\), but it is part of the probability of the observed count.
Combining separate observations without checking what was recorded. The likelihood for a vector of counts is a product of probabilities. A likelihood for the total count corresponds to observing only that total.

Key takeaway: A Poisson likelihood needs the observed count and the exposure-adjusted mean; a binomial likelihood needs the number of successes and the fixed trial count. In either case, identify the observation and the unknown parameter first, then use the probability mass function for the data actually recorded.

Check Your Understanding

For each problem, identify the model and write the likelihood for the stated data. Treat each requested numerical value as an exact-count probability under the model.

A hypothetical repair portfolio records 2 claims over 5 policy-years. Under a Poisson model with frequency \(\lambda\) per policy-year, write the likelihood and evaluate it at \(\lambda=0.30\). (Answer: \(L(\lambda)=e^{-5\lambda}(5\lambda)^2/2!\); \(L(0.30)=0.2510\).)
Among 8 independent commercial buildings, exactly 1 has a claim during a specified year. Under a binomial model with common annual claim probability \(p\), write the likelihood. (Answer: \(L(p)=\binom{8}{1}p(1-p)^7=8p(1-p)^7\).)
Two independent periods have a common Poisson frequency parameter \(\lambda\). The first has exposure 1.5 and count 0; the second has exposure 2.5 and count 3. Write the likelihood for the pair of observed counts. (Answer: \(L(\lambda)=e^{-4\lambda}(2.5\lambda)^3/3!=\frac{125}{48}\lambda^3e^{-4\lambda}\).)
In a fixed set of 20 independent trials, 5 are claim-positive. Write the binomial likelihood and evaluate the exact-count probability at \(p=0.10\), rounding to four decimal places. (Answer: \(L(p)=\binom{20}{5}p^5(1-p)^{15}\); \(L(0.10)=0.0319\).)
Two independent Poisson segments have means \(2\lambda\) and \(6\lambda\), and their separately observed counts are 1 and 4. Write the likelihood for the separate counts, then state the mean of the total count if only the total is observed. (Answer: \(L(\lambda)=108\lambda^5e^{-8\lambda}\); the total is Poisson with mean \(8\lambda\).)

What You'll Learn