Poisson Likelihood For Policy Claim Counts

Why the Likelihood Matters for Claim Counts

A frequency model assigns a probability distribution to the number of claims observed for a policy. Its fitted mean is not merely a predicted number: under a Poisson model, it is the policy’s expected claim count, conditional on the information used to make the prediction. The likelihood then describes how compatible the observed portfolio counts are with specified fitted means.

Constructing the likelihood correctly requires keeping track of two things for every policy: the observed count and the mean assigned to that policy. A zero count still contributes information, and a count of two or more requires a factorial term. These details matter whether the likelihood is being evaluated for specified model parameters or used to estimate them.

Definition: Suppose policy $i$ has observed claim count $y_i$, where $y_i$ is a nonnegative integer, and a Poisson model assigns it mean $\mu_i>0$. Its probability mass, viewed as a function of $\mu_i$, is $$ P(Y_i=y_i\mid \mu_i)=\frac{e^{-\mu_i}\mu_i^{y_i}}{y_i!}. $$ If policy counts are independent conditional on the model information, the likelihood for $n$ policies is $$ L(\mu_1,\ldots,\mu_n)=\prod_{i=1}^{n}\frac{e^{-\mu_i}\mu_i^{y_i}}{y_i!}. $$

Here $y_i!$ is the factorial of the observed count, with $0!=1$. The likelihood is the joint probability of the observed count vector, expressed as a function of the model means or parameters. Once the observations are fixed, it is not a probability distribution over possible parameter values.

Model Assumptions and the Meaning of the Mean

For the policy-level likelihood above, assume that each policy count is conditionally Poisson given its rating information and any known exposure information used by the model. The assigned mean must be finite and positive. Also assume that counts from different policies are conditionally independent. Conditional independence is what allows the joint probability to be written as a product of policy-level probabilities.

For a Poisson random variable, the conditional mean and conditional variance are both $\mu_i$:

E[Y_i\mid \text{model information}]=\mu_i, \qquad \operatorname{Var}(Y_i\mid \text{model information})=\mu_i.

Thus a fitted mean of $0.7$ means an expected $0.7$ claims for that policy over the period represented by the model. It does not mean that the policy is predicted to have exactly $0.7$ of a claim. The realized count remains an integer, often zero when the mean is below one.

The Poisson likelihood is not appropriate simply because the response is a count. The conditional Poisson assumption and independence assumption must be reasonable for the application. For example, dependence between policies or materially greater conditional variability than the Poisson model allows may call for a different model. Those issues do not change how the Poisson likelihood is calculated; they limit whether it is a suitable description of the data.

From Policy Contributions to the Portfolio Likelihood

For a single policy, the likelihood contribution is the probability mass at the observed count. If the count is zero, the contribution is $e^{-\mu_i}$. If it is one, the contribution is $e^{-\mu_i}\mu_i$. For a count of two, the contribution is $e^{-\mu_i}\mu_i^2/2$. For multiple independent policies, multiply the contributions.

A convenient equivalent calculation uses the log-likelihood. Taking the logarithm turns the product into a sum:

\ell(\mu_1,\ldots,\mu_n) =\log L =\sum_{i=1}^{n}\left(-\mu_i+y_i\log\mu_i-\log(y_i!)\right).

The log-likelihood is easier to work with algebraically and numerically. Because the logarithm is strictly increasing, the parameter values that maximize $L$ also maximize $\ell$. The factorial term depends on the observed data, not on the means. It can therefore be omitted when finding a maximum over model parameters, but it must be included when reporting the actual likelihood value.

Worked Example: Calculating Contributions for Three Policies

A hypothetical small portfolio has three independent policy claim counts. The observed counts are $0,1,2$, and their fitted Poisson means are $0.5,1.0,1.5$, respectively. Calculate each policy’s contribution and the joint likelihood. Treat the fitted means as fixed, and assume the conditional Poisson and independence assumptions hold.

Policy 1. The observed count is zero and the mean is $0.5$:

L_1=\frac{e^{-0.5}(0.5)^0}{0!}=e^{-0.5}=0.606531.

Policy 2. The observed count is one and the mean is $1.0$:

L_2=\frac{e^{-1}(1.0)^1}{1!}=e^{-1}=0.367879.

Policy 3. The observed count is two and the mean is $1.5$. The factorial is $2!=2$:

L_3=\frac{e^{-1.5}(1.5)^2}{2!} =e^{-1.5}\frac{2.25}{2} =e^{-1.5}(1.125) =0.251021.

By conditional independence, multiply the three contributions. Equivalently, collect the exponential terms and the remaining factors:

L=L_1L_2L_3 =e^{-0.5}e^{-1}e^{-1.5}(1)(1)(1.125) =e^{-3}(1.125) =0.056010.

Check. The exponents combine to $-3$, the only non-unit power/factorial term is $1.5^2/2!=1.125$, and the resulting joint probability is between zero and one. The common mistake here is to treat the zero count as a zero likelihood contribution. Its contribution is $e^{-0.5}$, not zero.

Why the Likelihood Has This Form

The Poisson probability mass for one policy is the probability of observing exactly $y_i$ events when the mean count is $\mu_i$. Under conditional independence, the probability of observing the full vector of policy counts is the product of those individual probabilities. Substituting the Poisson mass for each policy gives:

P(Y_1=y_1,\ldots,Y_n=y_n\mid \mu_1,\ldots,\mu_n) =\prod_{i=1}^{n}P(Y_i=y_i\mid\mu_i) =\prod_{i=1}^{n}\frac{e^{-\mu_i}\mu_i^{y_i}}{y_i!}.

This joint probability, considered as a function of the means after the observed counts have been fixed, is the likelihood. Taking logs gives the sum form because $\log(ab)=\log a+\log b$. These steps explain both the product across policies and the factorial term within each policy’s contribution.

The same reasoning applies when each mean comes from a model with shared parameters. In that case, the means are functions of those parameters, and the likelihood becomes a function of the parameters through the $\mu_i$'s. The likelihood still has one count contribution per policy; it does not replace the observed policy-level counts with an average count.

Worked Example: Computing Means Before Contributions

For a hypothetical portfolio, suppose each policy’s fitted mean follows $\mu_i=\exp(\beta_0+\beta_1x_i)$, where $x_i$ is a given rating indicator. Take $\beta_0=\log(0.4)$ and $\beta_1=\log(2)$. Four independent policies have $(x_i,y_i)$ values $(0,0),(0,1),(1,1),(1,2)$. Calculate their fitted means and the likelihood of the observed counts.

Policies with $x_i=0$. For either of the first two policies, $\mu_i=\exp(\log(0.4))=0.4$. For the policy with count zero:

L_1=e^{-0.4}=0.670320.

For the policy with count one:

L_2=e^{-0.4}(0.4)=0.268128.

Policies with $x_i=1$. Their means are $\exp(\log(0.4)+\log(2))=0.8$. The count-one contribution is $L_3=e^{-0.8}(0.8)=0.359463$. The count-two contribution includes $2!$:

L_4=\frac{e^{-0.8}(0.8)^2}{2!} =e^{-0.8}(0.32) =0.143785.

The joint likelihood can be checked by combining powers and exponentials. The sum of the fitted means is $0.4+0.4+0.8+0.8=2.4$; the count powers contribute $0.4(0.8)^3$, and the count-two factorial divides this by two:

L=L_1L_2L_3L_4 =e^{-2.4}\frac{0.4(0.8)^3}{2} =e^{-2.4}(0.1024) =0.009290.

Check. The means are $0.4,0.4,0.8,0.8$, and the observed counts sum to four. The count of two contributes $0.8^2/2!$, not $0.8^2$. Omitting that factorial would double the computed likelihood.

Estimating a Common Poisson Mean

A simple likelihood calculation also shows how maximum likelihood estimation works. Suppose $n$ independent policy counts share one unknown mean $\theta>0$, so each count has a Poisson distribution with mean $\theta$. The likelihood is

L(\theta)=\prod_{i=1}^{n}\frac{e^{-\theta}\theta^{y_i}}{y_i!} =e^{-n\theta}\theta^{\sum_i y_i}\prod_{i=1}^{n}\frac{1}{y_i!}.

Taking logarithms and differentiating with respect to $\theta$ gives the score equation. If the total observed count is positive, setting the derivative equal to zero yields the maximum likelihood estimate:

\ell(\theta)=-n\theta+\left(\sum_i y_i\right)\log\theta-\sum_i\log(y_i!), \qquad \frac{d\ell}{d\theta}=-n+\frac{\sum_i y_i}{\theta}=0, \qquad \widehat{\theta}=\frac{\sum_i y_i}{n}.

This estimate is the arithmetic mean of the policy counts because all policies in this particular model have the same mean. The second derivative is $-(\sum_i y_i)/\theta^2<0$ when the total count is positive, confirming that the stationary point is a maximum. If all observed counts are zero, the likelihood is $e^{-n\theta}$, which decreases as $\theta$ increases; its supremum is approached as $\theta$ tends to zero. There is then no positive interior maximum.

Worked Example: Finding a Common Mean by Likelihood

Four independent policies in a hypothetical book have observed counts $0,1,3,2$. Assume all four counts have the same unknown Poisson mean $\theta>0$. Find the maximum likelihood estimate and evaluate the likelihood at that estimate.

The total observed count is $0+1+3+2=6$, and the number of policies is four. By the common-mean likelihood result,

\widehat{\theta}=\frac{6}{4}=1.5.

To evaluate the likelihood, retain the factorial terms for the observed counts. The individual factorials are $0!=1$, $1!=1$, $3!=6$, and $2!=2$, whose product is $12$. Thus

L(\theta)=\frac{e^{-4\theta}\theta^{0+1+3+2}}{1\cdot1\cdot6\cdot2} =\frac{e^{-4\theta}\theta^6}{12}.

At $\widehat{\theta}=1.5$, the exponential term is $e^{-6}$, and $1.5^6=11.390625$. Therefore

L(1.5)=e^{-6}\frac{11.390625}{12} =e^{-6}(0.94921875) \approx 0.002353.

The log-likelihood at this estimate is $\ell(1.5)=-6+6\log(1.5)-\log(12)\approx-6.052116$. Check. The estimated common mean equals the observed average count, $6/4=1.5$. A frequent mistake is to divide the total count by the total count plus the number of policies; that is not the Poisson maximum likelihood estimate.

Common Mistakes and Practical Checks

Dropping a zero-count policy. When $y_i=0$, the contribution is $e^{-\mu_i}$, which generally is not one. Policies with no claims still affect the likelihood.
Forgetting the factorial. For a count of two, divide by $2!$; for a count of three, divide by $3!$. The factorial is part of the probability mass even if it does not affect estimation of parameters.
Adding rather than multiplying contributions. Under conditional independence, the joint likelihood is the product of the policy-level likelihoods. The log-likelihood is the quantity that adds.
Confusing fitted mean with realized count. A mean may be non-integer; the observed number of claims cannot be. The mean describes the distribution, not a fractional claim outcome.
Using the independence product without justification. If policy counts are dependent conditional on the information in the model, their joint likelihood need not be the product of separate Poisson masses.
Reporting a log-likelihood as a likelihood. A log-likelihood can be negative. The likelihood itself is a nonnegative joint probability mass, while its logarithm is a transformed scale used for calculation and optimization.

Key takeaway: Specify the conditional Poisson model, calculate each policy contribution as $e^{-\mu_i}\mu_i^{y_i}/y_i!$, and multiply across policies only when conditional independence is assumed. The fitted mean is the expected claim count; the likelihood measures how well the specified means assign probability to the observed count vector.

Check Your Understanding

For each problem, assume the stated policy counts are conditionally independent and Poisson. Round likelihoods to six decimal places when needed.

Two policies have observed counts $0$ and $2$, with fitted means $0.25$ and $1.25$. Calculate the joint likelihood, including the factorial for the second count. (Answer: $0.174320$)
Two policies have counts $1$ and $0$, and fitted means $0.6$ and $1.2$. Calculate the joint likelihood. (Answer: $0.099179$)
Three independent policies have observed counts $1,1,2$, and are assumed to share an unknown Poisson mean. Find its maximum likelihood estimate. (Answer: $\widehat{\theta}=4/3$)
For two policies, use $\mu_i=\exp(\log(0.3)+x_i\log(3))$. Their values are $(x_1,y_1)=(0,2)$ and $(x_2,y_2)=(1,1)$. Find both fitted means and the joint likelihood. (Answer: Means $0.3$ and $0.9$; likelihood $0.012198$)
Two independent policies have counts $0$ and $2$, with fitted means $0.5$ and $2.0$. Calculate both the likelihood and log-likelihood, and state which one is negative. (Answer: Likelihood $2e^{-2.5}=0.164170$; log-likelihood $\log(2)-2.5=-1.806853$; the log-likelihood is negative.)

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