Worked Frequency Problems With Binomial And Poisson Models

Choosing a Claim-Count Model

A frequency problem asks how many claims occur, not how large those claims are. The first task is to identify what generates the count. If there is a fixed number of independent opportunities, each with the same probability of a claim, a binomial model is appropriate. If claims occur over an amount of exposure at a constant rate, a Poisson model is often appropriate. These assumptions determine the model; choosing a distribution simply because its probabilities are convenient is not a sound method.

The examples use hypothetical insurance arrangements. In each case, keep the count’s unit and time period explicit. A probability per policy-year cannot be applied to a number of policies without also accounting for how many years each policy is exposed.

Definition: If \(N\) is the number of claims in a specified period, then \(N\sim\operatorname{Bin}(n,p)\) when there are exactly \(n\) independent claim opportunities and each opportunity has probability \(p\) of producing one claim. Its mean and variance are \(E[N]=np\) and \(\operatorname{Var}(N)=np(1-p)\).

Definition: If \(N\sim\operatorname{Poisson}(\lambda)\), then \(N\) is a count over a specified exposure with expected count \(\lambda\), and \(P(N=k)=e^{-\lambda}\lambda^k/k!\) for \(k=0,1,2,\ldots\). Its mean and variance are both \(\lambda\).

The binomial model assumes the number of opportunities is fixed, the opportunities are independent, and the claim probability is the same for each. It also allows at most one claim per opportunity. If those conditions fail, the binomial formula may not describe the count. The Poisson model assumes a constant claim rate per unit of exposure and independent counts in disjoint exposure intervals. In the usual Poisson-process interpretation, claims occur singly and the count over exposure \(e\) has mean \(\lambda=re\), where \(r\) is the rate per exposure unit.

For either model, the mean gives a useful check on the scale of the count, while the variance is a separate check on the model’s implications. A probability calculation can be arithmetically correct but based on the wrong parameter or wrong assumptions.

Worked Binomial Calculations

Worked Example: Exactly Two Claims From Fixed Opportunities

A hypothetical warranty portfolio covers 10 identical machines for one season. For each machine, the probability of at least one covered claim during the season is \(0.20\), independently of the other machines. Let \(N\) be the number of machines with a claim. Find \(P(N=2)\), \(P(N\leq 1)\), and the mean and variance of \(N\).

Method and model. There are exactly \(n=10\) independent machine-season opportunities, with the same claim probability \(p=0.20\). Each opportunity contributes zero or one to this count, so \(N\sim\operatorname{Bin}(10,0.20)\).

Exactly two claims. Choose which two of the 10 machines have a claim; the other eight have none:

P(N=2)=\binom{10}{2}(0.20)^2(0.80)^8 =45(0.04)(0.16777216) =0.301989888 \approx 0.3020.

At most one claim. This event consists of zero claims or one claim. The probability of zero is \((0.80)^{10}=0.1073741824\). The probability of one is \(10(0.20)(0.80)^9=2(0.134217728)=0.268435456\). Therefore,

P(N\leq 1)=0.1073741824+0.268435456 =0.3758096384\approx 0.3758.

Moment check. The mean is \(np=10(0.20)=2\) claims. The variance is \(np(1-p)=10(0.20)(0.80)=1.6\) claims squared. A mean of 2 is consistent with a substantial probability at two claims, and the variance is less than the mean as expected for this binomial model.

Check. The event probabilities are valid: \(P(N=2)=0.3020\) and \(P(N\leq1)=0.3758\), so their sum is less than 1. The common mistake this method avoids is treating “at most one” as only “exactly one” and omitting the zero-claim probability.

Poisson Counts and Exposure

For a Poisson model, exposure determines the mean count. If the claim rate is \(r\) claims per policy-year and the portfolio contributes \(e\) policy-years, then \(\lambda=re\). For \(m\) identical policies each observed for \(t\) years, \(e=mt\) policy-years and \(\lambda=rmt\). The time unit in the rate and the time unit in the exposure must agree.

Exposure rule: Under a constant Poisson claim rate \(r\) per exposure unit, total exposure \(e\) gives \(N\sim\operatorname{Poisson}(re)\). If independent groups have Poisson counts with means \(\lambda_1,\ldots,\lambda_j\), their combined count is Poisson with mean \(\lambda_1+\cdots+\lambda_j\).

The additivity result explains why the exposure rule works across independent periods. For two independent counts \(N_1\sim\operatorname{Poisson}(\lambda_1)\) and \(N_2\sim\operatorname{Poisson}(\lambda_2)\), the probability of a total of \(k\) is obtained by summing over all possible splits of that total:

\begin{aligned} P(N_1+N_2=k) &=\sum_{j=0}^{k}P(N_1=j)P(N_2=k-j)\\ &=e^{-(\lambda_1+\lambda_2)} \sum_{j=0}^{k}\frac{\lambda_1^j\lambda_2^{k-j}}{j!(k-j)!}\\ &=e^{-(\lambda_1+\lambda_2)} \frac{(\lambda_1+\lambda_2)^k}{k!}. \end{aligned}

The last equality follows from the binomial theorem after multiplying the sum by \(k!/k!\). This is the probability mass function of a Poisson random variable with mean \(\lambda_1+\lambda_2\). Thus, independent time periods or groups can be combined by adding their expected counts, provided the Poisson assumptions hold.

Worked Example: Adjusting a Poisson Rate for Partial-Year Exposure

A hypothetical insurer models claims for a particular type of equipment at \(0.06\) claims per equipment-year. It observes 25 pieces of equipment for six months each. Assuming a constant rate and independent Poisson claim counts, find the probability of at least two claims and the mean and variance of the total count \(N\).

Method and exposure conversion. Six months is \(0.5\) year, so total exposure is \(25(0.5)=12.5\) equipment-years. The Poisson mean is \(\lambda=(0.06)(12.5)=0.75\). Hence \(N\sim\operatorname{Poisson}(0.75)\).

Probability calculation. Use the complement of zero or one claim:

\begin{aligned} P(N\geq2) &=1-P(N=0)-P(N=1)\\ &=1-e^{-0.75}-e^{-0.75}(0.75)\\ &=1-e^{-0.75}(1.75). \end{aligned}

Using \(e^{-0.75}=0.4723665527\) to the displayed precision, \(e^{-0.75}(1.75)=0.8266414672\). Therefore, \(P(N\geq2)=1-0.8266414672=0.1733585328\), or \(0.1734\) to four decimal places.

Moment check. The Poisson mean and variance are both \(0.75\). The probability of at least two claims, approximately \(0.1734\), is plausible for a count whose expected value is less than one; most outcomes are zero or one.

Check. The probability is between 0 and 1, and the complement calculation includes both excluded counts, zero and one. The common mistake avoided is using 25 as the exposure, which would incorrectly treat each piece of equipment as observed for a full year.

Worked Example: A Poisson Probability and Its Moment Check

A hypothetical roadside-assistance program covers 20 vehicles for one year. Assume each vehicle’s claims follow a Poisson model with rate \(0.15\) claims per vehicle-year, independently across vehicles. Find the probability of exactly four claims in total, and verify the count’s mean and variance.

Method and parameter. Total exposure is \(20(1)=20\) vehicle-years. Therefore the total count is Poisson with \(\lambda=(0.15)(20)=3\).

Exactly four claims.

\begin{aligned} P(N=4) &=e^{-3}\frac{3^4}{4!}\\ &=e^{-3}\frac{81}{24}\\ &=e^{-3}(3.375)\\ &=0.1680313557\ldots\\ &\approx 0.1680. \end{aligned}

Moment check. Since \(N\sim\operatorname{Poisson}(3)\), \(E[N]=3\) and \(\operatorname{Var}(N)=3\). Four claims is one above the expected count, so a probability near \(0.1680\) is reasonable. As an additional arithmetic check, the Poisson probabilities at 3 and 4 satisfy \(P(N=4)=P(N=3)\lambda/4\), because the ratio of successive Poisson probabilities is \(\lambda/(k+1)\). Here \(P(N=3)=e^{-3}(27/6)=0.2240418077\ldots\), and multiplying by \(3/4\) gives \(0.1680313557\ldots\).

Check. The probability is below 1, and the adjacent-probability ratio agrees with the direct calculation. The common mistake avoided is using \(4!\) as \(4\), rather than \(24\), or rounding an intermediate value too early.

Model Choice and Common Traps

A fixed number of policy periods does not automatically imply a binomial model. The binomial model requires a fixed number of opportunities with at most one claim per opportunity and the same probability on every opportunity. A Poisson model is appropriate when claims are modeled as events over exposure at a constant rate. For example, an opportunity defined as “a policy has any claim during the year” is binary, whereas a count of all claims during policy-years can exceed one and may be modeled as Poisson.

The models also make different moment predictions. For a binomial count, \(\operatorname{Var}(N)=np(1-p)<np=E[N]\) when \(0<p<1\). For a Poisson count, \(\operatorname{Var}(N)=E[N]=\lambda\). These relationships are useful checks, not permission to choose a model solely by comparing a single observed count with a mean.

Exam trap: Do not multiply a per-year rate by the number of contracts alone if each contract has less or more than one year of exposure. First calculate total exposure in the rate’s time unit, then multiply by the rate. Also distinguish “exactly \(k\)” from “at least \(k\)”: for a tail probability, add the relevant probabilities or use a correct complement.

Define the count.
State what qualifies as one claim and specify the observation period and exposure unit.

Match assumptions to the model.
Use binomial for a fixed number of independent binary opportunities with common probability; use Poisson for a constant event rate over exposure under the stated Poisson assumptions.

Calculate the parameter.
For binomial, identify \(n\) and \(p\). For Poisson, convert all exposure to matching units and calculate \(\lambda=re\).

Compute and check.
Evaluate the requested probability, then verify the result is between 0 and 1 and that the model’s mean and variance match the parameterization.

Check Your Understanding

For each hypothetical situation, identify the model, calculate the requested quantity, and use the stated moments or exposure to check the parameter.

A device portfolio has 12 independent units, each with probability \(0.10\) of one or more claims during the coverage period. If \(N\) is the number of units with claims, find \(P(N=0)\), \(E[N]\), and \(\operatorname{Var}(N)\). (Answer: \(N\sim\operatorname{Bin}(12,0.10)\); \(P(N=0)=0.2824\), \(E[N]=1.2\), \(\operatorname{Var}(N)=1.08\).)
A hypothetical repair program has a Poisson rate of \(0.08\) claims per machine-year and covers 15 machines for six months each. Find the probability of at least one claim. (Answer: \(\lambda=0.08(15)(0.5)=0.6\); \(P(N\geq1)=1-e^{-0.6}=0.4512\).)
Eight independent service contracts each have probability \(0.25\) of a claim in the term, with at most one claim counted per contract. Find the probability of exactly one contract having a claim. (Answer: \(N\sim\operatorname{Bin}(8,0.25)\); \(P(N=1)=8(0.25)(0.75)^7=0.2670\).)
A hypothetical portfolio’s claim count is Poisson with mean \(2.4\). Find the probability of exactly two claims and state its mean and variance. (Answer: \(P(N=2)=e^{-2.4}(2.4^2/2)=0.2613\); mean \(=2.4\), variance \(=2.4\).)
A claim process has a constant rate of \(0.12\) claims per device-year. Ten devices are observed for three months each. Find the probability of no claims and give the count’s mean and variance. (Answer: \(\lambda=0.12(10)(0.25)=0.3\); \(P(N=0)=e^{-0.3}=0.7408\); mean \(=0.3\), variance \(=0.3\).)

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