Poisson Point Probabilities

Why Poisson Point Probabilities Matter

A frequency model describes the number of claims in a specified period or other unit of exposure. When a Poisson model is appropriate, it gives a direct way to calculate the probability of a particular count: for example, the probability of no claims during a coverage period or exactly two claims in a portfolio. The essential first step is to identify the model parameter for the exposure in the question; a rate by itself is not yet that parameter.

This tutorial focuses on probabilities for specified claim counts, as well as the closely related probabilities of no claims and at least one claim. It is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Definition: A claim count $N$ has a Poisson distribution with parameter $\lambda$, written $N\sim\operatorname{Poisson}(\lambda)$, if $N$ takes values in the nonnegative integers. For $\lambda>0$, its probability mass function is $$ P(N=k)=e^{-\lambda}\frac{\lambda^k}{k!},\qquad k=0,1,2,\ldots. $$ For $\lambda=0$, the distribution is degenerate at zero: $P(N=0)=1$, and $P(N=k)=0$ for $k>0$. Equivalently, the formula may be used at $\lambda=0$ with the convention $0^0=1$. The parameter $\lambda$ is the expected count over the specified exposure.

The count $k$ must be a nonnegative integer. The factorial $k!$ means the product $1\cdot2\cdots k$, with $0!=1$. Thus, when $\lambda>0$, the first point probabilities are

P(N=0)=e^{-\lambda},\qquad P(N=1)=\lambda e^{-\lambda},\qquad P(N=2)=\frac{\lambda^2}{2}e^{-\lambda}.

The probability of at least one claim follows by taking the complement of zero claims:

P(N\geq 1)=1-P(N=0)=1-e^{-\lambda}.

This complement is useful because it requires only one point probability. In contrast, a question asking for a general range of counts involves adding the relevant point probabilities or using an appropriate complement; that is a separate calculation from a single point probability.

Setting the Parameter From Rate and Exposure

Suppose claims occur according to a homogeneous Poisson process with constant rate $r$ per unit of exposure. The standard assumptions are that claim occurrences in disjoint exposure intervals are independent and that the expected number of claims in an interval is proportional to its length, with proportionality constant $r$. For exposure $t$, the Poisson parameter is

\lambda=rt.

The units in $r$ and $t$ must match. For example, a rate stated per policy-year must be multiplied by the number of policy-years, not by an unconverted number of months. A Poisson model can also be used directly when a problem supplies $\lambda$ as the expected count for the specified period.

Important distinction: The rate $r$ describes expected claims per unit of exposure; $\lambda$ is the expected count for the particular exposure being studied. Use $\lambda=rt$ only after expressing $t$ in the same exposure units as the rate.

Why the Poisson Formula Has This Form

One justification comes from the binomial model for a large number of small, independent opportunities for a claim. Suppose there are $n$ opportunities, each with claim probability $p=\lambda/n$, so the expected count $np$ remains $\lambda$. For a fixed nonnegative integer $k$, the binomial probability is

P(N=k)=\binom{n}{k}\left(\frac{\lambda}{n}\right)^k \left(1-\frac{\lambda}{n}\right)^{n-k}.

As $n$ grows, the first factors satisfy $\binom{n}{k}(\lambda/n)^k\to\lambda^k/k!$, while $(1-\lambda/n)^{n-k}\to e^{-\lambda}$. Therefore the limiting probability is $e^{-\lambda}\lambda^k/k!$, the Poisson point probability. This argument explains why the Poisson model is associated with many small, independent opportunities and a fixed expected count. It does not mean that every claim process is Poisson: the model is inappropriate when the relevant assumptions are not reasonable.

Calculating Point Probabilities

For a specified count, substitute the parameter and the count into the probability mass function. Keep the factorial attached to the count, and use the same $\lambda$ for every point probability associated with the same exposure. The examples show the steps for direct parameters, rate-based parameters, and independent groups.

Worked Example: Zero, One, or More Claims

A hypothetical block of policies has a Poisson claim count $N$ over a defined coverage period with parameter $\lambda=2.4$. Calculate the probabilities of no claims, exactly one claim, and at least one claim. Give probabilities to four decimal places.

Method. The specified parameter already applies to the full coverage period, so no rate conversion is needed. First compute $e^{-2.4}=0.0907179533\ldots$.

For no claims, use $k=0$ and $0!=1$:

P(N=0)=e^{-2.4}\frac{2.4^0}{0!} =0.0907179533\ldots \approx 0.0907.

For exactly one claim, use $k=1$ and $1!=1$:

P(N=1)=e^{-2.4}\frac{2.4^1}{1!} =0.0907179533\ldots\times2.4 =0.2177230879\ldots \approx 0.2177.

At least one claim is the complement of no claims:

P(N\geq1)=1-P(N=0) =1-0.0907179533\ldots =0.9092820467\ldots \approx 0.9093.

Check. The no-claims and at-least-one probabilities sum to $0.0907179533\ldots+0.9092820467\ldots=1$. The exactly-one probability is less than the at-least-one probability, as it must be because exactly one is included among the outcomes with at least one claim. A common mistake is to treat “at least one” as “exactly one”; the complement calculation correctly includes every positive count.

Worked Example: Converting Monthly Exposure

A hypothetical warranty portfolio has a constant Poisson claim rate of $0.30$ claims per covered contract-month. A contract is observed for 10 months. Calculate the probability of no claims and the probability of exactly two claims during that contract’s observation period. Give probabilities to four decimal places.

Method. The rate is per contract-month, so multiply it by the 10 contract-months:

\lambda=rt=0.30\times10=3.00.

The period count is therefore modeled as $N\sim\operatorname{Poisson}(3)$. For no claims,

P(N=0)=e^{-3}\frac{3^0}{0!} =e^{-3} =0.0497870684\ldots \approx 0.0498.

For exactly two claims, $2!=2$, and $3^2=9$:

P(N=2)=e^{-3}\frac{3^2}{2!} =e^{-3}\frac{9}{2} =0.0497870684\ldots\times4.5 =0.2240418077\ldots \approx 0.2240.

Check. Both results use the 10-month parameter $\lambda=3$, not the one-month rate $0.30$. The probability of exactly two claims is $4.5$ times the probability of no claims, consistent with the ratio $P(N=2)/P(N=0)=3^2/2!=4.5$. A common error is to substitute the rate $0.30$ directly into the formula for a 10-month count, which would answer a different question.

Worked Example: Combining Independent Claim Groups

Two hypothetical, independent service regions are observed for one year. Region A has a Poisson claim count with parameter $0.6$; Region B has a Poisson claim count with parameter $0.8$. Calculate the probability that the combined regions have exactly two claims and the probability they have at least one claim. Give probabilities to four decimal places.

Method. The sum of independent Poisson counts is Poisson with parameter equal to the sum of their parameters. If $N_A$ and $N_B$ are the regional counts, then

N=N_A+N_B\sim\operatorname{Poisson}(0.6+0.8) =\operatorname{Poisson}(1.4).

For exactly two claims, $e^{-1.4}=0.2465969639\ldots$, $1.4^2=1.96$, and $2!=2$:

P(N=2)=e^{-1.4}\frac{1.4^2}{2!} =0.2465969639\ldots\times0.98 =0.2416650246\ldots \approx 0.2417.

For at least one claim, calculate the zero-claim probability for the combined count and subtract it from one:

P(N\geq1)=1-e^{-1.4} =1-0.2465969639\ldots =0.7534030361\ldots \approx 0.7534.

Check. The combined parameter is $1.4$, not either region’s parameter alone. Also, independence is needed for the sum to have this Poisson distribution; merely knowing the two marginal parameters is not enough if the regional counts may be dependent. A common mistake is to calculate a point probability for each region separately and treat one of those as the combined portfolio probability.

Common Exam Traps

Using a rate instead of a period parameter. Convert $r$ to $\lambda=rt$ for the stated exposure, keeping the time units consistent.
Dropping the factorial. For $k=2$, the probability includes division by $2!$; for $k=3$, it includes division by $3!$. The factorial is part of the point-probability formula.
Confusing “exactly” and “at least.” $P(N=1)$ is one point probability. $P(N\geq1)$ includes all positive counts and equals $1-P(N=0)$.
Ignoring the zero-parameter case. When $\lambda=0$, there are zero claims with probability one and every positive count has probability zero. Do not describe positive counts as possible with positive probability in this case.
Adding parameters without checking independence. Independent Poisson counts have a Poisson sum with the parameters added. Dependence can invalidate that result.

Key takeaway: Identify the count and its exposure, determine the corresponding $\lambda$, then apply $P(N=k)=e^{-\lambda}\lambda^k/k!$ for $\lambda>0$. Handle $\lambda=0$ as a degenerate count, and use $1-e^{-\lambda}$ for the probability of at least one claim.

Check Your Understanding

For each question, identify the relevant Poisson parameter before calculating. Round probabilities to four decimal places.

A hypothetical portfolio has a Poisson claim count with parameter $1.5$ for a specified period. Find the probability of no claims and the probability of exactly one claim. (Answer: $P(N=0)=0.2231$; $P(N=1)=0.3347$.)
A device has a constant Poisson claim rate of $0.25$ claims per month and is observed for eight months. Find the probability of exactly two claims. (Answer: $0.2707$.)
A hypothetical policy group has $N\sim\operatorname{Poisson}(4.5)$. Find the probabilities of zero claims and exactly one claim. (Answer: $P(N=0)=0.0111$; $P(N=1)=0.0500$.)
Two independent hypothetical groups have Poisson claim counts with parameters $0.7$ and $1.3$. Find the probability of exactly four claims in the combined groups. (Answer: $0.0902$.)
A hypothetical coverage block has a constant claim rate of two claims per 30 days and is observed for 18 days. Find the probability of at least one claim. (Answer: $0.6988$.)

What You'll Learn