Poisson Count Examples In Insurance

From a Claim Rate to Claim Probabilities

A claim frequency model turns an expected count into probabilities for possible claim counts. In ratemaking, the calculation must match the exposure being priced: a rate stated per policy-year cannot be applied directly to a six-month exposure, and the expected number of claims in a portfolio is not itself the probability that the portfolio has a claim.

The Poisson distribution is used for a nonnegative integer claim count when events are modeled as occurring independently at a constant rate over the specified exposure. For a single policy or other portfolio unit, write $N$ for the number of claims and $\lambda$ for its expected claim count during the exposure period. The parameter $\lambda$ is nonnegative and is measured in expected claims for that period.

Definition: If $N\sim\operatorname{Poisson}(\lambda)$, then for $k=0,1,2,\ldots$, $$ P(N=k)=e^{-\lambda}\frac{\lambda^k}{k!}. $$ The mean is $E[N]=\lambda$. Under a constant claim rate $r$ per exposure-year and exposure $t$ exposure-years, the parameter is $\lambda=rt$.

The rate-exposure relationship assumes the same modeled rate applies throughout the stated exposure and that expected counts scale proportionally with exposure. For example, if the rate is $0.20$ claims per policy-year, then half a policy-year has expected count $0.20(0.5)=0.10$. This is a frequency calculation only; interest and mortality assumptions are not applicable.

Useful Poisson Probabilities

Substituting $k=0$, $k=1$, or $k=2$ into the probability formula gives the most common count probabilities. The probability of at least one claim follows by taking the complement of zero claims.

\begin{aligned} P(N=0)&=e^{-\lambda},\\ P(N=1)&=\lambda e^{-\lambda},\\ P(N=2)&=\frac{\lambda^2}{2}e^{-\lambda},\\ P(N\geq 1)&=1-e^{-\lambda}. \end{aligned}

More generally, to find the probability of at least $m$ claims, subtract the probabilities below $m$ from one:

P(N\geq m)=1-\sum_{k=0}^{m-1}e^{-\lambda}\frac{\lambda^k}{k!}.

The complement formula is often the quickest way to calculate “at least one.” It also makes a useful reasonableness check: as $\lambda$ increases, $e^{-\lambda}$ decreases, so the probability of at least one claim increases.

Why Exposure and Portfolio Aggregation Work

If the modeled claim rate is $r$ per exposure-year and the unit has $t$ exposure-years, the expected count is $rt$. The Poisson model then uses this expected count as its parameter. This is not a claim that every unit will have $rt$ claims; individual outcomes are integers, while $rt$ is their probability-weighted average.

Independent Poisson counts can also be added. If $N_1$ and $N_2$ are independent Poisson variables with parameters $\lambda_1$ and $\lambda_2$, then their total is Poisson with parameter $\lambda_1+\lambda_2$. For a total of $k$ claims, the convolution is

\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 factoring out $1/k!$, the sum has terms $\binom{k}{j}\lambda_1^j\lambda_2^{k-j}$. The result is the Poisson probability formula with parameter $\lambda_1+\lambda_2$. Repeating this argument shows that the sum of any finite collection of independent Poisson counts is Poisson with parameter equal to the sum of their parameters.

This aggregation result requires independent Poisson counts. If dependence or a different count distribution is present, the result cannot be applied automatically. The impact of count variation beyond the Poisson model is a separate issue.

Worked Example: Zero, One, Two, or More Claims

A hypothetical policy has $0.8$ policy-years of exposure. Its constant modeled claim rate is $1.20$ claims per policy-year. Find the expected claim count and the probabilities of zero, one, two, and at least three claims. Use a Poisson count model. Interest and mortality assumptions are not applicable.

Step 1: Find the Poisson parameter. The exposure-adjusted parameter is

\lambda=rt=(1.20)(0.8)=0.96.

Thus $N\sim\operatorname{Poisson}(0.96)$, and $E[N]=0.96$ claims.

Step 2: Calculate zero claims.

P(N=0)=e^{-0.96}=0.382892886\ldots\approx 0.382893.

Step 3: Calculate exactly one claim.

P(N=1)=e^{-0.96}\frac{0.96^1}{1!} =(0.382892886\ldots)(0.96) =0.367577171\ldots\approx 0.367577.

Step 4: Calculate exactly two claims.

\begin{aligned} P(N=2) &=e^{-0.96}\frac{0.96^2}{2!}\\ &=(0.382892886\ldots)\frac{0.9216}{2}\\ &=(0.382892886\ldots)(0.4608)\\ &=0.176437042\ldots\approx 0.176437. \end{aligned}

Step 5: Find at least three claims by a complement. The probabilities for zero, one, and two claims sum to $0.382893+0.367577+0.176437=0.926907$, using the displayed rounded values. Using unrounded values,

\begin{aligned} P(N\geq 3) &=1-P(N=0)-P(N=1)-P(N=2)\\ &=1-0.382892886\ldots-0.367577171\ldots-0.176437042\ldots\\ &=0.073092902\ldots\approx 0.073093. \end{aligned}

The four probabilities add to $1.000000$ at six decimal places, as required. The method avoids treating the expected count $0.96$ as a probability; it is the distribution parameter and mean, not the chance of a claim.

Worked Example: A Portfolio With Six Months of Exposure

A hypothetical portfolio contains 40 independent policies, each with six months of exposure. The modeled rate for each policy is $0.15$ claims per policy-year. Find the portfolio’s expected claim count, the probability of no claims, the probability of exactly one claim, and the probability of at least one claim. Assume the policy counts are independent Poisson variables. Interest and mortality assumptions are not applicable.

Step 1: Calculate the parameter for one policy. Six months is $0.5$ policy-year, so each policy has parameter

\lambda_{\text{policy}}=(0.15)(0.5)=0.075.

Step 2: Aggregate the parameters. By independence and the Poisson sum result, the portfolio count is Poisson with parameter

\lambda_{\text{portfolio}}=40(0.075)=3.00.

E[N_{\text{portfolio}}]=3.00\text{ claims}.

Step 3: Calculate the requested probabilities.

\begin{aligned} P(N_{\text{portfolio}}=0)&=e^{-3}=0.049787068\ldots\approx 0.0498,\\ P(N_{\text{portfolio}}=1)&=3e^{-3}=0.149361205\ldots\approx 0.1494,\\ P(N_{\text{portfolio}}\geq 1)&=1-e^{-3}=0.950212932\ldots\approx 0.9502. \end{aligned}

The expected count is 3.00, while the probability of at least one claim is approximately 0.9502. The common exposure trap is to use the full annual rate for each six-month policy. The correction is to multiply $0.15$ by $0.5$ for each policy before aggregating.

Worked Example: Combining Two Risk Groups

A hypothetical equipment program has two independent groups. Group A contains 12 units, each with $0.5$ exposure-year at a rate of $0.20$ claims per exposure-year. Group B contains 8 units, each with $1.0$ exposure-year at a rate of $0.10$ claims per exposure-year. Find the expected total number of claims, the probability of at least one claim, and the probability of exactly two claims. Assume independent Poisson counts for all units. Interest and mortality assumptions are not applicable.

Step 1: Calculate each group’s parameter. Group A has parameter

\lambda_A=12(0.5)(0.20)=1.20.

Group B has parameter

\lambda_B=8(1.0)(0.10)=0.80.

Step 2: Aggregate the independent counts.

\lambda_{\text{total}}=\lambda_A+\lambda_B=1.20+0.80=2.00, \qquad E[N_{\text{total}}]=2.00.

Step 3: Calculate the probabilities. The total count is Poisson with parameter $2.00$, so

\begin{aligned} P(N_{\text{total}}\geq 1) &=1-e^{-2}=0.864664717\ldots\approx 0.8647,\\ P(N_{\text{total}}=2) &=e^{-2}\frac{2^2}{2!} =2e^{-2}\\ &=0.270670566\ldots\approx 0.2707. \end{aligned}

The calculation checks because the group parameters sum to the expected total count, and the probability of at least one claim is the complement of the zero-claim probability. Do not average the two group rates without accounting for each group’s size and exposure.

Common Exam Traps and Checks

The most frequent errors arise before the Poisson formula is used: misreading the exposure period, confusing an expected count with a probability, or aggregating group rates instead of group parameters. Apply the rate to the appropriate exposure first, then use the resulting parameter in the probability calculation.

For example, for 40 policies each exposed for six months at $0.15$ claims per policy-year, the incorrect calculation $40(0.15)=6.00$ treats every policy as having a full year of exposure. The corrected calculation is

$$ 40(0.15)(0.5)=3.00. $$

Another error is to report $P(N\geq1)=\lambda$. For $\lambda=3$, this would give 3, which cannot be a probability. The correct result is $1-e^{-3}=0.9502$. As a final check, probabilities must be between 0 and 1, and probabilities for all possible count values must sum to 1.

Key takeaway: Convert each rate to an exposure-adjusted Poisson parameter, add parameters only for independent Poisson counts, and use the Poisson probability formula for exact counts. For at least one claim, use $1-e^{-\lambda}$.

Check Your Understanding

For each problem, use the stated Poisson model and retain full precision until the final probability, which should be reported to four decimals.

A hypothetical policy has $0.5$ exposure-year at a rate of $1.00$ claim per exposure-year. Find its expected count, probability of zero claims, probability of exactly one claim, and probability of at least one claim. (Answer: $E[N]=0.50$, $P(N=0)=0.6065$, $P(N=1)=0.3033$, $P(N\geq1)=0.3935$.)
A hypothetical policy has five months of exposure at a rate of $0.30$ claims per policy-year. Find the Poisson parameter and the probability of zero claims. (Answer: $\lambda=0.125$, $P(N=0)=0.8825$.)
There are 15 independent hypothetical units, each with $0.4$ exposure-year at a rate of $0.50$ claims per exposure-year. Find the expected portfolio count and the probability of at least one claim. (Answer: $E[N]=3.00$, $P(N\geq1)=0.9502$.)
Two independent hypothetical groups have parameters obtained as follows: 6 units each have $1.5$ exposure-years at rate $0.20$, and 4 units each have $2.0$ exposure-years at rate $0.30$. Find the total parameter and probability of no claims. (Answer: $\lambda=4.20$, $P(N=0)=e^{-4.2}=0.0150$.)
Twenty independent hypothetical policies each have three months of exposure at a rate of $0.25$ claims per policy-year. Find the probability of exactly one claim and the probability of at least two claims. (Answer: $\lambda=1.25$, $P(N=1)=0.3581$, $P(N\geq2)=1-e^{-1.25}(1+1.25)=0.3554$.)

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