Poisson Thinning Across Claim Categories

Why Thinning Matters

A portfolio may have a Poisson-distributed total number of claims, while an actuary also needs counts for particular claim categories. For example, a hypothetical insurer might distinguish claims reported through a mobile application from other claims, or classify claims by a mutually exclusive cause. Poisson thinning describes how the category counts behave when each claim is assigned independently to a category with a fixed probability.

The key result is simple: if the total count is Poisson with mean \(\lambda\), and each claim independently belongs to a specified category with probability \(p\), then the count in that category is Poisson with mean \(\lambda p\). This result concerns counts only; it does not require or describe the amount paid on a claim.

Definition: Let \(N\sim\operatorname{Poisson}(\lambda)\) be the total number of claims in a fixed observation period. Suppose each claim is independently assigned to a specified category with the same probability \(p\), where \(0\le p\le1\), independently of \(N\) and of the assignments of other claims. If \(K\) is the number of claims assigned to that category, then \(K\) is the thinned count.

Deriving the Category Count Distribution

Conditional on there being \(n\) total claims, each claim is an independent category/non-category trial with category probability \(p\). Therefore \(K\mid N=n\sim\operatorname{Binomial}(n,p)\). The unconditional distribution is obtained by averaging that conditional probability over the possible values of \(N\).

\begin{aligned} P(K=k) &=\sum_{n=k}^{\infty}P(K=k\mid N=n)P(N=n)\\ &=\sum_{n=k}^{\infty}\binom{n}{k}p^k(1-p)^{n-k} e^{-\lambda}\frac{\lambda^n}{n!}\\ &=e^{-\lambda}\frac{(\lambda p)^k}{k!} \sum_{m=0}^{\infty}\frac{[\lambda(1-p)]^m}{m!}\\ &=e^{-\lambda}\frac{(\lambda p)^k}{k!} e^{\lambda(1-p)} =e^{-\lambda p}\frac{(\lambda p)^k}{k!}, \qquad k=0,1,2,\ldots \end{aligned}

In the third line, set \(m=n-k\), and use \(\binom{n}{k}/n!=1/[k!(n-k)!]\). The sum is the exponential series. The resulting probability function is that of a Poisson random variable with mean \(\lambda p\), which proves the thinning result.

Theorem: Under the assumptions above, \(K\sim\operatorname{Poisson}(\lambda p)\). In particular, \(E[K]=\operatorname{Var}(K)=\lambda p\), and \(P(K=k)=e^{-\lambda p}(\lambda p)^k/k!\) for nonnegative integers \(k\).

The mean also has a direct interpretation. If there are \(N\) total claims, the conditional expected category count is \(Np\). Averaging over \(N\) gives \(E[K]=pE[N]=\lambda p\). The variance equals the mean as well. By the law of total variance, \(\operatorname{Var}(K)=E[Np(1-p)]+\operatorname{Var}(Np)\). Since \(E[N]=\operatorname{Var}(N)=\lambda\), this becomes \(\lambda p(1-p)+\lambda p^2=\lambda p\).

Several Mutually Exclusive Categories

Suppose every claim is assigned to exactly one of \(r\) mutually exclusive and exhaustive categories. Let category \(j\) have assignment probability \(p_j\), with \(\sum_{j=1}^r p_j=1\), and let \(N_j\) be its claim count. Applying thinning to each category gives \(N_j\sim\operatorname{Poisson}(\lambda p_j)\).

More strongly, the category counts are mutually independent. To see why, for nonnegative integers \(n_1,\ldots,n_r\), the probability of those category counts is

\begin{aligned} P(N_1=n_1,\ldots,N_r=n_r) &=e^{-\lambda}\frac{\lambda^{n_1+\cdots+n_r}} {(n_1+\cdots+n_r)!} \frac{(n_1+\cdots+n_r)!}{\prod_{j=1}^r n_j!} \prod_{j=1}^r p_j^{n_j}\\ &=\prod_{j=1}^r \left(e^{-\lambda p_j}\frac{(\lambda p_j)^{n_j}}{n_j!}\right). \end{aligned}

The first factor in the first line is the probability of the total count; the next factor is the conditional multinomial probability of the category counts given that total. The final expression factors into the separate Poisson probability functions, establishing independence.

This independence result requires the stated partition: every claim receives one and only one category label, and the assignment probabilities are fixed. If labels can overlap—for example, a claim can be both “large” and “weather-related”—the counts need not be independent. The marginal count for either label may still be Poisson under independent assignment, but the joint independence conclusion does not follow.

Worked Example: Claims in a Specified Category

In a hypothetical portfolio, the total number of claims in a quarter is \(N\sim\operatorname{Poisson}(12)\). Independently for each claim, the probability that it is assigned to a specified review category is \(0.25\). Find the category count’s distribution, its probability of being exactly two, and its probability of being at least one.

Method and assumptions. The total count is Poisson with parameter \(\lambda=12\), and each claim is independently assigned to the category with fixed probability \(p=0.25\). By thinning, the category count \(K\) is Poisson with parameter \(\lambda p=12(0.25)=3\).

Thus \(K\sim\operatorname{Poisson}(3)\). The probability of exactly two category claims is

P(K=2)=e^{-3}\frac{3^2}{2!} =e^{-3}\frac{9}{2} =4.5e^{-3} =0.2240.

The probability of at least one is the complement of zero claims:

P(K\ge1)=1-P(K=0) =1-e^{-3}\frac{3^0}{0!} =1-e^{-3} =0.9502.

The category mean is 3, which also equals its variance. Check: the zero-count probability is \(e^{-3}\), so the at-least-one probability is approximately \(1-0.0498=0.9502\). This method avoids treating the total number of claims as fixed.

Worked Example: Counts in Two Categories

A hypothetical service program has \(N\sim\operatorname{Poisson}(10)\) claims in a month. Each claim is assigned to exactly one of categories A, B, or C, with probabilities \(0.20\), \(0.30\), and \(0.50\), respectively. Find the mean count in each category and the probability that category A has no claims while category C has exactly two, with no restriction on B.

Method. Apply thinning to each category. The three Poisson parameters are \(10(0.20)=2\), \(10(0.30)=3\), and \(10(0.50)=5\). Hence the category means are 2, 3, and 5, and the counts are mutually independent because the categories are an exhaustive, mutually exclusive partition.

The required probability is therefore the product of the probability that A is zero and the probability that C is two. Category B is unrestricted, so its probability contribution is 1.

\begin{aligned} P(N_A=0,N_C=2) &=P(N_A=0)P(N_C=2)\\ &=e^{-2}\left(e^{-5}\frac{5^2}{2!}\right)\\ &=12.5e^{-7} =0.0114. \end{aligned}

Check: the category means sum to \(2+3+5=10\), the original total mean. The independence step is valid because each claim is assigned to exactly one of the three categories.

Worked Example: Exactly Two Target Claims and at Least One Other Claim

A hypothetical portfolio has \(N\sim\operatorname{Poisson}(5)\) claims in a period. Each claim independently has probability \(0.40\) of being assigned to a target category; all other claims form the complementary category. Find the probability of exactly two target claims and at least one claim outside the target category.

Method. The target count \(K\) is Poisson with mean \(5(0.40)=2\). The complementary count \(L\) is Poisson with mean \(5(0.60)=3\). These are independent, since the two labels partition all claims.

First calculate the target-count probability:

P(K=2)=e^{-2}\frac{2^2}{2!}=2e^{-2}=0.2707.

Next calculate the probability of at least one complementary claim:

P(L\ge1)=1-P(L=0)=1-e^{-3}=0.9502.

Independence allows the two probabilities to be multiplied:

P(K=2,L\ge1) =(2e^{-2})(1-e^{-3}) =0.2572.

Check: the complementary mean is 3, not \(5-2\) by coincidence alone; it follows from \(5(1-0.40)=3\). This solution uses the category probabilities rather than assuming a fixed total of five claims.

Common Exam Trap: Treating the Total as Fixed

A frequent error is to use a binomial distribution with the Poisson mean as though it were a fixed number of claims. In the first example, this incorrect approach would treat \(N=12\) with certainty and calculate the probability of no category claims as \(0.75^{12}=0.0317\). But 12 is the mean of the random total, not a fixed total. The correct calculation is \(P(K=0)=e^{-12(0.25)}=e^{-3}=0.0498\).

A binomial model is appropriate for the category count conditional on a known fixed total \(N=n\). When that total itself is Poisson, average over its possible values; the result is the thinned Poisson distribution. Do not replace the random total by its mean.

Key takeaway: For a Poisson total count with mean \(\lambda\), independent assignment to a category with fixed probability \(p\) produces a Poisson category count with mean \(\lambda p\). For an exhaustive set of mutually exclusive categories, the resulting counts are independent Poisson variables with means \(\lambda p_j\).

Check Your Understanding

For each problem, use the assumptions given and round probabilities to four decimal places.

A hypothetical count is Poisson with mean 16. Each claim independently has probability \(0.25\) of receiving a particular label. What is the probability of no claims with that label? (Answer: \(e^{-4}=0.0183\))
A total count is Poisson with mean 9, and each claim independently enters category A with probability \(2/3\). Find the probability of exactly four category A claims. (Answer: \(e^{-6}6^4/4!=54e^{-6}=0.1339\))
A Poisson total has mean 14 and is partitioned into three categories with probabilities \(0.10\), \(0.40\), and \(0.50\). Give the three category means. (Answer: \(1.4,\ 5.6,\ 7.0\))
A total count is Poisson with mean 8. Each claim is assigned to category A with probability \(0.25\), and otherwise to category B. Find the probability of at least one A claim and no B claims. (Answer: \(e^{-6}(1-e^{-2})=0.0021\))
A Poisson total has mean 12 and is partitioned into categories A, B, and C with probabilities \(0.25\), \(0.50\), and \(0.25\). Find the probability of exactly two A claims and exactly four B claims, with any number of C claims. (Answer: \((e^{-3}3^2/2!)(e^{-6}6^4/4!)=243e^{-9}=0.0300\))

What You'll Learn