Poisson Thinning Under Independent Claim Retention

Why Independent Retention Changes the Frequency Model

A Poisson model may describe the number of claims arising in a period, but an insurer may retain only some of them. For example, a hypothetical insurer might independently decide for each claim whether it is retained or ceded under a reinsurance arrangement. The question is not just how many claims occur: it is also how the total count is divided between the two categories.

When the original count is Poisson and each claim is independently retained with the same probability, the retained and ceded counts are each Poisson. A less immediate result is that they are independent without conditioning on the original total. Conditioning on that total produces a different relationship: except in degenerate cases, the two counts are dependent because they must add to the known total.

The results below concern claim counts only. They do not require assumptions about claim amounts. This frequency calculation is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Definition: Suppose the total claim count \(N\) in a fixed period has a Poisson distribution with parameter \(\lambda>0\). Independently for each claim, retain it with probability \(p\), where \(0\le p\le1\), and cede it with probability \(1-p\). Let \(R\) be the retained count and \(C\) the ceded count. Then \(N=R+C\). The retention decisions are assumed independent of one another and of \(N\), and the retention probability is the same for every claim.

The Thinning Result

Given \(N=n\), each of the \(n\) claims is classified as retained or ceded independently. Therefore \(R\), conditional on \(N=n\), has a binomial distribution with \(n\) trials and success probability \(p\), and \(C=n-R\). To obtain the unconditional distribution of the pair, combine that binomial split with the Poisson probability of the total.

\begin{aligned} P(R=r,C=c) &=P(N=r+c)\,P(R=r\mid N=r+c)\\ &=\frac{e^{-\lambda}\lambda^{r+c}}{(r+c)!} \binom{r+c}{r}p^r(1-p)^c\\ &=\left(\frac{e^{-\lambda p}(\lambda p)^r}{r!}\right) \left(\frac{e^{-\lambda(1-p)}(\lambda(1-p))^c}{c!}\right), \qquad r,c=0,1,2,\ldots . \end{aligned}

The factorization follows because \(\binom{r+c}{r}=(r+c)!/(r!c!)\), and \(e^{-\lambda}=e^{-\lambda p}e^{-\lambda(1-p)}\). Each factor is a Poisson probability mass function. The joint mass therefore factors into a function of \(r\) times a function of \(c\), establishing both the marginal distributions and their independence.

Theorem: Under the assumptions above, \(R\sim\operatorname{Poisson}(\lambda p)\) and \(C\sim\operatorname{Poisson}(\lambda(1-p))\), and \(R\) and \(C\) are independent. In particular, \(E[R]=\operatorname{Var}(R)=\lambda p\) and \(E[C]=\operatorname{Var}(C)=\lambda(1-p)\).

The parameters add to the original Poisson parameter: \(\lambda p+\lambda(1-p)=\lambda\). This is consistent with \(N=R+C\). The theorem also covers the boundary cases \(p=0\) and \(p=1\): one count is then identically zero, and the other equals \(N\). A constant random variable is independent of every random variable.

Worked Example: Calculate Both Count Distributions and a Joint Probability

A hypothetical portfolio has a Poisson claim count with parameter \(\lambda=12\) for a specified period. Each claim is independently retained with probability \(p=0.75\). Find the retained and ceded count distributions and calculate the probability of exactly 8 retained claims and 4 ceded claims.

Method and assumptions. Use Poisson thinning: the original count is Poisson with parameter 12, and each claim has an independent, identical retention decision with probability 0.75. Thus \(\lambda p=12(0.75)=9\) and \(\lambda(1-p)=12(0.25)=3\).

R\sim\operatorname{Poisson}(9),\qquad C\sim\operatorname{Poisson}(3),\qquad R\ \text{and}\ C\ \text{are independent}.

By independence, the requested probability is the product of the two Poisson probabilities:

\begin{aligned} P(R=8,C=4) &=P(R=8)P(C=4)\\ &=\left(e^{-9}\frac{9^8}{8!}\right) \left(e^{-3}\frac{3^4}{4!}\right). \end{aligned}

The intermediate terms are \(9^8/8!=43{,}046{,}721/40{,}320\approx1067.6265\), so \(P(R=8)\approx0.1318\). Also, \(3^4/4!=81/24=3.375\), so \(P(C=4)\approx0.1680\). Their product, using unrounded terms, is approximately \(0.0221\). To four decimal places, the answer is 0.0221.

Check. The counts sum to 12, matching the specified total in this event. The calculation uses the unconditional independent marginals; the common mistake of treating the counts as independent after fixing the total is addressed below.

What Changes When the Total Is Fixed?

The unconditional independence result does not say that the counts remain independent after observing \(N\). If \(N=n\), then \(C=n-R\), so knowing one count determines the other. More formally, for \(n>0\) and \(0<p<1\), the conditional distribution of \(R\) is nonconstant and \(\operatorname{Cov}(R,C\mid N=n)=-np(1-p)<0\). This nonzero conditional covariance rules out conditional independence.

There are important exceptions. If \(n=0\), both conditional counts equal zero. If \(p=0\), \(R=0\); if \(p=1\), \(C=0\). In each of these cases the conditional counts are constants and hence independent. These exceptions matter when stating the result precisely.

R\mid N=n\sim\operatorname{Binomial}(n,p),\qquad C=n-R, \qquad \operatorname{Cov}(R,C\mid N=n)=-np(1-p).

The covariance follows from \(C=n-R\), conditional on \(N=n\). Since adding a constant does not change variance and \(\operatorname{Var}(R\mid N=n)=np(1-p)\), \(\operatorname{Cov}(R,C\mid N=n)=\operatorname{Cov}(R,n-R\mid N=n)=-\operatorname{Var}(R\mid N=n)\).

Worked Example: Split a Known Total

A hypothetical reinsurance review establishes that exactly 10 claims occurred in a period. Each claim was independently retained with probability \(p=0.60\). Find the conditional probability that 7 claims were retained and 3 were ceded. Then assess whether the two counts are conditionally independent.

Method. Because the total is fixed at 10, use a binomial split for the retained count, rather than multiplying unconditional Poisson probabilities. The event of 7 retained claims automatically means 3 ceded claims.

\begin{aligned} P(R=7,C=3\mid N=10) &=\binom{10}{7}(0.60)^7(0.40)^3\\ &=120(0.0279936)(0.064)\\ &=120(0.0017915904)\\ &=0.214990848. \end{aligned}

To four decimal places, the conditional probability is 0.2150. For the dependence check, \(\operatorname{Cov}(R,C\mid N=10)=-10(0.60)(0.40)=-2.4\). Since \(10>0\) and \(0<0.60<1\), this is a nondegenerate case with negative conditional covariance; the counts are not conditionally independent.

Check. The binomial probabilities account for all 10 independent decisions, and the result is between 0 and 1. The conclusion does not apply to \(N=0\) or to retention probabilities 0 or 1, where the conditional counts are constants.

A Joint-Probability Check in Two Ways

A useful way to check an application is to compute a joint event in two ways. Unconditionally, use the independent Poisson marginals. Alternatively, first specify the total count \(N=r+c\), then multiply the Poisson probability of that total by the conditional binomial split probability. The two expressions must agree because the joint probability calculation in the derivation is exactly this decomposition.

Worked Example: Verify a Joint Probability by Conditioning

A hypothetical system has \(N\sim\operatorname{Poisson}(8)\) claims in a period. Each claim is independently retained with probability \(p=0.25\). Calculate the probability of exactly 2 retained claims and 5 ceded claims, first from the independent marginals and then by conditioning on the total.

Method 1: Independent Poisson counts. The retained parameter is \(8(0.25)=2\), and the ceded parameter is \(8(0.75)=6\). Thus

\begin{aligned} P(R=2,C=5) &=\left(e^{-2}\frac{2^2}{2!}\right) \left(e^{-6}\frac{6^5}{5!}\right)\\ &=(2e^{-2})(64.8e^{-6})\\ &\approx(0.2706706)(0.160623)\\ &\approx0.0435. \end{aligned}

Method 2: Condition on the total. The event requires \(N=2+5=7\). The probability of that total is \(P(N=7)=e^{-8}8^7/7!\approx0.1396\). Given \(N=7\), the probability of 2 retained claims is \(\binom{7}{2}(0.25)^2(0.75)^5 =21(0.0625)(0.2373046875)\approx0.3115\). Therefore,

P(R=2,C=5) =P(N=7)P(R=2\mid N=7) \approx(0.1396)(0.3115)\approx0.0435.

Both methods give 0.0435 to four decimal places. The second method uses rounded intermediate values for display; the final value is calculated from the unrounded expressions. The agreement checks both the parameters and the conditioning logic.

Assumptions and Exam Traps

Poisson thinning is a model result, not a rule that applies to every claim allocation. Its assumptions should be matched to the scenario before using it.

One common retention probability. Every claim must have the same probability \(p\) of retention. A count-dependent or claim-specific retention rule need not produce this pair of Poisson distributions.
Independent decisions. Retention decisions for different claims must be independent and independent of the original count. A shared decision affecting all claims can create dependence and invalidate the factorization.
Unconditional versus conditional question. If the original Poisson total is not fixed, the retained and ceded counts are independent. If the total is fixed at \(n>0\) and \(0<p<1\), they are dependent.
Count parameter versus retention probability. The retained Poisson parameter is \(\lambda p\), not \(p\) alone; the ceded parameter is \(\lambda(1-p)\).

Common mistake. A candidate is told that \(N=10\) and \(p=0.60\), then computes \(P(R=7)P(C=3)\) using independent Poisson counts. That calculation incorrectly treats a known total as an unconditioned Poisson model. The correction is \(P(R=7,C=3\mid N=10)=\binom{10}{7}(0.60)^7(0.40)^3=0.2150\). By contrast, if the total were not fixed but \(N\sim\operatorname{Poisson}(\lambda)\), the independent Poisson marginals would be appropriate.

Key takeaway: A Poisson count split by independent, identical claim-level decisions produces independent Poisson counts with parameters \(\lambda p\) and \(\lambda(1-p)\). Once the total is fixed, the split is binomial, and the counts are generally dependent; conditional constant-count cases must be treated separately.

Check Your Understanding

Use the stated assumptions in each problem, and distinguish a fixed total from an unconditioned Poisson total.

A hypothetical portfolio has \(N\sim\operatorname{Poisson}(9)\), with independent retention probability \(p=1/3\). Give the distributions of \(R\) and \(C\), including their parameters. (Answer: \(R\sim\operatorname{Poisson}(3)\), \(C\sim\operatorname{Poisson}(6)\), independent.)
Suppose \(N\sim\operatorname{Poisson}(4)\) and \(p=0.5\), with independent claim-level decisions. Find \(P(R=0,C=0)\), to four decimal places. (Answer: \(e^{-4}\approx0.0183\).)
Exactly 12 claims are observed, and each is independently retained with probability \(0.25\). Find the probability of exactly 3 retained and 9 ceded claims, to four decimal places. (Answer: \(\binom{12}{3}(0.25)^3(0.75)^9\approx0.2581\).)
For an unconditioned model \(N\sim\operatorname{Poisson}(5)\) with \(p=0.20\), find \(P(R=0,C=1)\), to four decimal places. (Answer: \(e^{-1}\left(e^{-4}4\right)=4e^{-5}\approx0.0270\).)
Given \(N=6\) and \(p=0.5\), state the conditional covariance of \(R\) and \(C\), and whether they are conditionally independent. (Answer: \(\operatorname{Cov}(R,C\mid N=6)=-6(0.5)(0.5)=-1.5\); they are not conditionally independent.)

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