Poisson Thinning

Study guide: Exam MAS-I

Poisson Thinning (splitting) is the result that if each event of a Poisson Process with rate $\lambda$ is independently classified as type 1 with probability $p$ and type 2 with probability $1-p$ , the two type counts are themselves independent Poisson processes with rates $\lambda p$ and $\lambda(1-p)$ .

$N_1(t) \sim \text{Poi}(\lambda p\, t)$

$N_2(t) \sim \text{Poi}\big(\lambda (1-p)\, t\big)$

$N_1(t)$ and $N_2(t)$ are independent of each other, and each has independent, stationary increments. With $k$ types of probabilities $p_1, \ldots, p_k$ , the result gives $k$ independent Poisson processes with rates $\lambda p_i$ .
Independence is the surprising part. Learning that many small claims occurred says nothing about the number of large ones. Contrast conditioning on the total: given $N(t) = n$ , $N_1(t)$ is binomial $(n, p)$ and the two counts are negatively dependent.
Time-dependent classification (Ross): if an event at time $s$ is type $i$ with probability $P_i(s)$ , the $N_i(t)$ are independent Poisson with means $\lambda \int_0^t P_i(s)\,ds$ . Reporting-lag and IBNR counts come from this version.
Thinning with a probability $p(t)$ that varies over time produces a Nonhomogeneous Poisson Process with intensity $\lambda p(t)$ — which is also how such a process is simulated.
Standard uses: claims that pierce a Deductible ( $p = P(X > d)$ ), claims by cause of loss or coverage, losses in a reinsurance layer. The reverse operation is Poisson Superposition, and each type's aggregate amount is its own Compound Poisson Process.

Claims That Pierce a Deductible Losses on a book occur as a Poisson process at $\lambda = 10$ per year. Loss amounts are exponential with mean $\$5{,}000$ , independent of the arrival process, and every policy carries a $\$2{,}000$ deductible.

(a) At what rate do paid claims arrive? (b) What is the probability of no paid claims in a year? (c) This year $4$ losses below the deductible have been recorded. What is the expected number of paid claims this year?
Answer
 $\begin{align*} p &= P(X > 2{,}000) \\ &= e^{-2{,}000/5{,}000} \\ &= 0.67032 \\ \lambda p &= 10 \times 0.67032 \\ &= 6.7032 \end{align*}$ 
(a) Paid claims form a Poisson process at $6.70$ per year.

(b) $P(N_1(1) = 0) = e^{-6.7032} = 0.00123$ .

(c) Still $6.70$ . The small-loss count $N_2$ is independent of $N_1$ , so the four losses under the deductible carry no information about the number of paid claims.

Counting IBNR Claims from Reporting Lags Claims occur as a Poisson process at $\lambda = 20$ per month starting at time $0$ . Each claim's reporting delay is exponential with mean $2$ months, independent of everything else. At $t = 6$ months, find the expected numbers of incurred-but-not-reported and reported claims.
Answer A claim occurring at time $s \leq 6$ is still unreported at $t = 6$ with probability $e^{-(6-s)/2}$ . By time-dependent thinning:
 $\begin{align*} E[N_{\text{IBNR}}(6)] &= 20 \int_0^6 e^{-(6-s)/2}\,ds \\ &= 20 \times 2\left(1 - e^{-3}\right) \\ &= 40 \times 0.95021 \\ &= 38.01 \end{align*}$ 
The reported count has mean $20 \times 6 - 38.01 = 81.99$ . Both are Poisson and independent — under this model, the number of claims already reported says nothing about how many are still to come.