Poisson Superposition

Study guide: Exam MAS-I

Poisson Superposition is the result that the sum of independent Poisson processes is itself a Poisson process whose rate is the sum of the individual rates — merging independent claim streams produces one Poisson stream.

$N(t) = N_1(t) + N_2(t) + \cdots + N_k(t)$

$\lambda = \lambda_1 + \lambda_2 + \cdots + \lambda_k$

The component processes must be independent. The merged process inherits independent increments, and a sum of independent Poisson counts is Poisson with the summed mean.
Labelling: each event of the merged process came from stream $i$ with probability $\lambda_i/\lambda$ , independently of the other events and of the event times. Superposition run backwards is Poisson Thinning.
First arrival: the wait for the first event of any type is exponential with rate $\lambda$ (mean $1/\lambda$ ), and that event is of type $i$ with probability $\lambda_i/\lambda$ — the competing-exponentials result on Interarrival Time.
Given $N(t) = n$ merged events, the number from stream $i$ is binomial $(n, \lambda_i/\lambda)$ .
$n$ of type 1 before $m$ of type 2: this happens exactly when at least $n$ of the first $n+m-1$ merged events are type 1, so with $p = \lambda_1/(\lambda_1 + \lambda_2)$ :

$\sum_{k=n}^{n+m-1} \binom{n+m-1}{k} p^k (1-p)^{n+m-1-k}$

Streams driven by a common shock (two lines hit by the same storm) are not independent, and their sum need not be Poisson. For amounts, merged compound Poisson streams stay compound Poisson with a rate-weighted mixture of severities.

Merging Three Auto Coverages On a block of auto policies, collision claims arrive at $3$ per month, comprehensive at $2$ per month and liability at $1$ per month, as independent Poisson processes.

(a) Find the probability of no claims of any kind in a week ( $t = 1/4$ month). (b) Find the probability that the next claim is a liability claim. (c) Given $8$ claims in a month, find the distribution and mean of the number of collision claims.

Answer The merged process is Poisson with $\lambda = 3 + 2 + 1 = 6$ per month.

(a) $P(N(1/4) = 0) = e^{-6/4} = e^{-1.5} = 0.2231$ .

(b) $\lambda_{\text{liab}}/\lambda = 1/6 = 0.1667$ .

(c) Each of the $8$ claims is collision with probability $3/6 = 0.5$ , independently, so the collision count is $\text{Binomial}(8, 0.5)$ with mean $4$ .

Which Large-Loss Stream Hits Twice First? Large property losses arrive at $3$ per year and large casualty losses at $1$ per year, as independent Poisson processes. What is the probability that the insurer sees its second large property loss before its second large casualty loss?
Answer Each merged large loss is property with probability $p = 3/(3+1) = 0.75$ . Two property before two casualty means at least $2$ property among the first $n + m - 1 = 3$ large losses:
 $\begin{align*} P &= \binom{3}{2}(0.75)^2(0.25) + \binom{3}{3}(0.75)^3 \\ &= 0.421875 + 0.421875 \\ &= 0.84375 \end{align*}$ 
There is about an $84\%$ chance the property stream reaches two losses first.