Compound Poisson Mixtures

Why a Compound Poisson Mixture Is Useful

An aggregate loss depends on two sources of randomness: how many claims occur and how large each claim is. When the claim count is Poisson and severities have finite support, conditioning on the count separates these sources. For each possible count, calculate the probability of the required severity sum; then weight it by the probability of that count.

This approach is especially practical for exact aggregate probabilities and cumulative probabilities on a lattice. It also makes the zero-claim outcome explicit, so it is easier to see whether a probability at zero has been included or omitted.

Definition: Let $N\sim\operatorname{Poisson}(\lambda)$, where $\lambda>0$, and let $X_1,X_2,\ldots$ be independent, identically distributed, nonnegative claim severities, independent of $N$. The compound Poisson aggregate loss is $$ S=\sum_{j=1}^{N}X_j. $$ When $N=0$, the sum is defined to be zero. Conditional on $N=n$, the aggregate is the sum of $n$ independent severities; for $n=0$, it is zero with probability 1.

The Poisson-Weighted Mixture

The Poisson probability mass function is $P(N=n)=e^{-\lambda}\lambda^n/n!$, for $n=0,1,2,\ldots$. Write $q_n(s)$ for the conditional probability that the sum of $n$ severities equals $s$:

q_n(s)=P\left(\sum_{j=1}^{n}X_j=s\right),\qquad q_0(s)= \begin{cases} 1,&s=0,\\ 0,&s\ne 0. \end{cases}

Conditioning on the claim count gives the mixture formula

P(S=s)=\sum_{n=0}^{\infty}P(N=n)q_n(s) =e^{-\lambda}\sum_{n=0}^{\infty}\frac{\lambda^n}{n!}q_n(s).

The same argument applies to a cumulative probability. If $r_n(x)=P(\sum_{j=1}^{n}X_j\le x)$, with the empty sum treated as zero, then

P(S\le x)=e^{-\lambda}\sum_{n=0}^{\infty}\frac{\lambda^n}{n!}r_n(x).

These formulas are an application of the law of total probability: the events $N=0,N=1,\ldots$ are disjoint and cover all possibilities. Within each event $N=n$, the relevant aggregate probability is the conditional severity-sum probability. The Poisson weight is therefore applied to the entire conditional probability for that count.

Zero-claim probability: If severities are strictly positive, then $S=0$ exactly when $N=0$, so $P(S=0)=e^{-\lambda}$. If severity can be zero, positive claim counts may also produce $S=0$; in that case the zero-aggregate probability is not just $e^{-\lambda}$.

When Only Finitely Many Counts Matter

Suppose every claim severity is at least $m>0$. Then $n$ claims produce an aggregate of at least $nm$. For an exact target $s$, counts $n>s/m$ cannot contribute to $P(S=s)$. For a cumulative event $S\le x$, counts $n>x/m$ cannot contribute either. Thus, although the Poisson mixture is written as an infinite sum, only finitely many terms need to be evaluated for these events.

This finite cutoff depends on the positive lower bound for severity. If zero severity is possible, arbitrarily many claims may still have a sum below a fixed threshold, so the relevant mixture may retain infinitely many terms. A finite-support severity distribution by itself does not rule out zero severity.

Identify the conditional event.
For each possible count $n$, determine which combinations of severities give the required exact total or fall within the cumulative threshold.

Find feasible counts.
Use the smallest possible severity to rule out counts that cannot produce the target or satisfy the threshold.

Apply the Poisson weights.
Multiply each conditional probability by $e^{-\lambda}\lambda^n/n!$, including the $n=0$ term whenever it belongs to the event.

Add and check.
Sum the weighted terms, verify that no feasible count was omitted, and round only the final probability.

Worked Applications

Worked Example: An Exact Aggregate Probability

A hypothetical portfolio has $N\sim\operatorname{Poisson}(1)$ claims. Independently, each claim has severity 1 or 2, each with probability $1/2$. Find $P(S=3)$, where $S$ is the aggregate claim amount.

Method. Condition on $N$. Since each claim is at least 1, three or more claims could reach 3 only when there are exactly three claims, and four or more claims are impossible. One claim cannot produce 3.

For $N=2$, the total is 3 if the severities are 1 and 2 in either order. Thus $q_2(3)=2(1/2)(1/2)=1/2$. Its Poisson-weighted contribution is $P(N=2)q_2(3)=\bigl(e^{-1}/2\bigr)(1/2)=e^{-1}/4$.

For $N=3$, the total is 3 only if all three severities are 1. Thus $q_3(3)=(1/2)^3=1/8$. Its contribution is $P(N=3)q_3(3)=\bigl(e^{-1}/6\bigr)(1/8)=e^{-1}/48$.

Adding the feasible contributions, $P(S=3)=e^{-1}(1/4+1/48)=13/(48e)\approx0.099634$. Rounded to four decimal places, the answer is 0.0996.

Check. The two conditional cases cover all feasible counts, and the probability is less than 1. A common mistake is to omit the Poisson factorials; that would give $e^{-1}(1/2+1/8)\approx0.2299$, which is not the correct mixture. The factors $1/2!$ and $1/3!$ are part of the Poisson probabilities.

Worked Example: A Cumulative Probability

For a hypothetical portfolio, let $N\sim\operatorname{Poisson}(1.5)$. Each claim has severity 1 or 2 with probability $1/2$ each, independently of all other claims and of $N$. Calculate $P(S\le2)$.

Method. Condition on the count and use the positive minimum severity of 1. If $N\ge3$, then $S\ge3$, so only counts 0, 1, and 2 contribute.

If $N=0$, the aggregate is 0, so the conditional probability is 1. If $N=1$, either severity is at most 2, so the conditional probability is also 1. If $N=2$, the sum is at most 2 only when both severities are 1; this has probability $(1/2)^2=1/4$.

The weighted sum is therefore $e^{-1.5}\left[1+1.5+(1.5^2/2)(1/4)\right]$. Since $1.5^2/2=1.125$, the bracket is $1+1.5+1.125(0.25)=2.5+0.28125=2.78125$. Thus $P(S\le2)=2.78125e^{-1.5}\approx0.6205808$, or 0.6206 to four decimal places.

Check. The $N=0$ and $N=1$ cases contribute $e^{-1.5}(1+1.5)$, and the only qualifying two-claim outcome contributes $e^{-1.5}(0.28125)$. No count of three or more can meet the threshold.

Worked Example: Enumerating Counts for an Exact Total

Suppose $N\sim\operatorname{Poisson}(2)$ and each hypothetical claim has severity 1, 2, or 3 with probabilities 0.5, 0.3, and 0.2, respectively. Find $P(S=4)$.

Method. Enumerate the feasible counts and the severity combinations within each count. Because the smallest severity is 1, counts above 4 cannot produce a total of 4. A single claim cannot equal 4 because severity 4 is not in the support.

For $N=2$, the total 4 comes from severities 1 and 3 in either order, or from 2 and 2. Therefore $q_2(4)=2(0.5)(0.2)+(0.3)^2=0.20+0.09=0.29$. The weighted contribution is $(2^2/2!)e^{-2}(0.29)=2e^{-2}(0.29)$.

For $N=3$, the only way to total 4 is to have two severities of 1 and one severity of 2. There are three positions for the severity 2, so $q_3(4)=3(0.5)^2(0.3)=0.225$. The weighted contribution is $(2^3/3!)e^{-2}(0.225)=(8/6)e^{-2}(0.225)$.

For $N=4$, all four severities must be 1, giving $q_4(4)=(0.5)^4=0.0625$. The weighted contribution is $(2^4/4!)e^{-2}(0.0625)=(16/24)e^{-2}(0.0625)$. Counts $N\ge5$ have aggregate at least 5, so they cannot contribute.

Combining the terms gives $P(S=4)=e^{-2}[2(0.29)+(8/6)(0.225)+(16/24)(0.0625)]$. Inside the brackets, the terms are $0.58$, $0.30$, and $0.0416667$, for a total of $0.9216667=553/600$. Hence $P(S=4)=(553/600)e^{-2}\approx0.1247340$, or 0.1247 to four decimal places.

Check. The count-specific contributions before multiplication by $e^{-2}$ are $0.58$, $0.30$, and $0.0416667$. Their sum is $0.9216667$; retaining the $N=4$ term is essential because four minimum-sized claims also produce a total of 4.

Common Traps and a Final Check

The mixture is simple in principle, but errors often arise from confusing conditional probabilities with unconditional probabilities. The conditional probability $q_n(s)$ describes severity combinations given $N=n$; it is not itself the probability that the aggregate equals $s$. It must be multiplied by $P(N=n)$.

Dropping the factorial. The weight for $n$ claims is $e^{-\lambda}\lambda^n/n!$, not $e^{-\lambda}\lambda^n$. The factorial matters even when the conditional severity probability is straightforward.
Forgetting the zero-count term. For a cumulative event $S\le x$ with $x\ge0$, $N=0$ qualifies. Its contribution is $e^{-\lambda}$. For an exact positive total, it does not qualify.
Including impossible counts. If each severity is at least $m>0$, an aggregate no greater than $x$ cannot arise from more than $x/m$ claims.
Using the wrong conditional event. For a cumulative probability, count all severity combinations below or equal to the threshold; for an exact probability, count only combinations equal to the target.

This conditioning method is useful because it converts one random-sum calculation into a sequence of ordinary finite-support severity calculations. Keep the model assumptions visible, list feasible counts before calculating, and attach the correct Poisson weight to every conditional probability.

Check Your Understanding

For each problem, condition on the claim count, identify feasible counts, and apply the Poisson weights. Answers are rounded only where indicated.

A hypothetical portfolio has $N\sim\operatorname{Poisson}(0.5)$, and each claim is 1 or 2 with probability $1/2$ each. Find $P(S=2)$. (Answer: $0.1706$, to four decimal places.)
Let $N\sim\operatorname{Poisson}(1)$. Claim severities are 1 with probability 0.6 and 3 with probability 0.4. Find $P(S=3)$. (Answer: $0.1604$, to four decimal places.)
Let $N\sim\operatorname{Poisson}(2)$, with severities 1 and 2 having probabilities 0.25 and 0.75. Find $P(S\le2)$. (Answer: $3.125e^{-2}\approx0.4229$, to four decimal places.)
Let $N\sim\operatorname{Poisson}(1.5)$, with severities 1, 2, and 4 having probabilities 0.5, 0.25, and 0.25. Find $P(S=4)$. (Answer: $0.56396484375e^{-1.5}\approx0.1258$, to four decimal places.)
In a model with $N\sim\operatorname{Poisson}(2)$ and strictly positive severities, what is $P(S=0)$? (Answer: $e^{-2}\approx0.1353$, to four decimal places.)

What You'll Learn