Exact Aggregate Probabilities By Count Mixtures

Why Condition on the Claim Count?

An aggregate loss is a random sum: the number of claims is random, and each claim has a random severity. A specified aggregate amount may arise in more than one way. For example, a total of two units could result from one claim of size two or two claims of size one. To calculate an exact aggregate probability, identify every claim count that could produce the amount, find the conditional probability for each such count, and weight it by the probability of that count.

This procedure is useful for both finite and infinite claim-count models. A binomial count has a finite set of possible counts. A Poisson count has infinitely many possible counts, but positive minimum claim severity can make only finitely many counts relevant to a fixed aggregate amount.

Definition: Let $N$ be a nonnegative integer-valued claim count, and let $X_1,X_2,\ldots$ be independent, identically distributed claim severities that are independent of $N$. The aggregate loss is $S=\sum_{j=1}^{N}X_j$, with $S=0$ when $N=0$. For any set of aggregate amounts $B$, $$ P(S\in B)=\sum_{n=0}^{\infty}P(N=n)\,P\left(\sum_{j=1}^{n}X_j\in B\right). $$ For $n=0$, the sum of severities is defined to be zero. A term contributes only when its count probability and its conditional severity-sum probability are both nonzero.

Finding the Counts That Can Contribute

For a point probability $P(S=s)$, the main efficiency improvement is to determine feasible counts before calculating conditional probabilities. Suppose each severity takes values in a finite set with smallest value $m>0$ and largest value $M$. Given $N=n$, the aggregate must lie between $nm$ and $nM$. Thus a necessary condition for $S=s$ is $nm\leq s\leq nM$. For an interval event $a\leq S\leq b$, a necessary condition is $nm\leq b$ and $nM\geq a$. These conditions can rule out counts, though they do not guarantee that the target is attainable on the severity support.

A useful distinction is that the count distribution and the severity support impose separate limits. If $N$ is binomial, counts above its maximum are impossible regardless of severity. If $N$ is Poisson, there is no maximum count from the frequency model; a positive minimum severity may nevertheless rule out all counts above a bound for a fixed target. If zero severity is possible, arbitrarily many claims might still contribute to the same total, so that bound need not hold.

List possible counts.
Use the support of $N$, then apply minimum- and maximum-severity bounds to remove counts that cannot reach the specified amount or range.

Find conditional probabilities.
For each remaining $n$, calculate the probability that the sum of $n$ severities satisfies the event. For a point probability, enumerate combinations or use the relevant severity-sum distribution.

Weight and add.
Multiply each conditional probability by $P(N=n)$, then sum over all contributing counts. Do not add conditional probabilities without their count weights.

Worked Point Probabilities

Worked Example: A Binomial Count With Two Severity Sizes

A hypothetical portfolio has $N\sim\operatorname{Binomial}(4,0.25)$. Given a claim, its severity is one unit or two units, each with probability $0.5$, independently of other claims and of $N$. Find $P(S=2)$. The method is to condition on each feasible count; it avoids overlooking a one-claim way to reach the target.

Each claim is at least one unit. Therefore, $n\geq 3$ implies $S\geq 3$, so only $n=0,1,2$ need consideration. With no claims, $S=0$, which cannot equal two.

For one claim, the severity must be two. The count probability and conditional probability are $$ P(N=1)=\binom{4}{1}(0.25)(0.75)^3=\frac{27}{64}, \qquad P(S=2\mid N=1)=0.5. $$ This contribution is $\frac{27}{64}(0.5)=\frac{27}{128}$.

For two claims, both severities must be one. The count probability and conditional probability are $$ P(N=2)=\binom{4}{2}(0.25)^2(0.75)^2=\frac{27}{128}, \qquad P(S=2\mid N=2)=(0.5)^2=\frac14. $$ This contribution is $\frac{27}{128}(\frac14)=\frac{27}{512}$. Adding all feasible contributions gives $$ P(S=2)=\frac{27}{128}+\frac{27}{512} =\frac{108+27}{512} =\frac{135}{512} =0.263671875. $$ The checked probability, rounded to four decimal places, is $0.2637$.

Common mistake. Using only $N=2$ gives $27/512$, but omits the valid case $N=1$ with severity two. The support check shows why counts must be examined one by one: both one and two claims can produce the target.

Worked Example: A Poisson Count With an Infinite Support

For a hypothetical portfolio, $N\sim\operatorname{Poisson}(1)$. Each severity is one unit or three units, each with probability $0.5$, independently. Find $P(S=3)$. Conditioning on count makes this exact calculation finite even though a Poisson count has no finite upper bound.

With one claim, the severity must be three, so $P(S=3\mid N=1)=0.5$. With two claims, possible sums are two, four, and six, so the conditional probability is zero. With three claims, the sum is three only if all three severities are one, giving $P(S=3\mid N=3)=(0.5)^3=1/8$. Counts $n\geq4$ have aggregate at least four, and $n=0$ gives zero. Thus only $n=1$ and $n=3$ contribute.

For a Poisson random variable with mean one, $P(N=n)=e^{-1}/n!$. The two weighted contributions are $$ P(N=1)P(S=3\mid N=1)=e^{-1}\left(\frac12\right), $$ and $$ P(N=3)P(S=3\mid N=3)=\frac{e^{-1}}{3!}\left(\frac18\right)=\frac{e^{-1}}{48}. $$ Therefore, $$ P(S=3)=e^{-1}\left(\frac12+\frac1{48}\right) =\frac{25}{48e} \approx 0.1916. $$ The answer is rounded to four decimal places. The count truncation here is exact: no omitted Poisson count can produce three because every severity is at least one.

Common mistake. Summing over every Poisson count as though each count might contribute is unnecessary, while stopping at an arbitrary count is not justified. The minimum severity supplies the valid cutoff $n\leq3$ for this point probability.

Events Covering a Range of Aggregate Amounts

The same conditioning formula applies to a range event, but the conditional probability must include every severity combination in that range. It is not enough to calculate the probability of one particular aggregate amount within the range. The next example illustrates both the count weighting and this distinction.

Worked Example: An Aggregate Interval With a Finite Count

A hypothetical portfolio has $N\sim\operatorname{Binomial}(3,0.4)$. Each claim has severity one unit with probability $0.75$ and four units with probability $0.25$, independently. Find $P(4\leq S\leq5)$. The method is to calculate the conditional probability of the whole interval for every feasible count.

For $n=0$, the aggregate is zero. For $n=1$, it lies in the interval only when the severity is four, so the conditional probability is $0.25$. For $n=2$, the possible sums are two, five, and eight; the interval event occurs when one severity is one and the other is four. Its conditional probability is $$ 2(0.75)(0.25)=0.375. $$ For $n=3$, the possible sums are three, six, nine, and twelve, so none is in the interval. The binomial count cannot exceed three.

The relevant count probabilities are $$ P(N=1)=\binom31(0.4)(0.6)^2=0.432, \qquad P(N=2)=\binom32(0.4)^2(0.6)=0.288. $$ Weighting the conditional probabilities gives $$ P(4\leq S\leq5) =(0.432)(0.25)+(0.288)(0.375) =0.108+0.108 =0.216. $$ The result is $0.2160$ to four decimal places.

Common mistake. For two claims, counting only the ordered outcomes “one then four” or “four then one” without using their probabilities can lead to a missing factor or a mistaken conditional probability. The factor of two in $2(0.75)(0.25)$ accounts for both mutually exclusive orders.

Why the Count Mixture Formula Holds

The count values partition the sample space: exactly one of $N=0,N=1,N=2,\ldots$ occurs. The event $S\in B$ can therefore be separated into disjoint events according to the count. Adding their probabilities gives the law of total probability. Independence of $N$ and the severities permits the conditional severity-sum distribution given $N=n$ to use the stated severity model without further adjustment.

Justification: The events $\{N=n\}$ are mutually exclusive and exhaustive, so $$ P(S\in B)=\sum_{n=0}^{\infty}P(S\in B,N=n) =\sum_{n=0}^{\infty}P(N=n)P(S\in B\mid N=n). $$ Given $N=n$, $S$ is the sum of $n$ severities (zero if $n=0$). Substituting this conditional distribution yields the count-mixture formula. For a point event $B=\{s\}$, it gives the exact probability $P(S=s)$.

Checks That Prevent Count-Truncation Errors

A correct calculation has two parts: the conditional severity-sum probabilities must be correct, and the list of counts being weighted must be complete. The severity support is especially useful for checking both. For a target $s$, if the severities are positive integers with minimum one, counts above $s$ cannot contribute. But a count below $s$ is not automatically feasible, nor does one feasible count imply that it is the only one.

For lattice severities, conditional sums must land on attainable lattice values. A numerical range may contain values that cannot occur at all. For instance, if every severity is either two or five, a total of six is impossible: no sum of those values equals six. This support check should happen before any arithmetic with the count probabilities.

Enumerate the event, not just a convenient case. A target can arise from different counts and different severity combinations.
Apply the count probability once per count. The conditional severity probability is not itself an aggregate probability until it is weighted by $P(N=n)$.
Use an exact cutoff. A finite count model has its own upper bound; an infinite count model needs a support-based reason for omitting terms.
Check point versus interval events. For an interval, include all attainable sums in the interval, not only its endpoints or one selected value.

These calculations are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus because they combine a claim-count model with a severity-sum event. Clear conditioning and a documented feasible-count list make the calculation easier to verify and reduce the risk of silently dropping a contributing case.

Check Your Understanding

For each problem, condition on the claim count and check which severity sums are attainable before adding contributions.

A hypothetical portfolio has $N\sim\operatorname{Binomial}(2,0.5)$. Each severity is two or five with probability $0.5$ each. Find $P(S=7)$. (Answer: $0.1250$)
A hypothetical portfolio has $N\sim\operatorname{Poisson}(2)$. Each severity is one or two with probability $0.5$ each. Find $P(S=2)$. (Answer: $1.5e^{-2}\approx0.2030$)
A hypothetical portfolio has $N\sim\operatorname{Binomial}(4,0.3)$. Each severity is one with probability $0.6$ and three with probability $0.4$. Find $P(S=4)$. (Answer: $0.1281$)
A hypothetical portfolio has $N\sim\operatorname{Binomial}(3,0.2)$. Each severity is one or two with probability $0.5$ each. Find $P(3\leq S\leq4)$. (Answer: $0.0760$)
A hypothetical portfolio has $N\sim\operatorname{Poisson}(1.5)$. Each severity is two or five with probability $0.5$ each. Find $P(S=10)$. (Answer: approximately $0.0632$)

What You'll Learn