Aggregate Loss Distributions by Conditioning

Why Condition on the Claim Count?

An aggregate loss combines two sources of randomness: how many claims occur and how large each claim is. Knowing the frequency and severity models separately does not, by itself, give the probability of an aggregate event such as “total payments do not exceed $7.” Conditioning on the claim count makes the calculation manageable: for each possible count, determine the probability of the aggregate event given that count, then weight it by the probability of that count.

This approach is useful when conditional sums of severities can be worked out directly, especially for discrete severities with a small number of possible values. It also gives a general expression for the aggregate distribution, whether the severities are discrete or continuous. It is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Definition: Let $N$ be the claim count, and let $X_1,X_2,\ldots$ be claim severities. Assume the severities are independent and identically distributed as $X$, and that the entire severity sequence is independent of $N$. The aggregate loss is $$ S=\sum_{j=1}^{N}X_j, $$ with $S=0$ when $N=0$. The distribution of $S$ is called a compound distribution: a random number of severity amounts are summed.

Here, $X_j$ represents the amount of the $j$th claim included in the aggregate. If a problem defines the aggregate using payments after a coverage modification, the relevant severity distribution is the distribution of those payments, not automatically the distribution of the original losses. The conditioning formula requires that the chosen severity amounts satisfy the stated independence and identical-distribution assumptions.

The Conditioning Formula

For any event $B$ involving the aggregate, partition the event according to the possible values of $N$. The events $N=0,N=1,\ldots$ are mutually exclusive and cover all possible claim counts. Therefore, the law of total probability gives the aggregate event probability as a weighted sum of conditional probabilities.

P(S\in B)=\sum_{n=0}^{\infty}P(N=n)\,P(S\in B\mid N=n).

In particular, for a cumulative probability at $s$, define $F_{S\mid N=n}(s)=P(S\le s\mid N=n)$. When $N=n$, aggregate loss is the sum of $n$ independent severities. For $n=0$, it is the constant zero, so $F_{S\mid N=0}(s)=1$ if $s\ge0$, and $0$ if $s<0$. Thus

P(S\le s)=\sum_{n=0}^{\infty}P(N=n)F_{S\mid N=n}(s).

The same reasoning applies to an exact point probability:

P(S=s)=\sum_{n=0}^{\infty}P(N=n)P(S=s\mid N=n).

These expressions are not an instruction to use the frequency probabilities alone. Each frequency probability must be multiplied by the appropriate conditional aggregate probability. If the frequency model has an unbounded number of possible claims, the sum is formally infinite. A calculation may still be finite when the event is impossible for sufficiently large $n$, for example when every severity is at least a positive amount and the aggregate is bounded above.

Worked Examples

Worked Example: A Cumulative Probability with Poisson Frequency

A hypothetical portfolio has claim count $N\sim\operatorname{Poisson}(1)$. Each claim severity is independently $1 or $3, each with probability $0.5$, and severities are independent of $N$. Find $P(S\le3)$. Give the probability to four decimal places.

Method. Condition on $N$, and retain every claim count that could produce an aggregate no greater than 3. The smallest possible severity is 1, so any $N\ge4$ makes $S\ge4$; those counts contribute zero.

For $N=0$, $S=0$, so $P(S\le3\mid N=0)=1$. For $N=1$, either severity is at most 3, so the conditional probability is 1. For $N=2$, the sum is at most 3 only when both severities are 1, giving conditional probability $(0.5)^2=0.25$. For $N=3$, the sum is at most 3 only when all three severities are 1, giving $(0.5)^3=0.125$.

The Poisson probabilities are $P(N=n)=e^{-1}/n!$. Weighting each conditional probability gives

\begin{aligned} P(S\le3) &=P(N=0)(1)+P(N=1)(1)+P(N=2)(0.25)+P(N=3)(0.125)\\ &=e^{-1}\left(1+1+\frac{0.25}{2}+\frac{0.125}{6}\right)\\ &=e^{-1}\left(2+\frac18+\frac1{48}\right) =e^{-1}\frac{103}{48}\\ &\approx 0.7894. \end{aligned}

Check. The conditional contributions inside the brackets are $1$, $1$, $0.125$, and approximately $0.0208$, respectively. Multiplying their sum by $e^{-1}\approx0.3679$ gives approximately $0.7894$, a probability between 0 and 1.

Common mistake avoided. Replacing the event $S\le3$ with $N\le3$ would give $P(N\le3)=e^{-1}(1+1+1/2+1/6)\approx0.9810$. That includes every outcome with two or three claims, even when their severities make the aggregate exceed 3. The conditional severity probabilities are essential.

Worked Example: Binomial Frequency and a Severity Threshold

A hypothetical repair plan has $N\sim\operatorname{Binomial}(3,0.4)$. Claim severities are independent: each is $2 with probability $0.75$ and $5 with probability $0.25$, independently of $N$. Find $P(S\le7)$.

Method. Calculate the conditional probability for each possible count $n=0,1,2,3$, then multiply by the binomial probability of that count. For $n=0$, the aggregate is zero; for $n=1$, either severity is at most 7, so both conditional probabilities are 1.

For $n=2$, the possible sums are 4, 7, and 10. The aggregate exceeds 7 only when both claims are $5, which has probability $(0.25)^2=0.0625$. Therefore $P(S\le7\mid N=2)=1-0.0625=0.9375$. For $n=3$, the aggregate is at most 7 only when all three severities are $2: any $5 severity would make the sum at least $5+2+2=9$. Hence $P(S\le7\mid N=3)=(0.75)^3=0.421875$.

The frequency probabilities are $P(N=0)=(0.6)^3=0.216$, $P(N=1)=3(0.4)(0.6)^2=0.432$, $P(N=2)=3(0.4)^2(0.6)=0.288$, and $P(N=3)=(0.4)^3=0.064$.

\begin{aligned} P(S\le7) &=(0.216)(1)+(0.432)(1)+(0.288)(0.9375)+(0.064)(0.421875)\\ &=0.216+0.432+0.270+0.027\\ &=0.945. \end{aligned}

Check. The four frequency probabilities sum to $0.216+0.432+0.288+0.064=1$. The aggregate probability is $0.945$, so the probability of exceeding $7 is $1-0.945=0.055$.

Common mistake avoided. It would be incorrect to add $1+1+0.9375+0.421875$ as though all four claim counts were equally likely. The conditional probabilities must be weighted by their respective binomial probabilities.

Worked Example: An Exact Aggregate Point Probability

A hypothetical book has $N\sim\operatorname{Poisson}(1)$. Independent severities, also independent of $N$, are $1 with probability $0.6$ and $2 with probability $0.4$. Find $P(S=4)$, to four decimal places.

Method. Condition on the claim count and identify which severity combinations sum exactly to 4. Since each claim is at least $1, $N>4$ cannot produce an aggregate of 4. Counts 0 and 1 cannot produce it either.

For $N=2$, the sum is 4 only if both severities are $2, with probability $(0.4)^2=0.16$. For $N=3$, the sum is 4 only if one severity is $2 and the other two are $1. There are $\binom31=3$ choices for the $2 claim, so this conditional probability is $3(0.4)(0.6)^2=0.432$. For $N=4$, all four severities must be $1, with probability $(0.6)^4=0.1296$.

Using $P(N=n)=e^{-1}/n!$,

\begin{aligned} P(S=4) &=P(N=2)(0.16)+P(N=3)(0.432)+P(N=4)(0.1296)\\ &=e^{-1}\left(\frac{0.16}{2!}+\frac{0.432}{3!}+\frac{0.1296}{4!}\right)\\ &=e^{-1}(0.08+0.072+0.0054)\\ &=e^{-1}(0.1574)\approx0.0579. \end{aligned}

Check. The three conditional probabilities are each at most 1, and all other claim counts have conditional probability zero for $S=4$. The weighted result is approximately $0.0579$.

Common mistake avoided. The number of ways to place the single $2 severity among three claims matters when $N=3$. Using only $(0.4)(0.6)^2=0.144$ omits the three possible positions and understates that conditional probability.

How to Organize a Conditioning Calculation

A reliable calculation separates the frequency model from the conditional severity-sum calculation. The following steps make it easier to see whether any possible claim count has been omitted or assigned the wrong conditional probability.

State the model.
Identify the distribution of $N$, the severity distribution, and the assumption that severities are independent of one another and of $N$.

List relevant claim counts.
For a cumulative or point event, identify which counts can contribute. Use zero contributions for impossible counts rather than silently dropping them.

Find conditional probabilities.
For each relevant $n$, compute the event probability for a sum of $n$ severities. Handle $n=0$ using the convention $S=0$.

Weight and sum.
Multiply each conditional probability by $P(N=n)$, add the contributions, and check that the result lies between 0 and 1.

Assumptions and Exam Traps

Conditioning is a probability identity, but a chosen conditional calculation still depends on the model. In the compound model above, given $N=n$, the aggregate is a sum of $n$ identically distributed, independent severities. If claim sizes depend on the count, if claims are dependent, or if the portfolio contains materially different claim types without a specified mixture model, the simple conditional severity calculations may not be valid as stated. The conditioning identity remains true, but the conditional distribution must match the actual assumptions.

A frequent error is to confuse a claim-count event with an aggregate-loss event. For example, $S\le s$ is not generally equivalent to $N\le n$: even one claim may exceed $s$, and several small claims may remain below it. Another error is omitting the $N=0$ contribution. When $s\ge0$, this contributes $P(N=0)$ to $P(S\le s)$; when finding $P(S=s)$, it contributes only if $s=0$.

For an unbounded frequency distribution, do not assume that a finite list of counts is exhaustive. It is exhaustive only if the event is impossible for all omitted counts, or if a separately justified approximation or truncation is being used. For example, with strictly positive severities bounded below by $c>0$, $S\le s$ is impossible when $n>s/c$. Without such a condition, larger counts may still contribute.

Key takeaway: An aggregate probability is a frequency-weighted average of conditional severity-sum probabilities. First determine what the severity sums can do at each claim count; then apply the frequency probabilities. This is the central safeguard against treating claim count alone as aggregate loss.

Check Your Understanding

For each problem, condition on the claim count and include every claim-count contribution that can affect the requested event.

$N\sim\operatorname{Poisson}(0.5)$. Independent severities are $1 or $2, each with probability $0.5$. Find $P(S\le2)$. (Answer: 0.9288)
$N\sim\operatorname{Binomial}(2,0.3)$. Independent severities are $1 with probability $0.8$ and $4 with probability $0.2$. Find $P(S\le4)$. (Answer: 0.9676)
$N\sim\operatorname{Poisson}(1.5)$. Independent severities are $2 or $5, each with probability $0.5$. Find $P(S\le5)$. (Answer: 0.6206)
$N\sim\operatorname{Poisson}(2)$. Independent severities are $1 with probability $0.7$ and $3 with probability $0.3$. Find the exact point probability $P(S=5)$. (Answer: 0.0856)
A frequency model has $P(N=0)=0.2$, $P(N=1)=0.3$, $P(N=2)=0.3$, and $P(N=3)=0.2$. Suppose the conditional probabilities $P(S\le s\mid N=n)$, for $n=0,1,2,3$, are respectively $1,0.8,0.5,0.1$. Find $P(S\le s)$. (Answer: 0.6100)

What You'll Learn