Aggregate Distributions by Conditioning on Claim Count

Why Condition on the Claim Count?

An aggregate loss is a random sum: the number of claims can vary, and each claim amount is also uncertain. Conditioning on the number of claims separates these two sources of variation. Once the count is fixed at $n$, the aggregate is a sum of exactly $n$ severity amounts. The resulting conditional distribution can then be weighted by the probability of that count.

This approach gives a direct way to express aggregate probabilities, whether the count distribution has finite support or continues indefinitely. It is also useful for interpreting calculations made by other methods: every aggregate probability must account for the possible claim counts that can produce the event.

Definition: Let $N$ be a nonnegative integer-valued claim count. Let $X_1,X_2,\ldots$ be claim severities, and define the aggregate loss by $S=\sum_{j=1}^{N}X_j$, with $S=0$ when $N=0$. In the standard compound model, $N$ is independent of the severities, and the $X_j$ are independent and identically distributed with severity distribution $F_X$. Write $p_n=P(N=n)$. The distribution of the sum of $n$ independent severities is the $n$-fold convolution $F_X^{*n}$; $F_X^{*0}$ is the point mass at zero.

The independence and identical-distribution assumptions matter. They ensure that, conditional on $N=n$, the sum has distribution $F_X^{*n}$. If severities depend on the count, or if the claim amounts are dependent, the conditional sum distribution may differ; the general conditioning identity still holds, but the ordinary severity convolutions may not.

The Conditioning Formula

For any set of aggregate-loss values $B$, the law of total probability gives

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

Under the standard compound-model assumptions, $S\mid(N=n)$ has the $n$-fold severity convolution. Therefore,

P(S\in B)=\sum_{n=0}^{\infty}p_n\,F_X^{*n}(B).

In particular, for a cumulative probability,

P(S\le s)=\sum_{n=0}^{\infty}p_n F_X^{*n}(s).

For discrete severity with probability mass function $f_X(x)=P(X=x)$, the conditional probability at a point $s$ is

P(S=s)=\sum_{n=0}^{\infty}p_n f_X^{*n}(s),

where $f_X^{*0}(0)=1$ and $f_X^{*0}(s)=0$ for $s\ne0$. The $n$-fold convolution for $n\ge1$ sums over all $n$-claim combinations whose total is $s$. For example, with discrete severities,

f_X^{*2}(s)=\sum_x f_X(x)f_X(s-x).

Important distinction: If each claim severity is strictly positive, $S=0$ occurs exactly when $N=0$, so $P(S=0)=p_0$. If zero-valued severities are possible, positive claim counts can also contribute to $P(S=0)$; in that case the full mixture at zero must be used.

For continuous severity, the aggregate distribution generally has an atom at zero of size $p_0$, as well as a continuous part. When the severity has density $f_X$, the density of the continuous part at $s>0$, where the convolutions have densities, is $\sum_{n\ge1}p_n f_X^{*n}(s)$. The atom at zero is not represented by that density. The conditioning identity itself applies to discrete, continuous, or mixed distributions.

Why the Weighted Convolution Formula Holds

The justification is a partition argument. The events $\{N=n\}$, for $n=0,1,\ldots$, are mutually exclusive and together cover all possible outcomes. Intersecting the event $\{S\in B\}$ with this partition and adding the probabilities gives

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).

When $N=n$, the aggregate is $X_1+\cdots+X_n$. Independence and the common severity distribution imply that its distribution is the $n$-fold convolution, which supplies the final form of the formula. For $n=0$, the sum is empty and equals zero; this is why the zero-count term is a point mass at zero rather than an ordinary severity convolution.

List the possible counts.
Identify the values of $n$ with nonzero $p_n$, or determine which counts could contribute to the requested aggregate event.

Find the conditional aggregate probability.
For each relevant count $n$, calculate the probability using the $n$-fold severity convolution.

Weight and add.
Multiply each conditional probability by $p_n$, then sum the contributions. Include the $n=0$ term when the event contains zero.

Worked Examples

Worked Example: Constructing a Finite Aggregate Distribution

A hypothetical portfolio has claim count probabilities $P(N=0)=0.4$, $P(N=1)=0.4$, and $P(N=2)=0.2$. Given a claim, its severity is \$100 with probability 0.7 and \$200 with probability 0.3. Claim amounts are independent and identically distributed and independent of $N$. Find the aggregate distribution and $P(S\le \$200)$.

Method. Calculate each conditional distribution and weight it by the corresponding claim-count probability. This avoids treating a conditional severity probability as though it were already an aggregate probability.

For $N=0$, $S=0$ with probability 1. Its weighted contribution is $P(S=0,N=0)=0.4$.

For $N=1$, the conditional aggregate equals the single severity: \$100 has probability 0.7 and \$200 has probability 0.3. After weighting by $P(N=1)=0.4$, the contributions are $0.4(0.7)=0.28$ at \$100 and $0.4(0.3)=0.12$ at \$200.

For $N=2$, the possible sums and their conditional probabilities are:

\begin{aligned} P(S=200\mid N=2)&=0.7(0.7)=0.49,\\ P(S=300\mid N=2)&=0.7(0.3)+0.3(0.7)=0.42,\\ P(S=400\mid N=2)&=0.3(0.3)=0.09. \end{aligned}

Weight these probabilities by $P(N=2)=0.2$. The resulting contributions are $0.098$ at \$200, $0.084$ at \$300, and $0.018$ at \$400. At \$200, both one-claim and two-claim cases contribute, so the total probability there is $0.12+0.098=0.218$.

Thus the aggregate distribution is $P(S=0)=0.4$, $P(S=100)=0.28$, $P(S=200)=0.218$, $P(S=300)=0.084$, and $P(S=400)=0.018$. Therefore,

P(S\le 200)=0.4+0.28+0.218=0.898.

Check. The probabilities sum to $0.4+0.28+0.218+0.084+0.018=1$. The calculation includes all counts and correctly combines the different counts that can produce a \$200 aggregate.

Worked Example: A Poisson Count and an Exact Aggregate Amount

A hypothetical portfolio has $N\sim\operatorname{Poisson}(0.5)$. Each claim is independently \$1 or \$2, each with probability 0.5, independent of $N$. Find the probability that the aggregate is exactly \$2. Give the probability to four decimal places.

Method. Condition on the count values that can produce a total of \$2. Since each claim is at least \$1, only $N=1$ and $N=2$ can produce this amount; $N=0$ produces zero and $N\ge3$ produces at least \$3.

For a Poisson random variable with mean $\lambda=0.5$, $P(N=n)=e^{-0.5}(0.5)^n/n!$. With one claim, the severity must be \$2:

P(N=1)P(S=2\mid N=1) =\left(0.5e^{-0.5}\right)(0.5) =0.25e^{-0.5}.

With two claims, both severities must be \$1. The count probability is $P(N=2)=e^{-0.5}(0.5)^2/2=0.125e^{-0.5}$, and the conditional probability of two \$1 claims is $0.5^2=0.25$. Thus this contribution is $0.03125e^{-0.5}$.

Adding the two disjoint count cases gives

P(S=2)=(0.25+0.03125)e^{-0.5} =0.28125e^{-0.5} \approx 0.1706.

Check. The $N=1$ contribution is approximately $0.1516$, and the $N=2$ contribution is approximately $0.0190$; their sum is $0.1706$. Counts above two cannot contribute because every claim is at least \$1.

Worked Example: Conditioning for an Aggregate Tail Probability

A hypothetical contract has $P(N=0)=0.1$, $P(N=1)=0.3$, and $P(N=2)=0.6$. Each claim severity is \$1, \$2, or \$3 with probabilities 0.5, 0.3, and 0.2, respectively. Claim severities are independent and independent of $N$. Find $P(S\ge \$4)$.

Method. Evaluate the tail event conditional on each possible count, then weight each result. For $N=0$, the aggregate is zero, so the conditional tail probability is 0. For $N=1$, the largest possible severity is \$3, so it is also 0.

For $N=2$, the ordered severity pairs whose total is at least \$4 are $(1,3),(3,1),(2,2),(2,3),(3,2),(3,3)$. Their total conditional probability is

\begin{aligned} P(S\ge4\mid N=2) &=2(0.5)(0.2)+(0.3)(0.3)+2(0.3)(0.2)+(0.2)(0.2)\\ &=0.20+0.09+0.12+0.04\\ &=0.45. \end{aligned}

Only the two-claim case contributes, so

P(S\ge4)=0.1(0)+0.3(0)+0.6(0.45)=0.27.

Check. The conditional probability $0.45$ is less than 1, and weighting by the probability $0.6$ of two claims gives an unconditional tail probability of $0.27$. The one-claim case cannot reach the threshold.

Interpreting and Checking the Mixture

The count weights $p_n$ are not optional multipliers: a conditional probability such as $P(S=s\mid N=n)$ describes only the cases with that count. It becomes an unconditional contribution only after multiplication by $P(N=n)$. Conversely, if a problem already gives an unconditional joint probability for a count and an aggregate event, do not multiply by the count probability a second time.

A finite count distribution makes the mixture a finite sum. For an unbounded count distribution, a particular event may still permit a finite calculation, as in the Poisson example where positive severities bounded the relevant counts. Otherwise the mixture is an infinite sum. If it is truncated at $n=K$, the omitted contribution to any event probability is at most $P(N>K)$, because a conditional event probability cannot exceed 1.

Several checks are useful. The count probabilities should sum to 1. For a complete aggregate probability mass function, the aggregate probabilities should also sum to 1. Every conditional probability must lie between 0 and 1, and its weighted contribution cannot exceed the corresponding count probability. For a tail threshold, check which counts are even capable of reaching it before doing convolution arithmetic.

Common Exam Trap

A frequent error is to use a conditional severity-convolution probability directly as an unconditional aggregate probability. For example, in the first worked example, the probability of two \$100 claims conditional on $N=2$ is $0.7^2=0.49$. Reporting $P(S=200)=0.49$ is wrong: it ignores both the chance of having two claims and the one-claim route to a \$200 total. The corrected calculation is

$$ P(S=200)=P(N=1)(0.3)+P(N=2)(0.7^2) =0.4(0.3)+0.2(0.49)=0.218. $$

The reverse trap is to omit $N=0$ when the requested event includes zero aggregate loss. When severities are strictly positive, $P(S=0)=P(N=0)$; when severity may be zero, the zero-aggregate probability can receive contributions from positive claim counts as well.

Key takeaway: Under the standard independent and identically distributed severity assumptions, an aggregate probability is a count-weighted sum of conditional severity-convolution probabilities. Identify the contributing counts, calculate each conditional probability, multiply by its count probability, and check that no count case has been omitted or counted twice. This treatment is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, condition on the claim count and keep conditional probabilities distinct from their unconditional weighted contributions.

A hypothetical portfolio has $P(N=0)=0.3$, $P(N=1)=0.4$, and $P(N=2)=0.3$. Each severity is \$10 or \$20 with equal probability, independently. Find $P(S=20)$. (Answer: 0.2750)
Let $N\sim\operatorname{Poisson}(0.8)$, and suppose every claim has strictly positive severity. What is $P(S=0)$? Give the answer to four decimal places. (Answer: 0.4493)
A claim count has probabilities $P(N=0)=0.2$, $P(N=1)=0.5$, and $P(N=2)=0.3$. Each severity is 1 with probability 0.7 or 3 with probability 0.3, independently. Find $P(S=4)$. (Answer: 0.1260)
Let $N\sim\operatorname{Binomial}(3,0.2)$. Claim severities are independently 1 or 2, each with probability 0.5, independent of $N$. Find $P(S=3)$. (Answer: 0.0490)
Let $N\sim\operatorname{Poisson}(1)$, and let each independent claim severity be 1 or 2 with probability 0.5 each. Find $P(S\le2)$, to four decimal places. (Answer: 0.7817)

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