Conditioning Formula For Aggregate Distributions

Why Condition on Claim Count?

An aggregate loss combines two sources of uncertainty: the number of claims and the size of each claim. Even when the frequency and severity models are individually simple, the distribution of their total can be less direct to calculate. Conditioning on the number of claims separates the calculation into manageable pieces: first find the distribution of the total for a fixed count, then weight that distribution by the probability of the count.

This tutorial focuses on selected probabilities for discrete aggregate losses. The method applies across frequency models, including Poisson, binomial, and geometric models, provided the count and severities satisfy the independence assumptions stated below. It is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Definition: Let $N$ be the number of claims, and let $X_1,X_2,\ldots$ be independent, identically distributed claim severities that are independent of $N$. The aggregate loss is $S=\sum_{i=1}^{N}X_i$, with $S=0$ when $N=0$. For a discrete severity, define $q_n(s)=P(X_1+\cdots+X_n=s)$, with $q_0(0)=1$ and $q_0(s)=0$ for $s\ne0$. If $p_n=P(N=n)$, then $$ P(S=s)=\sum_{n=0}^{\infty}p_nq_n(s). $$

The quantity $q_n(s)$ is the probability of an aggregate amount $s$ conditional on exactly $n$ claims. The frequency probability $p_n$ is the weight assigned to that conditional probability. The sum is therefore a probability mixture, not an average of conditional probabilities with equal weights.

Repeated Severity Convolution

For integer-valued severities, write $g(x)=P(X_i=x)$. Once $q_{n-1}$ is known, the distribution for $n$ claims is found by adding one independent severity:

q_n(s)=\sum_x g(x)q_{n-1}(s-x).

The sum runs over the severity values with positive probability. In particular, $q_1(s)=g(s)$. For two claims, $q_2(s)=\sum_x g(x)g(s-x)$. Each additional convolution adds another severity to the conditional total. This is a direct calculation for selected values; it is not a frequency recursion.

List feasible claim counts.
Use the severity support and the target aggregate amount to determine which values of $n$ could contribute.

Find conditional probabilities.
For each feasible $n$, calculate $q_n(s)$ by enumerating severity combinations or applying repeated convolution.

Weight by frequency.
Multiply each $q_n(s)$ by $p_n=P(N=n)$, then add the weighted terms.

Check the result.
Confirm that omitted claim counts cannot produce the target and that the final probability lies between 0 and 1.

A useful shortcut is available when severities are strictly positive integers and their smallest possible value is $m$. With $n$ claims, the aggregate is at least $nm$. Thus, for a target $s$, counts above $\lfloor s/m\rfloor$ cannot contribute. If severity can be zero, this finite cutoff need not apply: arbitrarily many claims could still sum to a fixed amount if some of them have zero severity.

Why the Mixture Formula Holds

The events $\{N=n\}$, for $n=0,1,\ldots$, are mutually exclusive and together cover every possible claim count. The event $\{S=s\}$ can therefore be partitioned according to the value of $N$. Applying the law of total probability gives

\begin{aligned} P(S=s) &=\sum_{n=0}^{\infty}P(S=s,N=n)\\ &=\sum_{n=0}^{\infty}P(S=s\mid N=n)P(N=n)\\ &=\sum_{n=0}^{\infty}q_n(s)p_n. \end{aligned}

The last equality uses the model assumption that, conditional on $N=n$, the aggregate consists of $n$ independent severities with the specified common distribution. Independence of $N$ and the severity sequence ensures that this conditional severity distribution does not change with the claim count. If claim sizes depend on the count, the same total-probability partition still holds, but $q_n(s)$ must use the appropriate count-conditional severity model.

Worked Example: A Poisson Count and a Target Aggregate

A hypothetical portfolio has $N\sim\operatorname{Poisson}(2)$. Each claim severity is independently 1 or 3 monetary units, each with probability $1/2$, and severities are independent of $N$. Calculate $P(S=3)$.

Method. Condition on the claim count and use the severity support to find feasible counts. Each claim is at least 1, so $n\ge4$ cannot produce a total of 3. With one claim, the severity must be 3, giving $q_1(3)=1/2$. With two claims, possible totals are 2, 4, and 6, so $q_2(3)=0$. With three claims, all three severities must be 1, giving $q_3(3)=(1/2)^3=1/8$. Zero claims cannot produce 3.

The relevant Poisson probabilities are

P(N=1)=e^{-2}\frac{2^1}{1!}=2e^{-2}, \qquad P(N=3)=e^{-2}\frac{2^3}{3!}=\frac{4}{3}e^{-2}.

Weight each conditional probability by its frequency probability:

\begin{aligned} P(S=3) &=(2e^{-2})\left(\frac12\right) +\left(\frac43e^{-2}\right)\left(\frac18\right)\\ &=e^{-2}+\frac16e^{-2} =\frac76e^{-2}\\ &\approx 0.157891. \end{aligned}

To four decimal places, $P(S=3)=0.1579$. As a check, the result is between 0 and 1, and the only nonzero contributions come from one or three claims. A common error would be to add $1/2$ and $1/8$ without weighting them by $P(N=1)$ and $P(N=3)$; those are conditional probabilities, not probabilities of the joint events.

Frequency Weights Are Not Equal Weights

The conditional distribution answers, “What is the probability of this aggregate if the claim count is fixed?” The frequency model answers, “How likely is that claim count?” Both are needed. This distinction is particularly important when several counts can produce the same aggregate amount: each count contributes its own conditional probability, multiplied by the probability of that count.

Worked Example: A Binomial Count With Multiple Routes to the Same Total

A hypothetical coverage has $N\sim\operatorname{Binomial}(3,0.4)$. Each claim independently has severity 1 or 2, each with probability $1/2$, independently of $N$. Find $P(S=3)$.

For one claim, severity 3 is impossible, so $q_1(3)=0$. For two claims, the total is 3 precisely when the severities are 1 and 2 in either order. Therefore $q_2(3)=2(1/2)(1/2)=1/2$. For three claims, the only way to total 3 is for all three severities to equal 1, so $q_3(3)=(1/2)^3=1/8$. Zero claims cannot produce a positive total.

The required frequency probabilities are

P(N=2)=\binom32(0.4)^2(0.6)=0.288, \qquad P(N=3)=\binom33(0.4)^3=0.064.

Thus,

\begin{aligned} P(S=3) &=(0.288)\left(\frac12\right)+(0.064)\left(\frac18\right)\\ &=0.144+0.008\\ &=0.152. \end{aligned}

The probability is 0.1520 to four decimal places. Check that the total includes both feasible counts and no others. The exam trap is to report $1/2+1/8=0.625$, which adds conditional probabilities as though the count were equally likely to be 2 or 3. The corrected calculation weights each term by the relevant binomial probability.

Infinite-Support Frequency Models

A frequency distribution can have infinitely many possible counts even when a selected aggregate value can be reached in only a few ways. When severities have a strictly positive minimum, use that minimum to identify a finite set of possible counts before calculating. This avoids unnecessary work and makes it easier to verify that no term has been missed.

The frequency parameterization also matters. For example, a geometric count may include zero claims or may be defined as the number of failures before a success. The model must state its support and probability formula. The following example uses a zero-based geometric distribution, so $P(N=n)=p(1-p)^n$ for $n=0,1,\ldots$.

Worked Example: A Zero-Based Geometric Count

A hypothetical portfolio has $P(N=n)=0.5(0.5)^n$ for $n=0,1,\ldots$. Severities are independent of $N$, with $P(X_i=2)=P(X_i=3)=1/2$. Calculate $P(S=6)$.

Each severity is at least 2. One claim cannot produce 6, and four or more claims have aggregate at least 8. For two claims, a total of 6 requires both severities to equal 3, so $q_2(6)=(1/2)^2=1/4$. For three claims, a total of 6 requires all three severities to equal 2, so $q_3(6)=(1/2)^3=1/8$. No other claim count contributes.

The frequency weights are

P(N=2)=0.5(0.5)^2=0.125, \qquad P(N=3)=0.5(0.5)^3=0.0625.

Therefore,

\begin{aligned} P(S=6) &=(0.125)\left(\frac14\right)+(0.0625)\left(\frac18\right)\\ &=0.03125+0.0078125\\ &=0.0390625. \end{aligned}

To four decimal places, $P(S=6)=0.0391$. The terms correspond to two distinct ways of reaching the target under different claim counts. The calculation also checks the geometric convention explicitly: using a geometric model supported on positive integers instead would give different frequency weights.

Practical Checks and Common Traps

Conditioning is useful because it makes the structure of an aggregate probability visible. It also provides a disciplined way to check a calculation before accepting a numerical result.

Omitting frequency weights. A value $q_n(s)$ is conditional on $N=n$. It must be multiplied by $P(N=n)$ before it contributes to $P(S=s)$.
Including impossible counts. If severities have a positive minimum, use the target amount to bound the possible number of claims. Also check whether the target can be formed from the available severity values for each remaining count.
Using the wrong count support. State the frequency model’s support and parameterization, especially for geometric and negative binomial models.
Changing the severity model across counts without justification. Under the standard independent, identically distributed severity model, each claim has the same severity distribution regardless of $N$. If the model says otherwise, the conditional distribution must reflect that dependence.
Confusing an exact amount with an exceedance event. $P(S=s)$ concerns one aggregate value. For $P(S\leq s)$ or $P(S>s)$, first obtain the relevant conditional cumulative probabilities, then apply the same frequency weighting.

Key takeaway: For a selected aggregate amount, calculate the conditional severity probability for each feasible claim count, multiply it by that count’s probability, and sum. The mixture is exact when the frequency and independent, identically distributed severity assumptions are correctly represented.

Check Your Understanding

For each problem, condition on claim count and apply the severity probabilities. Probabilities in the answers are rounded to four decimal places where needed.

A hypothetical policy has one possible claim, with $P(N=1)=0.3$ and $P(N=0)=0.7$. Given a claim, its severity is 2 with probability 0.4 and 4 with probability 0.6. Find $P(S=4)$. (Answer: 0.1800)
Let $N\sim\operatorname{Poisson}(0.5)$, and let each severity be 1 or 2 with equal probability, independently of $N$. Find $P(S=2)$. (Answer: 0.1706)
Let $N\sim\operatorname{Binomial}(4,0.25)$. Each severity is 2 with probability 0.7 and 5 with probability 0.3, independently of $N$. Find $P(S=10)$. (Answer: 0.0190)
Let $P(N=n)=0.4(0.6)^n$ for $n=0,1,\ldots$. Each severity is 1 with probability 0.25 and 3 with probability 0.75, independently of $N$. Find $P(S=3)$. (Answer: 0.1814)

What You'll Learn