Calculating Discrete Aggregate Loss Distributions

Why Aggregate Loss Requires Two Distributions

An insurer’s total loss over a period depends both on how many claims occur and on how large each claim is. When the claim count and claim severities have discrete distributions, the aggregate loss also has a discrete distribution. Constructing it means combining the two sources of uncertainty without mixing up their probabilities.

Let $N$ be the number of claims and let $Y_1,Y_2,\ldots$ be the claim severities. The aggregate loss is $S=\sum_{i=1}^{N}Y_i$, with $S=0$ when $N=0$. Throughout this tutorial, unless a problem states otherwise, assume the severities are independent and identically distributed, and independent of $N$. These assumptions allow the same conditional severity distribution to be used for every claim count.

Definition: The discrete aggregate-loss distribution gives $P(S=s)$, the probability that total loss equals $s$, for every possible aggregate amount $s$. Given the claim-count probabilities $P(N=n)$ and the conditional aggregate probabilities $P(S=s\mid N=n)$, the distribution is $$ P(S=s)=\sum_{n=0}^{\infty}P(N=n)P(S=s\mid N=n). $$ For $n=0$, $P(S=0\mid N=0)=1$, and $P(S=s\mid N=0)=0$ for $s\ne0$.

The sum over $n$ is the law of total probability: the events $N=0,N=1,\ldots$ are mutually exclusive and cover all possible claim counts. For a count distribution with a finite maximum, only its possible counts need to be included. For an unbounded count distribution, the sum may have infinitely many terms; a numerical calculation then needs a suitable truncation and an assessment of the omitted probability.

Convolving Discrete Severities

For one claim, write $p_Y(y)=P(Y_i=y)$ for the severity probability mass function. If $N=n$, aggregate loss is a sum of $n$ severities. The probability of a particular sum is found by adding the probabilities of all combinations of severities that produce that sum. This operation is called convolution.

Result: Let $p_n(s)=P(Y_1+\cdots+Y_n=s)$, with $p_0(0)=1$ and $p_0(s)=0$ for $s\ne0$. Under the independent, identically distributed severity assumption, $$ p_n(s)=\sum_y p_{n-1}(s-y)p_Y(y),\qquad n\ge1. $$ The aggregate probability is then $P(S=s)=\sum_n P(N=n)p_n(s)$.

To see why the convolution formula holds, condition on the last claim severity being $y$. For the total of $n$ claims to equal $s$, the first $n-1$ claims must total $s-y$. Independence makes the probability of those two events $p_{n-1}(s-y)p_Y(y)$. Summing over every possible last severity counts all ways to reach $s$, with no overlap between the different last-severity values.

In practice, make a conditional distribution for each possible claim count, then weight each row by $P(N=n)$. When different combinations produce the same aggregate amount, add their probabilities. The following examples use dollar amounts and finite supports so every possible outcome can be listed.

Worked Example: Two Possible Claims with Two Severity Amounts

A hypothetical coverage has $N\sim\operatorname{Binomial}(2,0.4)$. Conditional on a claim, its severity is \$2 with probability $0.75$ and \$5 with probability $0.25$. Claim severities are independent of each other and of $N$. Construct the aggregate distribution and calculate $P(S=7)$.

Method. First calculate the count probabilities: $P(N=0)=0.6^2=0.36$, $P(N=1)=2(0.4)(0.6)=0.48$, and $P(N=2)=0.4^2=0.16$. These sum to $0.36+0.48+0.16=1$.

If $N=0$, then $S=0$ with conditional probability 1, so $P(S=0)=0.36(1)=0.36$.

If $N=1$, the conditional outcomes are \$2 with probability $0.75$ and \$5 with probability $0.25$. Weighting by $P(N=1)$, the aggregate masses are $P(S=2)=0.48(0.75)=0.36$ and $P(S=5)=0.48(0.25)=0.12$.

If $N=2$, convolve the severity probabilities. The possible sums are \$4, \$7, and \$10: $P(S=4\mid N=2)=0.75^2=0.5625$; $P(S=7\mid N=2)=2(0.75)(0.25)=0.375$; and $P(S=10\mid N=2)=0.25^2=0.0625$. After weighting by $0.16$, these contribute $0.09$, $0.06$, and $0.01$, respectively.

Thus the complete distribution is $P(S=0)=0.36$, $P(S=2)=0.36$, $P(S=4)=0.09$, $P(S=5)=0.12$, $P(S=7)=0.06$, and $P(S=10)=0.01$. Therefore $P(S=7)=0.06$. As a check, the probabilities sum to $0.36+0.36+0.09+0.12+0.06+0.01=1.00$. The method avoids treating the conditional probability $0.375$ of a \$7 total given two claims as the unconditional aggregate probability; it must also be multiplied by $P(N=2)=0.16$.

Combining Conditional Distributions

A useful calculation table keeps conditional probabilities separate from the count weights. In particular, each row conditional on $N=n$ should sum to 1 before it is weighted. The weighted contributions across all rows should then sum to 1 in the aggregate distribution.

List possible counts.
Record $P(N=n)$ for each count that can occur, including zero if it is possible.

Find the conditional sum distribution.
For each count, add severities directly or use convolution to calculate $P(S=s\mid N=n)$.

Weight and combine.
Multiply each conditional probability by $P(N=n)$; add contributions that lead to the same aggregate amount.

Check the result.
Confirm that the aggregate probabilities sum to 1 and, when practical, compare the aggregate mean with $E[N]E[Y]$ under the independence assumptions.

The mean check follows from conditioning on the count: given $N=n$, the expected total is $nE[Y]$. Averaging over the count gives $E[S]=\sum_n P(N=n)nE[Y]=E[N]E[Y]$. This is a check on a distribution, not a substitute for calculating its probabilities: different distributions can have the same mean.

Worked Example: A Specified Count Distribution and a Tail Probability

For a hypothetical portfolio, $P(N=0)=0.50$, $P(N=1)=0.30$, and $P(N=2)=0.20$. Each claim is \$1 with probability $0.60$ or \$4 with probability $0.40$, independently of other severities and of $N$. Find the aggregate distribution and $P(S>4)$.

Method. For $N=0$, all probability $0.50$ is at $S=0$. For $N=1$, the weighted contributions are $P(S=1)=0.30(0.60)=0.18$ and $P(S=4)=0.30(0.40)=0.12$.

For $N=2$, the conditional sums are \$2, \$5, and \$8. Their conditional probabilities are $0.60^2=0.36$, $2(0.60)(0.40)=0.48$, and $0.40^2=0.16$, respectively. Weighting by $P(N=2)=0.20$ gives contributions $0.072$ at \$2, $0.096$ at \$5, and $0.032$ at \$8.

The aggregate distribution is therefore $P(S=0)=0.50$, $P(S=1)=0.18$, $P(S=2)=0.072$, $P(S=4)=0.12$, $P(S=5)=0.096$, and $P(S=8)=0.032$. The requested probability is $P(S>4)=P(S=5)+P(S=8)=0.096+0.032=0.128$. The total probability is $0.50+0.18+0.072+0.12+0.096+0.032=1.000$.

A mean check gives $E[N]=0(0.50)+1(0.30)+2(0.20)=0.70$, and $E[Y]=1(0.60)+4(0.40)=2.20$, so $E[S]=0.70(2.20)=1.54$. Summing loss times mass from the distribution gives $1(0.18)+2(0.072)+4(0.12)+5(0.096)+8(0.032)=1.54$. The check confirms the weighted distribution’s mean.

When Different Combinations Produce the Same Total

Convolution requires adding probabilities for all severity combinations that produce the same sum. For example, with two claims that can each be \$2 or \$4, a total of \$6 can occur in two orders: first \$2 and then \$4, or first \$4 and then \$2. If the severities are independent and each amount has probability $0.5$, each order has probability $0.5(0.5)=0.25$, so the conditional probability of \$6 given two claims is $0.25+0.25=0.50$.

Worked Example: Aggregating Coincident Severity Combinations

A hypothetical contract has $P(N=0)=0.20$, $P(N=1)=0.50$, and $P(N=2)=0.30$. Each claim is \$2 with probability $0.25$ and \$4 with probability $0.75$. Severities are independent of each other and of $N$. Construct the aggregate distribution and check its mean.

Method. With no claims, $P(S=0)=0.20$. With one claim, the weighted masses are $P(S=2)=0.50(0.25)=0.125$ and $P(S=4)=0.50(0.75)=0.375$.

With two claims, the conditional sums and probabilities are: \$4 from two \$2 claims, probability $0.25^2=0.0625$; \$6 from either order of \$2 and \$4, probability $2(0.25)(0.75)=0.375$; and \$8 from two \$4 claims, probability $0.75^2=0.5625$. Weighting by $0.30$ gives contributions $0.01875$ at \$4, $0.1125$ at \$6, and $0.16875$ at \$8.

The total at \$4 has contributions from both $N=1$ and $N=2$, so combine them: $P(S=4)=0.375+0.01875=0.39375$. The complete distribution is $P(S=0)=0.20$, $P(S=2)=0.125$, $P(S=4)=0.39375$, $P(S=6)=0.1125$, and $P(S=8)=0.16875$. The probabilities sum to $1.00000$.

For the mean check, $E[N]=0(0.20)+1(0.50)+2(0.30)=1.10$, and $E[Y]=2(0.25)+4(0.75)=3.50$. Thus $E[S]=1.10(3.50)=3.85$. The distribution gives $2(0.125)+4(0.39375)+6(0.1125)+8(0.16875)=0.25+1.575+0.675+1.35=3.85$. The agreement checks both the combination at \$4 and the final distribution.

Common Mistake: Leaving Out the Count Weight

A conditional probability is not automatically an aggregate probability. In the first example, $P(S=7\mid N=2)=0.375$, but the unconditional probability is $P(N=2)P(S=7\mid N=2)=0.16(0.375)=0.06$. Reporting $0.375$ as $P(S=7)$ wrongly assumes that two claims occur with certainty.

Another frequent error is to count only one ordering of a mixed pair of severities. When claims are independent but the severity amounts differ, the orderings are distinct outcomes and their probabilities must be added if they yield the same total. Finally, check whether an amount can be reached at more than one claim count: a total of \$4, for example, might arise from one larger claim or from multiple smaller claims. Contributions from every possible count belong in $P(S=4)$.

Key takeaway: Calculate severity sums conditional on each claim count, multiply by the probability of that count, and combine equal aggregate amounts. A valid probability distribution sums to 1; when the independence assumptions hold, its mean also agrees with $E[N]E[Y]$. This calculation skill is aligned with the CAS Exam 9 (Risk Management for Actuaries) syllabus.

Check Your Understanding

For each problem, construct or use the conditional distributions carefully, and check that the aggregate probability calculation accounts for all relevant claim counts.

A hypothetical policy has $P(N=0)=0.60$ and $P(N=1)=0.40$. A claim is \$3 with probability $0.40$ and \$7 with probability $0.60$. Find the full aggregate distribution. (Answer: $P(S=0)=0.60,\ P(S=3)=0.16,\ P(S=7)=0.24$.)
A portfolio has $P(N=0)=0.40,\ P(N=1)=0.40,\ P(N=2)=0.20$. Each severity is \$1 or \$3 with probability $0.50$ each. Find $P(S\ge4)$ and give the full aggregate distribution. (Answer: $P(S\ge4)=0.15$; masses are $0.40$ at 0, $0.20$ at 1, $0.05$ at 2, $0.20$ at 3, $0.10$ at 4, and $0.05$ at 6.)
$N\sim\operatorname{Binomial}(2,0.25)$. Each claim is \$2 or \$5 with probability $0.50$ each, independently. Calculate the aggregate distribution and $P(S>5)$. (Answer: masses are $0.5625$ at 0, $0.1875$ at 2, $0.1875$ at 5, $0.015625$ at 4, $0.03125$ at 7, and $0.015625$ at 10; $P(S>5)=0.046875$.)
A hypothetical portfolio has $P(N=0)=0.10,\ P(N=1)=0.60,\ P(N=2)=0.30$. Each claim is \$1 with probability $0.20$ and \$4 with probability $0.80$. Find the full aggregate distribution, $P(S\ge5)$, and verify the mean using $E[S]=E[N]E[Y]$. (Answer: masses are $0.10$ at 0, $0.12$ at 1, $0.012$ at 2, $0.48$ at 4, $0.096$ at 5, and $0.192$ at 8; $P(S\ge5)=0.288$, and $E[S]=1.20(3.40)=\$4.08$.)

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