Small Discrete Compound Models By Hand

Why Small Compound Models Are Best Built in Layers

An aggregate loss is a sum of a random number of claim amounts. Even when both the claim count and individual claim amounts have only a few possible values, it is easy to lose track of which combinations produce which aggregate totals. The reliable hand-calculation method is to work in two layers: first find the distribution of the total conditional on each possible claim count, then weight those conditional distributions by the probabilities of the claim counts.

This approach makes the source of every entry visible. It also helps identify impossible totals: a total that cannot be formed from one claim must not receive a one-claim contribution, regardless of whether it can be formed from two claims.

Definition: Let $N$ be a nonnegative integer-valued claim count, and let $X_1,X_2,\ldots$ be independent, identically distributed claim severities, independent of $N$. The aggregate loss is $S=\sum_{j=1}^{N}X_j$, with $S=0$ when $N=0$. For a possible total $s$, $$ P(S=s)=\sum_{n=0}^{\infty}P(N=n)P(S=s\mid N=n). $$ The conditional probability $P(S=s\mid N=n)$ is the probability that the sum of $n$ independent severities equals $s$. For $n=0$, that conditional distribution assigns probability 1 to $s=0$ and 0 to every other total.

For a small model, only finitely many claim counts may have positive probability. If the severity is supported on a few amounts, each conditional distribution can be found by listing the possible combinations, or by repeatedly convolving the severity probabilities. A convolution adds amounts and multiplies probabilities; when different combinations produce the same total, their probabilities are added.

Conditional Convolution and Frequency Mixtures

Suppose a single claim has possible amounts $x_1,\ldots,x_k$, with probabilities $q_1,\ldots,q_k$. For two claims, the probability of total $s$ is obtained from every ordered pair of severities that sums to $s$:

P(X_1+X_2=s)=\sum_{i=1}^{k}\sum_{j=1}^{k}q_iq_j\,\mathbf{1}\{x_i+x_j=s\}.

Here $\mathbf{1}\{\cdot\}$ equals 1 when its condition is true and 0 otherwise. Ordered pairs matter: if the amounts differ, the outcomes $(x_i,x_j)$ and $(x_j,x_i)$ are distinct possibilities. For three or more claims, one can continue the same process, convolving the existing conditional table with the one-claim distribution.

After finding a conditional table for each possible $n$, multiply every entry in that table by $P(N=n)$. Then add the weighted contributions at each aggregate total. The result is an unconditional probability table for $S$.

List the frequency probabilities.
Record each possible claim count $n$ and its probability $P(N=n)$, including $n=0$ if it is possible.

Build each conditional severity sum.
For each $n$, find the possible totals and their probabilities assuming exactly $n$ claims. Combine different claim combinations that give the same total.

Apply the mixture weights.
Multiply each conditional probability for count $n$ by $P(N=n)$, and add contributions that reach the same total.

Check the table.
Confirm that every listed total is attainable and that the unconditional probabilities sum to 1.

Worked Tables

Worked Example: A Two-Point Severity and a Correctly Indexed Total

A hypothetical portfolio has claim count $N$ with $P(N=0)=0.25$, $P(N=1)=0.625$, and $P(N=2)=0.125$. Each claim independently costs either 1 or 2 units, each with probability 0.5. Find the aggregate distribution. The method is to calculate the conditional distribution for each claim count, then apply the frequency mixture weights.

Zero claims. If $N=0$, the aggregate is 0 with probability 1. Its unconditional contribution is $0.25$ at total 0.

One claim. Conditional on $N=1$, totals 1 and 2 each have probability 0.5. Weighting by $P(N=1)=0.625$ gives contributions $0.625(0.5)=0.3125$ at total 1 and $0.3125$ at total 2.

Two claims. The ordered severity pairs are $(1,1),(1,2),(2,1),(2,2)$, each with probability $0.5(0.5)=0.25$. Thus the conditional probabilities at totals 2, 3, and 4 are $0.25,0.50,0.25$, respectively. Weighting by $P(N=2)=0.125$ gives contributions $0.03125,0.0625,0.03125$ at those totals.

Combine equal totals. Total 2 receives both a one-claim and a two-claim contribution. Total 3 can only come from two claims: one claim can produce only 1 or 2. Therefore the one-claim contribution at total 3 is 0, while the two-claim contribution is $0.125(0.50)=0.0625$.

\begin{aligned} P(S=0)&=0.25,\\ P(S=1)&=0.3125,\\ P(S=2)&=0.3125+0.03125=0.34375,\\ P(S=3)&=0+0.0625=0.0625,\\ P(S=4)&=0.03125. \end{aligned}

Check. The probabilities sum to $0.25+0.3125+0.34375+0.0625+0.03125=1$. The table contains only attainable totals. This calculation avoids the common mistake of carrying a one-claim probability at total 2 over to total 3.

Worked Example: Separating Overlapping Totals From Distinct Totals

For a hypothetical policy group, let $P(N=0)=0.20$, $P(N=1)=0.50$, and $P(N=2)=0.30$. Given a claim, its amount is 2 units with probability 0.75 or 5 units with probability 0.25. Find the aggregate distribution and $P(S=7)$. The method is to form the two-claim convolution before weighting it.

Zero claims. The aggregate is 0, contributing $0.20$.

One claim. The conditional totals are 2 with probability 0.75 and 5 with probability 0.25. After weighting by 0.50, the contributions are $0.50(0.75)=0.375$ at total 2 and $0.50(0.25)=0.125$ at total 5.

Two claims. Both claims are 2 with probability $0.75^2=0.5625$, giving total 4. Exactly one claim is 2 and one is 5 in two ordered ways, so total 7 has probability $2(0.75)(0.25)=0.375$. Both claims are 5 with probability $0.25^2=0.0625$, giving total 10. The conditional probabilities sum to $0.5625+0.375+0.0625=1$. Weighting by 0.30 gives contributions 0.16875 at 4, 0.1125 at 7, and 0.01875 at 10.

Combine and answer. The one-claim total 5 and the two-claim total 5 do not overlap in this model: two claims can total 4, 7, or 10, but not 5. Thus:

\begin{array}{c|rrrrrrr} s & 0 & 2 & 4 & 5 & 7 & 10\\ \hline P(S=s) & 0.20 & 0.375 & 0.16875 & 0.125 & 0.1125 & 0.01875 \end{array}

Therefore $P(S=7)=0.1125$. As a check, the table sums to $0.20+0.375+0.16875+0.125+0.1125+0.01875=1$. The key trap avoided is treating the probability of one particular ordering, 2 followed by 5, as the entire probability of total 7; the reverse ordering also produces 7.

Worked Example: A Three-Point Severity With Several Coincident Sums

Consider a hypothetical aggregate model with $P(N=0)=0.10$, $P(N=1)=0.60$, and $P(N=2)=0.30$. Each claim is independently 1 unit with probability 0.50, 2 units with probability 0.25, or 3 units with probability 0.25. Find the full aggregate distribution. The method is to enumerate the two-claim ordered pairs, combine equal sums, and then apply the count probabilities.

One claim. Weighting its severity probabilities by 0.60 gives contributions 0.30 at total 1, 0.15 at total 2, and 0.15 at total 3.

Two claims. The ordered pairs and their probabilities are:

Total 2: $(1,1)$, with probability $0.50(0.50)=0.25$.
Total 3: $(1,2)$ and $(2,1)$, with probability $2(0.50)(0.25)=0.25$.
Total 4: $(1,3)$, $(3,1)$, and $(2,2)$, with probability $2(0.50)(0.25)+(0.25)(0.25)=0.3125$.
Total 5: $(2,3)$ and $(3,2)$, with probability $2(0.25)(0.25)=0.125$.
Total 6: $(3,3)$, with probability $0.25(0.25)=0.0625$.

These conditional probabilities sum to $0.25+0.25+0.3125+0.125+0.0625=1$. Multiplying by $P(N=2)=0.30$ gives contributions of 0.075, 0.075, 0.09375, 0.0375, and 0.01875 at totals 2 through 6, respectively.

Combine with the other claim counts. Total 2 also receives a one-claim contribution; total 3 does as well. Totals 4, 5, and 6 can only arise from two claims here.

\begin{array}{c|rrrrrrr} s & 0 & 1 & 2 & 3 & 4 & 5 & 6\\ \hline P(S=s) & 0.10 & 0.30 & 0.225 & 0.225 & 0.09375 & 0.0375 & 0.01875 \end{array}

Check. The probabilities sum to $0.10+0.30+0.225+0.225+0.09375+0.0375+0.01875=1$. For example, total 4 has probability 0.09375, not the probability of only $(1,3)$ and $(3,1)$: the $(2,2)$ pair must also be included.

Why the Mixture Formula Holds

The claim count takes one value at a time. The events $N=0,N=1,\ldots$ are mutually exclusive and cover all possibilities for $N$. For any fixed total $s$, the events $\{S=s,N=n\}$ are therefore disjoint across $n$, and their union is $\{S=s\}$. The addition rule for disjoint events and the definition of conditional probability give the mixture formula:

\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). \end{aligned}

This argument does not require a particular named frequency distribution. It applies whenever the claim count is nonnegative integer-valued and the conditional aggregate probabilities are defined. Independence and identical distribution of severities make the conditional convolution table especially simple; if severities are dependent or have different distributions, the conditional sum must instead be calculated using their actual joint model.

Checks and Common Traps

A small table is manageable enough to verify directly. Use these checks before accepting a result:

Check the support by count. With positive severities bounded between $m$ and $M$, exactly $n$ claims can only total between $nm$ and $nM$. More specifically, only sums formed from the listed severity values are possible.
Check ordered combinations. For distinct severities, both orders contribute to a two-claim total unless the model treats the claims differently.
Check frequency weights. A conditional probability must be multiplied by $P(N=n)$ exactly once. Do not add conditional distributions across claim counts without their weights.
Check overlapping totals. If different claim counts can produce the same aggregate amount, add their weighted contributions. If a count cannot produce that amount, its contribution is zero.
Check the total probability. Conditional probabilities for each fixed count should sum to 1, and the final unconditional table should also sum to 1.

A probability table can sum to 1 and still be wrong if a contribution is assigned to an impossible total and another entry is omitted. Check both the arithmetic and the combinations that support each total. These hand calculations are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each model, calculate the requested aggregate probability by conditioning on the claim count and forming the needed severity sums.

A hypothetical model has $P(N=0)=0.30$, $P(N=1)=0.50$, and $P(N=2)=0.20$. Each claim is 2 or 4 units with probability 0.50 each. Find $P(S=4)$. (Answer: 0.3000)
Let $P(N=0)=0.20$, $P(N=1)=0.50$, and $P(N=2)=0.30$. A claim is 1 unit with probability 0.75 and 3 units with probability 0.25. Find $P(S=3)$. (Answer: 0.1250)
A claim count has a binomial distribution with parameters 2 and 0.5. Each claim is independently 1 or 2 units with probability 0.50 each. Find $P(S=2)$. (Answer: 0.3125)
Let $P(N=0)=0.10$, $P(N=1)=0.60$, and $P(N=2)=0.30$. Each claim is 1 unit with probability 0.50, 2 units with probability 0.25, or 3 units with probability 0.25. Find $P(S=4)$. (Answer: 0.09375)
In the model in Question 4, find $P(S\geq 5)$. (Answer: 0.05625)

What You'll Learn