Compound Binomial Models

Why a Binomial Count Changes an Aggregate Model

An aggregate loss is the total payment from a group of claims. When there are a fixed number of potential claimants or exposures, and each independently produces a claim with the same probability, the number of claims can be modeled by a binomial random variable. The aggregate is then a mixture: its value depends both on how many claims occur and on the severity of each claim.

This tutorial develops that model for small trial counts, where exact probabilities can be found by conditioning on the number of claims. The key is to keep the two stages distinct: first determine the probability of each possible claim count, then determine the aggregate severity distribution conditional on that count.

Definition: Suppose there are $n$ independent potential claim opportunities, each producing a claim with probability $r$. Let $N\sim\operatorname{Bin}(n,r)$ be the number of claims. Let $X_1,X_2,\ldots$ be independent, identically distributed claim severities, independent of $N$, with $E[X_i]=\mu$ and $\operatorname{Var}(X_i)=\sigma^2$. The aggregate loss is $$ S=\sum_{j=1}^{N}X_j, $$ where an empty sum is defined to be zero. Thus $S=0$ when $N=0$.

The binomial count probabilities are $$ P(N=k)=\binom{n}{k}r^k(1-r)^{n-k},\qquad k=0,1,\ldots,n. $$ For $k$ claims, the conditional aggregate is a sum of $k$ severities. Write $S_k=X_1+\cdots+X_k$, with $S_0=0$. Conditioning on $N$ gives the general probability formula $$ P(S\in B)=\sum_{k=0}^{n}P(N=k)P(S_k\in B) $$ for any set of aggregate values $B$. For an exact value $s$, take $B=\{s\}$. This formula applies whether severities are discrete or continuous, though exact-value probabilities are most directly useful for discrete or lattice-valued severities.

Calculating Aggregate Probabilities by Conditioning

For small $n$, list the possible claim counts and identify which severity combinations can produce the target event at each count. Counts that cannot produce the event contribute zero. The conditional probability for a count must then be multiplied by that count's binomial probability.

Find the count probabilities.
Calculate $P(N=k)$ for the possible values $k=0,\ldots,n$, checking that they sum to 1.

Analyze the severities at each count.
For each $k$, determine whether $k$ severities can satisfy the event and calculate the corresponding conditional probability.

Weight and add.
Multiply each conditional probability by $P(N=k)$, then sum the contributions over all counts.

Worked Example: A Target Total From One, Two, or Three Claims

Hypothetical model. A portfolio has three independent claim opportunities. Each produces a claim with probability $r=0.25$. Conditional on a claim, its payment is $\$1$ or $\$2$, each with probability $0.5$. Claim payments are independent. Find the probability that the aggregate payment is $\$3$. Use conditioning on the claim count, and state the calculation that avoids confusing the two-claim and three-claim cases.

Method: condition on $N$. Here $N\sim\operatorname{Bin}(3,0.25)$. Its probabilities are $$ P(N=0)=0.75^3=0.421875,\quad P(N=1)=3(0.25)(0.75)^2=0.421875, $$ $$ P(N=2)=3(0.25)^2(0.75)=0.140625,\quad P(N=3)=0.25^3=0.015625. $$ They sum to $1.000000$.

With zero or one claim, the aggregate cannot be $\$3$, because zero claims pay $\$0$ and one claim pays at most $\$2$. Given two claims, the sum is $\$3$ when the payments are $(1,2)$ or $(2,1)$. These are two of the four equally likely ordered pairs, so $P(S=3\mid N=2)=2/4=0.5$.

Given three claims, the aggregate is $\$3$ only when all three payments are $\$1$. By independence, $P(S=3\mid N=3)=0.5^3=0.125$. Therefore, $$ P(S=3)=0.140625(0.5)+0.015625(0.125) =0.0703125+0.001953125 =0.072265625. $$ Rounded to four decimal places, $P(S=3)=0.0723$.

Check and common mistake. The only contributing counts are two and three, and their contributions sum to less than $P(N=2)+P(N=3)=0.15625$, as they should. A wrong calculation might use $P(S=3\mid N=2)=0.5$ but then multiply the three-claim probability by $0.5^2$, giving $0.005859375$ for that contribution. Three claims require three severity outcomes, so the correct conditional probability is $0.5^3$, and the correct total is $0.072265625$.

The same conditioning principle works for cumulative probabilities. In a discrete model, $P(S\leq s)$ is found by calculating $P(S_k\leq s)$ for each count and weighting those probabilities. The conditional event changes with $k$: a threshold that includes every one-claim outcome may include only some two-claim outcomes, and no outcomes at a larger count.

Worked Example: A Threshold With Different Contributions by Count

Hypothetical model. Three independent opportunities each have claim probability $r=0.40$. Each claim pays $\$2$ with probability $0.75$ or $\$5$ with probability $0.25$, independently of all other claims and the count. Find $P(S\leq \$5)$.

Method: sum conditional threshold probabilities. The count probabilities are $$ P(N=0)=0.6^3=0.216,\quad P(N=1)=3(0.4)(0.6)^2=0.432, $$ $$ P(N=2)=3(0.4)^2(0.6)=0.288,\quad P(N=3)=0.4^3=0.064. $$ These add to $1.000$.

If $N=0$, the aggregate is $\$0$, so $P(S\leq5\mid N=0)=1$. If $N=1$, either possible payment is at most $\$5$, so $P(S\leq5\mid N=1)=1$. If $N=2$, the total is at most $\$5$ only when both claims pay $\$2$; the conditional probability is $0.75^2=0.5625$. If $N=3$, even the smallest possible total is $3(2)=\$6$, so the conditional probability is zero.

Weighting these results gives $$ P(S\leq5)=0.216(1)+0.432(1)+0.288(0.5625)+0.064(0) =0.216+0.432+0.162=0.810. $$ Thus $P(S\leq \$5)=0.8100$ to four decimal places.

Check and common mistake. The count probabilities sum to $1$, and the event excludes every three-claim outcome. A common error is to count all two-claim outcomes as satisfying the threshold. Only the $(2,2)$ severity pair has total at most $\$5$; pairs involving a $\$5$ payment exceed it.

Aggregate Expectation and Variance

The same count-conditioning setup gives the aggregate moments. Under the assumptions in the definition, conditional on $N=k$, the aggregate is the sum of $k$ independent severities. Its conditional expectation is $k\mu$, and its conditional variance is $k\sigma^2$. The binomial count has $E[N]=nr$ and $\operatorname{Var}(N)=nr(1-r)$.

Result: For $S=\sum_{j=1}^{N}X_j$, with $N\sim\operatorname{Bin}(n,r)$ independent of iid severities with mean $\mu$ and variance $\sigma^2$, $$ E[S]=nr\mu $$ and $$ \operatorname{Var}(S)=nr\sigma^2+nr(1-r)\mu^2. $$ Equivalently, if $E[X^2]=\sigma^2+\mu^2$, then $\operatorname{Var}(S)=nrE[X^2]-nr^2\mu^2$.

To justify these formulas, first condition on the count. Taking the expectation of the conditional mean gives $$ E[S]=E\bigl[E[S\mid N]\bigr]=E[N\mu]=\mu nr. $$ For variance, the law of total variance separates the average conditional variance from the variation in the conditional mean: $$ \operatorname{Var}(S)=E[\operatorname{Var}(S\mid N)] +\operatorname{Var}(E[S\mid N]). $$ Since $\operatorname{Var}(S\mid N)=N\sigma^2$ and $E[S\mid N]=N\mu$, the first term is $\sigma^2 E[N]=nr\sigma^2$, and the second is $\mu^2\operatorname{Var}(N)=nr(1-r)\mu^2$. Adding the terms proves the variance formula. The first term reflects severity variability among claims; the second reflects variability in the number of claims.

Worked Example: Probability and Moments in a Two-Opportunity Portfolio

Hypothetical model. There are two independent opportunities, each with claim probability $r=0.30$. Each claim pays $\$1$ or $\$4$, each with probability $0.5$, independently. Find $P(S=\$4)$, $E[S]$, and $\operatorname{Var}(S)$. Give the probability to four decimals and the monetary moments in dollars and squared dollars.

Method: condition for the probability, then use the moment formulas. The count probabilities are $P(N=0)=0.7^2=0.49$, $P(N=1)=2(0.3)(0.7)=0.42$, and $P(N=2)=0.3^2=0.09$. Given one claim, a $\$4$ total occurs with probability $0.5$. Given two claims, the possible totals are $\$2,\$5,\$8$, so none equals $\$4$. Hence $$ P(S=4)=0.49(0)+0.42(0.5)+0.09(0)=0.2100. $$

For one claim, the severity mean and second moment are $$ \mu=1(0.5)+4(0.5)=2.5,\qquad E[X^2]=1^2(0.5)+4^2(0.5)=8.5. $$ Thus $\sigma^2=8.5-(2.5)^2=8.5-6.25=2.25$. The count has $E[N]=2(0.30)=0.6$ and $\operatorname{Var}(N)=2(0.30)(0.70)=0.42$. Therefore, $$ E[S]=0.6(2.5)=1.50\text{ dollars}, $$ $$ \operatorname{Var}(S)=0.6(2.25)+0.42(2.5)^2 =1.35+0.42(6.25)=1.35+2.625=3.975\text{ dollars}^2. $$

Check and common mistake. The possible aggregate values have probabilities $P(S=0)=0.49$, $P(S=1)=P(S=4)=0.21$, $P(S=2)=0.0225$, $P(S=5)=0.045$, and $P(S=8)=0.0225$; these sum to $1$. Their mean is $1.50$, and their second moment is $6.225$, giving variance $6.225-(1.50)^2=3.975$. A common error is to use only $nr\sigma^2=1.35$ as the aggregate variance. That omits the additional $2.625$ from the random claim count.

Zero Claims, Assumptions, and Exam Traps

When every claim severity is strictly positive, $S=0$ happens exactly when $N=0$. In that case, $$ P(S=0)=P(N=0)=(1-r)^n. $$ If a claim can have zero severity, however, positive claim counts may also produce zero aggregate loss, so this equality need not hold. The event of no claims and the event of no payment are not always the same.

The binomial model requires a fixed number $n$ of opportunities, the same claim probability $r$ for each opportunity, and independence among claim occurrences. The moment formulas additionally require severity distributions that are identical across claims, independent of the count, and independent of one another. If claim probabilities or severity distributions differ, these formulas may not apply as stated.

A useful check for exact calculations is to ask whether the target total is possible at each count before multiplying probabilities. Also verify that the severity probabilities sum to one and that the count probabilities sum to one. These checks often reveal a missing severity combination, an incorrect exponent, or a count included even though its minimum possible aggregate exceeds the target.

Key takeaway: A compound binomial aggregate is calculated by weighting each conditional severity distribution with its binomial count probability. For moments, $E[S]=nr\mu$, while the variance contains both severity variation and claim-count variation.

Check Your Understanding

For each problem, use the stated independence assumptions and condition on the claim count where needed. Probabilities are rounded to four decimal places.

A hypothetical portfolio has two independent opportunities, each with claim probability $0.20$. Every claim pays exactly $\$3$. Find the probability of an aggregate payment of $\$3$. (Answer: $0.3200$)
Three independent opportunities each have claim probability $0.50$. Each claim pays $\$1$ or $\$2$ with equal probability, independently. Find $P(S=2)$. (Answer: $0.2813$)
Two independent opportunities each have claim probability $0.40$. A claim pays $\$2$ with probability $0.75$ and $\$6$ with probability $0.25$. Find $P(S\leq \$6)$. (Answer: $0.9300$)
Four independent opportunities each have claim probability $0.25$, and every claim pays exactly $\$4$. Find $E[S]$ and $\operatorname{Var}(S)$. (Answer: $E[S]=\$4$, $\operatorname{Var}(S)=12\text{ dollars}^2$)
Two independent opportunities each have claim probability $0.30$. Each claim pays $\$1$ or $\$3$ with equal probability. Find $P(S=\$3)$ and $E[S]$. (Answer: $P(S=3)=0.2100$, $E[S]=\$1.20$)

What You'll Learn