Compound Models With Binomial and Negative Binomial Frequencies

Why the Frequency Model Matters

An aggregate loss is the total payment from a random number of claims. Its distribution depends on both the claim count and the payment amounts. If the count is binomial, it is bounded by a fixed number of opportunities. If it is negative binomial, the count can be unbounded and may exhibit more variability than a Poisson count with the same mean. These differences affect both aggregate moments and exact probabilities.

This tutorial considers compound models in which claim payments are independent and identically distributed and are independent of the claim count. Those assumptions allow the aggregate to be analyzed by conditioning on the count. They must be checked before using the formulas: if payments depend on the count, or if payments are dependent on one another, the standard results need not hold. The methods here are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Definition: Let $N$ be a claim count, and let $X_1,X_2,\ldots$ be independent and identically distributed claim payments, independent of $N$. The aggregate payment is $$ S=\sum_{j=1}^{N}X_j, $$ where the sum is defined as zero when $N=0$. Write $m=E[X]$ and $\sigma_X^2=\operatorname{Var}(X)$ for the mean and variance of one payment. Payments may include zero values if the model counts events that result in no payment.

Frequency Assumptions and Aggregate Moments

For a binomial frequency model, there are $n$ independent opportunities for a claim, each with probability $p$. Thus $N\sim\operatorname{Bin}(n,p)$, with $E[N]=np$ and $\operatorname{Var}(N)=np(1-p)$. The count cannot exceed $n$.

For a negative binomial model, the parameterization must be made explicit. Here, $N$ is the number of failures before the $r$th success in independent Bernoulli trials, where each trial has success probability $p$. Write $q=1-p$. Then $P(N=k)=\binom{k+r-1}{k}p^r q^k$, for $k=0,1,\ldots$, and $E[N]=rq/p$, $\operatorname{Var}(N)=rq/p^2$. Other negative binomial conventions exist, so a formula should not be used until the meaning of its parameters is established.

Theorem: For a compound model with $N$ independent of iid payments $X_j$, provided the moments are finite, $$ E[S]=E[N]E[X], \qquad \operatorname{Var}(S)=E[N]\operatorname{Var}(X)+\operatorname{Var}(N)(E[X])^2. $$ Consequently, for $N\sim\operatorname{Bin}(n,p)$, $$ E[S]=np\,m,\qquad \operatorname{Var}(S)=np\,\sigma_X^2+np(1-p)m^2, $$ and for the negative binomial parameterization above, $$ E[S]=\frac{rq}{p}m,\qquad \operatorname{Var}(S)=\frac{rq}{p}\sigma_X^2+\frac{rq}{p^2}m^2. $$

The variance formula separates two sources of uncertainty. The term $E[N]\operatorname{Var}(X)$ reflects variation in payment amounts when the count is held fixed. The term $\operatorname{Var}(N)(E[X])^2$ reflects variation in how many payments occur. A variable payment amount does not replace the frequency-variation term.

Why the Moment Formula Holds

Conditional on $N=n$, the aggregate is a sum of $n$ iid payments, so $E[S\mid N=n]=nm$ and $\operatorname{Var}(S\mid N=n)=n\sigma_X^2$. Taking expectations gives the mean. Applying the law of total variance gives the variance:

E[S]=E\{E[S\mid N]\}=E[Nm]=E[N]m, $$ $$ \operatorname{Var}(S) =E\{\operatorname{Var}(S\mid N)\} +\operatorname{Var}\{E[S\mid N]\} =E[N]\sigma_X^2+\operatorname{Var}(N)m^2.

This argument does not require $N$ to be Poisson. It does require the conditional sum to have the stated mean and variance, which follows from the iid payment assumption and independence from $N$. If payment amounts change with the number of claims, the conditional calculations must be reconsidered.

Worked Example: Binomial Opportunities With Zero Payments

A hypothetical device has $n=3$ independent one-year failure opportunities, each with failure probability $p=0.5$. For each failure, the insurer's payment $X$, after the contract terms have been applied, is \$0 with probability $0.2$, \$2 with probability $0.5$, and \$5 with probability $0.3$. Payments are independent across failures and of the failure indicators. Find the aggregate mean and variance, and calculate $P(S=5)$.

Method and assumptions. The number of failures is $N\sim\operatorname{Bin}(3,0.5)$. Treat each failure opportunity's payment as an iid payment with the given distribution; a zero payment is possible even when a failure occurs.

First calculate the payment moments: $$ E[X]=0(0.2)+2(0.5)+5(0.3)=0+1+1.5=2.5, $$ $$ E[X^2]=0^2(0.2)+2^2(0.5)+5^2(0.3)=0+2+7.5=9.5, $$ $$ \operatorname{Var}(X)=9.5-(2.5)^2=9.5-6.25=3.25. $$

For the binomial count, $E[N]=3(0.5)=1.5$ and $\operatorname{Var}(N)=3(0.5)(0.5)=0.75$. Therefore, $$ E[S]=1.5(2.5)=3.75, \qquad \operatorname{Var}(S)=1.5(3.25)+0.75(2.5)^2=4.875+4.6875=9.5625. $$

For the probability, each of the three opportunities has aggregate-contribution probabilities $P(0)=0.5+0.5(0.2)=0.6$, $P(2)=0.5(0.5)=0.25$, and $P(5)=0.5(0.3)=0.15$. A total of 5 occurs only when one opportunity contributes 5 and the other two contribute 0. There are three choices for the opportunity contributing 5, so $$ P(S=5)=3(0.15)(0.6)^2=0.162. $$ The result is a mean of \$3.75, a variance of $9.5625$ dollars squared, and probability $0.1620$ to four decimal places. This check also confirms that zero payments after failures are included rather than mistaken for no failure.

Exact Probabilities by Conditioning

For a target aggregate value or range, condition on $N=k$. When payments are discrete, the conditional aggregate distribution is the $k$-fold convolution of the payment distribution; for $k=0$, all probability is at $S=0$. Therefore, for an event $B$, $$ P(S\in B)=\sum_{k=0}^{\infty}P(N=k)P(X_1+\cdots+X_k\in B). $$ For a binomial count the sum stops at $n$, while for the negative binomial count it is generally infinite. In practice, terms can often be ruled out by the payment support or by the event being evaluated.

For integer-valued payments, probability generating functions offer a compact expression for the same conditioning calculation. If $G_X(z)=E[z^X]$, then independence gives $G_S(z)=E[(G_X(z))^N]$. Substituting the frequency generating function yields:

G_S(z)=[1-p+pG_X(z)]^n \quad\text{for }N\sim\operatorname{Bin}(n,p), $$ $$ G_S(z)=\left(\frac{p}{1-qG_X(z)}\right)^r \quad\text{for }N\sim\operatorname{NegBin}(r,p) \text{ under the stated failures-before-}r\text{-successes convention.}

The coefficient of $z^s$ in $G_S(z)$ is $P(S=s)$. These formulas encode convolution; they do not make the frequency model Poisson. In particular, Poisson-specific thinning identities or recursions should not be applied just because a model is compound. Use a method whose assumptions match the actual count distribution.

Worked Example: Negative Binomial Count and a Target Aggregate

A hypothetical portfolio has $N\sim\operatorname{NegBin}(r=2,p=0.5)$, where $N$ counts failures before the second success. Each claim payment is independently \$2 or \$4, each with probability $0.5$, and payments are independent of $N$. Find $E[S]$, $\operatorname{Var}(S)$, and $P(S=4)$.

Method and parameterization. Here $q=1-p=0.5$, so $E[N]=rq/p=2$ and $\operatorname{Var}(N)=rq/p^2=4$. The probability mass function is $P(N=k)=\binom{k+1}{k}(0.5)^2(0.5)^k$.

The payment moments are $$ E[X]=2(0.5)+4(0.5)=3,\qquad E[X^2]=2^2(0.5)+4^2(0.5)=10, $$ $$ \operatorname{Var}(X)=10-3^2=1. $$ Thus $$ E[S]=2(3)=6,\qquad \operatorname{Var}(S)=2(1)+4(3)^2=2+36=38. $$

To obtain $P(S=4)$, condition on the count. If $N=1$, the payment must be 4, with conditional probability $0.5$. If $N=2$, both payments must be 2, with conditional probability $0.5^2=0.25$. Counts of zero or at least three cannot produce a total of 4: zero claims give 0, while three or more claims pay at least 6. The relevant count probabilities are $$ P(N=1)=\binom{2}{1}(0.5)^2(0.5)=0.25,\qquad P(N=2)=\binom{3}{2}(0.5)^2(0.5)^2=0.1875. $$ Therefore, $$ P(S=4)=0.25(0.5)+0.1875(0.25)=0.125+0.046875=0.171875. $$ The probability is $0.1719$ to four decimal places. Conditioning avoids overlooking the two distinct ways to reach 4: one payment of 4 or two payments of 2.

Worked Example: Binomial Frequency and an Aggregate Threshold

A hypothetical program covers at most two independent events. Each event occurs with probability $0.4$. Given an event, its payment is \$1 or \$3 with equal probability, independently of the other event. Calculate the aggregate mean and variance and find the probability that aggregate payment is at least \$4.

Method and assumptions. Let $N\sim\operatorname{Bin}(2,0.4)$, independent of iid payments. For one payment, $$ E[X]=1(0.5)+3(0.5)=2,\qquad E[X^2]=1^2(0.5)+3^2(0.5)=5, $$ $$ \operatorname{Var}(X)=5-2^2=1. $$ The count moments are $E[N]=2(0.4)=0.8$ and $\operatorname{Var}(N)=2(0.4)(0.6)=0.48$. Hence $$ E[S]=0.8(2)=1.6,\qquad \operatorname{Var}(S)=0.8(1)+0.48(2)^2=0.8+1.92=2.72. $$

A total of at least 4 requires two events: with zero or one event, the aggregate is at most 3. Given two events, the four equally likely payment pairs are $(1,1),(1,3),(3,1),(3,3)$. Exactly three pairs have total at least 4. Since $P(N=2)=0.4^2=0.16$, $$ P(S\ge4)=P(N=2)\frac{3}{4}=0.16(0.75)=0.12. $$ The probability is $0.1200$ to four decimal places. The finite maximum of two claims is important: no omitted count above two can contribute.

Common Traps and Model Checks

Using the wrong negative binomial convention. The formulas $rq/p$ and $rq/p^2$ here apply when $N$ counts failures before $r$ successes. If a problem defines $N$ differently, derive or identify the matching moments before proceeding.
Omitting frequency variation. Using only $E[N]\operatorname{Var}(X)$ understates aggregate variance whenever $\operatorname{Var}(N)>0$ and $E[X]\ne0$. The missing term is $\operatorname{Var}(N)(E[X])^2$.
Treating a zero payment as necessarily a zero claim count. A claim or event can occur and still produce no payment. When computing an aggregate probability, include every way the count and payment amounts can combine to the target.
Applying a Poisson shortcut without checking its assumptions. Binomial and negative binomial counts have different probability structures. A shortcut derived for Poisson frequency does not transfer merely because the aggregate is a sum of random payments.
Ignoring support restrictions. A binomial count cannot exceed $n$, and payment bounds can rule out many conditional cases. Conversely, a negative binomial count has no finite upper bound, so a sum over counts must be truncated only when the event or payment support justifies it.

Key takeaway: Establish the frequency parameterization and independence assumptions first. Use conditioning to obtain aggregate moments, and use the count distribution together with severity convolution to obtain exact probabilities. Match any shortcut to the model rather than assuming Poisson properties.

Check Your Understanding

For each problem, state the relevant frequency model and use conditioning or the compound moment formulas as appropriate.

A hypothetical fund has $N\sim\operatorname{Bin}(4,0.25)$, and every claim pays \$2. Find the aggregate mean and the probability of an aggregate payment of \$4. (Answer: Mean $=\$2.00$; probability $=0.2109$.)
A count is negative binomial with $r=2$, $p=0.5$, counting failures before the second success. Every claim pays \$1. Find the probability of no aggregate payment. (Answer: $0.2500$.)
A portfolio has $N\sim\operatorname{Bin}(2,0.5)$. Each payment is \$1 or \$3 with equal probability, independently. Find the aggregate mean, variance, and probability that the aggregate equals \$4. (Answer: Mean $=\$2.00$; variance $=3.00$ dollars squared; probability $=0.1250$.)
A hypothetical service has $N\sim\operatorname{NegBin}(r=1,p=0.5)$, counting failures before the first success. Each payment is \$1 or \$2 with equal probability. Find the probability that aggregate payment is \$2. (Answer: $0.1563$.)
A program has $N\sim\operatorname{Bin}(3,0.4)$, and each event pays exactly \$5. Find the probability that aggregate payment is at least \$10. (Answer: $0.3520$.)

What You'll Learn