Compound Binomial Aggregate Distributions

Why the Binomial Frequency Model Changes the Aggregate

A binomial claim count is appropriate when there are a fixed number of claim opportunities and each opportunity can produce at most one claim. For example, a hypothetical insurer might cover a fixed group of $n$ devices during a period, with each device independently producing a claim with the same probability $q$. The fixed number of opportunities matters: unlike a Poisson count, the binomial count cannot exceed $n$.

Let $N$ be the number of claims and let $X_1,X_2,\ldots$ be independent claim severities, independent of $N$, with common distribution $X$. The aggregate loss is $S=\sum_{j=1}^{N}X_j$, with $S=0$ when $N=0$. Unless stated otherwise, the examples below use undiscounted losses at the same valuation date; interest and mortality adjustments are not part of this model.

Definition: A compound binomial aggregate has $N\sim\operatorname{Bin}(n,q)$, where $n$ is a fixed number of independent claim opportunities and $q$ is the probability of a claim at each opportunity. Conditional on $N$, the claim severities are independent copies of $X$, and they are independent of $N$. The aggregate loss is $S=\sum_{j=1}^{N}X_j$.

The identical claim probability and independence assumptions are substantive. If opportunities have different claim probabilities, or their claim indicators are dependent, $N$ need not be binomial. If one opportunity can generate multiple claims, this model does not describe that opportunity without further modeling.

Generating Functions for a Compound Binomial

For nonnegative integer-valued severity $X$, its probability generating function (PGF) is $G_X(z)=E[z^X]$. The aggregate PGF is obtained by first conditioning on the number of claims. Given $N=k$, the PGF of the sum of $k$ independent severities is $G_X(z)^k$.

G_S(z)=E[z^S]=E\!\left[G_X(z)^N\right] =\left(1-q+qG_X(z)\right)^n.

The last equality uses the PGF of a binomial random variable: $E[t^N]=(1-q+qt)^n$. Another way to see the result is to consider each of the $n$ opportunities separately. Its contribution is zero with probability $1-q$, and, with probability $q$, it contributes a severity with PGF $G_X(z)$. Its contribution PGF is therefore $1-q+qG_X(z)$. Independence across the $n$ opportunities means these PGFs multiply.

For a severity distribution that is not integer-valued, the same conditioning argument gives a moment generating function (when it exists):

M_S(t)=E[e^{tS}] =\left(1-q+qM_X(t)\right)^n.

The PGF is particularly useful for exact probabilities when severities take discrete integer values. The coefficient of $z^s$ in $G_S(z)$ is $P(S=s)$. In particular, if claim severities are strictly positive, then $G_X(0)=0$ and $P(S=0)=(1-q)^n$. If zero-valued severities are possible, the probability of zero aggregate can also include cases where claims occur but their severities are zero.

Aggregate Mean and Variance

Write $\mu_X=E[X]$ and $\sigma_X^2=\operatorname{Var}(X)$, assuming the severity has a finite second moment. The binomial count has $E[N]=nq$ and $\operatorname{Var}(N)=nq(1-q)$. Applying conditional expectation gives the aggregate mean; applying the conditional variance formula gives its variance.

E[S]=nqE[X], \qquad \operatorname{Var}(S)=nq\operatorname{Var}(X)+nq(1-q)\bigl(E[X]\bigr)^2.

An equivalent variance formula, using $E[X^2]=\operatorname{Var}(X)+(E[X])^2$, is:

\operatorname{Var}(S)=nqE[X^2]-nq^2\bigl(E[X]\bigr)^2.

To justify the formulas, conditional on $N$, the aggregate is a sum of $N$ independent severities. Thus $E[S\mid N]=N E[X]$ and $\operatorname{Var}(S\mid N)=N\operatorname{Var}(X)$. Taking expectations yields $E[S]=E[N]E[X]$. For the variance, the conditional variance identity gives:

\begin{aligned} \operatorname{Var}(S) &=E[\operatorname{Var}(S\mid N)]+\operatorname{Var}(E[S\mid N])\\ &=E[N]\operatorname{Var}(X)+\operatorname{Var}(N)\bigl(E[X]\bigr)^2\\ &=nq\operatorname{Var}(X)+nq(1-q)\bigl(E[X]\bigr)^2. \end{aligned}

The second term reflects variation in the number of claims. Its factor $1-q$ is a consequence of binomial frequency variation; replacing the binomial count by a Poisson count with mean $nq$ would instead use $\operatorname{Var}(N)=nq$. The two models therefore do not generally have the same aggregate variance.

Worked Example: Aggregate Moments and the Binomial Adjustment

A hypothetical portfolio has $n=4$ independent claim opportunities. Each produces a claim with probability $q=0.20$. Conditional on a claim, the undiscounted severity $X$, in thousands of dollars, is 1 with probability 0.75 and 3 with probability 0.25. Find the aggregate mean and variance. No interest or mortality adjustment applies.

Method and severity moments. The claim count is $N\sim\operatorname{Bin}(4,0.20)$. The severity distribution gives $E[X]=1(0.75)+3(0.25)=1.50$ thousand dollars, and $E[X^2]=1^2(0.75)+3^2(0.25)=0.75+2.25=3.00$ thousand dollars squared. Therefore, $\operatorname{Var}(X)=3.00-(1.50)^2=0.75$ thousand dollars squared.

Mean. Using $E[S]=nqE[X]$, $E[S]=4(0.20)(1.50)=1.20$ thousand dollars, or $1,200.00.

Variance. Using the conditional-variance form, $\operatorname{Var}(S)=4(0.20)(0.75)+4(0.20)(0.80)(1.50)^2 =0.60+1.44=2.04$ thousand dollars squared.

Check. The equivalent second-moment expression gives $4(0.20)(3.00)-4(0.20)^2(1.50)^2=2.40-0.36=2.04$, agreeing with the result. The variance is in thousands of dollars squared; the mean is in thousands of dollars.

Common mistake avoided. Using the compound Poisson variance $nqE[X^2]$ would give $2.40$, not $2.04$. That calculation incorrectly treats the count variance as $nq$; the binomial variance is $nq(1-q)$.

Calculating Exact Probabilities from a Discrete Severity

For a small discrete-severity model, the aggregate PGF provides an exact probability calculation. Form the PGF for the contribution from one opportunity, raise it to the $n$th power, and identify which combinations of contributions produce the requested aggregate. This is a coefficient calculation, not a new frequency model.

Worked Example: Finding an Aggregate Probability by Coefficients

A hypothetical set of three independent equipment contracts has one claim opportunity per contract. Each opportunity produces a claim with probability $q=0.50$. Given a claim, the undiscounted severity is 1 or 2 thousand dollars, each with probability 0.50. Find $P(S=2)$. No interest or mortality adjustment applies.

Method: form the aggregate PGF. The severity PGF is $G_X(z)=0.50z+0.50z^2$. With $n=3$, the aggregate PGF is $G_S(z)=[0.50+0.50G_X(z)]^3=[0.50+0.25z+0.25z^2]^3$. Each factor represents one contract: no claim contributes zero, while a claim contributes 1 or 2.

Identify the ways to total 2. One contract can contribute 2 while the other two contribute zero. There are 3 choices for that contract, giving probability $3(0.25)(0.50)^2=0.1875$. Alternatively, two contracts can each contribute 1 and the remaining contract contributes zero. There are 3 choices for the zero-contribution contract, giving probability $3(0.25)^2(0.50)=0.09375$.

Add the disjoint cases. Thus $P(S=2)=0.1875+0.09375=0.28125$, or 0.2813 to four decimal places.

Check. These cases correspond to the $z^2$ terms in the expanded PGF: choosing one $0.25z^2$ and two constant terms, or choosing two $0.25z$ terms and one constant term. A claim count of zero or one cannot produce aggregate 2 except for the one-claim severity-2 case already included; a claim count of three has minimum aggregate 3. The cases are exhaustive and mutually exclusive.

Common mistake avoided. The severity probabilities are conditional on a claim. The unconditional probabilities for one contract are $P(0)=0.50$, $P(1)=0.50(0.50)=0.25$, and $P(2)=0.50(0.50)=0.25$. Applying the claim probability $q$ a second time after using these unconditional probabilities would understate the aggregate probability.

Worked Example: A Target Aggregate with Two Possible Claim Counts

A hypothetical pool has $n=5$ independent claim opportunities, each with claim probability $q=0.40$. Conditional on a claim, severity $X$, in thousands of dollars, is 2 with probability 0.60 and 5 with probability 0.40. Find $P(S=10)$. Losses are undiscounted; interest and mortality are not modeled.

Method: use the one-opportunity PGF. The severity PGF is $G_X(z)=0.60z^2+0.40z^5$, so the aggregate PGF is $G_S(z)=[0.60+0.24z^2+0.16z^5]^5$. The three terms inside the brackets represent no claim, a claim of 2, and a claim of 5, respectively.

Find all configurations totaling 10. If $a$ opportunities contribute 2 and $b$ contribute 5, then $2a+5b=10$, with $a+b\leq 5$. For nonnegative integers, the only solutions are $(a,b)=(5,0)$ and $(a,b)=(0,2)$. The first is five claims of severity 2. The second is two claims of severity 5 and three no-claim outcomes.

Calculate the probabilities. Five claims of severity 2 have probability $(0.24)^5=0.0007962624$. For two claims of severity 5 and three no claims, choose the two opportunities in $\binom{5}{2}=10$ ways, giving $10(0.16)^2(0.60)^3=10(0.0256)(0.216)=0.055296$.

Combine and check. The two cases are disjoint, so $P(S=10)=0.0007962624+0.055296=0.0560922624$, or 0.0561 to four decimal places. The coefficient of $z^{10}$ in the PGF consists of exactly these two terms, confirming that no other combination contributes.

Common mistake avoided. Counting only two claims of severity 5 misses the five-claim outcome with five severities of 2. The equation $2a+5b=10$, together with the fixed limit of five opportunities, makes the complete set of contributing configurations explicit.

Interpretation, Limits, and Exam Traps

The binomial model is useful when the maximum number of claims is known from the exposure structure and each opportunity can produce no more than one claim. Its frequency variance, $nq(1-q)$, is smaller than its mean $nq$ when $0<q<1$. This distinction carries through to aggregate variance. For a fixed mean claim count, the binomial model’s frequency component is lower than the Poisson model’s, although the severity component and the full aggregate distributions must still be considered.

The model relies on a common claim probability and independent opportunities. If these assumptions do not fit the situation, simply using $nq$ as the expected count does not make the compound binomial distribution appropriate. Similarly, the aggregate PGF formula assumes severities are independent copies of $X$ and independent of the count. If severity depends on which opportunities claim, a single common severity PGF may not describe the aggregate.

Using a Poisson variance formula: the compound Poisson expression $nqE[X^2]$ is not the compound binomial variance. Use $nqE[X^2]-nq^2(E[X])^2$, or the equivalent conditional-variance formula.
Confusing conditional and unconditional severity probabilities: $P(X=x)$ describes severity given that a claim occurs. The probability that an opportunity contributes $x$ is $qP(X=x)$; its no-claim probability is $1-q$.
Ignoring the fixed opportunity limit: the number of claims cannot exceed $n$, so a proposed claim-count case with more than $n$ claims is impossible.
Stopping before checking all coefficient cases: a target aggregate can arise from different claim counts and different severity combinations. Write the contribution equation and enumerate its feasible integer solutions.

Key takeaway: Under independent, identical claim opportunities and independent identically distributed severities, $G_S(z)=[1-q+qG_X(z)]^n$, $E[S]=nqE[X]$, and $\operatorname{Var}(S)=nq\operatorname{Var}(X)+nq(1-q)(E[X])^2$. The binomial adjustment in the variance and the fixed limit on claim count are central features of the model.

Check Your Understanding

For each problem, treat losses as undiscounted at the same valuation date, with no interest or mortality adjustment. Give probabilities to four decimal places where appropriate.

A hypothetical portfolio has four independent claim opportunities, each with claim probability 0.30. Every claim has a fixed severity of $2. Find the aggregate mean and variance. (Answer: mean $2.40; variance $3.36.)
There are three independent opportunities with claim probability 0.20. Given a claim, severity is 1 or 3, each with probability 0.50. Find the aggregate mean and variance. (Answer: mean 1.20; variance 2.52.)
For two opportunities with claim probability 0.40, severity given a claim is 1 or 2 with equal probability. Find the aggregate PGF and $P(S=2)$. (Answer: $G_S(z)=(0.60+0.20z+0.20z^2)^2$; $P(S=2)=0.2800$.)
Four independent opportunities each have claim probability 0.50. Given a claim, severity is 2 with probability 0.75 and 4 with probability 0.25. Find the probability of aggregate severity 4. (Answer: $0.2734$.)
Eight independent opportunities each have claim probability 0.10, and each claim has fixed severity $2,500. Find the probability of zero aggregate loss and the aggregate mean. (Answer: probability $0.4305$; mean $2,000.00.)

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