Compound Binomial Distributions

Why a Binomially Counted Sum Is a Mixture

Suppose a fixed number of opportunities are available, each independently producing an event with probability $p$. The number of events is random. If each event carries a random severity, amount, or claim size, the total is a sum with a random number of terms. Its distribution must account for both sources of randomness: how many terms occur and the values of those terms.

The key step is to condition on the event count. If the count is $k$, the total has the ordinary distribution of a sum of $k$ independent severities. The overall law is therefore a finite mixture of these sum distributions, with binomial mixture weights. The term $k=0$ is essential: it contributes a point mass at zero, even when every individual severity has a continuous distribution.

Definition: Let $M$ be a fixed nonnegative integer, let $N\sim\operatorname{Bin}(M,p)$, and let $X_1,X_2,\ldots$ be iid random variables independent of $N$. The compound binomial sum is $$ S=\sum_{i=1}^{N}X_i, $$ where the sum is defined to be $0$ when $N=0$. The $X_i$ are called the severity variables. For $k\geq 1$, write $F^{*k}$ for the distribution of a sum of $k$ independent variables with the same distribution $F$ as $X_1$. Set $F^{*0}=\delta_0$, the distribution placing probability one at zero.

Here $M$ is the number of opportunities, while $N$ is the number that produce a severity. The independence assumption matters: it ensures that conditioning on $N=k$ does not change the distribution of the severities. The convolution distributions $F^{*k}$ are those of iid sums; the $k=0$ convention lets the same mixture expression include the empty sum.

The Conditional Mixture Formula

For every Borel set $A\subseteq\mathbb R$, the law of $S$ is

P(S\in A) = \sum_{k=0}^{M} \binom{M}{k}p^k(1-p)^{M-k}F^{*k}(A).

Equivalently, its distribution function is the binomially weighted sum of the distribution functions of the conditional sums. If $F^{*k}(x)$ denotes the probability that a sum of $k$ iid severities is at most $x$, then

P(S\leq x) = \sum_{k=0}^{M} \binom{M}{k}p^k(1-p)^{M-k}F^{*k}(x).

Theorem: Under the definition above, the distribution of the compound binomial sum is the finite mixture of convolution laws with weights $P(N=k)=\binom{M}{k}p^k(1-p)^{M-k}$, for $k=0,\ldots,M$.

Proof. The events $\{N=k\}$, $k=0,\ldots,M$, partition the sample space. For any Borel set $A$, the law of total probability gives

P(S\in A) = \sum_{k=0}^{M}P(S\in A\mid N=k)P(N=k).

Because $N$ is independent of the sequence of severities, conditional on $N=k$ the variables $X_1,\ldots,X_k$ retain their iid distribution. Thus $P(S\in A\mid N=k)=F^{*k}(A)$. For $k=0$, $S=0$, so this conditional distribution is $\delta_0$, exactly the convention $F^{*0}$. Substituting the binomial probability for $P(N=k)$ proves the formula. $\square$

When the severities are discrete, the formula gives a probability mass function directly. If $f^{*k}(s)$ is the probability that a sum of $k$ iid severities equals $s$, and $f^{*0}(s)$ is one at $s=0$ and zero elsewhere, then

P(S=s) = \sum_{k=0}^{M} \binom{M}{k}p^k(1-p)^{M-k}f^{*k}(s).

Each $f^{*k}$ can be found by repeated discrete convolution. Since there are only $M+1$ possible counts, the mixture is finite, even if the severity distribution has infinite support.

Worked Example: A Discrete Severity Distribution

Let $N\sim\operatorname{Bin}(3,0.2)$. Independently, let the severities be iid with $P(X_i=1)=0.75$ and $P(X_i=2)=0.25$. Find the distribution of $S=\sum_{i=1}^{N}X_i$.

The count probabilities are

P(N=0)=0.512,\quad P(N=1)=0.384,\quad P(N=2)=0.096,\quad P(N=3)=0.008.

For one severity, the probabilities at 1 and 2 are $0.75$ and $0.25$. For two severities, convolution gives probabilities $0.75^2=0.5625$ at 2, $2(0.75)(0.25)=0.375$ at 3, and $0.25^2=0.0625$ at 4. For three severities, the probabilities at 3, 4, 5, and 6 are respectively $0.75^3=0.421875$, $3(0.75)^2(0.25)=0.421875$, $3(0.75)(0.25)^2=0.140625$, and $0.25^3=0.015625$.

Weight each conditional probability by its count probability, and add contributions when different counts produce the same total. This gives

\begin{aligned} P(S=0)&=0.512,\\ P(S=1)&=0.384(0.75)=0.288,\\ P(S=2)&=0.384(0.25)+0.096(0.5625)=0.150,\\ P(S=3)&=0.096(0.375)+0.008(0.421875)=0.039375,\\ P(S=4)&=0.096(0.0625)+0.008(0.421875)=0.009375,\\ P(S=5)&=0.008(0.140625)=0.001125,\\ P(S=6)&=0.008(0.015625)=0.000125. \end{aligned}

The probabilities sum to $0.512+0.288+0.150+0.039375+0.009375+0.001125+0.000125=1$, as required. In particular, $S=0$ occurs only when $N=0$, because every severity is strictly positive.

The Probability Mass at Zero

The zero-count case always contributes $P(N=0)=(1-p)^M$ to $P(S=0)$. It is the entire probability at zero if severities are strictly positive almost surely. But if a severity can itself be zero, a positive number of events can also produce a total of zero. In general, the exact expression is

P(S=0) = \sum_{k=0}^{M} \binom{M}{k}p^k(1-p)^{M-k}P(X_1+\cdots+X_k=0),

where the $k=0$ probability in the sum is defined as one. This formula also handles signed severities, for which positive and negative terms might cancel to zero.

Worked Example: Zero Severity as Well as Zero Count

Let $N\sim\operatorname{Bin}(5,0.3)$, and let each independent severity be Bernoulli with $P(X_i=1)=0.4$ and $P(X_i=0)=0.6$. Find $P(S=0)$ and $P(S=1)$.

For each of the five opportunities, a nonzero contribution occurs only if it is activated and its severity equals one. The probability of this is $0.3(0.4)=0.12$, independently across opportunities. Thus $S\sim\operatorname{Bin}(5,0.12)$. In particular,

$$ P(S=0)=(1-0.12)^5=0.88^5=0.5277319168. $$

The count itself is zero with probability $P(N=0)=0.7^5=0.16807$. These probabilities differ because $S=0$ can also occur when $N>0$ but every realized severity is zero. For exactly one unit of total severity,

P(S=1) = \binom{5}{1}(0.12)(0.88)^4 = 5(0.12)(0.59969536) = 0.359817216.

The example shows why the zero-count weight should not automatically be reported as the full mass at zero.

Probability Generating Functions for Integer Severities

When $X_i$ takes values in the nonnegative integers, define its probability generating function by $G_X(z)=E[z^{X_i}]$, for $|z|\leq 1$. Let $G_S$ be the corresponding function for $S$. Conditioning on $N$ yields a compact expression for the entire compound binomial law.

Theorem: If $X_i$ are iid nonnegative-integer-valued random variables, independent of $N\sim\operatorname{Bin}(M,p)$, then for $|z|\leq 1$, $$ G_S(z)=\bigl(1-p+pG_X(z)\bigr)^M. $$

Proof. Conditional on $N=k$, independence of the severities gives the generating function $G_X(z)^k$ for their sum; when $k=0$, this is $1$, as required for the empty sum. Therefore

\begin{aligned} G_S(z) &=E\bigl[E[z^S\mid N]\bigr]\\ &=\sum_{k=0}^{M}\binom{M}{k}p^k(1-p)^{M-k}G_X(z)^k\\ &=\bigl((1-p)+pG_X(z)\bigr)^M, \end{aligned}

where the last equality is the binomial theorem. These expectations and the finite sum are well-defined for $|z|\leq1$, because $|z^S|\leq1$ for nonnegative integer $S$. $\square$

Worked Example: Recovering a Compound Binomial Law from Its Generating Function

Suppose $N\sim\operatorname{Bin}(2,0.4)$, and each severity is geometric on $\{1,2,\ldots\}$, with $P(X_i=j)=0.5^j$. Its generating function is $G_X(z)=z/(2-z)$. Find the generating function of $S$ and its zero mass.

The compound binomial formula gives

G_S(z) = \left(0.6+0.4\frac{z}{2-z}\right)^2 = \left(\frac{1.2-0.2z}{2-z}\right)^2.

At $z=0$, $G_S(0)=P(S=0)$ because $S$ is nonnegative integer-valued. Since every geometric severity is at least one, the total is zero only when $N=0$; accordingly,

$$ P(S=0)=G_S(0)=(0.6)^2=0.36. $$

The generating function packages the count mixture and the severity convolution in one expression. Its coefficients are the probabilities $P(S=s)$; extracting them is useful when the explicit convolution sums become cumbersome.

Continuous Severities and an Atom at Zero

A compound binomial distribution need not have a probability mass function or a density alone. For example, if every severity is strictly positive and continuous, the event $N=0$ still puts positive probability at exactly zero. Conditional on $N\geq1$, the total may have a density, so the unconditional distribution combines a point mass and a continuous part.

Worked Example: Exponential Severities with Two Opportunities

Let $N\sim\operatorname{Bin}(2,0.4)$, independently of iid $X_i\sim\operatorname{Exp}(2)$, where 2 is the rate. Find the mass at zero and the density for $s>0$.

The count probabilities are $P(N=0)=0.6^2=0.36$, $P(N=1)=2(0.4)(0.6)=0.48$, and $P(N=2)=0.4^2=0.16$. With one term, the total has density $2e^{-2s}$ for $s>0$. With two terms, convolution gives the gamma density $4se^{-2s}$ for $s>0$. Hence the total has an atom of size $0.36$ at zero and, for $s>0$, density

f_S(s)=0.48(2e^{-2s})+0.16(4se^{-2s}) =(0.96+0.64s)e^{-2s}.

The continuous component integrates to $0.48+0.16=0.64$: the one-term density has total weight $0.48$, and the two-term density has total weight $0.16$. Together with the atom, the total probability is $0.36+0.64=1$. For $t\geq0$, the distribution function is

P(S\leq t) = 0.36+0.48(1-e^{-2t}) +0.16\bigl(1-e^{-2t}(1+2t)\bigr).

The first term is the zero-count event. The remaining terms are the weighted probabilities for one and two exponential severities.

Common Pitfall and Practical Checks

A frequent mistake is to write $P(S=0)=(1-p)^M$ without checking whether severities can equal zero or cancel. That expression is correct when each severity is strictly positive almost surely, since then a positive count forces a positive total. It is not generally correct for severities that can be zero or can take both positive and negative values. Use the full mixture formula for $P(S=0)$ unless strict positivity has been established.

A second useful check is normalization. Each conditional law $F^{*k}$ has total probability one, and the binomial weights sum to one, so the mixture is a probability distribution. For discrete calculations, adding the resulting masses should give one. For mixed or continuous cases, add the atom probabilities to the integrated continuous components.

Specify the model.
State $M$, $p$, the severity law, and the independence assumptions.

Condition on the count.
For each $k=0,\ldots,M$, identify the law of the sum of $k$ severities, including the empty sum at $k=0$.

Weight and combine.
Multiply each conditional law by the binomial probability of $N=k$, then add contributions at shared values or across the relevant set.

Check zero and total probability.
Separate the zero-count contribution from any other ways the total can equal zero, and verify that the full law has probability one.

Key takeaway: A compound binomial law is the finite mixture of the iid-sum laws for counts $0$ through $M$. Its zero-count contribution is always $(1-p)^M$, but that is the full mass at zero only when no positive count can produce a total of zero.

Check Your Understanding

Use conditioning on the binomial count to answer these questions.

Write the mixture formula for the distribution of $S$ when $N\sim\operatorname{Bin}(M,p)$, and explain the role of the $k=0$ term.
If every severity is strictly positive almost surely, why is $P(S=0)=(1-p)^M$?
If severities can equal zero, what additional conditional probabilities must be included when calculating $P(S=0)$?
For nonnegative-integer severities, state the probability generating function of $S$ in terms of $G_X$, $M$, and $p$.
Explain why a continuous severity distribution can still yield a compound binomial distribution with an atom at zero.