Compound Poisson Distribution Theory

What Makes a Sum Compound Poisson?

A compound Poisson model describes a total accumulated from a random number of independent contributions. For example, a random number of claims may arrive during a fixed period, with each claim having a random size. The total depends both on how many contributions occur and on their individual values. Conditioning on the count separates these two sources of randomness and leads to concise formulas for the distribution and its moments.

Let \(N\sim\operatorname{Poisson}(\lambda)\), where \(\lambda>0\), and let \(Y_1,Y_2,\ldots\) be independent and identically distributed random variables, independent of \(N\). Define the total \(S\) as the sum of the first \(N\) contributions, with an empty sum equal to zero.

S=\sum_{j=1}^{N}Y_j,\qquad S=0\ \text{when }N=0.

Definition: The distribution of \(S\) is a compound Poisson distribution with rate \(\lambda\) and summand distribution equal to the distribution of \(Y_1\). The count \(N\) is Poisson, and the summands are iid and independent of \(N\). Write \(\phi_Y(t)=E[e^{itY_1}]\) for the characteristic function of one summand and \(\phi_S(t)=E[e^{itS}]\) for that of the total.

The value \(S=0\) is possible whenever the count is zero, no matter what distribution the summands have. If summands themselves can equal zero, a positive count can also produce a zero total. The distinction matters both for calculation and interpretation.

Characteristic Function by Conditioning

For a fixed count \(n\), the total is the sum of \(n\) independent copies of \(Y_1\). Its conditional characteristic function is therefore \(\phi_Y(t)^n\); when \(n=0\), the empty sum has characteristic function \(1\). Averaging over the Poisson count gives the compound Poisson transform.

Theorem: For any real-valued summand \(Y_1\), with no moment assumption required, the compound Poisson total has characteristic function

\phi_S(t)=\exp\!\left\{\lambda\bigl(\phi_Y(t)-1\bigr)\right\},\qquad t\in\mathbb{R}.

Proof. Since \(\lvert e^{itS}\rvert=1\), the conditional expectation and the countable sum over \(N\) are well-defined. Conditional on \(N=n\), independence of the summands gives \(E[e^{itS}\mid N=n]=\phi_Y(t)^n\). Thus, using the Poisson probability mass function,

\begin{aligned} \phi_S(t) &=E\!\left[E[e^{itS}\mid N]\right]\\ &=\sum_{n=0}^{\infty} e^{-\lambda}\frac{\lambda^n}{n!}\phi_Y(t)^n\\ &=e^{-\lambda}\exp\!\bigl(\lambda\phi_Y(t)\bigr)\\ &=\exp\!\left\{\lambda\bigl(\phi_Y(t)-1\bigr)\right\}. \end{aligned}

The series is absolutely convergent because \(\lvert\phi_Y(t)\rvert\leq 1\), and its \(n=0\) term is the contribution from the empty sum. \(\square\)

Mean and Variance

The moments of a compound Poisson total depend on moments of a single summand. Write \(\mu=E[Y_1]\) and \(\sigma^2=\operatorname{Var}(Y_1)\). The mean formula requires \(E[|Y_1|]<\infty\); the variance formula below requires \(E[Y_1^2]<\infty\).

Theorem: If \(E[|Y_1|]<\infty\), then \(E[S]=\lambda\mu\). If \(E[Y_1^2]<\infty\), then

\operatorname{Var}(S)=\lambda E[Y_1^2] =\lambda\bigl(\sigma^2+\mu^2\bigr).

Proof. Conditional on \(N=n\), linearity of expectation gives \(E[S\mid N=n]=n\mu\). Since a Poisson random variable with rate \(\lambda\) has expectation \(\lambda\), iterated expectation yields

E[S]=E[E[S\mid N]]=E[N\mu]=\lambda\mu.

For the variance, conditional independence gives \(\operatorname{Var}(S\mid N=n)=n\sigma^2\). The law of total variance applies because the second moment is finite. Its two terms are

\begin{aligned} E[\operatorname{Var}(S\mid N)]&=E[N\sigma^2]=\lambda\sigma^2,\\ \operatorname{Var}(E[S\mid N])&=\operatorname{Var}(N\mu) =\mu^2\operatorname{Var}(N)=\lambda\mu^2. \end{aligned}

Adding these terms gives \(\operatorname{Var}(S)=\lambda(\sigma^2+\mu^2)=\lambda E[Y_1^2]\), since \(E[Y_1^2]=\sigma^2+\mu^2\). The term involving \(\mu^2\) reflects variation in the random count, even when each summand is constant. \(\square\)

Worked Example: Bernoulli Contributions and Poisson Thinning

Suppose \(N\sim\operatorname{Poisson}(3)\), and each contribution independently equals \(1\) with probability \(0.4\) and \(0\) with probability \(0.6\). Find the characteristic function, mean, variance, and probability that the total is zero.

For one contribution, \(\phi_Y(t)=0.6+0.4e^{it}\), \(E[Y_1]=0.4\), and \(E[Y_1^2]=0.4\). Substitution in the compound Poisson characteristic function gives

\phi_S(t)=\exp\{3[(0.6+0.4e^{it})-1]\} =\exp\{1.2(e^{it}-1)\}.

This is the characteristic function of a \(\operatorname{Poisson}(1.2)\) random variable, so \(S\) has that distribution. The moment formulas also give \(E[S]=3(0.4)=1.2\) and \(\operatorname{Var}(S)=3(0.4)=1.2\).

The contributions are nonnegative and \(P(Y_1=0)=0.6\). Hence \(S=0\) exactly when every contribution is zero, including the possibility of no contributions at all:

P(S=0)=\exp\{-3(1-0.6)\}=e^{-1.2}.

The zero-count event alone has probability \(P(N=0)=e^{-3}\). The larger probability \(e^{-1.2}\) also includes cases with \(N>0\) in which every contribution equals zero.

Probability of a Zero Total

For nonnegative summands, a positive contribution cannot be cancelled by another contribution. This makes the event \(S=0\) especially simple: conditional on \(N=n\), all \(n\) summands must be zero. Let \(p_0=P(Y_1=0)\).

Theorem: If \(Y_1\geq 0\) almost surely, then

P(S=0)=\exp\{-\lambda(1-p_0)\}.

The zero-count event contributes \(e^{-\lambda}\). When \(p_0>0\), positive counts whose summands are all zero contribute an additional probability.

Proof. Given \(N=n\), nonnegativity implies that the sum is zero if and only if each summand is zero. By independence, \(P(S=0\mid N=n)=p_0^n\), including \(p_0^0=1\) when \(n=0\). Therefore

\begin{aligned} P(S=0) &=\sum_{n=0}^{\infty}e^{-\lambda}\frac{\lambda^n}{n!}p_0^n\\ &=e^{-\lambda}\exp(\lambda p_0) =\exp\{-\lambda(1-p_0)\}. \end{aligned}

The \(n=0\) term is exactly \(e^{-\lambda}\). The sum of the terms with \(n\geq 1\), representing zero-valued summands at positive counts, is consequently \(\exp\{-\lambda(1-p_0)\}-e^{-\lambda}\). \(\square\)

If \(Y_1>0\) almost surely, then \(p_0=0\), so only the zero-count event can produce a zero total and \(P(S=0)=e^{-\lambda}\). If summands can be negative, the theorem's equivalence fails: nonzero summands can cancel. In that case the zero-total probability must be determined from the particular summand distribution rather than from \(P(Y_1=0)\) alone.

Worked Example: A Fixed Positive Contribution

Let \(N\sim\operatorname{Poisson}(2.5)\), and suppose every contribution equals \(4\). Then \(S=4N\). Find its characteristic function, mean, variance, and zero-total probability.

A single contribution has \(\phi_Y(t)=e^{4it}\), \(E[Y_1]=4\), and \(E[Y_1^2]=16\). Therefore

\phi_S(t)=\exp\{2.5(e^{4it}-1)\},\qquad E[S]=2.5(4)=10,\qquad \operatorname{Var}(S)=2.5(16)=40.

Here \(p_0=0\), because each contribution is strictly positive. Thus \(S=0\) if and only if \(N=0\), and

P(S=0)=e^{-2.5}\approx 0.0821.

The variance is \(40\), not zero. Although the contribution size has no variability, the Poisson count does; applying \(\lambda E[Y_1^2]\) accounts for that count variation.

Worked Example: Zero-Valued Contributions Add to the Zero Probability

Let \(N\sim\operatorname{Poisson}(2)\). Independently, each contribution equals \(0\), \(2\), or \(5\) with probabilities \(0.2\), \(0.5\), and \(0.3\), respectively. Calculate the characteristic function, mean, variance, and probability of a zero total, and isolate the zero-count part.

For one contribution,

\begin{aligned} \phi_Y(t)&=0.2+0.5e^{2it}+0.3e^{5it},\\ E[Y_1]&=0(0.2)+2(0.5)+5(0.3)=2.5,\\ E[Y_1^2]&=0^2(0.2)+2^2(0.5)+5^2(0.3)=2+7.5=9.5. \end{aligned}

The compound Poisson characteristic function and moments are

\begin{aligned} \phi_S(t) &=\exp\{2[(0.2+0.5e^{2it}+0.3e^{5it})-1]\},\\ E[S]&=2(2.5)=5,\qquad \operatorname{Var}(S)=2(9.5)=19. \end{aligned}

The summands are nonnegative, and \(p_0=0.2\). Consequently

P(S=0)=e^{-2(1-0.2)}=e^{-1.6}\approx 0.2019.

The zero-count contribution is \(e^{-2}\approx 0.1353\). The contribution from positive counts with all summands zero is

P(S=0,N\geq 1) =e^{-2}\sum_{n=1}^{\infty}\frac{(2\cdot 0.2)^n}{n!} =e^{-2}(e^{0.4}-1) =e^{-1.6}-e^{-2} \approx 0.0666.

Adding the two contributions gives \(0.1353+0.0666\approx0.2019\), as required.

Interpretation and a Common Pitfall

The characteristic function summarizes how the single-contribution distribution and the Poisson rate combine: the rate multiplies \(\phi_Y(t)-1\) in the exponent. The moment formulas show a related distinction. The mean scales with the average contribution, but the variance scales with its second moment, not merely with its variance. This makes the compound Poisson model useful when a total reflects both a variable number of events and variable event sizes.

Wrong reasoning: “The variance is \(\lambda\operatorname{Var}(Y_1)\), because there are on average \(\lambda\) independent contributions.” This omits variation in the Poisson count.
Correction: Conditioning on the count gives \(\lambda\operatorname{Var}(Y_1)\) from within-count variation and \(\lambda\{E[Y_1]\}^2\) from count variation. Their sum is \(\lambda E[Y_1^2]\).
Wrong reasoning: “The probability of a zero total is always \(e^{-\lambda}\).” That is only the zero-count probability.
Correction: For nonnegative summands with \(P(Y_1=0)=p_0\), positive counts can also yield zero, and \(P(S=0)=e^{-\lambda(1-p_0)}\). With signed summands, cancellation is an additional possibility.

Key takeaway: Condition on the Poisson count to derive the transform and moments. For zero totals, first check whether summands are nonnegative and whether a single summand can equal zero; the empty sum always contributes probability \(e^{-\lambda}\).

Check Your Understanding

Use the independence assumptions and formulas developed above to answer these questions.

For a compound Poisson total, why is the characteristic function conditional on \(N=n\) equal to \(\phi_Y(t)^n\), including when \(n=0\)?
If \(E[Y_1]=3\), \(\operatorname{Var}(Y_1)=4\), and \(\lambda=2\), calculate \(E[S]\) and \(\operatorname{Var}(S)\).
Suppose the summands are nonnegative and \(P(Y_1=0)=0.25\). Write the formula for \(P(S=0)\) and identify the zero-count contribution.
For signed summands, explain why knowing \(P(Y_1=0)\) is not enough to determine \(P(S=0)\).
If every summand is a fixed positive constant, what event is equivalent to \(S=0\), and what is its probability?