Compound Poisson Distributions from Jump Laws

From a Random Count to a Compound Sum

A compound Poisson variable describes the total contribution from a random number of events. The event count is Poisson, and each event contributes a random jump. The count rate and the distribution of one jump together determine the distribution of the total. Conditioning on the count makes this structure explicit: with $k$ events, the total is a sum of exactly $k$ independent jumps.

Definition: Let $N\sim\operatorname{Poisson}(\lambda)$, where $\lambda\geq 0$, and let $Y_1,Y_2,\ldots$ be independent and identically distributed real-valued random variables, independent of $N$. The compound Poisson sum is $$ S=\sum_{j=1}^{N}Y_j, $$ with the empty sum defined as $0$ when $N=0$. The variables $Y_j$ are the jumps, and their common distribution is the jump law.

The jump law need not be supported on positive values: jumps may be zero, positive, or negative. The type of transform available depends on the jump law. A probability-generating function (PGF) is appropriate when the jumps are nonnegative integers. A moment-generating function (MGF) is available only at arguments where the jump MGF is finite. The characteristic function (CF) exists for every real-valued jump law.

The Transform Formulas

Write $G_Y(z)=E[z^Y]$ when $Y$ is nonnegative integer-valued, $M_Y(t)=E[e^{tY}]$ when this expectation is finite, and $\varphi_Y(u)=E[e^{iuY}]$ for real $u$. The compound-sum formulas follow by first fixing the count.

Theorem: For a compound Poisson sum with rate $\lambda$:

If the jumps are nonnegative integers, then for $0\leq z\leq 1$, $$ G_S(z)=\exp\{\lambda(G_Y(z)-1)\}. $$
For each real $t$ such that $M_Y(t)<\infty$, $$ M_S(t)=\exp\{\lambda(M_Y(t)-1)\}. $$
For every real $u$, with no moment assumption on $Y$, $$ \varphi_S(u)=\exp\{\lambda(\varphi_Y(u)-1)\}. $$

Proof. Conditional on $N=k$, the sum is $S=Y_1+\cdots+Y_k$, and independence gives the transform of the sum as the $k$th power of the one-jump transform. For the PGF, $0\leq z\leq1$ ensures the expectations exist, so conditioning and the Poisson probabilities give

G_S(z)=\sum_{k=0}^{\infty}P(N=k)G_Y(z)^k =e^{-\lambda}\sum_{k=0}^{\infty}\frac{(\lambda G_Y(z))^k}{k!} =\exp\{\lambda(G_Y(z)-1)\}.

For the MGF, $M_Y(t)$ is finite and nonnegative. Conditional expectation and summation of the nonnegative terms therefore yield

M_S(t)=\sum_{k=0}^{\infty}e^{-\lambda}\frac{\lambda^k}{k!}M_Y(t)^k =\exp\{\lambda(M_Y(t)-1)\}.

For the CF, $|\varphi_Y(u)|\leq1$, and conditioning gives an absolutely convergent series:

\varphi_S(u)=\sum_{k=0}^{\infty}e^{-\lambda}\frac{\lambda^k}{k!}\varphi_Y(u)^k =\exp\{\lambda(\varphi_Y(u)-1)\}.

Each identity uses the Poisson power series $e^x=\sum_{k=0}^{\infty}x^k/k!$. The CF formula requires no existence of moments or an MGF. $\square$

The Compound Law as a Poisson Mixture

The same conditioning argument describes the distribution itself, not just its transforms. Let $Q_k$ denote the distribution of a sum of $k$ independent jumps, with $Q_0$ the point mass at zero. Equivalently, $Q_k$ is the $k$-fold convolution of the jump law, and $Q_0=\delta_0$.

Theorem: For every Borel set $A\subseteq\mathbb{R}$, $$ P(S\in A)=\sum_{k=0}^{\infty}e^{-\lambda}\frac{\lambda^k}{k!}Q_k(A). $$ Thus the compound Poisson distribution is a Poisson-weighted mixture of the convolution powers of the jump law.

Proof. The events $\{N=k\}$, for $k=0,1,\ldots$, are disjoint and cover the sample space. By the law of total probability,

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

Given $N=k$, the sum has distribution $Q_k$, so $P(S\in A\mid N=k)=Q_k(A)$. Substituting $P(N=k)=e^{-\lambda}\lambda^k/k!$ proves the formula. The series is well-defined because its terms are nonnegative and their sum is at most $\sum_{k=0}^{\infty}P(N=k)=1$. $\square$

Worked Example: Bernoulli Jumps Produce a Poisson Total

Let $N\sim\operatorname{Poisson}(4)$, and let each jump equal $1$ with probability $0.3$ and $0$ with probability $0.7$, independently of all other jumps and of $N$. Find the PGF of $S$ and identify its distribution.

For one jump, $G_Y(z)=0.7+0.3z$. Therefore,

G_S(z)=\exp\{4[(0.7+0.3z)-1]\} =\exp\{1.2(z-1)\}.

The PGF of a $\operatorname{Poisson}(\mu)$ variable is $\exp\{\mu(z-1)\}$, so $S\sim\operatorname{Poisson}(1.2)$. For example, the probability of a total of two is

P(S=2)=e^{-1.2}\frac{1.2^2}{2!} =0.72e^{-1.2}\approx 0.2169.

This example also has a direct interpretation: zero-valued jumps add nothing, and the number of jumps equal to one is Poisson with rate $4(0.3)=1.2$. The PGF derivation establishes the distribution without needing to reason separately about which events produced nonzero jumps.

Worked Example: Exponential Jumps and the MGF

Let $N\sim\operatorname{Poisson}(3)$, and take independent exponential jumps with rate $2$, so the density of one jump is $2e^{-2y}$ for $y>0$. Find the MGF and CF of $S$, and describe its atom at zero.

For $t<2$, integrating the jump density gives

M_Y(t)=\int_0^\infty e^{ty}2e^{-2y}\,dy =2\int_0^\infty e^{-(2-t)y}\,dy =\frac{2}{2-t}.

The integral is finite exactly when $2-t>0$. Hence, for $t<2$,

M_S(t)=\exp\left\{3\left(\frac{2}{2-t}-1\right)\right\} =\exp\left(\frac{3t}{2-t}\right).

For real $u$, the one-jump CF is $2/(2-iu)$, so

\varphi_S(u)=\exp\left\{3\left(\frac{2}{2-iu}-1\right)\right\}.

Because every jump is strictly positive, $S=0$ precisely when $N=0$. Thus $P(S=0)=e^{-3}$. Conditional on $N=k\geq1$, the sum has a gamma distribution with shape $k$ and rate $2$; the full law places Poisson weight $e^{-3}3^k/k!$ on each such conditional distribution, as in the mixture formula.

Worked Example: Signed Jumps and the Characteristic Function

Let $N\sim\operatorname{Poisson}(2.5)$. Independently, each jump is $+1$ with probability $0.4$ and $-1$ with probability $0.6$. Find the CF of $S$, and compute the probability that the total is zero.

The one-jump CF is $\varphi_Y(u)=0.4e^{iu}+0.6e^{-iu}$. Therefore,

\varphi_S(u)=\exp\{2.5(0.4e^{iu}+0.6e^{-iu}-1)\}.

A zero total can occur with no jumps, or with positive and negative jumps that cancel. To calculate its probability, let $A$ count the $+1$ jumps and $B$ the $-1$ jumps. Poisson splitting gives independent counts $A\sim\operatorname{Poisson}(2.5\cdot0.4)=\operatorname{Poisson}(1)$ and $B\sim\operatorname{Poisson}(2.5\cdot0.6)=\operatorname{Poisson}(1.5)$. Since $S=A-B$, the event $S=0$ is $A=B$, and

P(S=0)=\sum_{k=0}^{\infty}P(A=k)P(B=k) =e^{-2.5}\sum_{k=0}^{\infty}\frac{1.5^k}{(k!)^2}.

In particular, this is strictly greater than $e^{-2.5}$, the probability of no jumps, because the $k=1$ term is positive. The CF remains valid even though this signed jump law is not a setting for a PGF.

Using the Formulas Carefully

The transform identities are useful because they replace a random-length sum by a simple expression involving the one-jump transform. They also distinguish what information is needed for different tasks: a PGF can identify an integer-valued total, an MGF can be used only on its finite domain, and a CF applies without a moment assumption. In each case the rate $\lambda$ appears outside the jump transform, while the term $-1$ ensures the transform equals $1$ at its neutral argument.

A frequent mistake is to equate “the sum is zero” with “there were no jumps.” This is true for strictly positive jumps, as in the exponential example, but not for jumps that may be zero or may cancel. For Bernoulli jumps, zero-valued events do not change the sum; for signed jumps, nonzero contributions can cancel. The mixture formula makes the distinction precise: the $k=0$ component contributes $e^{-\lambda}$ mass at zero, while other convolution components may contribute additional mass there.

Key takeaway: Condition on the Poisson count, take the $k$th power of the one-jump transform for $k$ independent jumps, and sum using the Poisson series. This yields $\exp\{\lambda(T_Y-1)\}$ for the PGF, MGF, or CF whenever the relevant transform is defined.

Check Your Understanding

Use the count rate and jump transform to answer each question.

If $N\sim\operatorname{Poisson}(\lambda)$ and every jump is the constant $c$, what are the MGF and CF of $S$? State the MGF domain.
Why does the CF formula remain valid when the jump distribution has no finite MGF?
For nonnegative integer-valued jumps, what condition on $z$ was used to ensure the PGF calculation is well-defined?
Explain why $P(S=0)$ can be greater than $P(N=0)$ when jumps can be zero or can cancel.
In the exponential-jump example, identify the distribution of $S$ conditional on $N=k$, for $k\geq1$, and state its rate parameter.

Compound Poisson Distributions from Jump Laws

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