Probability Laws of Compound Poisson Sums

Conditioning on the Event Count

A compound Poisson sum records the total contribution from a random number of events. Each event brings a random mark, such as a claim amount, batch size, or jump size. The count determines how many marks are added; conditioning on that count turns the random sum into an ordinary sum of a fixed number of independent marks. This gives the entire distribution, not just a summary such as its mean.

Work on a probability space $(\Omega,\mathcal{F},\mathbb{P})$. Let $(N_t)_{t\ge0}$ be a Poisson process of rate $\lambda\ge0$, and let $(Y_k)_{k\ge1}$ be independent, identically distributed real-valued random variables with common probability law $F$. Assume that the marks are independent of the Poisson process. For a fixed $t\ge0$, define the compound Poisson sum $S_t=\sum_{k=1}^{N_t}Y_k$, with an empty sum equal to zero. Write $\theta=\lambda t$; then $N_t$ has the Poisson distribution with parameter $\theta$.

Definition: For probability laws $F$ and $G$ on the real line, their convolution $F*G$ is the law of $U+V$ when $U$ and $V$ are independent with laws $F$ and $G$. Write $F^{*n}$ for the convolution of $n$ copies of $F$, and set $F^{*0}=\delta_0$, the point mass at zero.

The convention $F^{*0}=\delta_0$ matters: if no events occur, the sum is zero regardless of the mark distribution. Also, marks need not be positive for the convolution-mixture law or the characteristic function below to hold. Nonnegative integer-valued marks will be required for the probability generating function.

Theorem (Compound Poisson law and transforms): For every Borel set $A\subseteq\mathbb{R}$, $$ \mathbb{P}(S_t\in A) =e^{-\theta}\sum_{n=0}^{\infty}\frac{\theta^n}{n!}F^{*n}(A). $$ In particular, the term $n=0$ assigns mass $e^{-\theta}$ to zero. For any real-valued marks, define $\phi_Y(u)=\mathbb{E}[e^{iuY_1}]$ and $\phi_{S_t}(u)=\mathbb{E}[e^{iuS_t}]$; then $$ \phi_{S_t}(u)=\exp\!\bigl(\theta(\phi_Y(u)-1)\bigr),\qquad u\in\mathbb{R}. $$ If the marks take values in the nonnegative integers, define $G_Y(z)=\mathbb{E}[z^{Y_1}]$ and $G_{S_t}(z)=\mathbb{E}[z^{S_t}]$. For $|z|\le1$, $$ G_{S_t}(z)=\exp\!\bigl(\theta(G_Y(z)-1)\bigr). $$

Proof. Fix a Borel set $A$. On the event $N_t=n$, the sum $S_t$ consists of the first $n$ marks. Since the marks are independent and have common law $F$, their sum has law $F^{*n}$, including the case $n=0$. Independence of the marks and $N_t$ therefore gives $\mathbb{P}(S_t\in A\mid N_t=n)=F^{*n}(A)$ whenever $\mathbb{P}(N_t=n)>0$. If $\theta=0$, then $N_t=0$ almost surely and the claimed mixture reduces to $F^{*0}$; otherwise every count $n\ge0$ has positive probability. Summing the conditional probabilities with weights $\mathbb{P}(N_t=n)=e^{-\theta}\theta^n/n!$ proves the mixture formula. For the characteristic function, condition on $N_t=n$. Independence gives $\mathbb{E}[e^{iuS_t}\mid N_t=n]=\phi_Y(u)^n$, so $$ \phi_{S_t}(u)=e^{-\theta}\sum_{n=0}^{\infty}\frac{\theta^n}{n!}\phi_Y(u)^n =e^{-\theta}e^{\theta\phi_Y(u)} =\exp\!\bigl(\theta(\phi_Y(u)-1)\bigr). $$ The series is valid because $|\phi_Y(u)|\le1$, and the exponential series converges absolutely. For nonnegative integer-valued marks, the same conditioning gives $\mathbb{E}[z^{S_t}\mid N_t=n]=G_Y(z)^n$. Since $|G_Y(z)|\le1$ when $|z|\le1$, summing the Poisson series gives $G_{S_t}(z)=e^{-\theta}e^{\theta G_Y(z)}$, as claimed. $\square$

The characteristic function applies whether marks are continuous, discrete, or mixed, and whether they can be negative or zero. The probability generating function is more specialized: its exponents represent integer-valued sums, so it is appropriate only when marks are nonnegative integers. Both formulas retain the contribution from zero events through the subtraction of $1$ in the exponent.

Bernoulli Marks and Poisson Counts

Proposition: Suppose each mark is Bernoulli with parameter $p$, so it equals $1$ with probability $p$ and $0$ with probability $1-p$, where $0\le p\le1$. Then $S_t$ has a Poisson distribution with parameter $\theta p$. If $\theta p=0$, this means $S_t=0$ almost surely.

Proof. If $\theta=0$ or $p=0$, there are almost surely no events or no nonzero marks, respectively, so the sum is zero. Otherwise, for each integer $k\ge0$, condition on the count $N_t=n$. A sum of $n$ Bernoulli marks equals $k$ with probability $\binom{n}{k}p^k(1-p)^{n-k}$ for $n\ge k$, and with probability zero for $n<k$. Therefore $$ \begin{aligned} \mathbb{P}(S_t=k) &=e^{-\theta}\sum_{n=k}^{\infty}\frac{\theta^n}{n!}\binom{n}{k}p^k(1-p)^{n-k}\\ &=e^{-\theta}\frac{(\theta p)^k}{k!} \sum_{m=0}^{\infty}\frac{(\theta(1-p))^m}{m!}\\ &=e^{-\theta}\frac{(\theta p)^k}{k!}e^{\theta(1-p)} =e^{-\theta p}\frac{(\theta p)^k}{k!}. \end{aligned} $$ Here the second equality uses $m=n-k$ and $\binom{n}{k}/n!=1/(k!(n-k)!)$. The final expression is the Poisson probability mass function with parameter $\theta p$, including $p=1$ when the inner sum is $1$. $\square$

This result says that zero-valued marks can be discarded without changing the type of count: the number of nonzero contributions is Poisson with parameter $\theta p$. The next example computes the law directly from the mixture and checks that conclusion.

Worked Example: Bernoulli Marks with Exact Parameters

Take $\lambda=2$, $t=1$, and marks with $\mathbb{P}(Y_k=1)=\tfrac13$ and $\mathbb{P}(Y_k=0)=\tfrac23$. Thus $\theta=\lambda t=2$. Find the probability that the sum equals $k$, for any integer $k\ge0$.

Given $N_t=n$, exactly $k$ marks must equal $1$. The mixture calculation is

\begin{aligned} \mathbb{P}(S_t=k) &=e^{-2}\sum_{n=k}^{\infty}\frac{2^n}{n!} \binom{n}{k}\left(\frac13\right)^k\left(\frac23\right)^{n-k}\\ &=e^{-2}\frac{(2/3)^k}{k!} \sum_{m=0}^{\infty}\frac{(4/3)^m}{m!}\\ &=e^{-2}e^{4/3}\frac{(2/3)^k}{k!} =e^{-2/3}\frac{(2/3)^k}{k!}. \end{aligned}

For the second line, set $m=n-k$; the factor from $2^n(2/3)^{n-k}$ becomes $2^k(4/3)^m$, and $2^k(1/3)^k=(2/3)^k$. Thus $S_t$ is Poisson with parameter $\theta p=2/3$. Check: summing the displayed probabilities over $k\ge0$ gives $e^{-2/3}\sum_{k\ge0}(2/3)^k/k!=e^{-2/3}e^{2/3}=1$, as required.

Exponential Marks: An Atom and a Density

When marks are positive and continuous, the mixture usually has both an atom at zero and a density away from zero. The atom comes from $N_t=0$; conditional on at least one exponential mark, the sum has a gamma distribution. The mixture formula specifies the full law even when the resulting density is best left as a series.

Worked Example: Exponential Marks at a Fixed Time

Let $\lambda=1$, $t=1$, so $\theta=1$, and let the marks be independent exponential random variables with rate $\beta=2$. Find the atom at zero and the density of $S_t$ for $x>0$.

With zero marks, $S_t=0$, so $\mathbb{P}(S_t=0)=\mathbb{P}(N_t=0)=e^{-1}$. Given $N_t=n\ge1$, the sum has density $2^n x^{n-1}e^{-2x}/(n-1)!$ for $x>0$. To verify this conditional formula, it holds for $n=1$, the exponential density. If it holds for $n$, convolution with one more exponential density gives $$ \int_0^x \frac{2^n u^{n-1}e^{-2u}}{(n-1)!}\,2e^{-2(x-u)}\,du =\frac{2^{n+1}e^{-2x}}{(n-1)!}\int_0^x u^{n-1}\,du =\frac{2^{n+1}x^n e^{-2x}}{n!}. $$ This is the asserted formula with $n+1$ in place of $n$.

Weight each conditional density by $\mathbb{P}(N_t=n)=e^{-1}/n!$. The density away from zero is consequently $$ f_{S_t}(x)=e^{-1-2x}\sum_{n=1}^{\infty}\frac{2^n x^{n-1}}{n!(n-1)!},\qquad x>0. $$ The first three terms of the sum are $2$, $2x$, and $\tfrac23x^2$: for $n=1$ the coefficient is $2/1!=2$; for $n=2$ it is $2^2/(2!\,1!)=2$; for $n=3$ it is $2^3/(3!\,2!)=\tfrac23$.

Check the total probability: each conditional density integrates to $1$, so the density’s total mass is $e^{-1}\sum_{n=1}^{\infty}1/n!=e^{-1}(e-1)=1-e^{-1}$. Adding the atom $e^{-1}$ gives $1$. As a transform check, an exponential mark of rate $2$ has Laplace transform $\mathbb{E}[e^{-sY_1}]=2/(2+s)$ for $s\ge0$. At $s=2$, the compound sum formula gives $\mathbb{E}[e^{-2S_t}]=\exp(1(2/4-1))=e^{-1/2}$. Conditioning on the count gives $e^{-1}\sum_{n=0}^{\infty}(1/2)^n/n!=e^{-1}e^{1/2}=e^{-1/2}$, the same value.

Deterministic Marks and a Lattice-Valued Sum

Worked Example: Every Event Adds Three Units

Suppose $\lambda=2$, $t=1$, and each mark equals $3$ with probability $1$. Then $\theta=2$ and $S_t=3N_t$. Find the probability that the total is $6$, and give its probability generating function.

The sum equals $6$ exactly when $N_t=2$. Thus $$ \mathbb{P}(S_t=6)=\mathbb{P}(N_t=2) =e^{-2}\frac{2^2}{2!}=2e^{-2}. $$ The mark generating function is $G_Y(z)=z^3$, so the compound-sum formula gives $G_{S_t}(z)=\exp(2(z^3-1))$. Expanding it, $$ G_{S_t}(z)=e^{-2}\sum_{n=0}^{\infty}\frac{2^n z^{3n}}{n!}. $$ The coefficient of $z^6$ is the $n=2$ term, namely $e^{-2}2^2/2!=2e^{-2}$, agreeing with the direct count calculation. Check: at $z=1$, the generating function is $\exp(2(1-1))=1$, so its probabilities sum to one.

Reading the Formulas Correctly

The mixture law is often the most direct route when the marks have a simple sum distribution conditional on a fixed count. The transform formulas are compact alternatives: a mark characteristic function $\phi_Y(u)$ becomes $\exp(\theta(\phi_Y(u)-1))$, while a mark probability generating function $G_Y(z)$ becomes $\exp(\theta(G_Y(z)-1))$. The two transforms answer different questions and have different domains: the characteristic function is always defined for real $u$, whereas the probability generating function requires nonnegative integer-valued marks.

Common mistake. A tempting but incorrect formula for Bernoulli marks is $G_{S_t}(z)=\exp(\theta pz)$. It omits the zero-event correction. For the exact parameters $\theta=2$ and $p=\tfrac13$, this wrong expression has value $e^{2/3}$ at $z=1$, not $1$, so it cannot be a probability generating function. The mark generating function is $G_Y(z)=\tfrac23+\tfrac13z$; substituting it correctly gives $$ G_{S_t}(z)=\exp\!\left(2\left(\frac23+\frac13z-1\right)\right) =\exp\!\left(\frac23(z-1)\right), $$ which equals $1$ at $z=1$. The subtraction of $1$ accounts for the fact that each Poisson event contributes a mark, while no event contributes nothing.

Check Your Understanding

Use conditioning on the event count or the transform formulas, and check that each proposed probability law is normalized.

If $\theta=3$ and marks are Bernoulli with success probability $\tfrac12$, what is the distribution of $S_t$?
Why does the convolution mixture contain the term $e^{-\theta}\delta_0$, and what happens to that term when $\theta=0$?
For exponential marks with rate $\beta$, write the conditional density of the sum given $N_t=n\ge1$. How does the mixture weight enter the unconditional density?
If every mark equals $2$, express $\phi_{S_t}(u)$ in terms of $\theta$ and $u$. What values can $S_t$ take?
Explain why the probability generating function formula is not appropriate for marks that can take arbitrary real values.