How to Compute the MGF of a Compound Poisson Distribution

A compound Poisson random variable is a random sum: the number of terms is random and follows a Poisson distribution. To compute its moment-generating function (MGF), condition on the number of terms, calculate the MGF for that fixed number, and then average over the Poisson count.

The result is a simple formula in terms of the MGF of one jump. The method works for positive or negative jumps, provided the relevant MGF exists.

Set up the compound Poisson variable

Let $N \sim \operatorname{Poisson}(\lambda)$, and let $X_1, X_2, \ldots$ be independent and identically distributed random variables, independent of $N$. Define

$$S = \sum_{i=1}^{N} X_i,$$

with the empty sum set to zero when $N=0$. The $X_i$ are often called jumps, claims, or summands. Write their MGF as

$$M_X(t) = \mathbb{E}[e^{tX_1}].$$

Derive the MGF by conditioning

If the count is fixed at $N=n$, then $S=X_1+\cdots+X_n$. Independence gives

$$\mathbb{E}[e^{tS}\mid N=n] = \prod_{i=1}^{n}\mathbb{E}[e^{tX_i}] = M_X(t)^n.$$

For $n=0$, this is $1$, because $S=0$. Now average over the Poisson distribution:

$$\begin{aligned} M_S(t) &= \mathbb{E}[e^{tS}] \\ &= \sum_{n=0}^{\infty} \mathbb{E}[e^{tS}\mid N=n]\Pr(N=n) \\ &= \sum_{n=0}^{\infty} M_X(t)^n e^{-\lambda}\frac{\lambda^n}{n!} \\ &= e^{-\lambda}\exp\bigl(\lambda M_X(t)\bigr) \\ &= \exp\bigl(\lambda(M_X(t)-1)\bigr). \end{aligned}$$

Therefore,

$$\boxed{M_S(t)=\exp\bigl(\lambda(M_X(t)-1)\bigr).}$$

This is also the MGF version of the compound Poisson probability-generating-function identity. It incorporates both the distribution of a single jump and the random number of jumps.

Worked example: exponential jumps

Suppose $N\sim\operatorname{Poisson}(2)$ and each jump is independent with an exponential distribution of rate $3$. The jump MGF is

$$M_X(t)=\frac{3}{3-t}, \qquad t<3.$$

Substitute this into the compound Poisson formula:

$$\begin{aligned} M_S(t) &= \exp\left(2\left(\frac{3}{3-t}-1\right)\right) \\ &= \exp\left(\frac{2t}{3-t}\right), \qquad t<3. \end{aligned}$$

The domain $t<3$ comes from the jump MGF. In particular, the formula gives $M_S(0)=1$, as every MGF must.

You can check the mean and variance by differentiating, or use the standard compound Poisson moment formulas. An exponential variable of rate $3$ has $\mathbb{E}[X]=1/3$ and $\mathbb{E}[X^2]=2/9$. Thus

$$\mathbb{E}[S]=\lambda\mathbb{E}[X]=\frac{2}{3}, \qquad \operatorname{Var}(S)=\lambda\mathbb{E}[X^2]=\frac{4}{9}.$$

Moments from the MGF

The log MGF, also called the cumulant-generating function, is

$$K_S(t)=\log M_S(t)=\lambda(M_X(t)-1).$$

If the jump MGF is finite in a neighborhood of zero, differentiating at zero gives

$$\mathbb{E}[S]=\lambda\mathbb{E}[X], \qquad \operatorname{Var}(S)=\lambda\mathbb{E}[X^2].$$

These moment formulas also hold more generally whenever the corresponding jump moments are finite. The variance uses the second raw moment of a jump, not just its variance. This distinction reflects the extra variation caused by the random number of jumps:

$$\operatorname{Var}(S) =\mathbb{E}[N]\operatorname{Var}(X)+\operatorname{Var}(N)(\mathbb{E}[X])^2 =\lambda\bigl(\operatorname{Var}(X)+(\mathbb{E}[X])^2\bigr).$$

Common mistakes

• Leaving out the zero-jump case: The sum is zero when $N=0$. Its contribution to the MGF is $e^{t\cdot 0}=1$, which is already included by the $n=0$ term.
• Using $\lambda M_X(t)$ as the exponent: The correct exponent is $\lambda(M_X(t)-1)$. The minus one ensures that the MGF equals one at $t=0$.
• Using $\lambda\operatorname{Var}(X)$ for the variance: The correct formula is $\lambda\mathbb{E}[X^2]$; it accounts for variation in the Poisson count.
• Ignoring existence: For $\lambda>0$, if $M_X(t)$ is infinite, then $M_S(t)$ is infinite as well: the event of exactly one jump has positive probability. The MGF formula is finite on the domain where the jump MGF is finite. If the jump distribution has no MGF around zero, the compound distribution need not have one either.

The formula assumes the jumps are identically distributed and independent of the Poisson count. If those assumptions change, condition on the count using the actual conditional jump distributions rather than applying this expression without adjustment.