Compound Poisson Calculations With Exponential Severities

Set-Up: A Poisson Number of Exponential Severities

A compound Poisson total adds a random number of positive amounts. With exponential severities, the conditional law of the total given the count has a particularly simple form: it is gamma. The unconditional law, however, is not just one gamma distribution. It includes an atom at zero, and for positive totals it combines the gamma laws corresponding to all possible positive counts.

Let $N\sim\operatorname{Poisson}(\lambda)$, where $\lambda\geq 0$. Independently of $N$, let $X_1,X_2,\ldots$ be independent and identically distributed exponential random variables with rate $\beta>0$. Thus each severity has density $\beta e^{-\beta x}$ for $x>0$, mean $1/\beta$, and variance $1/\beta^2$. Define the compound total by $S=\sum_{i=1}^{N}X_i$, taking the empty sum to be zero.

Definition: A compound Poisson sum with exponential severities is $S=\sum_{i=1}^{N}X_i$, where $N\sim\operatorname{Poisson}(\lambda)$ is independent of the independent, identically distributed severities $X_i\sim\operatorname{Exp}(\beta)$. Here $\lambda\geq0$ is the Poisson mean and $\beta>0$ is the exponential rate.

The independence assumptions matter: conditional on $N=n$, the sum has the law of $n$ independent exponential variables. For $n=0$, it is exactly zero. For $n\geq1$, it is positive with probability one and has a continuous distribution.

The Conditional Distribution: Gamma Given the Count

For a positive integer $n$, the sum of $n$ independent exponential variables with rate $\beta$ has a gamma distribution with shape $n$ and rate $\beta$. Since the shape is a positive integer, this is also called an Erlang distribution. Its density is

f_{S\mid N=n}(s) = \frac{\beta^n s^{n-1}e^{-\beta s}}{(n-1)!}, \qquad s>0,\quad n=1,2,\ldots.

This formula can be derived directly by convolution. For $n=1$, it is the exponential density. Suppose it holds for $n$. The density of the sum of $n+1$ severities at $s>0$ is the convolution of the density for the first $n$ with the exponential density of the next severity:

\begin{aligned} f_{S\mid N=n+1}(s) &=\int_0^s \frac{\beta^n u^{n-1}e^{-\beta u}}{(n-1)!} \,\beta e^{-\beta(s-u)}\,du\\ &=\frac{\beta^{n+1}e^{-\beta s}}{(n-1)!} \int_0^s u^{n-1}\,du\\ &=\frac{\beta^{n+1}s^n e^{-\beta s}}{n!}. \end{aligned}

This is the stated formula with $n+1$ in place of $n$. Induction establishes the conditional density for every positive integer count. For $n=0$, there is no gamma density: the conditional distribution is a point mass at zero.

Worked Example: Conditional Densities and the First Mixture Terms

Suppose $N\sim\operatorname{Poisson}(2)$, and each severity is exponential with rate $\beta=4$. Find the conditional laws for counts zero, one, and two, and write the corresponding contributions to the unconditional law.

The count probabilities are $P(N=n)=e^{-2}2^n/n!$. In particular, $P(N=0)=e^{-2}$, $P(N=1)=2e^{-2}$, and $P(N=2)=2e^{-2}$. Given $N=0$, $S=0$. Given $N=1$, $S$ is exponential with rate 4, with density $4e^{-4s}$ for $s>0$. Given $N=2$, $S$ has density $16se^{-4s}$ for $s>0$.

Thus the atom at zero has mass $e^{-2}$. The $N=1$ component contributes $2e^{-2}\cdot4e^{-4s}=8e^{-2}e^{-4s}$ to the unconditional density for $s>0$. The $N=2$ component contributes $2e^{-2}\cdot16se^{-4s}=32e^{-2}se^{-4s}$. These are only two terms of the positive-part density; counts $3,4,\ldots$ contribute additional terms.

The Unconditional Law

The unconditional law combines the Poisson probabilities for the count with the corresponding conditional gamma laws. Its atom at zero follows from the fact that every severity is strictly positive with probability one. Therefore $S=0$ if and only if $N=0$, up to events of probability zero.

Theorem: If $\lambda>0$, the compound total has an atom of mass $P(S=0)=e^{-\lambda}$ at zero and a density on $s>0$ given by $$ f_S(s)=e^{-\lambda-\beta s} \sum_{n=1}^{\infty} \frac{\lambda^n\beta^n s^{n-1}}{n!(n-1)!}, \qquad s>0. $$ There is no other point mass. If $\lambda=0$, then $S=0$ almost surely, so its law is a point mass at zero.

To derive the result for $\lambda>0$, first use the Poisson probability of zero:

P(S=0)=P(N=0)=e^{-\lambda}.

For $s>0$, condition on the count. The event $N=0$ contributes no density there, while each $n\geq1$ contributes its Poisson weight times its conditional gamma density. Summing those contributions gives

\begin{aligned} f_S(s) &=\sum_{n=1}^{\infty}P(N=n)f_{S\mid N=n}(s)\\ &=\sum_{n=1}^{\infty} \left(e^{-\lambda}\frac{\lambda^n}{n!}\right) \left(\frac{\beta^n s^{n-1}e^{-\beta s}}{(n-1)!}\right)\\ &=e^{-\lambda-\beta s} \sum_{n=1}^{\infty} \frac{\lambda^n\beta^n s^{n-1}}{n!(n-1)!}. \end{aligned}

This is a mixture of densities with weights $P(N=n)$ for $n\geq1$; those weights sum to $1-e^{-\lambda}$. Consequently the density integrates to $1-e^{-\lambda}$, while the remaining probability $e^{-\lambda}$ is at zero. Conditional on any positive count, the sum has a density, so it assigns zero probability to every individual positive value. This also shows that there are no other atoms.

For $\lambda>0$, one may express the same density using the modified Bessel function $I_1$, defined by its power series $I_1(2\sqrt{z})=\sum_{j=0}^{\infty}z^{j+1/2}/(j!(j+1)!)$ for $z\geq0$. Re-indexing the density series with $j=n-1$ gives

f_S(s)=e^{-\lambda-\beta s} \frac{\lambda\beta I_1(2\sqrt{\lambda\beta s})}{\sqrt{\lambda\beta s}}, \qquad s>0.

The series form is often more useful for understanding the mixture: its $n$th term is exactly the contribution from count $n$. The Bessel form is a compact representation of the same sum.

Worked Example: An Event Probability From One Count Component

Let $N\sim\operatorname{Poisson}(0.5)$, with independent exponential severities of rate 1. Find $P(N=1,\;0<S\leq1)$.

On $N=1$, the total $S$ is just $X_1$, so $P(0<S\leq1\mid N=1)=1-e^{-1}$. Also $P(N=1)=e^{-0.5}(0.5)^1/1!=0.5e^{-0.5}$. Multiplying gives

\begin{aligned} P(N=1,\;0<S\leq1) &=P(N=1)P(0<S\leq1\mid N=1)\\ &=0.5e^{-0.5}(1-e^{-1})\\ &\approx 0.5(0.60653)(0.63212)\\ &\approx 0.1917. \end{aligned}

This is the probability from the one-severity component alone. It is not the full probability $P(0<S\leq1)$, which also includes cases with $N=2,3,\ldots$ whose totals fall in that interval.

Unconditional Mean and Variance

The moments can be obtained by conditioning on the count. Given $N=n$, a sum of $n$ independent exponential severities has mean $n/\beta$ and variance $n/\beta^2$. These formulas also hold for $n=0$, with both quantities equal to zero. Because a Poisson random variable with mean $\lambda$ has both mean and variance $\lambda$, conditioning gives the following result.

Theorem: For $N\sim\operatorname{Poisson}(\lambda)$ independent of $X_i\sim\operatorname{Exp}(\beta)$, with $\lambda\geq0$ and $\beta>0$, $$ E[S]=\frac{\lambda}{\beta}, \qquad \operatorname{Var}(S)=\frac{2\lambda}{\beta^2}. $$

To verify the mean, apply the law of total expectation:

E[S]=E\{E[S\mid N]\} =E\left[\frac{N}{\beta}\right] =\frac{\lambda}{\beta}.

For the variance, the law of total variance separates variability within a fixed count from variability across counts:

\begin{aligned} \operatorname{Var}(S) &=E\{\operatorname{Var}(S\mid N)\} +\operatorname{Var}\{E[S\mid N]\}\\ &=E\left[\frac{N}{\beta^2}\right] +\operatorname{Var}\left(\frac{N}{\beta}\right)\\ &=\frac{\lambda}{\beta^2}+\frac{\lambda}{\beta^2}\\ &=\frac{2\lambda}{\beta^2}. \end{aligned}

The first term is the expected conditional variance of the severities when the count is fixed. The second is the variance in the conditional mean as the count changes. Both contribute $\lambda/\beta^2$. The required moments exist because the exponential severity and Poisson count each have finite first and second moments.

Worked Example: Decomposing the Variance

Suppose $N\sim\operatorname{Poisson}(3)$ and the severities are independent exponentials with rate $\beta=2$. Calculate the unconditional mean and variance, showing the two variance contributions.

The conditional mean is $E[S\mid N]=N/2$, so

$$ E[S]=E[N/2]=3/2=1.5. $$

The expected conditional variance is $E[N/4]=3/4$. The variance of the conditional mean is $\operatorname{Var}(N/2)=3/4$, since $\operatorname{Var}(N)=3$. Therefore

\operatorname{Var}(S)=\frac{3}{4}+\frac{3}{4}=\frac{3}{2}=1.5.

As a check, the direct formula gives $2\lambda/\beta^2=2(3)/2^2=6/4=1.5$, in agreement with the decomposition.

Common Pitfall and Interpretation

A frequent error is to say that the unconditional law is simply gamma because the sum is gamma for every fixed positive count. That reasoning treats the count as fixed when it is actually random. The correction is to condition on $N$, use the gamma law for each $n\geq1$, and then mix those laws using the Poisson probabilities. The $n=0$ component must be included as an atom at zero.

For $\lambda>0$, the result is a mixed distribution: it has a point mass $e^{-\lambda}$ at zero and a continuous density on the positive half-line. At $\lambda=0$, there are no arrivals, so $S=0$ almost surely and the law is purely discrete. This boundary case also agrees with the moment formulas: both the mean and variance are zero.

Key takeaway: Keep the count and severity roles distinct. Given $N=n\geq1$, the total is gamma with shape $n$ and rate $\beta$; unconditionally, it is a Poisson-weighted mixture of those gamma laws plus an atom $e^{-\lambda}$ at zero when $\lambda>0$. Its mean is $\lambda/\beta$ and its variance is $2\lambda/\beta^2$.

Check Your Understanding

Try these questions using the conditional and unconditional calculations above.

For $N\sim\operatorname{Poisson}(1.2)$, what is $P(S=0)$, assuming independent exponential severities?
State the density of $S\mid N=3$ when each severity has rate $\beta$.
Why is $P(S=0)=P(N=0)$ under the stated assumptions?
If $\lambda=4$ and $\beta=2$, calculate $E[S]$ and $\operatorname{Var}(S)$.
Explain why the conditional gamma densities alone do not describe the whole unconditional law.