Compound Poisson Distribution with sum of exponential random variables

Question

I'm trying to find the distribution and parameters in a Compound Poisson $S=\displaystyle\sum_{j=1}^{N}Y_{j},$ where $Y_{j}$ are exponential random variables independent and distributed identically with parameter $\alpha$ and $N$ is Poisson random variable with parameter $\lambda.$

I'm stuck because my attempts are based in computing the density probability function and I find a hard series: $$e^{-\lambda}e^{-\alpha x}\displaystyle\sum_{n=1}^{\infty}\frac{(\alpha\lambda)^{n}}{n!}\frac{x^{n-1}}{(n-1)!}.$$

The expression above was obteined using theorem of conditional probability in the random variable $N.$

If I use generating function I get $M_{S}(t)=e^{\lambda((\frac{\alpha}{\alpha-t})-1)},$ but I don't know the distribution of $S.$ This expression was got using conditional expectation again in random variable $N.$

Any kind of help is thanked in advanced.

Answer 1

The Question

Let $X \sim \text{Exponential}(\alpha)$, and let $\{X_1, X_2,\dots, X_n\}$ denote an iid sample of size $n$, where the sample size $n$ (instead of being fixed) is itself a Poisson random variable $N=n$. The OP seeks the distribution of the sample sum:

$$S = X_1 + X_2 + \dots + X_n \quad \quad \text{where} \quad N \sim \text{Poisson}(\lambda)$$

As $N$ is Poisson, and the domain of support of a Poisson includes 0, it follows that the sample size $N$ can be 0, in which case $S = 0$. This is important, because it means that $P(S = 0)$ will have discrete mass.

Solution

To proceed, first note that the sum of $n$ independent identical $\text{Exponential}(\alpha)$ variables has a $\text{Gamma}(n,\alpha)$ distribution i.e. $S$ has pdf $f(s \; \big| \; N = n)$:

where parameter $N \sim \text{Poisson}(\lambda)$ with pmf $g(n)$:

We seek the parameter mixture distribution of $S$ and $N$.

Unconditional pdf of $S$

• Discrete Part: $S = 0$ iff $n = 0$. This occurs with probability $P(N=0)$:

• Continuous Part: The parameter-mix distribution, for $S>0$, is given by $\mathbb{E}_g[f]$:

where:

• I am using the Expect function from the mathStatica package for Mathematica
• Hypergeometric0F1Regularized denotes the confluent hypergeometric function

In summary, the unconditional pdf of $S$ is:

$$\text{pdf}(S) = \left\{ \begin{array}{cc} e^{-\lambda} & \text{ if } s = 0 \\ \text{sol} & \text{ if } s > 0 \\ \end{array}\right.$$

which is a mixed discrete-continuous distribution.

Answer 2

Let $\{Y_i\}_{i\geq 1}$ be a sequence of IID $\mathrm{Exp}(\alpha)$ random variables, and let $$ S_N=\sum_{i=1}^N Y_i, $$ in which $N\sim\mathrm{Poisson}(\lambda)$. We know that $S_N\mid N=n\sim\mathrm{Gamma}(n,\alpha)$, for $n \geq 1$. Hence,

\begin{align*} \Pr\{S_N\in B\} &= \mathrm{E}\left[\Pr\{S_N\in B\mid N\}\right] \\ &=e^{-\lambda}I_B(0)+\sum_{n=1}^\infty \left(\frac{e^{-\lambda}\lambda^n}{n!} \int_B \frac{\alpha^n}{(n-1)!}\,u^{n-1}e^{-\alpha u}\,I_{(0,\infty)}(u)\,du\right). \end{align*}

TOL <- 0.01
lambda <- 4
alpha <- 2

B <- c(0, 3.5)

prob <- 0
if (B[1] == 0) prob <- exp(-lambda)

n <- 1
repeat {
  next_term <- exp(-lambda+n*log(lambda)-lfactorial(n)) * 
                 (pgamma(B[2], shape = n, rate = alpha) - 
                  pgamma(B[1], shape = n, rate = alpha))
  if (next_term < TOL) break
  prob <- prob + next_term
  n <- n + 1
}

print(prob)

Answer 3

The cumulative distribution does not have a simple closed form expression, nor does the density. Note that there is an atom at $S = 0$ with mass $\mathrm{Pr}\{N = 0\} = e^{-\lambda}$, so the density is for $S \vert S > 0$.

The series in the density can be related to the Bessel functions $I_0(x)$ and $I_1(x)$. But since this is a special case of the compound Poisson-Gamma distribution which itself is a special case of the Tweedie distribution, usable computing tools can be found under this name.

EDIT To derive an expression of the density, consider the following series $$ R(y) := \sum_{n=0}^{\infty} \frac{1}{n!\,n!} \, y^{n}, \qquad r(y) := \sum_{n=1}^{\infty} \frac{1}{n!\,(n-1)!} \, y^{n-1} = R'(y). $$ Note that $R(y) = I_0(2 \sqrt{y})$ where $I_0(z)$ is the usual modified Bessel function, with derivative $I_0'(z) = I_1(z)$. So, using the expression given in the question and some simple algebra we get the density: $f(x) = p \,\delta(x) + (1 - p) \, f_1(x)$ where $\delta(x)$ (abusively) stands for a Dirac density, $p:= e^{-\lambda}$ and $$ f_1(x) = \frac{p}{1 -p} \, e^{ - \alpha x } \alpha \lambda \, \frac{I_1(2 \sqrt{\alpha \lambda x})}{\sqrt{\alpha \lambda x}} \qquad \text{for } x > 0, $$ which is the density of $S$ conditional on $S > 0$. This must be the same solution as that of the answer by @wolfies, up to the $\alpha \leftrightarrow 1/ \alpha$ change of notation therein. The Bessel functions $I_\nu(z)$ are widely available, e.g. in R using besselI.