Solved – 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.