Suppose we have a Compound Poisson Process with intensity $\lambda$ $$ S_t = \sum_{i=1}^{N_t} X_i. $$ We can compute the formula for the CDF as follows \begin{align*} F_{S_t}(x) = P(S_t \leq x) &= \sum_{k=0}^\infty P(S_t \leq x \mid N_t = k)P(N_t = k) \\ & = \sum_{k=0}^\infty P(\sum_{i=1}^k X_i \leq x)\frac{(\lambda t)^k}{k!}e^{-\lambda t} \\ & = \sum_{k=0}^\infty F_X^{*k}(x)\frac{(\lambda t)^k}{k!}e^{-\lambda t}. \\ \end{align*} My question is, Is there a known distribution for $X_i$ such that this formula can be explicitly computed? To me the exponential distribution seemed like a promising choice, but I wasn't able to get far with it.

  • $\begingroup$ Compound Poisson distributions are special cases of infinitely divisible distributions. Their characteristic functions can be written in neat form. $\endgroup$ Mar 28 at 11:55
  • $\begingroup$ That I know, but this question is more for application purposes. It is easier to compute probabilities with closed form distribution functions instad of approximating them. $\endgroup$ Mar 28 at 14:27
  • $\begingroup$ Note that $S_t = \sum_{i=1}^{N_t} X_i$ needs to be corrected as $S_t = \sum_{\color{blue}{i=0}}^{N_t} X_i$ $\endgroup$
    – Amir
    Apr 2 at 15:01
  • $\begingroup$ I'm used to treating a sum from 1 to 0 as an empty one, if that's not the norm I can change it. $\endgroup$ Apr 2 at 16:17
  • $\begingroup$ Did you mean that $X_0=0$? If Yes, then it results in that $S_t$ has an atom at $0$ with probability $P(N_t=0)=e^{-\lambda t}$, and $S_t$ becomes a mixed random variable (neither continuous nor discrete) if $X_1, X_2, \dots$ follow a continuous distribution. You need to clarify it in the OP. In my answer, I assumed that $X_0, X_1, \dots$ are iid. $\endgroup$
    – Amir
    Apr 2 at 20:56

2 Answers 2


The case where $X_i\sim x^{a-1}e^{-x}/\Gamma(a)$ is Gamma can be relatively explicit since $X_1+\cdots+X_n$ has density $x^{na-1}e^{-x}/\Gamma(na).$ The function $$f_a(z)=\sum_{n=1}^{\infty}\frac{z^n}{n!\Gamma(na)}$$ is sometimes called a function of Mittag-Leffler. The law of $X_1+\cdots X_{N(t)}$ is therefore $$e^{-\lambda t}\delta_0(dx)+\frac{e^{-\lambda t-x}}{x}f_a(\lambda t x)1_{(0,\infty)}(x)dx.$$

Of course rather artificial examples can be created, like $$\Pr(X_i=n)=\frac{1}{-\log(1-p)}\frac{p^n}{n}$$ for $n=1,2,\ldots$ where $X_1+\cdots X_{N(t)}$ is negative binomial, or $X_i$ Bernoulli.

  • $\begingroup$ So there is no semi-known distribution (A bit more complex than Bernoulli of course) where a CDF can be explicitly computed? $\endgroup$ Apr 2 at 12:34

The cdf should be very complex, and cannot be obtained explicitly even for exponential distribution. Below, I show how you can handle it numerically.

Using the law of total expectation, it can be straightforwardly shown that the mgf $M_{S_t}(a)$ of (note that we need to consider $X_0$ as $N_t$ can become $0$)

$$ S_t = \sum_{\color{blue}{i=0}}^{N_t} X_i, $$

is given by

$$\color{blue}{M_{S_t}(a)=M_{X}(a)e^{\lambda t \left(M_{X}(a)-1\right)}} \tag{1}$$

where $M_{X}(a)$ denotes the mgf of $X$ (the same procedure can be done for the characteristic function).

Then, the pdf of $S_t$ is given by

$$\color{blue}{f_{S_t}(x)=\mathcal L_s^{-1} \left [ M_{S_t}(-s) \right ]} .$$

When $\color{blue}{X \sim \mathcal E (1)}$, for $a<1$ we have:

$$M_{S_t}(a)=\frac{1}{1-a} e^{ \lambda t \left(\frac{1}{1-a} -1\right)}.$$

For $\color{blue}{\lambda=1}$ and $t=1$, the inverse Laplace can be obtained here, so for $x>0$ we have:

$$\color{blue}{f_{S_1}(x)=e^{-x - 1} \frac{I_1 \left (2 \sqrt{x} \right )}{\sqrt{x}} + e^{-x - 1} I_2 \left (2 \sqrt{x} \right) }$$

where $I_n$ denotes the modified Bessel function of the first Kind.

You can use the above pdf to compute the cdf of $S_1$ numerically (just change $2$ here with any other positive number you want).

To derive (1), we assume that $X_0,X_1,\dots \sim X$ are iid. If we assume $\color{blue}{X_0=0}$ and $X_1,X_2,\dots \sim X$ are iid, then we have

$$\color{blue}{M_{S_t}(a)=e^{-\lambda t} +e^{\lambda t \left(M_{X}(a)-1\right)}} \tag{2}.$$

This majorly changes the distribution. For the example studied above when $\color{blue}{X \sim \mathcal E (1)}$, using (2), the invers Laplace is obtained here, and we obtain the following mixed distribution for arbitrary $\lambda , t>0 $:

$$\color{blue}{\mathbb P(S_t=0)=e^{-\lambda t}}$$

$$\color{blue}{f_{S_t}(x)=e^{-x - \lambda t}\, \frac{\sqrt{\lambda t}\,I_1 \left (2 \sqrt{\lambda tx} \right )}{\sqrt{x}}, \, x>0}. $$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .