Show that Poisson processes are closed under finite superposition.

Question

Given a collection $\{\{N_i(t): t\geqslant 0\} : 1\leqslant i\leqslant m\}$ of mutually independent Poisson processes with respective rates $\lambda_i$, $1\leqslant i\leqslant m$, show that the superposition

$$ N(t) = \sum_{i=1}^m N_i(t) $$ is itself a Poisson process with rate $\sum_{i=1}^m \lambda_i$.

Answer 1

Let $\{N_1(t):t\geqslant0\}$ and $\{N_2(t):t\geqslant0\}$ be two independent Poisson processes with rates $\lambda_1$ and $\lambda_2$, respectively Let $N(t)=N_1(t)+N_2(t)$. Clearly \begin{align} \mathbb P(N(0)=0) &= \mathbb P(\{N_1(0)=0\}\cap \{N_2(0)=0\})\\ &= \mathbb P(N_1(0)=0)\mathbb P(N_2(0)=0)\\ &=0\cdot0\\ &=0. \end{align} Given times $0\leqslant t_1 \leq t_2 \leq \cdots \leq t_m$, the corresponding increments of $N_1$ and $N_2$ \begin{align} &N_1(t_1), N_1(t_2)-N_1(t_1),\ldots N_1(t_m)-N_1(t_{m-1})\\ &N_2(t_1), N_2(t_2)-N_2(t_1),\ldots N_2(t_m)-N_2(t_{m-1})\\ \end{align} are mutually independent, and hence the increments of $N$ $$ N(t_1), N(t_2)-N_1(t_1),\ldots N(t_m)-N(t_{m-1}) $$ are as well. Moreover, for each $2\leqslant j\leqslant m$, \begin{align} N_1(t_j)-N_1(t_{j-1}) &\sim\mathsf{Pois}(\lambda_1(t_j-t_{j-1}))\text{ and }\\ N_2(t_j)-N_2(t_{j-1}) &\sim\mathsf{Pois}(\lambda_2(t_j-t_{j-1})), \end{align} and hence \begin{align} N(t_j)-N(t_{j-1}) &= N_1(t_j)+N_2(t_j) - N_1(t_{j-1})+N_2(t_{j-1})\\ &= (N_1(t_j) - N_1(t_{j-1}) )+ (N_2(t_j) - N_2(t_{j-1}) )\\ &\sim \mathsf{Pois}((\lambda_1+\lambda_2)(t_j-t_{j-1})). \end{align} It follows that $\{N(t):t\geqslant 0\}$ is a Poisson processs with rate $\lambda_1+\lambda_2$, and by induction, the result holds the for the superposition of any finite collection of independent Poisson processes.

Answer 2

The desired result can be proven by invoking the Probability Generating Function (PGF) of a Poisson-distributed random variable (RV), and appealing to the following results:

1. The PGF of the sum of a sequence of discrete RVs, $X_n \sim f_X(x)$ is the product of the PGFs of the individual RVs.

$$ \Phi_{\sum_{k=1}^N X_k}(t) = \Pi_{n=1}^{N} \Phi_{X_n}(t) $$

2. If two discrete RVs have identical PGFs, they are guaranteed to have identical PMFs.

A sketch of a proof of the desired result goes as follows:

Let $Po(\lambda; n) = \frac{e^{-\lambda} \lambda^n}{n!}$, where $\lambda$ is called the rate of the poisson process. If $X \sim Po(\lambda; n)$, then it is easy to show that the PGF for $X$ is given by

$$\Phi_X(t) = e^{-\lambda} e^{\lambda t}.$$

Let $X_k \sim Po(\lambda_k; n)\,\, \forall\, k \in \{1, 2, ..., N\}.$ Then, by result 1, the PGF for the RV $S = \sum_{k=1}^N X_k$ is given by:

\begin{align} \Phi_S(t) &= \Pi_{n=1}^N e^{-\lambda_k} e^{\lambda_k t} \\ &= e^{-\sum_{k=1}^N \lambda_k} e^{(\sum_{k=1}^N \lambda_k) t}, \end{align}

the final expression manifestly being the PGF for an RV following the distribution $Po(\sum_{k=1}^N \lambda_k; n).$