# Is the sum of several Poisson processes a Poisson process?

I would like to write an application that emits events distributed by a Poisson process with some $\lambda$

I need to separate the generation of these events in multiple processes, but I only have a $\lambda$, so I need to come up with a proper $\lambda_p$ for each process.

Is the sum of $p$ poisson processes, each with $\lambda_p = \frac{\lambda}{p}$ a Poisson process with parameter $\lambda$?

• You can show this since the waiting time is exponential. The minimum of exponentials is exponential with rate equal to the total rates, which you can show using the cdf. Commented Oct 30, 2014 at 8:52
• @NeilG Your comment assumes independence without which your whole edifice falls to the ground. See the very first sentence in Glen_b's excellent answer: +1 to Glen for that. Commented Oct 30, 2014 at 14:08

## 1 Answer

If they're independent of each other, yes.

Indeed a more general result is that if there are $k$ independent Poisson processes with rate $\lambda_i, \, i=1,2,\ldots,k$, then the combined process (the superposition of the component processes) is a Poisson process with rate $\sum_i\lambda_i$.

It's really only necessary to show the result for $k=2$ since that result can be applied recursively.

One way is to show that the conditions for a process to be a Poisson process are satisfied by the superposition of two Poisson processes.

For example, if we take the definition here, then the properties of a Poisson process are satisfied by the superposition of two processes:

• N(0) = 0 $\quad$ (clearly satisfied if it's true for the components)
• Independent increments (the numbers of occurrences counted in disjoint intervals are independent of each other) $\quad$ (follows from the independence mentioned above)
• Stationary increments (the probability distribution of the number of occurrences counted in any time interval only depends on the length of the interval) $\quad$ (if it applies to the independent components it will apply to their superposition)
• The probability distribution of N(t) is a Poisson distribution $\quad$ (see here*)
• No counted occurrences are simultaneous $\quad$ (simultaneity is an event with probability 0: follows from continuity and independence)

* (though I'd regard this as a consequence of the other properties)

• How would one prove the last property, that is no simultaneous occurrences? Commented Jan 25, 2023 at 19:54
• Typically, one doesn't attempt to prove model assumptions (George Box's maxim applies) but here's an example of a case where it is generally argued to hold and another where it is not; that may clarify the issues. Some naturally occurring forms of atomic decay are held to be completely independent events -- such event and another have no influence on each other; they occur one-by-one and a sufficiently small division of time by an observer would in principle separate the events. By contrast imagine we were modelling arrival of people at a queue; people from one family may arrive as a group. Commented Jan 26, 2023 at 0:03