# Splitting of Poisson processes and Bernoulli trials

When a Poisson process is split according to the results of independent Bernoulli trials, two Poisson processes are obtained. It is trivial to prove that these two processes are independent. I was examining the converse. If a Poisson process is split in two Poisson processes, is it necessary that the split takes place according to the results of Bernoulli trials? I was able to prove this on condition that the two processes are independent. Is it possible to prove this without the previous condition?

• Are you thinking of a process such as a queue with arrival of men and women? – Cehhiro Dec 26 '16 at 0:59
• I am mainly interested in a theoretical result. I don't have anything particular in mind. Another relevant question is coming up with an example of a way to split a Poisson process that results in two processes that are not Poisson. – susami1996 Dec 26 '16 at 1:15
• Split them by a method that is not independent of the counts of prior arrivals. – Graham Kemp Dec 26 '16 at 9:31

I think the following serves as a counterexample:

Let $N$ be a rate 2 Poisson process on $\bf{R}^+$. Independently within each unit interval $I=[k,k+1)$, split the process into $N=N_1+N_2$ by uniformly randomly selecting $x$ of the $N(I)=n$ jumps to assign to $N_1$ with the remaining $n-x$ assigned to $N_2$ where $x$ is determined by a conditional pdf $$f(x|n) = Pr(N_1(I)=x|N(I)=n)$$ as given below.

Note that, if we were to take $f(x|n)$ to be the binomial distribution with parameters $n$ and $1/2$, we would have the "usual" independent Bernoulli split.

Instead, let's take: \begin{align} f(0|0) &= 1 \\ f(0|1) &= 1/4, f(1|1) = 3/4 \\ f(0|2) &= 1/2, f(1|2) = 1/2, f(2|2) = 0 \\ f(0|3) &= 1/8, f(1|3) = 0, f(2|3) = 3/4, f(3|3) = 1/8 \\ f(x|n) &= Binom(n,1/2) \hbox{ otherwise} \end{align}

It should be clear that the resulting split is not generated by independent Bernoullis.

I claim that $N_1$ and $N_2$ are Poisson processes of rate 1. I think it's clear that we need only establish, for an arbitrary interval $I=[k,k+1)$, that $N_1(I)$ and $N_2(I)$ are Poisson r.v.s of mean 1, given the nature of the rest of the construction (uniform random selection of points from $N(I)$ conditioned on the total $N(I)=n$, and independent construction within each interval $I$).

Write $g(x)$ for the marginal pdf of $N_1$, and $h(n)$ for the pdf of $N$ and note: $$g(0)=\sum_x f(0|n)h(n) = e^{-2} (1*1 + 1/4*2 + 1/2*4/2 + 1/8*8/3! + 1/16*16/4! + \cdots) = e^{-1}$$ It can similarly be shown that $g(1) = e^{-1}$, $g(2)=e^{-1}/2$, etc. and $N_1(I)$ is Poisson with mean 1. Similarly, it can be shown that $N_2(I)$ is Poisson with mean 1. So, that should do it.

If you're curious how I constructed $f(x|n)$ above, I considered the infinite matrix for the joint distribution $f(x,y)$ of $N_1(I)$ and $N_2(I)$ under the usual Bernoulli split and realized that I could permute the top left corner by added the 3x3 matrix: $$\begin{matrix} 0 & \epsilon & -\epsilon \\ -\epsilon & 0 & \epsilon \\ \epsilon & -\epsilon & 0 \end{matrix}$$ in such a way that the column and row totals (i.e., the marginal pdfs of $N_1(I)$ and $N_2(I)$) and all the secondary diagonals (those moving from lower left to upper right, corresponding to the pdf of $N(I)$) were unchanged.

• I assume your construction results in two processes which are not independent. Because, splitting a Poisson process with rate $\lambda$ in two independent ones with respective rates $\lambda_1, \lambda_2$ would give: $$P[N_1(t) = n_1|N(t) = n_1 + n_2] = \frac{P[N_1(t) = n_1]P[N_2(t) = n_2]}{P[N(t) = n_1 + n_2]} = \frac{e^{- \lambda_1 t} \frac{(\lambda_1 t)^{n_1}}{n_1!} e^{-\lambda_2t}\frac{(\lambda_2 t)^{n_2}}{n_2!}}{e^{-(\lambda_1 + \lambda_2)t} \frac{(\lambda t)^{n_1 + n_2}}{(n_1 + n_2)!}} = \frac{(n_1 + n_2)!}{n_1! n_2!} (\frac{\lambda_1}{\lambda})^{n_1} (\frac{\lambda_2}{\lambda})^{n_2}$$. – susami1996 Dec 26 '16 at 12:13
• Which implies that $N_1(t)|N(t) = n_1 + n_2$ ~ $bin(n_1 + n_2, \frac{\lambda_1}{\lambda})$, which is basically the proof relying on the independence condition I referred to in my original post. – susami1996 Dec 26 '16 at 12:18

If the two random variables $X_1 \sim Pois(rate = \lambda_1)$ and $X_2 \sim Pois(\lambda_2)$ are independent. Then $Y = X_1 + X_2 \sim Pois(\lambda_1 + \lambda_2).$

It matters that $X_1$ and $X_2$ are independent, but not the mechanism by which either arose. The proof using moment generating functions is easy.

Intuitively, suppose $X_1$ describes counts from a radioactive solution into a counter and $X_2$ describes counts from another solution. Pour the two solutions into the same beaker. Counts from that beaker will be described by $Y.$

Here is a simulation by demonstration in R statistical software, using a million realizations of each constituent random variable, where $\lambda_1 = 4$ and $\lambda_2 = 5:$

m = 10^6;  x1 = rpois(m, 4);  x2 = rpois(m, 5);  y = x1 + x2
mean(x1);  mean(x2);  mean(y)
## 4.003022  # aprx 4,  consistent with POIS(4)
## 4.997015  # aprx 5,  POIS(5)
## 9.000037  # aprx 9,  and POIS(9)
var(x1);  var(x2);  var(y)
## 4.004901  # aprx 4
## 4.995319  # aprx 5
## 9.011688  # aprx 9


In the figure below, histograms show simulated distributions, and dots atop histogram bars show exact Poisson probabilities. • Small detail: $Pois(\lambda_1 + \lambda_2)$ is missing the 1. – Cehhiro Dec 26 '16 at 2:34
• @O.VonSeckendorff: Thanks. Waited to proofread until simulation and plots were done. Hope all is OK now. – BruceET Dec 26 '16 at 2:36
• I don't see how this answers the original question. Unless there is some implicit step that I am missing, the connection between your answer (sum of independent Poisson RVs is Poisson) and the question (splitting of Poisson processes) is not quite clear to me. – angryavian Dec 26 '16 at 2:43
• Your poat has two question marks. The answers are no (Bernoulli split need not be the mechanism) and no (indep req'd, if that is 'prev cond'), as discussed in my first two paragraphs. As for the additional issue raised in your comment, I guess if the splitting mechanism were three events to process 1 then three to process 2, then three to process 1, and so on, then both processes have only counts divisible by 3, and so not Poisson for which the domain is all nonnegative integers. If the intuitive justification and the simulation thwart your theoretical mind set, just ignore them. – BruceET Dec 26 '16 at 4:01

Arrivals in a Poisson process occur independently of any other arrivals, at a constant average rate, say $\lambda$.   If the arrivals are split into two categories according to the result of independent Bernoulli trials, with constant rate $p$, then you will have two independent Poisson processes with rates $p\lambda$ and $(1-p)\lambda$.

This can be shown by establishing that, if $X,Y$ are the count of arrivals of the two categories, then the conditional distribution of arrivals in a category among given count of total arrivals, will be Binomially distributed.   (Why?)   If that is so, then:

\begin{align} \mathsf P(X=k, Y=h) ~& =~ \mathsf P(X+Y=k+h)~\mathsf P(X=k\mid X+Y=k+h) \\[1ex] & =~ \dfrac{\lambda^{k+h}~\mathsf e^{-\lambda}}{(k+h)!}\cdotp\dfrac{(k+h)!~p^k~(1-p)^h}{k!~h!} \\[1ex] & =~ \dfrac{\lambda^{k+h}~\mathsf e^{-\lambda}~p^k~(1-p)^h}{k!~h!} \\[2ex] \mathsf P(X=k) ~&=~ \sum_{n=k}^\infty \mathsf P(X+Y=n)~\mathsf P(X=k\mid X+Y=n) \\[1ex] ~&=~ \sum_{n=k}^\infty \dfrac{\lambda^n~\mathsf e^{-\lambda}}{n!}\cdotp\dfrac{n!~p^k~(1-p)^{n-k}}{k!~(n-k)!} \\[1ex] & =~ \dfrac{(p\lambda)^k~\mathsf e^{-p\lambda}}{k!} \\[2ex]\mathsf P(Y=h) ~&=~ \dfrac{\bigl((1-p)\lambda\bigr)^h~\mathsf e^{-(1-p)\lambda}}{h!} \\[3ex] \therefore~\mathsf P(X=k,Y=h)~&=~ \mathsf P(X=k)~\mathsf P(Y=h) \end{align}

And we conclude that the Poisson processes are thus independent.