Is a subset of a Poisson process also following a Poisson process?

Question

Say there is a Poisson arrival process with rate lambda. The realization of that process is a sequence $X$. Denote $x_t$ as the element of $X$ at time $t$.

Suppose I specify the following rule to select a subset $Y$ of that sequence: From $t=1$ to $t=\infty$, for each $x_t$ I draw a random value from a uniform distribution between 0 and 1. If the value greater than 0.5 then I have $A=A+1$ with initial $A=0$, and otherwise, $A=A-0.5$. I take that $x_t$ and put into $Y$ only when $A$ is an even number.

My question is: is $Y$ also generated by a Poisson process?

Answer 1

Well I'm not coming up with a mathematical prove but this simulation should be fine:

lambda <- 10
n <- 100000
A <- 0
Y <- vector()
X <- rpois(n, lambda)

for(i in 1:n)
{
  A <- ifelse(runif(1, 0, 1) > 0.5, A + 1, A - 0.5)
  if(!A%%2){Y[i] <- X[i]}
}

par(mfrow = c(1,2))
hist(X, breaks = 100, main = "Histogram of X (Poisson Distribution)")
hist(Y, breaks = 100, main = "Histogram of Y (Subset of X)")

with this result:

where round(var(Y, na.rm = T)) == lambda [1] TRUE

and round(mean(Y, na.rm = T)) == lambda [1] TRUE

So I would say yes it is still Poisson distributed if you choose a simulation length that is big enough, in my example 100000. Hope this helps

Answer 2

A contribution, not an answer.

First, I'd rephrase the description. The original $X$ is a Poisson process with arrival rate $\lambda$. To build $Y$ you have a separate $(0,1)$ state $A$ determined by flipping a fair coin at each $X$ arrival and switching $A$ when the coin comes up heads. You put the value of $X$ into $Y$ just when the state is $1$.

My intuition says this is memoryless, hence a Poisson process.

The simulation is @Codutie 's answer suggests that the rate is $\lambda/4$. If that's true it's probably a consequence of the fact that the expected number of tosses to get the next head is $2$.

What if the coin toss to switch states is biased, with probability $p$ for heads? Then the expected number of tosses for the next head is $1/p$. When $p$ is small you get long periods in which you accept $X$ and long periods when you don't. My intuition gives out trying to decide whether that bunching destroys the Poisson-ness. If it doesn't then I'd guess the rate to be $\lambda p/4$. Maybe that's the rate even if the process is no longer Poisson.

You could get more information by running the simulation for a biased coin with probability $p$ for heads.

Maybe someone will step up and prove something first.