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Assume we have two random variables $G_1$ and $G_2$ that both follow a geometric distribution of type B (so RVs can assume 0) with success probability $p = p_1 = p_2$ that describe the interarrival times of independent arrival processes.

From Kingman, we know that the superposition of poisson point processes (independent, memoryless, negative exponentially distributed interarrival times) again results in a poisson point process.

Since in the scenario above, the processes generated by $G_1$ and $G_2$ are both memoryless and independent and the geometric distribution is the discrete analogon to the continuous exponential distribution, why does the superposition of these processes not result in a process whose interarrival times follow the geometric distribution?

What property of the geometric distribution destroys this behavior?

Below is a quick R snippet that simulates the superposition of two geometrically distributed RVs.

nSamples <- 5e6
s1 <- cumsum(rgeom(nSamples, prob = 1/100))
s2 <- cumsum(rgeom(nSamples, prob = 1/100))

minMax <- min(max(s1), max(s2))

s1 <- s1[s1 < minMax]
s2 <- s2[s2 < minMax]

s3 <- diff(sort(c(s1, s2)))
data.table(x = s3)[, .(n = .N), by = x][order(x)][, pct := n / sum(n)][1:10]

The resulting distribution is not strictly monotonically decreasing.

    x      n        pct
 1: 0 149315 0.01493716
 2: 1 195471 0.01955451
 3: 2 191880 0.01919528
 4: 3 187642 0.01877131
 5: 4 183838 0.01839077
 6: 5 180202 0.01802703
 7: 6 177592 0.01776593
 8: 7 173753 0.01738189
 9: 8 170504 0.01705686
10: 9 166771 0.01668342
...
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A process with "geometrically distributed" inter-arrival times is a discrete-time Bernoulli random process (the random variables $X[n]$ are i.i.d. Bernoulli random variables), and an arrival is said to have occurred at time $n$ if $X[n]$ has value $1$. If $X[n]=1$, the waiting time till the next arrival is a geometrically distributed random variable with fixed parameter $p$. Note that this requires the definition of a geometric random variable as one that takes on positive integer values only, as contrasted with the other convention in which the random variable takes on nonnegative integer values (the type B that the OP favors). But, the distinction does not really matter for the key reason that makes the superposition (presumably meaning the sum) of two independent processes $\{X[n]\colon n \in \mathbb Z\}$ and $\{Y[m]\colon m \in \mathbb Z\}$ of this type of processes on this type not be a process of the same type.

$P(X[n] + Y[n] = 2) = p^2 > 0$

In contrast, for two independent Poisson processes, the event that arrivals in the two processes occur at the same time instant is an event of probability $0$ -- something that we cannot logically exclude as impossible but rare enough in practice that the event is assigned probability $0$ in theory. Thus, the sum of two independent Poisson processes is also a Poisson process while the sum of two independent Bernoulli processes is not a Bernoulli process.

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  • $\begingroup$ Why can't we use OP's definition of a geometric distribution, allowing waiting times of 0? If we do then a single process can easily have simultaneous arrivals, so it's not problematic that a superposition of two of them can have simultaneous arrivals. $\endgroup$ – fblundun Dec 16 '20 at 18:51
  • $\begingroup$ @fblundun The OP's definition of geometric distribution makes no sense in this context. Forget the two processes stuff; if even for just one process, if the inter-arrival time is allowed to take on value $0$, then it would be possible to have two arrivals (indeed, arbitrarily many arrivals) all occurring at time $n$ with inter-arrival time of $0$ between these arrivals. Does it make sense to talk of two or more arrivals in the same process all occurring at the same time $n$? What probability do we assign to this event? $\endgroup$ – Dilip Sarwate Dec 16 '20 at 21:57
  • $\begingroup$ For one process, if $p$ is our geometric distribution's success probability parameter, then the probability of two or more arrivals at time 0 is $p^2$. This is the probability that our first two wait times are both 0. Why doesn't it make sense to allow multiple simultaneous arrivals? Is it part of the standard definition of an arrival process? If so that just means we need to use a different definition for OP's question. $\endgroup$ – fblundun Dec 16 '20 at 22:25
  • $\begingroup$ @fblundun Never mind. I will stick with my model and ideas, and I wish you success in your use of your preferred model and notions. $\endgroup$ – Dilip Sarwate Dec 16 '20 at 22:34
  • $\begingroup$ I am sure that mathematics can cope with discrete-time processes featuring multiple simultaneous arrivals, as otherwise there would be no way to model events like "at least 2 people get off the next train". $\endgroup$ – fblundun Dec 17 '20 at 8:54
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You actually do get a (positive) geometric distribution out of your superposition if, instead of considering wait times between consecutive events, you consider wait times between consecutive times at which at least one event occurs.

But for your original question, I'll try to give some intuition about why the superposition of continuous-time memoryless Poisson processes leading to another memoryless Poisson process doesn't directly translate to the discrete case that you simulated.

If you calculate the ratios between your consecutive computer-generated frequencies, you will see that they are mostly the same - except the ratio of P(0)/P(1), which is smaller than the others. This suggests that it's only the behaviour of the distribution at 0 which makes it non-geometric.

Here's a similar discrepancy: the distribution of the absolute difference between two i.i.d. continuous uniform random variables is monotonically decreasing (in fact it's a straight line). But the same is not true of the absolute difference between two i.i.d. discrete uniform random variables. For example the table below shows the relative frequencies of the possible differences between the rolls of two fair 6-sided dice.

x  n
0  6
1  10
2  8
3  6
4  4
5  2

Here a difference of 1 is almost twice as likely as a difference of 0. This is because to get a difference of 0, the second die must have exactly the same value as the first; but to get a difference of 1, the second die may have a value either 1 more or 1 less than the first. This effect only occurs for a difference of exactly 0, so doesn't show up in the continuous case.

I think something similar is going on in your case. A non-rigorous argument: suppose that your $G_1$ process has an arrival at time 500. And suppose that your $G_2$ process generates exactly one arrival in the interval (450, 550). What might the absolute time difference between this arrival and the $G_1$ arrival at time 500? If the $G_2$ arrival is at 498 or 502, the difference is 2; if it is at time 499 or 501, the difference is 1; but only an arrival at time exactly 500 will produce a difference of 0. So we are more likely to get a difference of 1 or 2 than a difference of 0.

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