Why does the superposition of two processes with geometrically distributed interarrival times not result in a process with similar distribution?

Question

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
...

Answer 1

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.

Answer 2

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.