# Intuitive explanation of paradoxical interval times distribution

When I simulate a Poisson process on the interval [0,1], then

• the interval time between successive points follows an exponential distribution.

E.g. in the code below when I select x[2]-x[1] then it is exponential distributed.

However,

• the interval time between successive points, conditional on that interval containing a particular point $$x$$, follows an Erlang distribution.

E.g. in the code below when I select x[k]-x[k-1], where 'k' is chosen such that x[k] is the lowest value above 0.5, then it is Erlang distributed

I can argue that it should be a sum of exponential distributed variables and hence Erlang distributed: When, for example, the specific value is 0.5 then the distance between the value before it and after it are exponential distributed.

However, it feels paradoxical that the distribution of any interval is exponential and the distribution of a specific interval that crosses a certain specific point is different distributed.

How can I explain this more intuitively?

The code below might explain some more:

set.seed(1)

t = replicate(10^4, expr = {
n = rpois(1,40)
x = runif(n)
x = x[order(x)]
k = min(which(x>0.5))
x[k]-x[k-1]
#x[2]-x[1]
})

hist(t, breaks = seq(0,1,0.005),freq = 0, xlim = c(0,0.3), main = "histogram of interval crossing 0.5")

t = replicate(10^4, expr = {
n = rpois(1,40)
x = runif(n)
x = x[order(x)]
k = min(which(x>0.5))
#x[k]-x[k-1]
x[2]-x[1]
})

hist(t, breaks = seq(0,1,0.005),freq = 0, xlim = c(0,0.3), main = "histogram of the first interval")

• I don't really get the motivation. How, for instance, would your analysis go if the waiting times were constant and non-random? Say, unit waiting time, so the process consists of a sequence $\ldots, x-1, x, x+1, x+2, \ldots.$ Any interval surrounding a randomly chosen point will have length 2, not 1: so even in this simplest possible situation the intervals differ.
– whuber
Commented Apr 23 at 16:36
• @whuber "Any interval surrounding a randomly chosen point will have length 2" It would be 1 not? Since the interval lengths are fixed at 1. Commented Apr 23 at 16:43
• @whuber alternatively consider t = replicate(10^4, expr = {x = 0; while(x<1) {dt = rexp(1,40); x = x+dt}; dt}) we keep adding exponential distributed variables untill reaching the value 1, then return the length of the interval that made us pass that value of 1. Commented Apr 23 at 17:14
• @whuber I have changed it now into "interval containing". The length of an interval, conditional on that interval containing a specific value is Erlang distributed instead of exponential distributed. I can argue why it is the case, but it feels counterintuitive... Commented Apr 23 at 17:23
• You have rediscovered the inspection paradox in renewal theory. Commented Apr 23 at 20:08

Here is some intuitive viewpoint different from simply adding the two exponential distributions:

The point 0.5 was fixed, but we can also consider it as a randomly chosen point.

In that case we may wonder why an interval picked by using a randomly chosen point has a different length distribution in comparison to an interval picked by a random interval.

The reason for the difference is that larger intervals have more probability of being picked by a random point.

So we can consider the distribution of picking a certain interval time as proportional to the product of the length and frequency of the interval $$\propto t\cdot f(t)$$ and that relates to the Erlang distribution.

• And it ends up being $Gamma(2, 1)$ in this case is because the probability that the point lands in a particular interval is proportional to the length of that interval. Commented Apr 23 at 17:43
• +1 -- this answer helps me appreciate the question!
– whuber
Commented Apr 23 at 19:00
• @whuber my motivation was the several elements neccesary for this answer here: stats.stackexchange.com/a/645685 Commented Apr 23 at 22:17