I understand that inter-arrival times of a Poisson process are exponentially distributed and therefore the inter-arrival times are memoryless.

However, how about the waiting times of Poisson process i.e wait time till $k$th arrival where $k \geq 2$. This is an Erlang distribution and shouldn't be memoryless -- or is it?

If it is, can someone help how to show it?

In general, why is Poisson process memoryless? I understand that interarrival times or time to next arrival are but doesn't look like time till kth arrival is also memoryless .. or is it?


1 Answer 1


Memorylessness is a property of the following form:

$$\Pr(X>m+n \mid X > m)=\Pr(X>n)\ .$$

This property holds for $X_1=\ \text{time to the next event in a Poisson process}\ $, but it doesn't hold for $X_k=\ \text{time to the}\, k^\text{th}\, \text{event in a Poisson process}\ $ when $k>1$.

As for how to show it, you could try to do it from first principles.

If you can show that the essentially equivalent form $P(X>s+t)\neq P(X>s)P(X>t)$, (for $s, t>0\ $), that would be sufficient; you already know the distribution for $X_k$.

  • $\begingroup$ Thanks. You address confusion that i had.So why is poisson process called memoryless when its only X1 thats memoryless and this is not true for Xk with k>1? $\endgroup$
    – toing
    Oct 7, 2014 at 1:30
  • $\begingroup$ As an extension, let me also ask if poisson distribution( not poisson process ) is memoryless? $\endgroup$
    – toing
    Oct 7, 2014 at 4:54
  • 2
    $\begingroup$ What is memoryless in the Poisson process is inter-event intervals (usually inter-event times). The Poisson distribution itself is the distribution of counts per unit interval. So you have a set of counts, but not the times (or whatever you're measuring events over). You don't have inter-event intervals to be memoryless about. $\endgroup$
    – Glen_b
    Oct 7, 2014 at 5:03

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