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I have a question concerning the superposition of renewal processes.

Assume that we have $n$ independent renewal processes with the same interrenewal time density $f(x)$ and distribution $F(x)$, both of which are known. The superposition of the $n$ processes leads to another renewal process with density $g(x)$ and distribution $G(x)$. I know that $g(x)$ and $G(x)$ can be obtained from $f(x)$ and $F(x)$. The derivation of the relation between the component processes and the superimposed process is explained here:

Superposition of renewal processes: Variance of lifetimes

and is given by:

$g(x) = -\frac{\mathsf d}{\mathsf d x}\left[\bar F(x)\left(\frac1\mu\int_x^\infty\bar F(u)\ \mathsf d u\right)^{n-1}\right]$

It seems that this relation is not valid when the component processes have a non-zero mínimum interrenewal time. For example if each component has exponential interrenewal times with a mínimum $\tau$, such that $f(x) = \lambda e^{-\lambda(x-\tau)}$ and $F(x) = 1 - e^{-\lambda(x-\tau)}$, both valid for $x\geq\tau>0$, then the relation above does not give valid results, which I have verified with some numerical simulations.

My question is: How should the relation above be modified to be valid when there is a mínimum non-zero interrenewal time $\tau > 0$?

Edit: Let me show what I have done.

I assume that the inter-arrival time of each component has a density $f(x) = \lambda e^{-\lambda(x-\tau)}$, which is a shifted exponential, where $\tau > 0$ is the mínimum value that the inter-arrival time can take in a single component. The mean inter-arrival time for a single component is $\mu = \tau + \frac{1}{\lambda}$. The corresponding CDF for each component is $F(x) = 1 - e^{-\lambda(x-\tau)}$.

Now I apply the equation:

$g(x) = - \frac{d}{dx} \left[ e^{-\lambda(x-\tau)} \left( \frac{1}{\tau + 1/\lambda} \int_x^\infty e^{-\lambda(u-\tau)} du \right)^{n-1}\right]$

$g(x) = -\frac{d}{dx} \frac{e^{-\lambda n (x-\tau)}}{(1+\lambda\tau)^{n-1}} = \frac{\lambda n e^{-\lambda n (x - \tau)}}{(1+\lambda\tau)^{n-1}}$

This result cannot be correct because it cannot be a PDF since $\int_\tau^\infty g(x) dx \neq 1$ and when I compare its shape with the one obtained from simulation for $n=10$ they only match for $\tau=0$.

João
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  • It is not generally true that "The superposition of the $n$ processes leads to a another renewal process." There are issues of memory here. It would be true of the original renewal processes were independent Poisson. – Michael Oct 17 '18 at 21:07
  • The above formula on PDF should be valid for "non-arithmetic" type processes which have the property that, for large $n$, the $n$th arrival at system 1 comes at what looks like a "uniformly random time" in the timeline of system 2 (and hence all other systems), in the sense that the probability that the residual time for system 2 is larger than $x$ is close to the time average fraction of time that the residual time for system 2 is larger than $x$. In fact this indeed holds for shifted exponentials so I am surprised you find some issue. – Michael Oct 17 '18 at 21:35
  • I have simulated this with n=10 processes and the result of solving the integral in the expression above holds only when $\tau=0$. I need the exact expression for an arbitrary $n$ (not necessarily large). – João Oct 17 '18 at 22:10
  • The above formula only holds for large $n$ (i.e., $n\rightarrow\infty$), perhaps that is your discrepancy. Recall that the superposition process is not a renewal process so inter-arrivals of the superposition process are not iid. Getting an individual result for each $n$ seems difficult. – Michael Oct 17 '18 at 22:41
  • Thank you for your answers. I first read a paper by D R Cox from 1954 (On the superposition of renewal processes) which seems to be the first to obtain this result and in fact there is a section for the derivation of that analytical result and another section where it is discussed that it converges to a Poisson process as n tends to infinity, regardless of the individual $f(x)$ for each component so I assumed that the expression is valid for any $n$. – João Oct 17 '18 at 22:44
  • Your mistake in evaluating that integral is in assuming only the case $x>\tau$. For $x<\tau$ we have $\overline{F}(x)=1$. And so indeed we have $g(x)>0$ for $x>0$, not just for $x>\tau$. PS: I am sure that paper makes additional assumptions like "non-arithmetic" to avoid counter-examples like when the inter-arrival times are always positive integers (in which case the result does not hold). – Michael Oct 17 '18 at 23:02
  • Note: When I use the index $n$ I am refering to a particular renewal (the $n$th one) not $n$ as the number of renewal processes. When I said it only applies "for large $n$" I did not realize the variable $n$ was already being used. The result holds for an arbitrary number of renewal processes, but it needs the number of renewals to go to infinity. You can see my comment on $n$ was for "the $n$th arrival." – Michael Oct 17 '18 at 23:07

1 Answers1

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Note that the superposition of renewal processes is not necessarily a renewal process.

The formula $g(x) = -\frac{\mathsf d}{\mathsf d x}\left[\bar F(x)\left(\frac1\mu\int_x^\infty\bar F(u)\ \mathsf d u\right)^{n-1}\right]$ for $x>0$ applies when the process has certain "non-arithmetic" properties which avoids counter-examples such as when all inter-arrival times are positive integers. Specifically, the processes must have the property that if we look at the $k$th arrival from system 1, for large $k$, that time looks like a "uniformly random time" with respect to system 2, in the sense that the residual time for the next arrival from system $2$ is distributed according to the time average residual life of system 2. (And, since all systems are identical but independent, the same holds for system 3, 4, ..., $n$).

However, a shifted exponential distribution such as you are examining is indeed "non-arithmetic" so there should not be any discrepancy, provided you are simulating "large $k$" i.e. a large number of renewals.

You make a mistake in calculating your integral for $g(x)$ because you only consider $x>\tau$. But we have $g(x)>0$ also for $0<x<\tau$ and indeed $\overline{F}(x)=1$ for $x<\tau$.

Michael
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  • Note that in the comments above I accidentally used $n$ to represent a renewal index, even though it was already defined as the number of renewal systems. To avoid confusion, in this answer I use $k$ as that renewal index, and keep $n$ as defined in the question. – Michael Oct 17 '18 at 23:17
  • This was indeed the (silly) mistake I was making. This solves my issues, now all my results fit nicely. Thanks a lot! – João Oct 18 '18 at 20:41
  • Glad to hear it. Yes, I would expect the formula to work for shifted exponentials. It would likely not work if all inter-arrival times are positive integers. – Michael Oct 18 '18 at 20:57