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 $\tau=0$.
Best Answer
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 $x<\tau$.