This problem is a dual question of "Splitting a renewal process".

Assume we have a renewal process $P_1$ with inter-renewal distribution $p_1(x)$ and rate $\lambda_1 = \lim_{t \to \infty} \frac{N_1(t)}{t}$, where $N_1(t)$ is the total number of renewals of $P_1$ during $[0,t]$. We want to find another renewal process $P_2$ with rate $\lambda_2$ such that the super position of $P_1$ and $P_2$ is a renewal process.

If $P_1$ is Poisson, $P_2$ would be simply Poisson and the superposition of them is another Poisson and everything works. But here, $P_1$ is general, not only Poisson.

Also, if $P_1$ and $P_2$ are independent and $P_2$ is renewal, the superposition $P_1+P_2$ is renewal if and only if both $P_1$ and $P_2$ are Poisson, or have special distribution (discussed here).

The question remains for the general case. To solve that, I am thinking of considering building the superposition by as a renewal process with inter-renewal distribution $p_t(x)$ where $\frac{1}{\mathbb{E}[p_t(x)]}=\lambda_1+\lambda_2$. Now, we have $P_1(t)$ and $P_1(t)+P_2(t)$, I don't know how to find $P_2(t)$. Maybe we could take advantage of the fact that pdf of sum of two random variables is the convolution of their pdf. Any idea? or simpler method?

  • $\begingroup$ I suggest you consult some texts on renewal theory. In particular, Renewal Theory by Cox (1962) is a rather comprehensive treatment; Stochastic Processes by Ross (1996) and Introduction to Stochastic Processes by Çinlar (1975) are some more general texts that treat renewal theory, and Stochastic Processes: Theory for Applications by Gallagher (2013) is a more recent text that appears decent. $\endgroup$ – Math1000 Nov 24 '15 at 21:05
  • $\begingroup$ Thanks a lot. I will do that. Meanwhile, I wait for a response. $\endgroup$ – Susan_Math123 Nov 24 '15 at 21:07

Along the lines of my last comment in your previous question, you could do a reverse convolution type thingy:

Suppose we have a renewal process $N(t)$ with rate $\lambda>0$. Let $X$ be a random variable with the same distribution as the inter-arrival times of $N(t)$. Then $E[X]=1/\lambda$. Fix $\epsilon>0$. We want to add another process $A(t)$ (not necessarily renewal, and not necessarily independent of $N(t)$) so that the sum process $N(t)+A(t)$ is a renewal process of rate $\lambda + \epsilon$.

Fix a probability $p \in (0,1]$. Suppose $X$ can be viewed as a geometric sum of positive iid random variables $Y$, so that $X = \sum_{i=1}^G Y_i$ where $G$ is independent and geometrically distributed with success probability $p$. Thus, $E[X]=E[Y]/p$. The process $N(t)$ can be viewed as having arrivals defined only at the end of the last $Y_i$ interval in every geometric sum. Let $A(t)$ be the arrival process with arrivals at the end of all the other $Y_i$ intervals. Then $N(t)+A(t)$ is a renewal process with rate $1/E[Y]$. We can choose $p$ to make the rate difference equal to $\epsilon$:
$$ \epsilon = \frac{1}{E[Y]} - \frac{1}{E[X]} = \frac{1}{pE[X]} - \frac{1}{E[X]} = \frac{1/p-1}{E[X]} $$ So choose $p = \frac{1}{\epsilon E[X] + 1}$.

Now it may not be possible to make such a decomposition of $X$. We want $X=\sum_{i=1}^G Y_i$. The moment generating function for $X$ gives: $$ E[s^X] = \sum_{i=1}^{\infty} (1-p)^{i-1}pE[s^X|G=i] = \sum_{i=1}^{\infty} (1-p)^{i-1}pE[s^Y]^i = \frac{pE[s^Y]}{1-(1-p)E[s^Y]} $$ where I am not worrying about region of convergence issues for simplicity. So this can be used to solve for the moment generating function $E[s^Y]$ in terms of the moment generating function $E[s^X]$.

  • $\begingroup$ That is a nice. If you are near a river and observing the boats on it ($P_1$) and you want to add some boats ($P_2$) in such a way that the super-position process is renewal, then how can you use the pdf of $Y$ to do that? I think we cannot, since we have the causality problem, that is when you see the first boat and want to put some boats on the river, you need to know the time that the second boat arrives. $\endgroup$ – Susan_Math123 May 17 '16 at 20:14
  • $\begingroup$ @Susan , That can be done in the Poisson case by basic Poisson properties. I was arguing above that it can also be done for non-Poisson, provided that the inter-arrival time has a certain structure. $\endgroup$ – Michael Jun 13 '16 at 23:51
  • $\begingroup$ Thanks a lot. Your answer is correct actually. The problem I mentioned in my last comment is a bit different, it has the causality problem. $\endgroup$ – Susan_Math123 Jun 14 '16 at 3:55

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