When superposition of two renewal processes is another renewal process?

If you merge (superpose) two Poisson processes with parameters $\lambda_1$ and $\lambda_2$, the outcome is another Poisson process with parameter $\lambda_1+\lambda_2$. But, how is that for two general renewal processes? Is there a class of renewal processes that merging two of the members of it makes another renewal process? More generally, under what conditions we can make sure the merged process is a renewal process?


Let $\{S_n\}$ and $\{T_n\}$ be independent renewal processes with interrenewal distributions $F$ and $G$. Define $$N(t):=N_S(t) + N_T(t)=\sum_{n=1}^\infty \left[\mathsf 1_{(0,t]}(S_n) +\mathsf 1_{(0,t]}(T_n)\right]. $$ Then the sequence of jump times of $N(t)$, $$U_n = \inf\{t: N(t)=n\} $$ is not in general a renewal sequence, because the inter-jump times need not be i.i.d. For a counterexample, consider when $F$ and $G$ are constant distributions, e.g. $S_n=\{i, 2i, 3i, \ldots\}$ and $T_n=\{j, 2j, 3j, \ldots\}$, with $i\ne j$. Take $i=2$ and $j=3$, then $$ U_n-U_{n-1} = \begin{cases} 1,& n\equiv 1,2,3,5\pmod 6\\ 2,& n\equiv 0,4\pmod 6. \end{cases} $$ Further, we have that $$R(t) = \mathbb E[N(t)] \stackrel{t\to\infty}\longrightarrow \frac43, $$ but as $R(n)-R(n-1)=U_n-U_{n-1}$, it is clear that $\lim_{t\to\infty}R(t+1)-R(t)$ does not exist, and thus Blackwell's renewal theorem does not hold.

There are two remaining questions to consider - is there more we can say about $N(t)$ than that it is a counting process, and whether being stable under superposition is equivalent to having independent and stationary increments (i.e. being a Poisson process)? As for the first, we can describe the jump times $\{U_n\}$ by the transitions in a Markov renewal process on $$E=\{(X_n,t) : X_n\in \{S,T\}, t>0\}. $$ As for the second, the superposition of $\{S_n\}$ and $\{T_n\}$ is a renewal process iff one of the following holds:

(i) One of the processes, WLOG $\{S_n\}$ has multiple renewals and $\{T_n\}$ does not (i.e. $F(0)>0$ and $G(0)=0$), $F$ and $G$ are concentrated on a semi-lattice $\{0,\delta,2\delta,\ldots\}$ and either \begin{align} F(x) &= \left(1 - p^{\left\lfloor\frac x\delta\right\rfloor+1}\right)\mathsf 1_{[0,\infty)}(x), \quad 0<p<1\\ G(x) &= \mathbb 1_{[\delta,\infty)}(x) \end{align} or \begin{align} F(x) &= \left(1 - p^{\left\lfloor\frac x\delta\right\rfloor+1}\right)\mathsf 1_{[0,\infty)}(x), \quad 0<p<1\\ G(x) &= \left(1 - q^{\left\lfloor\frac x\delta\right\rfloor}\right)\mathsf 1_{[0,\infty)}(x), \quad 0<q<1.\\ \end{align}

(ii) Neither process has multiple renewals, and $F$ and $G$ are exponential and hence $\{S_n\}$, $\{U_n\}$, and their superposition are Poisson processes.

The proof is given by Pairs of renewal processes whose superposition is a renewal process by J.A. Ferreira (2000).

Note in particular this means that ordinary renewal processes with strictly positive inter-renewal times, stability under superposition is equivalent to the processes being Poisson.

  • $\begingroup$ Thanks a lot for that. Do you know under what conditions we can make sure that the merged process is a renewal? $\endgroup$ – Susan_Math123 Nov 19 '15 at 14:43
  • $\begingroup$ Somehow I did not find this until now, but it addresses precisely that question: ac.els-cdn.com/S0304414999000952/… $\endgroup$ – Math1000 Nov 19 '15 at 15:07
  • $\begingroup$ thanks a lot for your comment. $\endgroup$ – Susan_Math123 Nov 19 '15 at 15:41
  • $\begingroup$ Also see here if you are interested in the asymptotic distribution of the superposition process: math.stackexchange.com/questions/49279/… $\endgroup$ – Math1000 Nov 19 '15 at 16:43
  • $\begingroup$ @Susan If you don't have any further questions, could you accept this answer? $\endgroup$ – Math1000 Nov 21 '15 at 10:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.