# When Superposition of Two Renewal Processes is another Renewal Process?

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.

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