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?
 A: 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.
