Thinkenning a Renewal Process 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?
 A: 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 $E[s^X]$. 
