Splitting a renewal process This is a follow-up question of the question "When superposition of two renewal processes is another renewal process?".

How can we split a renewal process $P$ into a renewal process $P_1$ and another process $P_2$ (not necessarily renewal), where $P_1$ and $P_2$ can be independent or dependent to $P$?
If the rate (expected value of number of jumps in duration of time $t$) of $P$
is $\lambda$, we want to get the rate $\lambda'$ for $P_1$ for any $\lambda'<\lambda$.

My solution: Assume $x = [x_1 x_2 \dots x_n]$ is generated according to the renewal process $P$ in time interval $[0,t]$, where $x_i$ is the $i^{th}$ jump time that is generated according to the inter-renewal probability distribution, $p(x)$. I think if we randomly take $(\lambda-\lambda') t$ of the jumps and remove them, we obtain another renewal process with rate $\lambda'$ because: 1. since we choose the jumps randomly, the pdf of the jumps will remain identical 2. since each removing of a jump means replacing the two consequent jumps $x_i,x_{i+1}$ by their sum $x_i+x_{i+1}$ that is independent to other elements of $x$.
Is that correct? any idea what is the general answer?
 A: The original renewal process has independent and identically disributed (i.i.d.) inter-renewal times $\{T_1, T_2, T_3, ...\}$.  Thus, starting from time 0, renewals occur at times $\{T_1, T_1+T_2, T_1+T_2+T_3, ...\}$.  Now fix a probability $p >0$.  Independently place each renewal time to $P_1$ with prob $p$.  So we get new inter-renewal times $\{Z_1, Z_2, Z_3, ...\}$, where $Z_1 = \sum_{i=1}^GT_i$ and $G$ is an independent geometric random variable with success probability $p$, and $Z_2, Z_3, ...$ are i.i.d. That is because, when you generate them, you generate them independently but using the same probability law (so each is just a random sum of i.i.d. $T_i$ variables). 
If the rate of $P$ was $\lim_{t\rightarrow\infty} \frac{N(t)}{t}= \frac{1}{E[T_1]}=\lambda$ (with prob 1), the rate of $P_1$ is $\lim_{t\rightarrow\infty} \frac{N_1(t)}{t} = p\lambda$ (with prob 1). 
Here I am of course defining: 
\begin{align} 
&N(t) = \mbox{ Total number of renewals from $P$ during $[0,t]$}\\
&N_1(t) = \mbox{ Total number of renewals from $P_1$ during $[0,t]$}
\end{align} 
A simple example is when $T_i=1/\lambda$ for all $i \in \{1, 2, 3, ...\}$, for a given constant $\lambda>0$. So inter-arrival times are constant (and hence trivially i.i.d.).  Then $N(t)$ is a deterministic staircase function and $E[N(t)]=N(t)$ for all $t$, and indeed $$\lim_{t\rightarrow\infty} \frac{N(t)}{t} = \lim_{t\rightarrow\infty} \frac{E[N(t)]}{t} = \lambda $$
Probabilistically splitting this deterministic process $N(t)$ (of rate $\lambda$ arrivals/time) using a probability $p$ gives a random process $N_1(t)$ that has rate $p\lambda$ arrivals/time (and this new process indeed has i.i.d. inter-arrival times). 

On visualizing the renewal times: I imagine renewal times as if they are things that arrive to a system, like job arrivals in a queueing system. So the original renewal process
$P$ can be drawn over a timeline with spikes arising at the renewal times (the times $\{T_1, T_1+T_2, T_1+T_2+T_3, ...\}$).  Then $N(t)$ counts the spikes up to time $t$. We can "probabilistically thin" this spikey process by independently including spikes with prob $p$, and throwing the others away.  The thinned process $N_1(t)$ counts the number of included spikes, and so $N_1(t)\leq N(t)$ for all $t$.

As an interesting side note:  Consider $\{T_1, T_2, T_3, ...\}$ as any random sequence of inter-arrival times (not necessarily i.i.d.) and let $N(t)$ count the number of arrivals up to time $t$. Now  probabilistically thin this process to a new process $N_1(t)$ by independently including each arrival with probability $p$.   Then $E[N_1(t)] = pE[N(t)]$ for all $t\geq 0$ since: 
$$ E[N_1(t)] = E[E[N_1(t)|N(t)]] = E[pN(t)] = pE[N(t)] $$
For example, if $E[N(t)]=\lambda t$ for all $t\geq 0$, then $E[N_1(t)]=p\lambda t$ for all $t\geq 0$. An example of such a process $N(t)$ that is not a Poisson process is this: Choose $T_1$ uniformly over $[0,1]$, then define $T_i=1$ for all $i\in\{2,3,4,...\}$. 
