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?

  • $\begingroup$ What about using all the even renewals for $P_1$? $\endgroup$ – Michael Nov 24 '15 at 4:59
  • $\begingroup$ @Michael That sounds good. But I am not sure how is the proof. Can you see my solution above? $\endgroup$ – Susan_Math123 Nov 24 '15 at 5:00
  • $\begingroup$ Randomly choosing the renewal events to put to $P_1$ would also work, as long as every choice is made independently and with the same probability. $\endgroup$ – Michael Nov 24 '15 at 5:00
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    $\begingroup$ Your description of "randomly choosign $m<n$" is a bit ambiguous since a renewal process has an infinite number of renewals, and we do not know what $n$ means. $\endgroup$ – Michael Nov 24 '15 at 5:02
  • $\begingroup$ @Michael Can you show me the math? I want to find a solution which the number of removed jumps over the total jumps $\frac{m}{n}$ can be changed easily. $\endgroup$ – Susan_Math123 Nov 24 '15 at 5:03

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 $i\in\{2,3,4,...\}$.

  • $\begingroup$ One really interesting question, likewise here, if we add spikes to $P$ randomly and independently, for example according to Poisson with rate $\lambda'$, will we obtain another renewal process with rate $ \lambda + \lambda'$? $\endgroup$ – Susan_Math123 Nov 24 '15 at 15:34
  • $\begingroup$ To answer my question, the only way that it can happen is when $P$ is also Poisson. How about for a general $P$? How can we add another process with rate $\lambda'$ to obtain a process with rate $\lambda + \lambda'$? $\endgroup$ – Susan_Math123 Nov 24 '15 at 15:44
  • $\begingroup$ If you have two spikey processes, and the counting process for each is $A_1(t)$ and $A_2(t)$, and if $\lim_{t\rightarrow\infty} A_1(t)/t =\lambda_1$ and $\lim_{t\rightarrow\infty} A_2(t)/t = \lambda_2$, then super-imposing them gives $\lim_{t\rightarrow\infty} (A_1(t)+A_2(t))/t = \lambda_1 + \lambda_2$. Now, the sum process $A_1(t) + A_2(t)$ may not be a renewal process anymore, even if both $A_1(t)$ and $A_2(t)$ are independent renewal processes and even if one of them is Poisson. Independent Poisson processes are special since adding them gives a new Poisson process with the sum rate. $\endgroup$ – Michael Nov 24 '15 at 19:32
  • $\begingroup$ Thanks for your comment. I see your point. My question is: if $A_1(t)$ is renewal, then how can we build $A_2(t)$ with parameter $\lambda_2$ such that $A_1(t)+A_2(t)$ is renewal? $\endgroup$ – Susan_Math123 Nov 24 '15 at 19:35
  • $\begingroup$ Another way to scale the rate of a renewal process is this: Suppose $N(t)$ is a renewal process with rate $\lambda$. Fix $r>0$ and define $A(t) = N(rt)$. Then $A(t)$ is also a renewal process, and it has rate $r\lambda$. $\endgroup$ – Michael Nov 24 '15 at 19:38

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