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Original Question

Let $N(t)$ be a Poisson process with intensity $\lambda$. Let $T_1<T_2<...$ be the occurrence times. Let $T_0=0$. For any $t>0$, define the $age$ random variable to be

$A_t := t-T_{N(t)} $

the $residual$ $life$ $time$ random variable to be

$R_t:=T_{N(t)+1}-t$

In other words, $age$ is the time elapsed since the last occurrence and $residual$ $life$ $time$ is the time remaining until the next occurrence.

Using a similar logic to the first answer of Relationship between poisson and exponential distribution, one can show that $A_t\sim$ Exp($\lambda$). Below is the reproduction of the logic:

$A_t>a \Leftrightarrow T_{N(t)}<T_{N(t)}+a<t \Leftrightarrow N(T_{N(t)})=N(T_{N(t)}+a)=N(t)$

I understand that the second $\Leftarrow$ logic requires the last equation to ALWAYS hold.

Continue on with user28's proof

$P(A_t>a)=P[N(T_{N(t)}+a)-N(T_{N(t)})=0]=P[N(a)=0]=e^{-\lambda a}$

However if I admit that $A_t\sim$ Exp($\lambda$), I will quickly run into the following paradox:

Since the exponential distribution is memoryless, one can easily prove that $R_t\sim$ Exp($\lambda$) and is independent of $A_t$. This means that two independent Exp($\lambda$) exponential random variables $A_t$ and $R_t$ sum to a different Exp($\lambda$) exponential random variable -- the interarrival time of the Poisson process.

According to Wikipedia https://en.wikipedia.org/wiki/Exponential_distribution, the sum, however, should be Gamma(2,$\lambda$) (in the ($\alpha,\beta$) parameterization).

What did I do wrong here? Was I too ambitious claiming $T_{N(t)}<T_{N(t)}+a<t \Leftrightarrow N(T_{N(t)})=N(T_{N(t)}+a)=N(t)$ ?

If the $\Leftarrow$ can not be claimed, then I can only get $P(A_t>a)>e^{-\lambda a}$. How should I get the distribution for $A_t$ in this case?


First Edit

I take the Poisson process as a limiting process of the Bernoulli process and arrive at the same conclusion $A_t\sim$ Exp($\lambda$). The key steps are as follows:

Step 1: Review the Law of Small Numbers for Bernoulli processes enter image description here

Step 2: Let $t\in\mathbb{N}$ for the Bernoulli process $\{X_r^m\}_{r\in\mathbb{N}/m}$. Then by time $t$, there are a total of $mt$ trials, and $N^m(t)$ of them are successes. For $N^m(t)>0$, the time that the last success takes place, $T_{N^m(t)}$, takes any value between $\frac{1}{m},\frac{2}{m},...,t$. Let $k\in\mathbb{N}$ such that $0<k<mt$

$P(t-T_{N^m(t)}=\frac{k}{m})$

$=P($ the last $k+1$ trials of the total $mt$ trials are one success and $k$ failures $)$

$=p_m(1-p_m)^k$

When $k=mt$, $P(t-T_{N^m(t)}=\frac{k}{m})=$$P($ all $mt$ trials are failures $)=(1-p_m)^{mt}$

Therefore

\begin{split} P \Big(t-T_{N^m(t)}>\frac{k}{m} \Big) &= \sum_{i=k+1}^{mt-1}P\Big(t-T_{N^m(t)}=\frac{i}{m}\Big) + P\Big(t-T_{N^m(t)}=\frac{mt}{m}\Big) \\ & = p_m(1-p_m)^{k+1} + ... + p_m(1-p_m)^{mt-1} + (1-p_m)^{mt} \\ & = (1-p_m)^{k+1} \end{split}

Taking $x=\frac{k}{m}$ and the limit $\lim_{m\to\infty} mp_m = \lambda$ yields \begin{split} P (t-T_{N^m(t)}>x) = \Big(1-\frac{\lambda}{m}\Big)^{mx+1} \to e^{-\lambda x} \end{split}

This again leads to the conclusion that $A_t\sim$ Exp($\lambda$) and the same paradox in the original question.

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Yes you are correct in suspecting that here the if and only if condition does not hold i.e.,

$A_t>a \Rightarrow N(T_{N(t)})=N(T_{N(t)}+a)=N(t)$ holds but not the other way around.

This can be seen as follows. Take $a$ such that $t<T_{N(t)}+a<T_{N(t)+1}.$ Then we still have the event $N(T_{N(t)})=N(T_{N(t)}+a)=N(t),$ but now $A_t<a.$


Edit 1:

An attempt: Let $X_t = T_{N(t)+1}-T_{N(t)}.$ We know $X_t \sim Exp(\lambda).$ Also from the memoryless property of exponential distribution $R_t \sim Exp(\lambda).$ However we know

$$ A_t = X_t - R_t $$

Using chain rule of probability.

\begin{eqnarray*} P(A_t>a)&=& P(X_t-R_t>a)=\int_a^\infty P(R_t<x_t-a|X_t=x_t) dP(X_t = x_t)\\ &=& \int_a^\infty P(R_t<x_t-a|X_t=x_t) \lambda e^{-\lambda x_t} dx_t \end{eqnarray*}

Till this step we do not require any dependence condition. Now we need to figure out the distribution $P(R_t<x_t-a|X_t=x_t)$ since $X_t,R_t$ are not independent.


Edit 2: I think I understand the basic problem, but still need to think a bit more. The following statement is true, "Consider a poisson process on a line, then randomly choose a point in the line. Then the distribution of the time to both the next and previous arrival point will be $Exp(\lambda)$, i.e. the process looks poisson in both direction." This is called the "Hitchhiker's paradox". But actually this is not a paradox, and can be proven, because we are inherently conditioning on the fact that we randomly arrive at a point $T_{N(t)}<t<T_{N(t)+1},$ and the probability of arriving at a larger interval is more than arriving at a smaller interval. So we observe a distance biased distribution. Its slightly non-intuitive. I suspect a similar thing happening here.

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  • $\begingroup$ Thank you A. Ray. In order to use convolution though, Xt and Rt must be independent. In this case, cov(Xt, Rt) = cov(Rt+At, Rt)=var(Rt)>0. Therefore, Xt and Rt are dependent, and I can not use convolution. $\endgroup$ – Ye Tian May 24 '16 at 20:59
  • $\begingroup$ @YeTian Yes you are correct, X_t and R_t are not independent. My mistake. Let me go back and check my queuing theory notes as I try to figure out the answer :) $\endgroup$ – A. Ray May 25 '16 at 3:00
  • $\begingroup$ Thank you! I don't disagree with your assessment in the first part. However, is it possible that the probability of the other A_t<a case is zero? I now have another method pointing to the same conclusion. $\endgroup$ – Ye Tian May 26 '16 at 3:19
  • $\begingroup$ No it cannot be zero. It will of course depend on the value of $a.$ $P(A_t<a)=0$ for all $a$ is a very strong statement. Just consider $a \to \infty,$ then $P(A_t<a)=0$ would imply the arrivals never occur ! $\endgroup$ – A. Ray May 27 '16 at 3:09
  • $\begingroup$ Please check my Edit 2. $\endgroup$ – A. Ray May 27 '16 at 3:22

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