# “Paradox” of a Poisson process on $\mathbb R$

This question concerns the apparently peculiar behavior around zero of the Poisson process defined over the entire real line $\mathbb R$.

A Poisson process $N(A)$ with mean measure $\mu$ on a general topological measure space $(\Omega, \mathcal B, \mathbb P)$ is defined such that for all $A \in \mathcal B$ satisfying $\mu(A) < \infty$, $$\mathbb P(N(A) = n) = e^{-\mu(A)} \frac{(\mu(A))^n}{n!}$$ and for disjoint $A_1,\ldots,A_k$, the $N(A_i)$ are mutually independent.

If $\Omega = [0,\infty)$ and $\mu$ is proportional to Lebesgue measure, then we can construct the process $N$ via a sequence $\{T_i\}$ of iid $\mathrm{Exp}(\lambda)$ random variables with $X_n = T_1 + \cdots + T_n$ and $N(t) = \#\{n: X_n \leq t\}$.

If $\Omega = (-\infty,\infty)$, a simple modification of the above construction works. We take a second iid $\mathrm{Exp}(\lambda)$ sequence $\{U_i\}$ independent of the first and place points at $X_{-n} = -(U_1 + \cdots + U_n)$. It seems easy to see that this satisfies the general definition of a Poisson process. We only really need to worry if $A$ contains points on both sides of zero. So, if $A$ is any Borel set on $\mathbb R$, we decompose it as $A = (A \cap (-\infty,0)) \cup (A \cap [0,\infty))$ and then use the fact that the construction on each half-line was independent of one another and that sums of independent Poissons are Poisson.

"Paradox": Every interarrival time is $\mathrm{Exp}(\lambda)$ independently of each other except for the interarrival time that straddles zero, namely $X_1 - X_{-1}$, which is still independent of all other interarrival times, but is $\Gamma(2,\lambda)$ since it is the sum of two independent $\mathrm{Exp}(\lambda)$ random variables.

How do we explain away this peculiar behavior of the process around zero?

(I realize this is not a proper paradox, which explains the quotation marks.)

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The problem may be one of definition of a Poisson process. Many developments use $(0, \infty)$ rather than $[0,\infty)$ as you have done, time intervals of the form $(t_1, t_2]$, and so the first arrival time can also be viewed as a possible inter-arrival time if there was an arrival at $t = 0$. There is a similar phenomenon in the study of "telegraph waves" that take on values $\pm 1$, switching back and forth each time a Poisson arrival occurs. If $X_0 = +1$ (or $-1$) is fixed, it is a semirandom telegraph wave. The random telegraph wave has $P(X_0 = +1) = P(X_0 = -1) = 1/2$. –  Dilip Sarwate Oct 21 '11 at 1:13
@DilipSarwate: I don't believe the distinction between $[0,\infty)$ and $(0,\infty)$ will matter in the OP's construction since the probability of an arrival at $t = 0$ is zero. –  cardinal Oct 21 '11 at 1:53
@cardinal The OP constructs a process on $[0,\infty)$. If he extends the same idea on the negative axis, his construction gives a process on $(-\infty,0]$. He then says the union is a process on $(-\infty, \infty)$. But what happens at $0$ is ambiguous. In particular, the zero probability event that both constructions put an arrival at $t = 0$ makes the union have two arrivals at $t = 0$ in violation of the definition of a Poisson process. On the other hand, the union of processes on $(0,\infty)$ and $(-\infty,0)$ is not a process on $(-\infty,\infty)$. –  Dilip Sarwate Oct 21 '11 at 10:04
@DilipSarwate: Think, by way of analogy, of the standard construction of the Poisson process on $(0,\infty)$. There also exists probability of zero of two jumps at every epoch, namely if some of the $T_i$ were identically equal to zero. By countable subadditivity, a double jump at any epoch taken over the whole collection is still zero. So, I don't think the "issue" you've identified is really an issue. –  cardinal Oct 21 '11 at 10:15
@DilipSarwate. This construction is not mine. I got it from Kingman's 1993 monograph on Poisson processes. This case of the process doesn't seem to show up as widely as others. –  bpt Oct 24 '11 at 12:10

In an ordinary one-dimensional Poisson process, where the expected time from one occurrence to the next is $\lambda$, if you single out any particular time and ask what is the expected time from the most recent occurrence before that time until the next occurrence after that time, it will be exactly twice as big as the average time between occurrences. In other words, if buses arrive at bus stops at random times with an average time between buses of about 10 minutes, and you arrive at the bus stop at a random time to wait for a bus, then you're more likely to arrive at a time when the time from the last previous bus until the next bus is longer than average, than you are to arrive between two buses whose arrival times differ by only 30 seconds. (This is not so unrealistic as a description of the 16A bus route between downtown Minneapolis and downtown St. Paul during certain hours, which are about 11 miles from each other.)
(+1) I think this is right. Note that for a homogeneous Poisson process on $(0,\infty)$ with rate $\lambda$, the age process $A(t_0) = t_0 - S_{N(t_0)}$ and forward recurrence time $B(t_0) = S_{N(t_0)+1} - t_0$ are independent exponential random variables both with rate $\lambda$ for each fixed $t_0$ and so $S_{N(t_0)+1} - S_{N(t_0)}$ is $\Gamma(2,\lambda)$. –  cardinal Oct 21 '11 at 3:40
This is the explanation. But @cardinal, what you write holds only in the limit when $t_0\to+\infty$, otherwise the part $A(t_0)$ is distributed like $\min\{t_0,Z\}$ with $Z$ an exponential random variable with parameter $\lambda$ independent on $B(t_0)$ (with the customary choice $S_0=0$). Note that $A(t_0)\leqslant t_0$ almost surely and that, when $t_0\to\infty$, $A(t_0)$ converges in distribution to what you say. –  Did Oct 21 '11 at 5:25