"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.)
 A: I believe this is essentially an instance of the inspection paradox.  This is the reason why, if you choose a random student and look at the average size of the classes he is in, it will on average be larger than the average size of all classes.
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.)
