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.)