Take a Poisson process on $\mathbb{R}$ with intensity given by Lebesgue measure. Think of this as the measure $d\mu=\sum_{n} \delta(t-\xi_n )dt$ where $\xi_n$ are the points of the process. Now compute the Cauchy transform of $\mu$. Formally this is $$G(t)=\int_{-\infty}^\infty \frac{1}{t-s}d\mu(s),$$ however $\int \frac{1}{1+|t|} d\mu(t)=\infty$ a.s., so the integral on the right is not convergent. However, $\int \frac{1}{1+t^2} d\mu(t)<\infty$ a.s. so the differences $$G(t)-G(t'):=\int_{-\infty}^\infty \left [ \frac{1}{t-s}-\frac{1}{t'-s} \right ] d\mu(s)$$ are well defined. So take $$G(t) = C +\left (G(t)-G(0 \right )$$ where $C$ is an arbitrary real constant.

Now let us look at the points $y_n$, $n= \ldots,-1,0,1,\ldots$, which solve $$G(y_n)=0.$$ List them in increasing order. The collection of these points is a point process on the real line. Has it been studied before?

An alternative route to defining $\{y_n\}$ goes by considering an $N\times N$ matrix consisting of ones off the diagonal and i.i.d. random variables on the diagonal. The spectrum of this matrix, suitably rescaled, gives the process $\{y_n\}$ in the large $N$ limit. A bit of numerical experimentation with these matrices using SciLab suggests that the distribution of nearest neighbor gaps ($y_{n+1}-y_{n}$) is not exponential, but instead has an intensity that drops to zero near zero. However, the higher order gaps ($y_{n+k}-y_{n}$ for $k>1$) seem to have a distribution very close to the similar distribution for the Poisson process.