Let $N(\omega,A)$ denotes a Poisson random measure on $\mathbb{R}$, for $\omega$ in a probability space and $A\subset \mathbb{R}$, with intensity being the Lebesgue measure. Then, we define a random variable: $$ Z(\omega):=\int_{\mathbb{R}}\frac{1}{x}\,dN(\omega,x). $$ $Z$ represents the sum of $1/x$ for each $x$ in the Poisson process.
- I want to show that $Z$ is well-defined, in the sense that $\int_{[-M,M]}\frac{1}{x}\,dN(\omega,x)$ converges almost surely as $M\to\infty$.
- Also, I want to compute the characteristic function $\phi$ of $Z$, but the only results I have for doing so are only the integral of the form $\int_{\mathbb{R}}x\,dN(\omega,x)$.
I can't figure out how to do this. My guess is that $\frac{1}{x}\,dN(\omega,x)$ also defines another Poisson process, but I don't know how to proceed or show this.
Let $Y_1,Y_2,\dots$ be all the points of the Poisson process on the real line, ordered so that the sequence $S_n=|Y_n|$ is increasing. Let $S_0=0$. Basic properties of the Poisson process ensure that $\{S_n-S_{n-1}\}_{n \ge 1}$ are i.i.d. exponential variables of mean $1/2$, so that $$S_n/n \to 1/2 \tag{*} \quad \text{almost surely $ $ as} \quad n \to \infty \,.$$ Observe that $J_n=\text{sgn}(Y_n)$ are i.i.d. $\pm$ valued with mean zero, and are independent of $S=\{S_n\}$. The a.s. convergence of $\int_{[-M,M]}\frac{1}{x}\,dN(\omega,x)$ as $M\to\infty$, is equivalent to the a.s. convergence of the series $$\sum_n \frac1{Y_n}=\sum_n \frac{J_n}{S_n}\,. \tag{**}$$
If we condition on $S=\{S_n\}$, then by [1], the series $(**)$ converges a.s. iff $$\sum_n \frac{1}{S_n^2} <\infty \,.$$ This inequality holds by $(*)$.
[1] https://en.wikipedia.org/wiki/Kolmogorov%27s_two-series_theorem