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Suppose that $v$ is a Radon measure on $(0,\infty)$ and let $X$ be a Poisson Point Process on $(0,\infty)$ with intensity measure $v$. Let $Y:=\int xX(dx)$.

In Theorem 24.17 of Probability Theory by A. Klenke (3rd version), the author, in showing that

$$ P[Y<\infty]>0 \Leftrightarrow P[Y<\infty]=1 \Leftrightarrow \int v(dx) \min(1,x) <\infty $$

writes $Y = Y_0+Y_\infty$ where $Y_0:=\int_{(0,1)} x X(dx)$ and $Y_\infty:= \int_{[1,\infty)} x X(dx)$.

I do not understand why it should be "clear" that

$$P[Y_\infty<\infty]>0 \Leftrightarrow P[Y_\infty<\infty]=1 \Leftrightarrow v([1,\infty))<\infty$$

My guess is that this last equivalence chain is linked to the fact that $Y_\infty$ is calculating the expectation of Poisson random variable on $[1,\infty)$ with parameter $v([1,\infty])$.

Any help in understanding this is very appreciated. Let me know if more details are needed. Thanks.

Enrico
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1 Answers1

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If $v([1,\infty))=\infty$ then denoting $v_n=v([n,n+1))$ we have $\sum_{n=1}^{\infty}v_n=\infty.$ Denote by $Y_n$ the number of points in $[n,n+1)$ of the Poisson process. By definition the $Y_n$ are Poisson independent and $E(Y_n)=v_n.$ Therefore $$Y_{\infty}=\sum_{n=1}^{\infty} Y_n=\infty.$$ A quick proof of this last fact goes as follows. Denote $w_n=\Pr(Y_n>0)=1-e^{-v_n}.$ If $\sum_{n=1}^{\infty}w_n<\infty$ this implies that $v_n\to 0$ and therefore $w_n\sim v_n$ implying the contradiction $\sum_{n=1}^{\infty}v_n<\infty.$ Hence $\sum_{n=1}^{\infty}\Pr(Y_n>0)=\infty$ and from the Borel Cantelli lemma we deduce that an infinite number of $Y_n$ are not $0.$

Gérard Letac
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  • Thanks for the answer. What do you mean with $w_n \sim v_n$? – Enrico Jan 14 '24 at 11:18
  • I'm also thinking about my guess: here you considered multiple Poisson random variables. Could also be correct to consider only one Poisson rv on $[1,\infty)$ with parameter $\lambda:=v([1,\infty]$? The link with the equivalence chain would be $E[Y_\infty]=\lambda$. – Enrico Jan 14 '24 at 12:32
  • If $ \lambda =v[1,\infty)<\infty$ of course $Y_{\infty}$ is Poisson with mean $\lambda.$ About $w_n\sim v_n$ this means $\lim {n\to \infty}w_n/v_n=1.$ – Gérard Letac Jan 14 '24 at 16:14
  • Clear. Just to see if I understood correctly: to prove that $P[Y_\infty<\infty]>0$ implies $P[Y_\infty<\infty]=1$ , I could argue that there exists some $Y_\infty(\omega)$ which represent the expectations of $X$ on $[1,\infty)$ that are not finite. But this contradicts the hypothesis that $X$ is a Poisson Point Process. Is it the right argument? Thanks for the patience. – Enrico Jan 15 '24 at 10:59
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    The second part of the Borel Cantelli lemma is concerned with the independent events $A_n={Y_n>0}$ of probability $v_n$. It says that either $\sum v_n<\infty$ and then with pb 1 a finite number of $A_n$ are realized, or $\sum v_n=\infty$ and with probability 1 an infinite number of $A_n$ are realized. This is a zero-one law. – Gérard Letac Jan 15 '24 at 11:26
  • Ok, thanks, so my guesses are totally wrong. I need to study to understand why now. – Enrico Jan 15 '24 at 18:49
  • I think my guesses are wrong because $X$, being a Poisson Point process, implies that $X(A)$ is a Poi$_{v(A)}$ only if A is relatively compact, which is not the case for $[1,\infty)$. – Enrico Jan 15 '24 at 20:19