Prove that a sum containing a Poisson variable is finite with probability 1

Question

Let $Z = \sum_{r=1}^{\infty}rX_{r}$, where $X_{r}$ is of Poisson distribution with intensity rate $\displaystyle \frac{1}{r^{2}}$, $r \geq 1$. I need to prove that $Z$ is finite with probability $1$.

First off, I am not sure if this problem is asking me to 1) Prove that $Z$ is finite AND has probability $1$ or 2) Prove that $Z$ is finite and the probability that it is finite is $1$.

Secondly, I have seen Poisson random variables before, but I have never heard the expression "intensity rate" used before - after doing some research, I have since discovered that the "intensity rate" is simply the parameter $\lambda$ that appears in the probability function $\displaystyle p(y) = \frac{\lambda^{y}}{y!}e^{-\lambda}$, where $Y$ is a Poisson distributed random variable.

Now, the hint that I was given for this problem is to "consider the total number of nonzero $X_{r}$", and appears in my notes at the same place that the first Borel-Cantelli Lemma is stated:

For a sequence of events $A_{1},A_{2},\cdots ,$, $$ P\left( \limsup_{n \to \infty} A_{n} \right) = 0 \, \text{whenever}\, \sum_{i=1}^{\infty}P(A_{i}) < \infty. $$

Now, the problem I'm having though, is how to apply this to this particular problem. First of all, in order to determine the convergence of $Z = \sum_{r=1}^{\infty}rX_{r}$, how do I write $X_{r}$ in a more meaningful form that includes its $\displaystyle p(y) = \frac{\left(\frac{1}{r^{2}}\right)^{y}}{y!}e^{\displaystyle -\frac{1}{r^{2}}}$ form? If I take $$P(Z) = P(\sum_{r=1}^{\infty}rX_{r}) \leq \sum_{r=1}^{\infty}P\left(rX_{r} \right) = \sum_{r=1}^{\infty}rP\left(X_{r} \right) \\ = \displaystyle \sum_{r=1}^{\infty}r \left(\frac{\left( \frac{1}{r^{2}}\right)^{x_{r}}}{x_{r}!}e^{\displaystyle -\frac{1}{r^{2}} }\right) = \displaystyle \sum_{r=1}^{\infty} \frac{r \cdot \frac{1}{r^{2x_{r}}}}{x_{r}!}e^{\displaystyle -\frac{1}{r^{2}}} = \displaystyle \sum_{r=1}^{\infty} \frac{\frac{1}{r^{2x_{r}-1}}}{x_{r}!}e^{\displaystyle -\frac{1}{r^{2}}} $$

it starts to look like something, but I really don't know 1) how to proceed or 2) if this is even the way to approach this problem.

I am extremely lost and am in desperate need of help. I thank you for your time and patience.

Answer 1

You are being asked to prove the probability $Z$ is finite is $1$. $Z$ is a random variable and not an event so it doesn't make sense to say $Z$ itself has probability $1$.

I'll begin by getting a handle on how the $X_r$ behave. We have $\mathbb{P}(X_r = k) = \frac{r^{-2k}}{k!}e^{-1/r^2}$ for $k \in \mathbb{N}$. In particular we have $\mathbb{P}(X_r \neq 0) = 1 - e^{-1/r^2}$. Since $\sum_{r=1}^\infty 1-e^{-1/r^2} < \infty$, by Borel-Cantelli, we have that $\mathbb{P}( \lim \sup A_r ) = 0$ where $A_r$ is the event $\{X_r \neq 0 \}$. Now $\lim \sup A_r$ is the event that $A_r$ happens for infinitely many $r$ and as a result we have that $X_r = 0$ for all but finitely many $r$ with probability $1$. On this probability $1$ event, $Z$ is just a finite sum and hence if finite so $\mathbb{P}(Z < \infty) = 1$ as desired.

Answer 2

Hint

To the first question, they are asking you to prove that the probability $Z$ is finite equals one.

Using their hint, let $A_r = \{X_r \ne 0\},$ i.e. the event that the RV $X_r$ is nonzero. For a the poisson distribution with rate $\lambda$, the probability the variable is zero is $e^{-\lambda},$ so you have $P(A_r) = 1-P(X_r = 0) = 1-e^{-\frac{1}{r^2}}. $

Note when applying Borel Cantelli that $\limsup_n A_n$ is the event that infinitely many of the events $A_i$ occur. So if you can show that probability is zero you'll have shown that the probability that infinitely many of the $X_i$'s is nonzero is zero.