Adaptation of Resnick Exercise 4.9: $P(\bigcup_{n=1}^\infty A_n)=1$ iff $P(A_n i.o.)=1$

Question

$P(\bigcup_{n=1}^\infty A_n)=1$ iff $P(A_n i.o.)=1$

I have always struggled with the "infinitely often" and "all but finitely often" concepts for sequences. My first time through Resnick, I don't think I was equipped with the tools necessary for this exercise using only what I had learned in the text previously, and so I wanted to come back to a few problems.

I still am not sure how to proceed at all here.

I think I understand the question to basically state "The probability that a union of a sequence of events is 1 if and only if the probability of that sequence occurring infinitely often is also 1.

As per the comments below, an additional assumption is that each of the events has $P(A_i)<1$ and each event is independent.

So, to start out, I can define:

$\{A_n i.o.\}=\limsup_{n\to\infty}A_n=\bigcap_{n=1}^\infty\bigcup_{k=n}A_k$

but from here I am not sure where to go (or if I've even begun correctly).

Feel free to be as explicit as possible as these ideas ($\limsup$, etc.) have always seemed to elude me. Thank you for the help.

Answer 1

Hint Show that (1) $\iff$ (2) $\iff$ (3) $\iff$ (4) $\iff$ (5) $\iff$ (6), where:

(1) $\mathbb P\left(\bigcup\limits_{n\geqslant1}A_n\right)=1$. (2) $\mathbb P\left(\bigcap\limits_{n\geqslant1}(\Omega\setminus A_n)\right)=0$. (3) $\prod\limits_{n\geqslant1}(1-\mathbb P(A_n))=0$. (4) $\sum\limits_{n\geqslant1}\mathbb P(A_n)=+\infty$. (5) $\mathbb P\left(\limsup A_n\right)=1$. (6) $\mathbb P\left(A_n\ \text{i.o.}\right)=1$.

If a step is unclear, please say so.

Answer 2

While did's solution is correct and complete, I thought that it needs a little justification that why from the step $\prod_{n \ge 1} (1 - P(A_n))$ we can conclude that $P(lim sup A_n) = 1$. Here is my solution:

We need to show: $\{A_n i.o.\}=\limsup_{n\to\infty}A_n=\bigcap_{n=1}^\infty\bigcup_{k=n}A_k$

Consider $B_i = \bigcup_{k=i}A_k $. We need to show that $\forall i : p(B_i) = 1$ to do that we have:

Based on the assumption $P(\bigcup_{n=1}^\infty A_n)=1 \rightarrow P(\bigcup_{n=1}^{i} A_n \cup \bigcup_{n= i + 1}^\infty A_n)=1 \rightarrow \text{we expand the prob} \rightarrow P(\bigcup_{n=1}^{i} A_n) + P(\bigcup_{n= i + 1}^\infty A_n) - P(\bigcup_{n=1}^{i} A_n)P(\bigcup_{n= i + 1}^\infty A_n) \text{the third part is because events are independent} \rightarrow 1 - P(\bigcup_{n=1}^{i} A_n) = P(\bigcup_{n= i + 1}^\infty A_n)(1 - P(\bigcup_{n=1}^{i} A_n)) \rightarrow \text{not that since $P(A_i)<1$ then $P({A_i}^{c})\ne 1$ so $(1 - P(\bigcup_{n=1}^{i} A_n)) = P({(\bigcup_{n=1}^{i} A_n)}^c) \ne 0$} \rightarrow P(\bigcup_{n= i + 1}^\infty A_n) = p(B_i) = 1 \text{ $\forall i$}$

Now note that because of continuity from above of $p(B_i) = P(\bigcup_{n= i + 1}^\infty A_n)$, we know that $P(B_i) \downarrow p(lim sup A_n)$ and note that $P(B_i) = 1 \text{ $\forall i$}$ $\rightarrow P(\limsup_{n\to\infty}A_n) = 1 \rightarrow P(A_n i.o.) = 1$

Hope it helps.

Answer 3

You can check Did's solution on how to show from $3\Rightarrow 4$ on the following post https://math.stackexchange.com/a/951072/775759

However, I think Sam's argument on $(1 - P(\bigcup_{n=1}^{i} A_n)) = P({(\bigcup_{n=1}^{i} A_n)}^c) \ne 0$ is problematic.

I added my solution here (Basically the detailed version of what Did wrote)

By independence, $$\begin{aligned} 0=1-P(\cup_{n=1}^\infty A_n)&=P((\cup_{n=1}^\infty A_n)^c)\\ &=P(\cap_{n=1}^\infty A_n^c)\\ &=\prod_{i=1}^\infty P(A_n^c)\\ &=\prod_{i=1}^\infty (1-P(A_n))\\ &=\exp{\sum_{i=1}^\infty \log(1-P(A_n))} \end{aligned}$$ For the sake of contradiction, suppose $\sum_{n=1}^\infty P(A_n)<\infty$. Then $\lim_n P(A_n)=0$ and we can take a large enough $n_0$ such that $P(A_m)\leq \frac{1}{2}$ for $m\geq n_0$. Note that when $x\leq \frac{1}{2}$, $\log(1-x)\geq -2x$ and $$\prod_{i=n_0}^\infty (1-P(A_n))=\exp{\sum_{i=n_0}^\infty \log(1-P(A_n))}\geq \exp{(-2\sum_{i=n_0}^\infty P(A_n))}>0$$ When $m<n_0$, we have that $P(A_m)<1$ and $1-P(A_m)\neq 0$. Therefore, $$\prod_{i=1}^{\infty} (1-P(A_n))=\prod_{i=1}^{n_0-1} (1-P(A_n))\times \prod_{i=n_0}^\infty (1-P(A_n))\neq 0$$ which is a contradiction. Hence, $\sum_{n=1}^\infty P(A_n)=\infty$.