Suppose that $\{A_i\}$ is a sequence of independent events with $P\left(\bigcup_{i=1}^\infty A_i\right) = 1$ and $P(A_i)<1$ for all $i\in \mathbb{N}$. Show that $$ P(A_i \text{ occurs infinitely often})=1 $$

My attempt: We only need to show $P\left(\cap_{i=1}^{\infty} A_i^c \right)=0 \Longrightarrow P(A_i\ i.o.)=1$. Note that $$ \begin{aligned} P\left(\cap_{i=1}^{\infty} A_i^c \right)&= \prod_{i=1}^{\infty}P(A_i^c)&&\text{(independence)}\\ &= \prod_{i=1}^{\infty}(1-P(A_i)) \end{aligned} $$ For any $k$, we have \begin{aligned} P\left(\cap_{i=1}^{k} A_i^c \right)&= \prod_{i=1}^{k}P(A_i^c)\\ &= \prod_{i=1}^{k}(1-P(A_i))\\ &\leq \prod_{i=1}^k e^{-P(A_i)}\quad(1-x\leq e^{-x}) \\ &=e^{-\sum_{i=1}^kP(A_i)} \end{aligned} Let $k \to \infty$, then $0=P\left(\cap_{i=1}^{\infty} A_i^c \right)\leq e^{-\sum_{i=1}^{\infty}P(A_i)}$. If we can show $e^{-\sum_{i=1}^{\infty}P(A_i)}=0$, which implies that $\sum_{i=1}^{\infty}P(A_i)=\infty$, then the result follows by the second Borel-Cantelli Lemma. My question is how to show $e^{-\sum_{i=1}^{\infty}P(A_i)}=0$. If we cannot, is there any other way to prove this result? I would appreciate if you could explain in details.

  • $\begingroup$ $$\lim_{n\to\infty}\sum_{i=1}^nP(A_i)=\infty\implies \lim_{n\to \infty}e^{-\sum_{i=1}^nP(A_i)}=0.$$ This because of the fact $x\to \infty\implies e^{-x}\to 0+$. But note that there is no real number $r$ such that $e^{-r}=0$. $\endgroup$ Sep 13 '20 at 9:59
  • 1
    $\begingroup$ math.stackexchange.com/questions/359404/probability-of-limsup $\endgroup$
    – d.k.o.
    Sep 13 '20 at 10:28

One way of solving this is to show that $\mathbb{P}\left(\bigcup_{i=n}^\infty A_i\right) = 1$ for all $n\in\mathbb{N}$ (Let's refer to this statement as $(\star)$). If this has been shown, we can conclude $$\mathbb{P}\left( A_i \text{ infinitely often}\right) = \mathbb{P} \left(\bigcap_{n=1}^\infty \bigcup_{i=n}^\infty A_i\right) = \lim_{n\to \infty} \mathbb{P}\left(\bigcup_{i=n}^\infty A_i\right) = 1,$$ where we have used continuity from above in the second step. To show $(\star)$, we use the following statement.

If $A$ and $B$ are two independent events and $\mathbb{P}(A\cup B) = 1$, then $\mathbb{P}(A) = 1$ or $\mathbb{P}(B) = 1$.

This follows easily from $0 = 1 - \mathbb{P}(A\cup B) = \mathbb{P}\left(A^c \cap B^c\right) = \mathbb{P}(A^c) \mathbb{P}(B^c)$.

Now in our case, if $n\in\mathbb{N}$, then by assumption, $\mathbb{P}\left( \bigcup_{i=1}^{n-1}A_i \cup \bigcup_{i=n}^\infty A_i\right) = \mathbb{P}\left(\bigcup_{i=1}^\infty A_i \right) = 1$, so we only need to show that $\mathbb{P}\left(\bigcup_{i=1}^{n-1}A_i\right) < 1$. This, however, follows from the assumption that $\mathbb{P}(A_i) < 1$ and thus $\mathbb{P}(A_i^c) > 0$ for all $i\in \mathbb{N}$, since \begin{align*} \mathbb{P}\left(\bigcup_{i=1}^{n-1} A_i\right) &= 1 - \mathbb{P}\left(\bigcap_{i=1}^{n-1} A_i^c\right) = 1 - \underbrace{\prod_{i=1}^{n-1} \mathbb{P}(A_i^c)}_{>0} < 1. \end{align*}


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