3
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ $$\lim_{n\to\infty}\sum_{i=1}^nP(A_i)=\infty\implies \lim_{n\to \infty}e^{-\sum_{i=1}^nP(A_i)}=0.$$ 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
2
$\begingroup$

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*}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.