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Prove the following equivalence relationship: $$P(\lim\inf A_n)\leq \lim\inf P(A_n) \Leftrightarrow \ P(\lim \sup A_n) \geq \lim \sup P(A_n).$$

My attempt: I could not see how to prove this relationship yet, except for my hypothesis, based on the definition of $\lim \inf$ and $\lim \sup$ that we could rewrite $\lim \sup A_n = -\lim \inf(-A_n)$?? But then how to show that $P(-\lim \inf(-A_n)) = - P(\lim \sup (A_n))$?

Could anyone please tell me if I'm on the right track help me with this problem? Really appreciate your input.

Robert Z
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ghjk
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    You definitely are not on the right track, considering the $A_n$ are events in an arbitrary probability space, and therefore the expression $-A_n$ is meaningless. – Math1000 Sep 09 '16 at 04:51
  • @Math1000: thank you for pointing that out. Could you help with this problem? – ghjk Sep 09 '16 at 04:53

2 Answers2

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We show that the $\liminf$ inequality implies the $\limsup$ inequality (the other way is similar).

Consider the sequence $\{A_n^c\}_{n\geq 1}$. We have that by De Morgan's laws:

$$\liminf_{n\to \infty} A_n^c = \bigcup_{n \ge 1} \bigcap_{k\ge n} A_k^c=\bigcup_{n \ge 1}\left( \bigcup_{k\ge n} A_k\right)^c =\left(\bigcap_{n \ge 1} \bigcup_{k\ge n} A_k\right)^c=\left(\limsup_{n\to \infty} A_n\right)^c.$$ Therefore $$1-\limsup_{n\to \infty} P(A_n)=\liminf_{n\to \infty} (1-P(A_n))\\=\liminf_{n\to \infty} P(A_n^c)\geq P(\liminf_{n\to \infty} A_n^c)=1-P(\limsup_{n\to \infty} A_n)$$ which implies $$\limsup_{n\to \infty} P(A_n)\leq P(\limsup_{n\to \infty} A_n).$$

P.S. We include a proof of the $\liminf$ inequality (my first answer where a miss OP's request about equivalence).

By definition $$\liminf_{n\to \infty} A_n = \bigcup_{n \ge 1} \bigcap_{k\ge n} A_k.$$ Since for any $n$, $\cap_{k\ge n} A_k \subseteq A_n$, it follows that $P(\cap_{k\ge n} A_k) \le P(A_n)$. Hence

$$\liminf_{n\to \infty} P\left(\bigcap_{k \ge n} A_k\right) \le \liminf_{n\to \infty} P(A_n)\tag{2}$$

Moreover the sequence $\{\cap_{k\ge n} A_k\}_{n\geq 1}$ is increasing and by continuity of $P$ from below, $$P(\liminf A_n) = \lim_{n\to \infty} P\left(\bigcap_{k \ge n} A_k\right)=\liminf_{n\to \infty} P\left(\bigcap_{k \ge n} A_k\right).$$

Robert Z
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  • I think you miss the point. The problem is to assume 1 of those inequalities are correct, and getting the other one, not to prove every single one of them is correct. For the forward direction, I'm considering $B_n = {A_n}^c$ as you said. But using the same argument as above, I could only end up with $\lim \sup P(\cup_{k\geq n} ({A_n}^{c}))\geq \lim \sup P({A_n}^{c})$, but $P(\lim \sup \cap_{n\geq 1} \cup_{k\geq n} ({A_n}^{c})) =$ $\ lim_{n\rightarrow \infty} P(\cup_{k\geq n} ({A_n}^c) \leq \lim \sup P(\cup_{k\geq n} {A_n}^{c})$ – ghjk Sep 09 '16 at 05:31
  • @user177196 Now I included the equivalence proof in my answer. Sorry for the mess! – Robert Z Sep 09 '16 at 06:00
  • great proof! I finally got the same thing. Many thanks for your help. – ghjk Sep 09 '16 at 06:38
  • For the other direction would it be correct to say

    \begin{align} 1-\liminf\mathsf P(A_n) &=\limsup(1-\mathsf P(A_n))\\ &=\limsup\mathsf P(A_n^c)\\ &\leq \mathsf P(\limsup A_n^c)\\ &=1-\mathsf P(\liminf A_n) \end{align} implies that

    $$\liminf\mathsf P(A_n)\geq \mathsf P(\liminf A_n)$$

     – Remy Sep 03 '20 at 05:04
  • @Remy Yes, that is correct. – Robert Z Sep 03 '20 at 06:37
  • Thanks! The only other question I had is that you seem to be using $$\mathsf P(\liminf A_n) \leq \liminf \mathsf P(A_n) \Leftrightarrow \mathsf P(\liminf A_n^c) \geq \liminf \mathsf P(A_n^c)$$ but that doesn't seem obvious to me. – Remy Sep 03 '20 at 06:42
  • @Remy For some reason, my inequalities were reversed. Is it clear now? – Robert Z Sep 03 '20 at 07:04
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I think the correct statement should be $$P(\liminf A_n) \leq \liminf P(A_n) \Leftrightarrow P(\limsup A_n^c) \geq \limsup P(A_n^c) \tag{1}$$ instead of the one you stated.

To prove the $(1)$, it suffices to note that $$\left(\liminf A_n\right)^c = \limsup A_n^c,$$ which can be easily verified by definitions of upper/lower limit sets and De Morgan law.

Now the proof can be completed by just using $P(E^c) = 1 - P(E)$ for any event $E$ and $\liminf(1 - a_n) = 1 - \limsup a_n $ for any real sequence $\{a_n\}$.

For example, if you are given $P(\liminf A_n) \leq \liminf P(A_n)$, then it implies $$1 - P(\liminf A_n) \geq 1 - \liminf P(A_n),$$ which is equivalent to $$P((\liminf A_n)^c) \geq \limsup P(A_n^c),$$ or $$P(\limsup A_n^c) \geq \limsup P(A_n^c).$$

Zhanxiong
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  • Thank you very much for your help. I just used the same idea, but then for the forward direction, you need $\lim \inf (A_n^c) = (\lim \sup A_n)^c$. Afterwards, it's straightforward to arrive at $P(\lim \sup A_n) \geq \lim \sup P(A_n)$. – ghjk Sep 09 '16 at 06:36