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$P(\cdot)$ is the probability measure.

$S_n$ is a sequence of events.

$P(\liminf S_n)=1$ does it mean that $S_n$ always happen after some large n?

Can I say that it must be true that $\liminf P(S_n)=1$ given $P(\liminf S_n)=1$ ?

lulu
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1 Answers1

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Recall that $S = \liminf S_n = \bigcup_{k=1}^\infty \left(\bigcap_{n \ge k} S_n\right)$; saying that $P(S)=1$ means that (almost) all outcomes $\omega$ belong to $S$, that is belong to $\bigcap_{n \ge k} S_n$ for at least one $k$; in other terms, $\omega\in S_n$ for all sufficiently large $n$'s.

In short, this means that that all events $S_n$ happen, except for a finite number of them.

Clement C.
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  • So if I can prove that $P(\liminf S_n)=1$. Then it implies that $P(S_n)=1$ after some large $n$. Is it correct? – lulu Feb 02 '14 at 20:26
  • Not really. You have (assuming the $S_n$'s are subsets of a probability space $\Omega$) that $$\forall_{\rm{}a.s.}\omega\in\Omega\exists k\text{ s.t. }\forall n\geq k,\quad \omega\in S_n$$ What you just asked is $$\exists k\text{ s.t. } \forall_{\rm{}a.s.}\omega\in\Omega\forall n\geq k,\quad \omega\in S_n$$ – Clement C. Feb 02 '14 at 20:34
  • Can I say $\lim_{n\to\infty}P(S_n)=1$ ?(, given $P(\liminf S_n)=1$)?

    Since $P(\liminf S_n)=1$, that mean there are infinite consecutive $S_i$ happen together, I think $P(\liminf S_n)=1$ implies $\lim_{n\to\infty} P(s_n)=1$

     – lulu Feb 02 '14 at 21:47
  • This is an implication, yes. Since $T_k=\cap_{n \ge k} S_n$ is non-decreasing, $$P \left( \cup_{k \geq 1} \cap_{n \ge k} S_n \right) = \lim_k P \left( \cap_{n \ge k} S_n \right) \le \lim_k \inf_{n \ge k} P(S_n).$$ (can also be proven with Fatou's lemma) and since the LHS is $1$ by assumption, you get $ \lim_k \inf_{n \ge k} P(S_n) = 1$ which implies here that $P(S_n)\to 1$. – Clement C. Feb 02 '14 at 21:58
  • Fantastic! This is just what i need. Thanks – lulu Feb 02 '14 at 22:03