probability theoryself-learning
Consider a sequence of random variable $(X_n)$. Prove the following inequality:
$$
\mathbb P\left(\liminf\{X_n \leq x\}\right) \leq \liminf\mathbb P\left(\{X_n \leq x\}\right)\leq \limsup\mathbb P\left(\{X_n \leq x\}\right) \leq \mathbb P\left(\limsup\{X_n \leq x\}\right).
$$
I can see that the second $\leq$ is simply the result for real sequence. However, I do not see how to get the first and the last one. It seems that the two are similar. Could anyone help me, please? Thank you!
Important inequalities
Williams - Probability with Martingales
Deduced similarly:
(iii) If $\liminf x_n > z$, then
$ \ \ \ \ \ \ \ (x_n > z)$ eventually (that is, for infinitely many n)
(iv) If $\liminf x_n < z $, then
$ \ \ \ \ \ \ \ (x_n < z)$ infinitely often (that is, for infinitely many n)
$$P( \limsup [X_n > a] ) = 1 \to P( [ \limsup X_n ] > a) = 1$$
No. Consider $X_n = a + \frac{1}{n}$.
The converse, which is,
$$P( \limsup [X_n > a] ) = 1 \leftarrow P( [ \limsup X_n ] > a) = 1$$
is true (see above).
$$P( [ \limsup X_n ] > a) = 1 \to P( \liminf [X_n > a] ) = 1$$
No.
$$P( [ \limsup X_n ] > a) = 1 \to P( \limsup (X_n > a)) = 1$$
$$P( \limsup (X_n > a)) = 1 \nrightarrow P( \liminf [X_n > a] ) = 1$$
The converse of the latter, which is,
$$P( \liminf [X_n > a] ) = 1 \to P( \limsup (X_n > a)) = 1$$
is true (obviously?).
$$P(([\limsup X_n] > a)^c)=P( \liminf(X_n > a)^c)$$
No.
$$P(([\limsup X_n] > a)^c) = P([\limsup X_n] \le a)$$
$$P( \liminf[(X_n > a)^c]) = P( \liminf(X_n \le a))$$
$$P( [\liminf(X_n > a)]^c) = P( \limsup(X_n \le a))$$
$$P( \liminf(X_n > a)^c)=P([\liminf X_n]\le a)$$
No.
$$P( \liminf[(X_n > a)^c]) = P( \liminf(X_n \le a))$$
$$P( [\liminf(X_n > a)]^c) = P( \limsup(X_n \le a))$$
$$P([\liminf X_n]\le a) = P(([\liminf X_n] > a)^C)$$
$$P([\liminf X_n]\le a) = 0 \to P( \liminf [X_n > a] ) = 1$$
Yes. Contrapositive of 3
Since $\liminf_n X_n = \sup_n \inf_{k\geq n} X_k$, the sequence $\inf_{k\geq n} X_k$ is monotonically increasing. Hence $$ \{\liminf_n X_n \le 0\} = \bigcap_{n=1}^\infty \{\inf_{k\geq n} X_k< 0\} = \bigcap_{n=1}^\infty \bigcup_{k\geq n} \{X_k< 0\} $$ As for $\{\liminf_n X_n>0\}$: the condition $\liminf_n X_n>0$ means that there is an index $n$ such that $\inf_{k\geq n} X_k >0$, i.e. there is an index $n$ such that for any $k\geq n$ $X_k>0$. Writing in set-formalism yields $$\{\liminf_n X_n\ge0\} = \bigcup_{n=1}^\infty \bigcap_{k\geq n} \{X_k>0\}. $$ By definition, for a sequence of sets $A_n$ we denote by \begin{align*} \limsup_n A_n &= \bigcap_{n=1}^\infty \bigcup_{k\geq n} A_k \\ \liminf A_n &=\bigcup_{n=1}^\infty \bigcap_{k\geq n} A_k.\end{align*} Now denote by $A_k = \{X_k>0\}$ and by $B_k=\{X_k<0\}$, then \begin{align*} \liminf_n A_n &= \{\liminf_n X_n\ge0\} \\ \limsup_n B_n & =\{\liminf_n X_n\le0\} \end{align*} Update: Now as for $\{\liminf_n X_n>0\}$. This means that there exists $m\in \mathbb N$ and an $N\in \mathbb N$ s.t. for any $n\geq N$ $\{X_k\geq \frac1m\}$, i.e. $$ \{\liminf_n X_n>0\} = \bigcup_{m\geq 1} \bigcup_{n\geq1} \bigcap_{k\geq n} \{X_k\geq \frac1m\}. $$ $\{\liminf_n X_n <0\}$ can be handled similarily.
Best Answer
Let $A_n = \{X_n \le x\}$. Then
$$\liminf A_n = \bigcup_{n \ge 1} \bigcap_{k\ge n} A_k$$
The sequence $\{\cap_{k\ge n} A_k\}_{n = 1}^\infty$ is a sequence which increases to $\liminf A_n$,
$$P(\liminf A_n) = \lim_{n\to \infty} P\left(\bigcap_{k \ge n} A_k\right)\tag{1}$$
by continuity of $P$ from below. Since $\cap_{k\ge n} A_k \subseteq A_n$ for all $n$, $P(\cap_{k\ge n} A_k) \le P(A_n)$ for all $n$. Hence
$$\liminf_{n\to \infty} P\left(\bigcap_{k \ge n} A_k\right) \le \liminf_{n\to \infty} P(A_n)\tag{2}$$
By $(1)$ and $(2)$,
$$P(\liminf_{n\to \infty} A_n) \le \liminf_{n\to \infty} P(A_n).$$
For the last equality, let $\Omega$ be the underlying sample space. Apply the lim inf inequality to the sequence $B_n = \Omega \setminus A_n$ to get $P(\limsup A_n) \ge \limsup P(A_n)$.