liminf and limsup in probability

Question

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!

Answer 1

Let $A_n = \{X_n \le x\}$. Then

$$\liminf A_n = \bigcup_{n \ge 1} \bigcap_{k\ge n} A_k$$

Since $\{\cap_{k\ge n} A_k\}_{n = 1}^\infty$ 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 $P(\limsup A_n) \ge \limsup P(A_n)$.

Answer 2

I would like to present a slightly general version here.

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Let $(A_n)$ be a sequence of measurable sets. Then $$\mathbb P\left(\liminf A_n \right) \leq \liminf\mathbb P\left( A_n \right) \leq \limsup\mathbb P\left( A_n \right) \leq \mathbb P\left(\limsup A_n \right).$$

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The second inequality is clear from real analysis. Let's prove the first one. We have

$$ \begin{aligned} \mathbb P\left(\liminf A_n \right) &= \mathbb P\left(\bigcup_{n} \bigcap_{k \ge n} A_k \right) \\ &= \lim_{n} \mathbb P\left( \bigcap_{k \ge n} A_k \right) \\ &\le \lim_n \inf_{k\ge n} \mathbb P\left( A_k \right) \text{ because } \mathbb P\left( \bigcap_{k \ge n} A_k \right) \le \mathbb P\left( A_t \right) \text{ for all } t\ge n\\ &= \liminf\mathbb P\left( A_n \right). \end{aligned} $$

Similarly, we have

$$ \begin{aligned} \mathbb P\left(\limsup A_n \right) &= \mathbb P\left(\bigcap_{n} \bigcup_{k \ge n} A_k \right) \\ &= \lim_{n} \mathbb P\left( \bigcup_{k \ge n} A_k \right) \\ &\ge \lim_n \sup_{k\ge n} \mathbb P\left( A_k \right) \text{ because } \mathbb P\left( \bigcup_{k \ge n} A_k \right) \ge \mathbb P\left( A_t \right) \text{ for all } t\ge n\\ &= \limsup \mathbb P\left( A_n \right). \end{aligned} $$