Lim sup of Random Variables and Events

Question

I just read a problem in a book and they used this logic below. I'm trying to reason it out. Some help would be appreciated.

Let $ ( \Omega, \mathcal{F} )$ be a Measurable Space.

Suppose $X_n$ is a sequence of IID Random variables with $$ P (X_n > x ) = e^{-x},\quad x\ge 0. $$

By Borel Cantelli 1 & 2 , $$ P (X_n > \alpha \log(n) \text{ i.o} ) = \begin{cases} 0 & \alpha > 1, \\ 1 & \alpha \leq 1. \end{cases} $$

Let $L= \limsup_n \frac {X_n}{\log(n)} $.

The book then uses the fact that $$ \{\omega : X_n > \log(n) \text{ i.o} \} \subset \{ L \geq 1 \}.$$

This allows to infer that $P( L \geq 1) = 1$.

He later goes on to show $P ( L > 1 ) = 0 $ so that in fact $ P(L=1) = 1$.

Answer 1

This has nothing to do with probability. The question is essentially why, given a real sequence $\{a_n,n\ge 1\}$, the fact that $a_n>1$ infinitely often implies $\limsup_n a_n\ge 1$.

Well, $\limsup$ of a sequence is at least $\limsup$ of any subsequence. So if there is a subsequence $a_{n_k}\ge 1$ then $\limsup_n a_n \ge \limsup_{a_{n_k}} \ge 1$ (note that it is enough to have the inequality $a_{n_k}\ge 1$).

Answer 2

Now we have By Borel Cantelli 1 & 2 , let $A_n=\{X_n/\log(n) >\alpha \}$ $$ \mathbb{P} (A_n \text{ i.o.} ) = \begin{cases} 0 & \alpha > 1, \\ 1 & \alpha \leq 1. \end{cases} $$

i.e. $$ \mathbb{P} (\frac{X_n}{\log(n)} > \alpha \quad \text{ i.o.} ) = \begin{cases} 0 & \alpha > 1, \\ 1 & \alpha \leq 1. \end{cases} $$

Let $\alpha = 1 + \epsilon$

$$ \mathbb{P} (\frac{X_n}{\log(n)} > 1 + \epsilon \quad\text{ i.o.} ) =0, \quad\forall \epsilon>0.$$ Thus

$$\mathbb{P}(\limsup\frac{{X_n}}{\log(n)} \leq 1+\epsilon) =1 \quad \forall \epsilon>0$$

let $\epsilon \downarrow 0$

$$\mathbb{P}(\limsup\frac{{X_n}}{\log(n)} \leq 1) =1$$

while

$$\mathbb{P} ( \frac{X_n}{\log(n)} \geq 1 \text{ i.o.} )\geq \mathbb{P} (\frac{X_n}{\log(n)} > 1 \text{ i.o.} ) =1, \forall \epsilon>0.$$ which means that $$\mathbb{P}(\limsup \frac{X_n}{\log(n)} \geq 1) \geq \mathbb{P} ( \frac{X_n}{\log(n)} \geq 1 \text{ i.o.} ) = 1$$

Finally we have

$$\mathbb{P}(\limsup\frac{{X_n}}{\log(n)} = 1) =1$$.

NOTE

In general

$$\big\{\limsup X_n\geq 1\big\} \supseteq \big\{\limsup \{X_n\geq 1\}\big\} $$

and

$$\big\{\limsup X_n > 1\big\} \subseteq \big\{\limsup \{X_n> 1\}\big\}$$