I Prove that the tail of this distribution goes to zero

[CGandC]

Theorem: Let $X$ be a random variable. Then $\lim_{s \to \infty} P( |X| \geq s ) =0$

Proof from teacher assistant's notes: We'll show first that $\lim_{s \to \infty} P( X \geq s ) =0$ and $\lim_{s \to \infty} P( X \leq -s ) =0$:

Let $(s_n)_{n=1}^\infty$ be a monotonically increasing sequence with $\lim_{ n \to \infty } s_n = \infty$. The sequences $\{ X \geq s_n \}_{n=1}^\infty$ and $\{ X \leq -s_n \}_{n=1}^\infty$ are decreasing sequences with zero intersection:

$\bigcap_{n=1}^{\infty}\left\{X \leq-s_n\right\} = \bigcap_{n=1}^{\infty}\left\{X \geq s_n\right\} = \emptyset$,
hence from continuity of probability:

$\begin{aligned} &0=\mathbb{P}(\emptyset)=\mathbb{P}\left(\bigcap_{n=1}^{\infty}\left\{X \geq s_n\right\}\right)=\lim _{n \rightarrow \infty} \mathbb{P}\left(X \geq s_n\right) \\ &0=\mathbb{P}(\emptyset)=\mathbb{P}\left(\bigcap_{n=1}^{\infty}\left\{X \leq-s_n\right\}\right)=\lim _{n \rightarrow \infty} \mathbb{P}\left(X \leq-s_n\right) \end{aligned}$

Hence we'll deduce:

$\lim _{s \rightarrow \infty} \mathbb{P}(|X| \geq s)=\lim _{s \rightarrow \infty}(\mathbb{P}(X \geq s)+\mathbb{P}(X \leq-s))=0$
and we're finished.Questions:
1. I understand that the proof above is according to Heine's definition of limit, but if so I don't understand why we took $(s_n)_{n=1}^\infty$ to be a monotonically increasing sequence and not an arbitrary sequence? ( we'd also like to prove for sequences that do go to infinity but are not necessarily monotonically increasing ).
2. Why does the equation $\mathbb{P}\left(\bigcap_{n=1}^{\infty}\left\{X \geq s_n\right\}\right)=\lim _{n \rightarrow \infty} \mathbb{P}\left(X \geq s_n\right)$ hold? how did we go from the left side to the right side?Thanks in advance for any help!

 

[pasmith]

Questions:
1. I understand that the proof above is according to Heine's definition of limit, but if so I don't understand why we took $(s_n)_{n=1}^\infty$ to be a monotonically increasing sequence and not an arbitrary sequence? ( we'd also like to prove for sequences that do go to infinity but are not necessarily monotonicallyincreasing ).

2. Why does the equation $\mathbb{P}\left(\bigcap_{n=1}^{\infty}\left\{X \geq s_n\right\}\right)=\lim _{n \rightarrow \infty} \mathbb{P}\left(X \geq s_n\right)$ hold? how did we go from the left side to the right side?

What happens if s_n is not necessarily strictly increasing? Since the minimum of an intersection is the maximum of the minima, we have <br /> \bigcap_{n=1}^N \{X \geq s_n\} = \{X \geq \max_{1 \leq n \leq N} s_n\}. Define <br /> M_N = \max_{1 \leq n \leq N} s_n. Then M_n is an increasing sequence with M_n \to \infty. Then <br /> \begin{split}<br /> \mathbb{P}\left( \bigcap_{n=1}^\infty \{X &gt; s_n\}\right) &amp;= \lim_{N \to \infty} \mathbb{P}\left(\bigcap_{n=1}^N \{X \geq s_n\}\right) \\<br /> &amp;= \lim_{N \to \infty} \mathbb{P}(\{X \geq M_N\}).\end{split}

 

[CGandC]

Thanks alot! everything's crystal clear now!

I also found an answer to my second question which stems from the law of continuity of probability which says the following ( in case anyone's interested ):
Let there be a monotonically decreasing sequence of events $A_1 \supseteq A_2 \supseteq ...$ in probability space $( \Omega , \mathbb{P} )$. Then: $\mathbb{P}\left( \bigcap_{n=1}^\infty A_n \right) = \lim_{n \to \infty } \mathbb{P}( A_n)$