Proof for $\lim_{k \to \infty} P (|X| > k) = 0$ for a random variable X.

Question

I'm trying to show the following statement.

If $X$ is a random variable, then $\lim_{k \to \infty} P (|X| > k) = 0$

What I tried:

the above statement is equivalent to the following.

$$ X \text{ : R.V. } \Rightarrow \forall \epsilon>0, \exists k>0 : \Big( P(|X| > k) < \epsilon \Big) $$

I defined a sequence of events $a_n = \{|X| > n \}$ which is decreasing.

Since $\lim_{n \to \infty} a_n = \emptyset $, by the continuity of probability, $\lim P(a_n) = 0$

However, I now realize that $k>0$ is a real number, so this approach might not be valid.

Is there anyone to help me?

Answer 1

Setting $B_n:=\{n-1\leq|X|<n\}$ for $n=1,2,\dots$ we have: $$\bigcup_{n=1}^{\infty}B_n=\Omega$$and because the sets $B_n$ are disjoint we conclude:$$\sum_{n=1}^{\infty}P(B_n)=P(\Omega)=1$$ or equivalently $$\lim_{n\to\infty}\sum_{k=0}^nP(B_k)=1$$

so that $$\lim_{n\to\infty}\sum_{k=n+1}^{\infty}P(B_k)=\lim_{n\to\infty}1-\sum_{k=0}^nP(B_k)=1-1=0$$

Now observe that: $$\sum_{k=n+1}^{\infty}P(B_k)=P(|X|\geq n)$$

because $$\bigcup_{k=n+1}^{\infty}B_k=\{|X|\geq n\}$$ and again the sets $B_k$ are disjoint.

Answer 2

We have $$ P(|X|>k) = 1-P(|X|\leq k) = 1- P(-k\leq X\leq k) = 1-P(X\leq k)-P(X<-k) $$ Let $F(x)$ be the distribution function of $X$. Using elemental properties of this function $$ \lim_{k\to \infty}P(X\leq k) = \lim_{k\to\infty} F(k) = 1 $$ $$ \lim_{k\to \infty}P(X<-k) \leq \lim_{k\to \infty}P(X\leq -k) = \lim_{k\to \infty} F(-k)= 0 $$ Hence we have the result.

Answer 3

If $X$ has a density function $f$, we can use that $$ P(|X| > k) = \int_{\Bbb R\setminus[-k,k]}f = \int_{\Bbb R}f\chi_{\Bbb R\setminus[-k,k]} $$ ($\chi =$ characteristic function) and apply the Dominated Convergence Theorem.

Answer 4

If not, then there exists an $\varepsilon > 0$ for which $P(|X|>k) > \varepsilon$ infinitely often, meaning $P(|X| \leq k) < 1-\varepsilon$ infinitely often, contradicting the fact that $P(|X| \leq k) = F_{|X|}(k) \to 1$ as $k \to \infty$.