# How to prove $P(|X_n-X_m|>\epsilon)\leq P(|X_n-X|>\epsilon/2)+P(|X_m-X|>\epsilon/2)$?

Consider random variables $X, X_1, X_2, ...$ in a probability space $(\Omega, \mathcal F, P)$ such that

$X_n\stackrel{p}{\rightarrow} X$.

Let $n, m \in \mathbb N$. How can I prove that for some $\epsilon > 0$,

$P(|X_n-X_m|>\epsilon)\leq P(|X_n-X|>\epsilon/2)+P(|X_m-X|>\epsilon/2)$?

I was thinking something along the lines of using the triangle inequality:

$|X_n-X-(X_m-X)|>\epsilon \Rightarrow |X_n-X|+|X_m-X|>\epsilon$

Any help would be appreciated

• Are they really both $X_n$? – user198044 Feb 11 '16 at 22:31
• @JackBauer I've corrected it. Thanks – An old man in the sea. Feb 11 '16 at 22:38

You're on the right track. Now argue that if $|X_n-X|+|X_m-X|>\epsilon$, then $|X_n-X|>\epsilon/2$, or $|X_m-X|>\epsilon/2$. (Hint: Suppose not.) The desired result then follows from $P(A\cup B)\le P(A)+ P(B)$.

• Thanks! I now understand what I was doing wrong... I was trying to reach $\{|X_n-X|>\epsilon/2\} \cap \{|X_m-X|>\epsilon/2\}$, when I should have been trying to reach $\{|X_n-X|>\epsilon/2\} \cup \{|X_m-X|>\epsilon/2\}$. – An old man in the sea. Feb 11 '16 at 21:25

Let us denote events:

$A_{n,m} = \{|X_n - X_m| > \epsilon\} = \{|X_n - X + X - X_m| > \epsilon\}$

$A_{n} = \{|X_n - X| > \epsilon/2\}$

$A_{m} = \{|X_m - X| > \epsilon/2\}$

Then we need to show that $P(A_{n,m}) \le P(A_n) + P(A_m)$.

Proof:

Now that's that out of the way.

$P(A_{n,m}) = P(|X_n - X_m| > \epsilon) = P(|X_n - X + X - X_m| > \epsilon)$

$\le P(|X_n - X| + |X - X_m| > \epsilon)$

since $A_{n,m} \subseteq \{|X_n - X| + |X - X_m| > \epsilon\}$

$\le P(|X_n - X| > \epsilon/2 \cup |X - X_m| > \epsilon/2)$

since $\{|X_n - X| + |X - X_m| > \epsilon\} \subseteq \{|X_n - X| > \epsilon/2 \cup |X - X_m| > \epsilon/2\}$

$\le P(|X_n - X| > \epsilon/2) + P(|X - X_m| > \epsilon/2)$

$= P(A_n) + P(|X - X_m| > \epsilon/2)$

$= P(A_n) + P(|X_m - X| > \epsilon/2)$

$= P(A_n) + P(A_m)$

And we are done!