TITLE: How to see Ky Fan metric satisfies the triangle inequality?
QUESTION [5 upvotes]: Let $X$ and $Y$ be random variables. The Ky Fan metric is defined as:
$$d(X,Y):= \min \{\epsilon>0 :P(|X-Y|>\epsilon)\le\epsilon\} $$
I want to show it is indeed a metric, for which I need to show that it satisfies triangle inequality.
Set $d(X,Y)=\epsilon_1, d(Y,Z)=\epsilon_2$,and
$d(Z,X)=\epsilon_3$ .
I approached this by trying to show that $ (\epsilon_1+\epsilon_2)  \in \{\epsilon>0 :P(|X-Z|>\epsilon)\le\epsilon\} $. But to show that, I had to show $P(|X-Y|>\epsilon_1)+P(|Y-Z|>\epsilon_2) \ge P(|X-Z|>\epsilon_1+\epsilon_2)$, which I could not.
Can you help in showing this?

REPLY [2 votes]: If $|X-Y| \le \epsilon_1$ and $|Y-Z| \le \epsilon_2$, then we have $|X-Z|\le \epsilon_1 + \epsilon_2$.
Hence if $|X-Z| > \epsilon_1 + \epsilon_2$, then $|X-Y| > \epsilon_1$ or $|Y-Z|> \epsilon_2$.
$$\{ \omega:|X(\omega) - Z(\omega)| > \epsilon_1 + \epsilon_2\} \subseteq \{\omega:|X(\omega)-Y(\omega)| > \epsilon_1\} \cup \{\omega:|Y(\omega)-Z(\omega)| > \epsilon_2\}$$
Hence, by the union bound,
$$P(|X-Y| > \epsilon_1) + P(|Y-Z| > \epsilon_2) \ge P(|X-Z| > \epsilon_1 + \epsilon_2)$$