Let $F$ and $G$ be distribution functions and define the Lévy metric by,
$$
d_L(F, G) = \inf \{\epsilon > 0\ |\ G(x – \epsilon) – \epsilon \leq F(x) \leq G(x + \epsilon) + \epsilon {\rm\ for\ all\ } x\}
$$
I want to show that $d_L(F, G) = d_L(G, F) \geq 0$.
The book that I am reading says that this is clearly true but doesn't provide a proof. It doesn't seem obvious to me. I can go about proving it by defining the sets $A$ and $B$ below,
\begin{eqnarray}
A &=& \{\epsilon > 0\ |\ G(x – \epsilon) – \epsilon \leq F(x) \leq G(x + \epsilon) + \epsilon {\rm\ for\ all\ } x\} \\
B &=& \{\epsilon > 0\ |\ F(x – \epsilon) – \epsilon \leq G(x) \leq F(x + \epsilon) + \epsilon {\rm\ for\ all\ } x\}
\end{eqnarray}
and then showing that $A = B$ and hence that $d_L(F, G) = \inf A = \inf B = d_L(G, F)$. Is there something more obvious that I'm missing?
Best Answer
Your plan is good. Just a small observation which can be useful:
$G(x - \epsilon) - \epsilon \leq F(x)$ for all $x$ $\Longleftrightarrow$ $G(x) \leq F(x+ \epsilon) + \epsilon$ for all $x$.