From Wikipedia:
Let $F, G : \mathbb{R} \to [0, 1]$ be two cumulative distribution functions. Define the Lévy distance between them to be
:$$L(F, G) := \inf \{ \varepsilon > 0 | F(x – \varepsilon) – \varepsilon \leq G(x) \leq F(x + \varepsilon) + \varepsilon \mathrm{\,for\,all\,} x \in \mathbb{R} \}.$$Intuitively, if between the graphs of $F$ and $G$ one inscribes squares with sides parallel to the coordinate axes (at points of discontinuity of a graph vertical segments are added), then the side-length of the largest such square is equal to $L$($F$, $G$).
I was wondering how to picture "between the graphs of $F$ and $G$ one inscribes squares with sides parallel to the coordinate axes (at points of discontinuity of a graph vertical segments are added)"?
Why "the side-length of the largest such square is equal to $L$($F$, $G$)"?
How is the Levy metric a special case of the Lévy–Prokhorov metric?
Thanks and regards!
Best Answer
For the picture, (first assuming that both $F$ and $G$ are continuous) for every $x$, I would start out from the point $P_G(x):=(x,G(x))$ on the graph of $G$ and draw both lines from that with tangents $\pm45^\circ$. Where these lines meet the graph of $F$ will determine the square on the picture around $P_G(x)$ which meets the graph of $F$ on both left and right side.
With this property we roll these squares around $P_G(x)$ for each $x$ and now have to take the supremum of their side lengths, because that will be the infimum of the $\varepsilon$'s that satisfy the criterium for all $x$.
For the other question, a distribution function $F:\Bbb R\to [0,1]$ determines a Borel measure $\mu_F$ on $\Bbb R$, defined as $$\mu_F((a,b]):=F(b)-F(a) \ . $$