# How would you visualize Levy metric?

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!

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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 $$\mu_F((a,b]):=F(b)-F(a) \ .$$
L-P metric is a metric among the (probability) measures on a space (now on $\Bbb R$), while the Levy metric is a metric rather among the (distribution) functions. The connection is that $$d_{L\text{-}P}(\mu_F,\mu_G)=d_{Levy}(F,G)\ .$$ In that sense, as they are equal, of course generate the same topology. – Berci Feb 25 '13 at 3:35
For (1), you actually have to draw it. For each $x$ I imagined the least square with center $(x,G(x))$ such that both its left and right side segment meets the graph of $F$ (or something like that). Try to draw the situation $F(x - \varepsilon) - \varepsilon \leq G(x) \leq F(x + \varepsilon) + \varepsilon$ – Berci Feb 25 '13 at 10:50
Thanks! I imagine a square centered as $(x, G(x))$ meets $F(x-\epsilon)-\epsilon$ at its right-down corner and meets $F(x+\epsilon)+\epsilon$ at its left-up corner? Is that the only case of meeting? – Tim Feb 25 '13 at 18:06