From Wikipedia:

Let $F, G : \mathbb{R} \to [0, 1]$ be two cumulative distribution functions. Define the Lévy distance between them to be :$$L(F, G) := \inf \{ \varepsilon > 0 | F(x - \varepsilon) - \varepsilon \leq G(x) \leq F(x + \varepsilon) + \varepsilon \mathrm{\,for\,all\,} x \in \mathbb{R} \}.$$

I was wondering what is the role of $\varepsilon$ in the range values ? It appears weird to me to take the same $\varepsilon$ in the range and domain values.

Does the following definition still satisfies metric properties ?

$$L(F, G) := \inf \{ \varepsilon > 0 | F(x - \varepsilon) \leq G(x) \leq F(x + \varepsilon) \mathrm{\,for\,all\,} x \in \mathbb{R} \}.$$

Set

$$K(F,G) := \inf\{\varepsilon>0; G(x-\varepsilon) \leq F(x) \leq G(x+\varepsilon) \, \text{for all} \, x \in \mathbb{R}\}.$$

It is not difficult to see that $K$ is a metric. Let us consider the following easy example in order to see the differences between the Lévy metric $L$ and the metric $K$:

Define random variables $X:= 1_{[1/2,1]}$ and $X_n := \frac{1}{2} 1_{(0,1/n]}+1_{[1/2,1]}$ on the unit interval $[0,1]$ (endowed with the Lebesgue measure). The corresponding distribution functions are given by

$$G(x) := \mathbb{P}(X \leq x) = 1_{[1,\infty)}(x) \qquad \qquad G_n(x) := \mathbb{P}(X_n \leq x) = \frac{1}{n} 1_{[1/2,1)}(x) + 1_{[1,\infty)}(x).$$

Using the definition of $L$ and $K$ it is not difficult to see that $L(G_n,G) \to 0$ whereas $K(G_n,G)=1/2$. On the other hand, we know that $X_n \to X$ almost surely; that's why the Lévy metric $L$ is more natural: If the random variables converge almost surely, then we would expect that the distance of the corresponding distribution functions converges to $0$.

A typical $\varepsilon$-neighborhood of the distribution function $G$ with respect to the Lévy metric $L$ looks like that whereas a neighborhood with respect to $K$ is of the following form Remark In fact, one can show that $L(G_n,G) \to 0$ if and only if $X_n \to X$ in distribution for any random variables $(X_n)_n$ and $X$.

• Thanks a lot for your detailed answer. I understand that it is a matter of convergence. However I'm still interested in proving that the mapping K is a metric. Hereafter an attempt of the triangle inequality (the others are quiet obvious). We have to prove the following. $$L(F,H)\leq L(F,G)+L(G,H)$$Consider $L(F,G)\leq a$ and $L(G,H)\leq b$ We have: \begin{eqnarray} G(x-a)\leq &F(x)& \leq G(x+a), \forall x\\ H(x-b)\leq&G(x)&\leq H(x+b) \forall x\\ H(x-a-b) \leq &F(x)& \leq H(x+a+b), \forall x \end{eqnarray} Then $L(F,H)\leq a+b$ Apr 11 '14 at 8:04
• @user140328 You have to assume that $L(F,G) < a$ and $L(G,H)<b$ (because, by definition, the inequality $$G(x-a) \leq F(x) \leq G(x+a)$$ holds for any $a>L(F,G)$ but not necessarily for $a=L(F,G)$). Then letting $a \downarrow L(F,G)$ and $b \downarrow L(G,H)$ finishes the proof.
– saz
Apr 11 '14 at 8:30
• Right ! Thanks. Apr 11 '14 at 9:25