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