Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

From Wikipedia

Let $D+$ be the set of all probability distribution functions $F$ such that $F(0) = 0$: $F$ is a nondecreasing, left continuous mapping from the real numbers $\mathbb{R}$ into $[0, 1]$ such that $$ \sup_{x \in\mathbb{R}} F(x) = 1 $$

The ordered pair $(S,d)$ is said to be a probabilistic metric space if $S$ is a nonempty set and $$ d: S×S →D+ $$ In the following, $d(p, q)$ is denoted by $d_{p,q}$ and is a distribution function dp,q(x). The distance-distribution function satisfies the following conditions: $$ d_{u,v}(x) = 0 \text{ for all }x > 0 \Leftrightarrow u = v (u, v ∈ S). $$ $$ ... $$

I wonder if $d_{u,v}(x) = 0 \text{ for all }x > 0$, whether $d_{u,v}$ is still a probability distribution function, since $\sup_{x \in \mathbb{R}} d_{u,v}(x) =0$ not $1$?

Thanks and regards!

share|cite|improve this question
I think there is some error in the Wikipedia article. The line should read $d_{u,v}(x) = 1$, all $x > 0$ $\iff u = v$. – martini Apr 23 '12 at 20:34
Wikipedia articles that have no references or sources should be considered suspect. At least Planet Math has some references ... – GEdgar Apr 23 '12 at 20:57
... and PlanetMath says we need to have a 1 there, as their $e_0$ is $\chi_{(0,\infty)}$. – martini Apr 23 '12 at 21:08

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.