Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

Let $\{X_i\}$ be $n$ iid uniform(0, 1) random variables. How do I compute the probability that the difference between the second smallest value and the smallest value is at least $c$?

I've messed around with this numerically and have arrived at the conjecture that the answer is $(1-c)^n$, but I haven't been able to derive this.

I see that $(1-c)^n$ is the probability that all the values would be at least $c$, so perhaps this is related?

share|cite|improve this question
Thanks for the accept, but it would be a good idea to take it back and let the question be listed as "still open" for a day or so, to see if someone can find a more intuitive explanation. – Henning Makholm Mar 6 '12 at 1:40
up vote 4 down vote accepted

There's probably an elegant conceptual way to see this, but here is a brute-force approach.

Let our variables be $X_1$ through $X_n$, and consider the probability $P_1$ that $X_1$ is smallest and all the other variables are at least $c$ above it. The first part of this follows automatically from the last, so we must have $$P_1 = \int_0^{1-c}(1-c-t)^{n-1} dt$$ where the integration variable $t$ represents the value of $X_1$ and $(1-c-t)$ is the probability that $X_2$ etc satisfies the condition.

Since the situation is symmetric in the various variables, and two variables cannot be the least one at the same time, the total probability is simply $nP_1$, and we can calculate $$ n\int_0^{1-c}(1-c-t)^{n-1} dt = n\int_0^{1-c} u^{n-1} du = n\left[\frac1n u^n \right]_0^{1-c} = (1-c)^n $$

share|cite|improve this answer
More generally, if $X_{(j)}$ are the order statistics, the joint density of $X_{(i)}$ and $X_{(i+1)}$ is $f(u,v) = \frac{n!}{(i-1)!(n-i-1)!} u^{i-1} (1-v)^{n-i-1}$, so $$P(X_{(i+1)}-X_{(i)}>c) = \int_0^{1-c} du\ \int_{u+c}^1 dv \ f(u,v) = (1-c)^n$$ – Robert Israel Mar 6 '12 at 2:12

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.