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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?

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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
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up vote 3 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 $$ n\int_0^{1-c}(1-c-t)^{n-1} dt = n\int_0^{1-c} u^{n-1} du = n\left[\frac1n u^n \right]_0^{1-c} = (1-c)^n $$

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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
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