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Consider $X_1,\dots,X_n$ are i.i.d. from a $U[0,1]$. What is the probability that $|X_i-X_j|>d$ for some $d>0$ and for all $i,j \in \{1,\dots,n\}$. ($d$ is sufficiently small to ensure that the probability is positive)

So far, I have considered the following: If we sort the observations starting from the smallest one, $X_{[1]},\dots,X_{[n]}$, then should hold that $X_{[1]} \in [0,1-(n-1)d]$, $X_{[2]} \in [X_{[1]}+d,1-(n-2)d],\dots$$\dots, X_{[n]} \in [X_{[n-1]}+k,1]$. From these conditions one can try to calculate the integral, which even for a uniform distribution seems quite tricky.

My first question is whether there is a more clever way to proceed and my second question is whether this could be generalized in more dimensions.

Nikolas
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  • You might want to consider "order statistics" for this, so as to have at least a valid pdf for the lowest, second lowest, etc. Then you could define the minimum difference of adjacent order statistics and call that $Y,$ and find $P(Y>d).$ [Likely it would be a mess of an integral...] – coffeemath Mar 30 '16 at 13:58
  • I had tried to work a bit with order statistics, but it seemed to mess things even further. Thanks a lot for the help @coffeemath . – Nikolas Mar 31 '16 at 06:05

1 Answers1

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The complex integral just reduces (for all $X$ uniform in $[0, 1]$) to: $$ P(X_n>X_{n-1}+d,\dots,X_3>X_2+d,X_2>X_1+d)={1 \over n!} \left (1-(n-1) d \right )^n $$

N74
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