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Finding $E(N)$ in this question

suppose $X_1,X_2,\ldots$ is sequence of independent random variables of $U(0,1)$ if $N=\min\{n>0 :X_{(n:n)}-X_{(1:n)}>\alpha , 0<\alpha<1\}$ that $X_{(1:n)}$ is smallest order statistic and $X_{(n:n)}$ is largest order statistic. how can show $P(N>n)=P(X_{(n:n)}-X_{(1:n)}<\alpha)$

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    Actually, the question asked here was also asked by the OP in a follow-up comment on Didier's answer to the earlier question, and has been answered there by Didier.2012-08-01

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If $(Z_n)_{n\geqslant0}$ is a sequence of random variables such that $Z_n\leqslant Z_{n+1}$ for every $n\geqslant0$, then $N_a=\inf\{n\geqslant0\,;\,Z_n\gt a\}$ is such that, for every $n\geqslant0$, $ [N_a\gt n]=[Z_1\leqslant a,\ldots,Z_n\leqslant a]=[Z_n\leqslant a]. $ Note: No probability here, this an almost sure result (as probabilists like to say), that is, a deterministic result (as everybody else says).

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    As already explained on the other page. @hadisanji: What is this game you are playing at?2012-08-01