show $P(N>n)=P(X_{(n:n)}-X_{(1:n)}<\alpha)$ [duplicate]

Possible Duplicate:
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 $P(N>n)=P(X_{(n:n)}-X_{(1:n)}<\alpha)$

-
 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. – Dilip Sarwate Aug 1 '12 at 14:09

marked as duplicate by Dilip Sarwate, Did, t.b., Sasha, Ｊ. M.Aug 2 '12 at 5:14

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).