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Let $\{U_n\}_{n\ge 1}$ be an i.i,d. sequence of uniform r.v.s on [0,1]. For $i\ge j,$ let $M_{i,j}=\max(U_i,...,U_j).$ Let $Z_{i,j}=(j-i+1)(1-M_{i,j})$.
(1) Let $Y_n=Z_{1:2n}-2Z_{1:n}.$ Compute $\lim\sup_{n\to\infty}P(Y_n<-3)$.
(2) Does $\{Z_{1:n}\}_{n\ge 1}$ converges almost surely?

I find $Y_n=2\min(Z_{1:n},Z_{n+1:n})-2Z_{1:n}$, then $P(Y_n<-3)=P(2Z_{n+1:n}-2Z_{1:n}<-3,\min(Z_{1:n},Z_{n+1:n})=Z_{n+1:2n})$. But I don't know how to compute the limsup of this probability. And it seems we may use this result and Fatou's Lemma to get a contradiction, since we know $Z_{1:n}$ converges in distribution to exponential distribution with rate 1.

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    Not sure why you are overcomplicating this, instead of trying to get back to the most elementary formulation of the setting... Note that $$Y_n=-2(Z_{n+1:2n}-Z_{1:n})^+$$ where $x^+=\max\{x,0\}$. Since $Z_{n+1:2n}$ and $Z_{1:n}$ are independent and both converge in distribution to the standard exponential distribution, one has, for every positive $c$, $$P(Y_n<-2c)=P(Z_{n+1:2n}>Z_{1:n}+c)\to P(Z>Z'+c)$$ for $(Z,Z')$ i.i.d. standard exponential. Thus, $$P(Y_n<-2c)\to\tfrac12e^{-c}$$ Now, assuming that $Z_{1:n}$ converges almost surely, its limit $Z_\infty$ would be standard exponential. ... – 2017-02-27
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    ... One would get $Y_n\to-Z_\infty$ almost surely but $P(Y_n<-2c)\to\tfrac12e^{-c}\ne e^{-2c}=P(-Z_\infty<-2c)$ for, say, $c=3$, a contradiction. – 2017-02-27

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