Question on a random vector This relate to that paper: 
http://www.stat.purdue.edu/docs/research/tech-reports/1982/tr82-17.pdf
Let $U_1,...,Un$ be iid uniform on (0,1).
Set $L_n=\max_{i\leq n} U_i$. 
Also $S(n)= \inf\{i\leq n| U_i = L_n \}$ the time were the highest value is attained and
$Z(n)= \inf\{i\leq n| U_i = L_{S(n)-1} \} \vee 0$ the time were the highest value before the very highest is attained (not necessary the second highest globally!)
Proof that there are $V$ ,$V' \sim Exp(1)$ and $W$, $W' \sim U(0,1)$ all mutually independent, such that:
$(n(1-L_n),(S(n)-1)(1-\frac{L_{S(n)-1}}{L_n}),\frac{S(n)}{n},\frac{Z(n)}{S(n)-1})\overset{d}{\rightarrow}(V,V',W,W').$ 
A proof sketch will be sufficient.
I also posted the question here:https://math.stackexchange.com/questions/1844238/limit-of-a-probability-vector
 A: For $1 \leq s' < s \leq n$ and $t,t' \in \mathbb{R}$, one has
$$
n(1-L_n) > t, (S(n)-1)(1-\frac{L_{S(n)-1}}{L_n}) > t',S(n)=s, Z(n) = s'
$$
precisely when 
$$
U_1,\dots,U_{s'-1} < U_{s'},\\
U_{s'+1},\dots,U_{s-1} \leq U_{s'}, \\
U_{s+1},\dots,U_{n} \leq U_{s},\\
U_{s'} < \left(1 - \frac{t'_+}{s-1} \right)_+ U_s, \\
 U_{s} < \left(1 - \frac{t_+}{n} \right)_+ ,
$$
where $x_+ = \max(x,0)$. Setting $\alpha =\left(1 - \frac{t_+}{n} \right)_+ $ and $\alpha' =  \left(1 - \frac{t'_+}{s-1} \right)_+$, the probability of these events is 
$$
\mathbb{E}[U_{s'}^{s-2} \mathbb{1}_{U_{s'} \leq \alpha' U_s} U_{s}^{n-s} \mathbb{1}_{U_{s} \leq \alpha}  ] = \mathbb{E}[\frac{(\alpha' U_s)^{s-1}}{s-1}U_{s}^{n-s} \mathbb{1}_{U_{s} \leq \alpha}] \\
= \frac{\alpha'^{s-1} \alpha^n}{n(s-1)}  = \frac{1}{n(s-1)}\left(1 - \frac{t'_+}{s-1} \right)_+^{s-1} \left(1 - \frac{t_+}{n} \right)_+^n
$$
The conclusion follows from this computation and from the following fact : for any $w,w' \in ]0,1]$, one has
$$
\sum_{\substack{2\leq s \leq wn \\ 1 \leq s' \leq w'(s-1)}} \frac{1}{n(s-1)}\left(1 - \frac{t'_+}{s-1} \right)_+^{s-1} \left(1 - \frac{t_+}{n} \right)_+^n \longrightarrow w w' e^{-t'_+ - t_+} ,
$$
as $n \rightarrow + \infty$.
