Question on a random vector

Question

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

Answer 1

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 $n \rightarrow + \infty$.