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Henry.L
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Largest deviations for uniform order statistics

Let n >0

Let $X_1,...,X_n$ be i.i.d. uniform random variable on [0,1]. Denote by $X^{(1)}\leq X^{(2)} \leq \ldots \leq X^{(n)}$ their order statistics, and write $\Delta^{(i)} = \vert X^{(i)} - EX^{(i)} \vert $

It is classical than $X^{(i)}$ follows a beta, with parameters $ \beta(i, n-i+1)$

I would like to show that :

$$ \sup_{1 \leq i \leq n} \Delta^{(i)} \overset{\mathcal{P}}{\underset{n\to+\infty}{\longrightarrow}} 0 $$

where the subscript "$\mathcal{P}$ " denote the convergence in probability.

If ones tries to be brutal, it goes like this :

$$P(\sup_{1 \leq i \leq n} \Delta^{(i)} \geq x) \leq \sum_{i=1..n} P( \Delta^{(i)} \geq x) $$ $$\leq \frac{1}{x^2}\sum_{i=1..n} Var(X^{(i)}) $$

$$ \leq \frac{1}{x^2}\sum_{i=1..n} \frac{i(n-i+1)}{(n+1)^2(n+2)} = O(1)$$

But i need $o(1)$ hence, one has to do something a little more refined, but I can't make it work.. Any idea ?

PS : A dream would be to prove such property for a whole classe of rvs, say rvs admitting a "nice" density on [0,1]

Gericault
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