Let n >0$n >0$.
Let $X_1,...,X_n$$X_1,\ldots,X_n$ be i.i.d. uniform random variable on [0,1].$[0,1].$ Denote by $X^{(1)}\leq X^{(2)} \leq \ldots \leq X^{(n)}$$X^{(1)}\leq X^{(2)} \leq \cdots \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 :
$$P(\sup_{1 \leq i \leq n} \Delta^{(i)} \geq x) \leq \sum_{i=1..n} P( \Delta^{(i)} \geq x) $$$$P(\sup_{1 \leq i \leq n} \Delta^{(i)} \geq x) \leq \sum_{i=1\ldots n} P( \Delta^{(i)} \geq x) $$ $$\leq \frac{1}{x^2}\sum_{i=1..n} Var(X^{(i)}) $$$$\leq \frac{1}{x^2}\sum_{i=1 \ldots n} \operatorname{Var}(X^{(i)}) $$
$$ \leq \frac{1}{x^2}\sum_{i=1..n} \frac{i(n-i+1)}{(n+1)^2(n+2)} = O(1)$$$$ \leq \frac{1}{x^2}\sum_{i=1\ldots n} \frac{i(n-i+1)}{(n+1)^2(n+2)} = O(1)$$
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]$[0,1]$
