This is a problem that I encountered in my research and have no clues to fully
resolve it. Basically, I need large (or moderate) deviation bounds on the
difference between an order statistic of independent and identically
distributed (i.i.d.) random variables on the compact interval $\left[
0,1\right]  $ and the expectation of this order statistic.

Let $X_{1}%
,...,X_{n}$, $n\in\mathbb{N}$ be i.i.d. and uniformly distributed on $\left[
0,1\right]  $. Let their order statistics be $X_{\left(  1\right)  }\leq
X_{\left(  2\right)  }\leq...\leq X_{\left(  n\right)  }$, where we can ignore
the zero probability event that any two order statistics are equal. Let $E$
and $V$ denote respectively the expectation operator and variance operator.
Then each $X_{\left(  r\right)  }$ follows a Beta distribution,
$$
E\left[  X_{\left(  r\right)  }\right]  =\frac{r}{n+1},r=1,...,n
$$
and
$$
V\left[  X_{\left(  r\right)  }\right]  =\frac{r\left(  n-r+1\right)
}{\left(  n+1\right)  ^{2}\left(  n+2\right)  },r=1,...,n
$$


This implies two things:

(a) when $r=o\left(  n\right)  $ where the small $o$ notation means that
$\lim_{n\rightarrow\infty}\frac{r}{n}=0$, for any $\varepsilon>0,$
\begin{equation}
P\left(  \left\vert X_{\left(  r\right)  }-\frac{r}{n+1}\right\vert
>\varepsilon\right)  \leq\frac{o\left(  1\right)  }{n
\varepsilon^{2}},\label{eq1}%
\end{equation}
where $o\left(  1\right)  $ denotes a nonnegative sequence that converges to
$0$ as $n\rightarrow\infty$; in this case, $\varepsilon=n^{\alpha}$ with $0 \le \alpha < \frac{1}{2}$ can be
set, such that
\begin{equation}
P\left(  \left\vert X_{\left(  r\right)  }-\frac{r}{n+1}\right\vert >\frac
{1}{n}\right)  \le  \frac{o(1)}{n^{1-2\alpha}}\rightarrow0,\label{eq3}%
\end{equation}
where the $o\left(  1\right)  $ can be set to be no smaller in order than rate
$\frac{r}{n}$.

(b) when $r=O\left(  n\right)  $ where the big $O$ notation here means that
$\liminf_{n\rightarrow\infty}\frac{r}{n}>0$, for any $\varepsilon>0$,
\begin{equation}
P\left(  \left\vert X_{\left(  r\right)  }-\frac{r}{n+1}\right\vert
>\varepsilon\right)  \leq C\frac{1}{\left(  n+2\right)  \varepsilon^{2}%
}\label{eq2}%
\end{equation}
from some constant $C\leq2$; in this case $\varepsilon=o\left(  \sqrt
{n}\right)  $ can be set such that
\begin{equation}
P\left(  \left\vert X_{\left(  r\right)  }-\frac{r}{n+1}\right\vert >\frac
{1}{n^{\alpha}}\right)  \leq\frac{2}{n^{1-2\alpha}}\rightarrow0\label{eq4}%
\end{equation}
for any $0\leq\alpha<\frac{1}{2}$.


A simple conclusion from the above discussion is that, regardless of the value
of $r$, we have that $X_{\left(  r\right)  }$ for each $r$ converges to
$E\left[  X_{\left(  r\right)  }\right]  $ in probability as $n\rightarrow
\infty$. My question is "**are the deviation bounds given above
the best**?" **Very likely NOT**. 

Let $a_{n,r}$ be a positive sequence
that depends on $n$ and $r$ such that
$$
\lim_{n\rightarrow\infty}a_{n,r}=0
$$
for each $r=1,...,n$, what is the best result available on
$$
\beta_{n,r}=P\left(  \left\vert X_{\left(  r\right)  }-\frac{r}{n+1}%
\right\vert >a_{n,r}\right)  ?
$$
By this I mean, what is the bound $\beta_{n,r}$ corresponding to the sequence $a_{n,r}$ that converges to $0$ at relatively and possibly the fastest speed? 
Any pointers or hints would be greatly appreciated! Thanks!