For a <cumulative distribution function> $F$, its <quantile function> is the generalized inverse
$$
F^{-1}(u)=\inf\{x\in\mathbb R:F(x)\geq u\},
\qquad 0<u\leq1,
$$
with the infimum allowed to be $+\infty$. For observations $X_1,\ldots,X_n$, the <empirical distribution function> is
$$
F_n(x)=\frac1n\sum_{i=1}^n\mathbf1_{\{X_i\leq x\}}.
$$
If $X_{(1)}\leq\cdots\leq X_{(n)}$ are the <order statistic>[order statistics], then
$$
X_{(j)}=F_n^{-1}(j/n).
$$

The <Bennett inequality> says that if $Y_1,\ldots,Y_N$ are independent, $\mathbb EY_i=0$, $Y_i\leq b$ almost surely, and $\sum_i\mathbb EY_i^2\leq v$, then
$$
\mathbb P\left(\sum_iY_i\geq t\right)
\leq\exp\left[-\frac v{b^2}
h\left(\frac{bt}{v}\right)\right],
\qquad
h(u)=(1+u)\log(1+u)-u.
$$
To prove it, convexity of $e^{\lambda y}$ on $y\leq b$, followed by the power-series bound for centered $Y_i$, gives
$$
\mathbb Ee^{\lambda Y_i}
\leq\exp\left\{\frac{\mathbb EY_i^2}{b^2}
(e^{\lambda b}-1-\lambda b)\right\}.
$$
Independence and the <Chernoff bound> therefore yield
$$
\mathbb P\left(\sum_iY_i\geq t\right)
\leq\exp\left\{-\lambda t+\frac v{b^2}
(e^{\lambda b}-1-\lambda b)\right\}.
$$
The minimizing value satisfies $e^{\lambda b}=1+bt/v$, namely $\lambda=b^{-1}\log(1+bt/v)$. Substitution gives the stated exponent.

For independent $U_i\sim\operatorname{Unif}[0,1]$, the <uniform order statistic> satisfies
$$
U_{(j)}\sim\operatorname{Beta}(j,n-j+1),
\qquad
\mathbb EU_{(j)}=\frac j{n+1}.
$$
Set $p=j/(n+1)-x$. The event $U_{(j)}\leq p$ means that at least $j$ sample points lie in $[0,p]$. If this occurs, some $j$-element subset consists entirely of such points. The <union bound> gives
$$
\mathbb P(U_{(j)}\leq p)
\leq\binom njp^j
\leq\left(\frac{enp}{j}\right)^j.
$$
This is exactly the required inequality for $0\leq x<j/(n+1)$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 210 / 1 / Solution

For independent uniform random variables $U_1,\ldots,U_n$ on $[0,1]$, the $j$th order statistic has distribution $\operatorname{Beta}(j,n-j+1)$ and mean $j/(n+1)$.


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