Let $R_1,...,R_n$ be i.i.d. exponentially distributet with mean 1 and let $0 =: R_{(0)} < R_{(1)} <...<R_{(n)}$ be their order statistics, i.e. $R_{(j)}$ is the j-th smallest of the $R_i$.
Now I want to proof that if $R_{(1)}$ is particularly large, then the sample has a small range, meaning that $R_{(n)}-R_{(1)}$ is small, i.e. \begin{align} Cov(R_{(n)}-R_{(1)},R_{(1)}) \overset{!}{<} 0. \end{align} Now we know the joint distribution of the order statistic \begin{align} (R_{(1)},...,R_{(n)}) \overset{d}{=} (R_n/n,...,R_n/n + R_{n-1}/(n-1) +...+ R_1/1), \end{align} unfortunately I could not find satisfactorily results concerning the question at hand. As I have never dealt with order statistics before, does anyone have an approach to this? Thanks!