Variance of extreme order statistics Let $x_{i:n}$ be the $i$-th order statistic of $n$ i.i.d. draws from CDF $F$ over compact support $[0,1]$, i.e. $x_{1:n}\leq x_{2:n}\leq \cdots \leq x_{n:n}$. 
Do we always have $Var(x_{1:n})\leq Var(x_{2:n})$?
 A: Question: Is Var($1^{\text{st}}$ order statistic) $\leq$ Var($2^{\text{nd}}$  order statistic) for all $n\geq 2$, when the parent distribution has support on [0,1]?
Answer: The easiest way of showing that the result does NOT always hold is by counterexample, and the easiest counterexample I can think of is to take an upwards triangular pdf:
$$f(x)  = 2x  \quad \quad \text{ for } 0\leq x\leq 1$$

One can then calculate the pdf of the $1^{\text{st}}$ order statistic, and of the $2^{\text{nd}}$ order statistic, and so the variance of each, which yields that:
$$\text{Var}(X_{1:n}) = \frac{1}{n+1}-\frac{\pi  \Gamma (n+1)^2}{4 \Gamma \left(n+\frac{3}{2}\right)^2}$$
$$\text{Var}(X_{2:n}) = \frac{2}{n+1}-\frac{9 \pi  (n!)^2}{16 \Gamma \left(n+\frac{3}{2}\right)^2}$$
The following diagram plots $\text{Var}(X_{1:n})$ and $\text{Var}(X_{2:n})$ as a function of $n$:

and, as apparent, for this case, the variance of the $1^{\text{st}}$ order statistic is larger (not smaller) than the variance of the $2^{\text{nd}}$ order statistic.
