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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 $Var(x_{1:n})\leq Var(x_{2:n})$?

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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:

$$f(x) = 2x \quad \quad \text{ for } 0\leq x\leq 1$$

enter image description here

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 $n$:

enter image description here

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.

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