Max and Min have the same variance

Question

Let $X_i$ be i.i.d uniform variables between 0 and 1. Define

$Min_n = \min(X_1, \ldots, X_n)$

$Max_n = \max(X_1, \ldots, X_n)$

How can we show that the variance of $Min_n$ and $Max_n$ are the same? What is the intuition behind it?

Thank you.

Answer 1

The intuition is simply that when $X_i\overset{\text{iid}}{\sim }U(0,1)$, then by symmetry,

$$1-\max\{X_1,...,X_n\}$$ has the same distribution as $$\min\{X_1,...,X_n\},$$

so $\text{Var}(\min\{X_1,...,X_n\})=\text{Var}(1-\max\{X_1,...,X_n\})=\text{Var}(\max\{X_1,...,X_n\}).$

If you want to more explicitly convince yourself that the above objects have the same distribution (i.e. same CDF), look into some prelims of order statistics. To obtain the CDFs, it may help to recognize

$$\{X_1,...,X_n:\max\{X_1,...,X_n\}\leq x\}\quad = \quad \bigcap_{i=1}^n\{X_1,...,X_n: X_i\leq x\}\\ \{X_1,...,X_n:\min\{X_1,...,X_n\}\geq x\}\quad=\quad \bigcap_{i=1}^n\{X_1,...,X_n: X_i\geq x\}.$$