If $X,Y$ are i.i.d. standard Normal r.v.s, why is $\operatorname{Var}(\max(X,Y)) = \operatorname{Var}(\min(X,Y))$?

Question

Let $X,Y$ be i.i.d. standard Normal r.v.s. Let $M = \max(X,Y), L = \min(X,Y)$. Why is $\operatorname{Var}(M) = \operatorname{Var}(L)$? The answer key simply states: "By the symmetry of the Normal distribution ..." which isn't sufficient for me.

Answer 1

This result does not require normal distribution. It is true for any i.i.d. symmetric random variables $X$ and $Y$.

$-M=\min \{-X,-Y\}$ which has the same distribution as $\min \{X,Y\}=L$ since the joint distribution of $(-X,-Y)$ is same as that of $(X,Y)$ (by independence and the the fact that $X,-X$ have the same distribution; $Y,-Y$ have the same distribution).

Thus, $Var (M)=Var (-M)=Var L$.

Answer 2

Note that $\max(X, Y) = \frac{1}{2}(X + Y + |X - Y|), \min(X, Y) = \frac{1}{2}(X + Y - |X - Y|)$. In addition, when $X, Y \text{ i.i.d. } \sim N(\mu, \sigma^2)$, $X + Y$ and $X - Y$ are independent, implying that $X + Y$ and $|X - Y|$ are independent. Therefore: \begin{align} & \operatorname{Var}(\max(X, Y)) \\ =& \frac{1}{4}\operatorname{Var}(X + Y + |X - Y|) \\ =& \frac{1}{4}\operatorname{Var}(X + Y) + \frac{1}{4}\operatorname{Var}(|X - Y|) \\ =& \frac{1}{4}\operatorname{Var}(X + Y - |X - Y|) \\ =& \operatorname{Var}(\min(X, Y)). \end{align}