Why is $\text{Var(}\text{min}(X, Y)) = \text{Var(}\text{max}(X, Y))$ for $X, Y \stackrel{\text{i.i.d}}{\sim} \mathcal{N}(0,1)$?

Question

I intuitively get that there is some sort of symmetry at play that makes $\text{Var}(\text{max}(X, Y)) = \text{Var}(\text{min}(X, Y))$ for $X, Y \stackrel{\text{i.i.d}}{\sim} \mathcal{N}(0,1)$ since, generally, there is symmetry for two continuous, i.i.d random variables $A, B$ in that $P(A > B) = P(A < B)$.

What I don't quite understand is this justification:

By symmetry of the Normal, $(X, Y)$ has the same distribution as $(-X, -Y)$, so $\text{Var}(\text{max}(X, Y)) = \text{Var}(\text{min}(X, Y))$. (Source)

This post begins to tease this out a little, but I'm still missing why the fact that $-X = X$ and $-Y=Y$ means it's necessarily so that the dispersion of the minimum and maximum of two i.i.d standard normal random variables is the same.

Answer 1

Because $\max(x, y) = -\min(-x, -y)$, you have that $$\text{Var}(\max(X, Y)) = \text{Var}(-\min(-X, -Y))$$

But since $\text{Var}(-X) = \text{Var}(X)$, this simplifies to $\text{Var}(\min(-X, -Y)$. And then since the distribution of $X$ and $-X$ are the same, this simplifies to $\text{Var}(\min(X, -Y))$. Finally because the distribution of $Y$ and $-Y$ are the same, this simplifies to $\text{Var}(\min(X, Y))$.

In general, if $X$ and $Y$ are chosen from symmetric distributions (it doesn't necessarily have to be the same distribution for each), then $$\text{Var}(\max(X, Y)) = \text{Var}(\min(X, Y))$$

Answer 2

A slight variant on @VarunVejalla's approach: let $\Phi$ denote the $\mathcal{N}(0,\,1)$ CDF so$$P(\min(X,\,Y)\le z)=1-(1-\Phi(z))^2=1-\Phi(-z)^2$$(the last $=$ uses symmetry), and $P(\max(X,\,Y)\le z)=\Phi^2(z)$. So $\min(X,\,Y)$'s distribution is that of $-\max(X,\,Y)$, whence$$\operatorname{Var}\min(X,\,Y)=\operatorname{Var}[-\max(X,\,Y)]=\operatorname{Var}\max(X,\,Y).$$