Variance of Minimum and Maximum of 2 iid Normal

Question

Let $X$ and $Y$ be iid $\sim Normal(0,1)$

Let $A=max(X,Y)$ and $B=min(X,Y)$

What are $Var(A)$ and $Var(B)$?

From simulation, I get $Var(A)=Var(B)$ approximately 0.70.

How do I get this analytically?

Answer 1

If you can convince yourself that $$ \max(X,Y) \overset{d}{=} -\min(X,Y), $$ then taking the variance on both sides will give you your answer.

Regarding the other part, you'll probably have to integrate by hand.

Answer 2

Doing it out the long way, which generalizes to more than 2 iid Normals, here are the integral calculations in MAPLE:

$EA^2 = $

2*int(z^2*1/sqrt(2*Pi)*int(exp(-x^2/2),x=-infinity..z)*1/sqrt(2*Pi)*exp(-z^2/2),z=-infinity..infinity);

which equals 1.

$EA = $

2*int(z*1/sqrt(2*Pi)*int(exp(-x^2/2),x=-infinity..z)*1/sqrt(2*Pi)*exp(-z^2/2),z=-infinity..infinity);

which equals $1/\sqrt{\pi}$.

Therefore, Var(A) = $1-1/\pi = $0.68169... which agrees with my simulation.

Of course, Var(B) is identical.

Answer 3

Consider the standard normal case (since it's trivial to generalize). Let $Z = \max(X,Y)$.

$F_Z(z)=P(\max(X,Y)\leq z) = P(X\leq z,Y\leq z) = \Phi(z)^2$

hence obtain $f_Z(z)$ by differentiation.

As for expectation, note the following:

$\frac{d}{dx} \phi(x)\Phi(x) = -x\phi(x)\Phi(x) + \phi(x)^2$

Further note that $\phi(x)^2$ can be written in terms of $a\phi(bx)$ for some constants $a$ and $b$. From there you should be able to show that

$\int x\phi(x)\Phi(x)\,dx={\frac{1}{\sqrt{2}}}\frac{1}{\sqrt{2\pi}}\Phi(x\sqrt{2})-\phi(x)\Phi(x)+C$
(if not, show it by differentiation ...)

And by taking derivatives of $x\phi(x)\Phi(x)$ you should be able to use previous results to get to $E(Z^2)$.

.... Or just use the table of definite integrals here: https://en.wikipedia.org/wiki/List_of_integrals_of_Gaussian_functions#Definite_integrals

with a little manipulation, I think you can do the expectation and variance from there.