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?


3 Answers 3


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.

  • 1
    $\begingroup$ The expression you provided implies that Var(A)=Var(B) and with this, I can compute for the individual variances already from the equation $Var(A)+Var(B)=2-\frac{2}{\pi}$. I get 0.68 from this which I think is close enough to the simulated answer. $\endgroup$
    – user164144
    Jul 3, 2017 at 1:21
  • 1
    $\begingroup$ I also just read that the expression you provided holds generally as $max(f)=-min(-f)$. Just to clarify, the negative for -f is not relevant in my case since X and Y have mean 0, correct? $\endgroup$
    – user164144
    Jul 3, 2017 at 1:31
  • 2
    $\begingroup$ @user164144 yes that's right, but the second part is more than that. $P(-\min(X,Y)\le a) = P(\min(X,Y)\ge -a)$ $= P(X \ge -a, Y \ge -a) = P(X \ge -a)P(Y \ge -a) = P(X \le a) P(Y \le a)$ by algebra, logic, independence, symmetric-ness respectively. Then you can do something similar for the other guy, and you'll see the cdf is the same. $\endgroup$
    – Taylor
    Jul 3, 2017 at 2:03

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

$EA^2 = $


which equals 1.

$EA = $


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

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

Of course, Var(B) is identical.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.