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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?

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3 Answers 3

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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.

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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ @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
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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.

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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.

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