POTW Expected Value of a Maximum

[Euge]

If X and Y are independent standard Gaussian variables, find the expected value of the maximum of X and Y.

Note: This problem is due to @Office_Shredder.

 

[jbergman]

If X and Y are independent standard Gaussian variables, find the expected value of the maximum of X and Y.

Note: This problem is due to @Office_Shredder.

What does the word standard mean here? Mean 0 and standard deviation of 1?

 

[Euge]

What does the word standard mean here? Mean 0 and standard deviation of 1?

Yes.

 

[bob012345]

What does the word standard mean here? Mean 0 and standard deviation of 1?

What do you mean by the maximum of X and Y? The sum, the product, independent of each other?

 

[Office_Shredder]

What do you mean by the maximum of X and Y? The sum, the product, independent of each other?

$\mathbf{E}(\max(X,Y))$

 

[BWV]

Will take a stab

The definite integral for the expectation of a Gaussian is
$-exp(-x^2/2)/sqrt(2 π)$

For 2 draws of a standard normal, 50% of the outcomes will have a positive and negative number, the expectation in this case is
$1/sqrt(2 π)$ = 0.40
The integral from 0 to inf

The next case has (x,y) both positive
Using above, the expectation for the minimum of the two positive draws
is
$1/sqrt(2 π)$, then the expectation of the max is the integral from
$1/sqrt(2 π)$ to inf, which is
$exp(-0.4^2)/sqrt(2π)$ = 2.13

For x,y both negative, the max is the integral from
$-1/sqrt(2 π)$ to zero, which is
$-1/sqrt(2 π) +exp(-1/4π)/sqrt(2π)$ =-0.22

So .5(0.40)+.25(2.13) + .25*(-0.22) = 0.68

 

[Office_Shredder]

That's not the right number. When they have opposite signs I think you have the right answer, but when they're the same sign I think you're doing something wrong. I don't know how you get the expected value of the smaller one, and I'm skeptical you can compute the expected value of the larger one by just integrating from e.v. of smaller one to infinity (maybe that works but it's not obvious to me)

 

[ergospherical]

I put Z = max(X,Y) then considered P(Z < z) = P(X < z)P(Y < z), which is\begin{align*}
P(Z<z) = \frac{1}{2\pi} \left( \int_{-\infty}^{z} e^{-\frac{1}{2} x^2} dx \right)^2
\end{align*}differentiating this gives the distribution of Z\begin{align*}
f(z) = \frac{1}{\pi} e^{-\frac{1}{2}z^2} \int_{-\infty}^z e^{-\frac{1}{2}x^2} dx
\end{align*}then the expectation of Z would be\begin{align*}
E(Z) = \int_{-\infty}^{\infty} zf(z) dz &= \frac{1}{\pi} \int_{-\infty}^{\infty} ze^{-\frac{1}{2}z^2} \left( \int_{-\infty}^z e^{-\frac{1}{2}x^2} dx \right) dz \\
&= \frac{1}{\pi} \left[ -e^{-\frac{1}{2} z^2} \int_{-\infty}^z e^{-\frac{1}{2}x^2} dx \right]_{-\infty}^{\infty} + \frac{1}{\pi} \int_{-\infty}^{\infty} e^{-z^2} dz \\
&= 0 + \frac{1}{\sqrt{\pi}}
\end{align*}

 

[PeroK]

I put Z = max(X,Y) then considered P(Z < z) = P(X < z)P(Y < z), which is\begin{align*}
P(Z<z) = \frac{1}{2\pi} \left( \int_{-\infty}^{z} e^{-\frac{1}{2} x^2} dx \right)^2
\end{align*}differentiating this gives the distribution of Z\begin{align*}
f(z) = \frac{1}{\pi} e^{-\frac{1}{2}z^2} \int_{-\infty}^z e^{-\frac{1}{2}x^2} dx
\end{align*}then the expectation of Z would be\begin{align*}
E(Z) = \int_{-\infty}^{\infty} zf(z) dz &= \frac{1}{\pi} \int_{-\infty}^{\infty} ze^{-\frac{1}{2}z^2} \left( \int_{-\infty}^z e^{-\frac{1}{2}x^2} dx \right) dz \\
&= \frac{1}{\pi} \left[ -e^{-\frac{1}{2} z^2} \int_{-\infty}^z e^{-\frac{1}{2}x^2} dx \right]_{-\infty}^{\infty} + \frac{1}{\pi} \int_{-\infty}^{\infty} e^{-z^2} dz \\
&= 0 + \frac{1}{\sqrt{\pi}}
\end{align*}

I think that's excellent, but you'll get zero marks from @Office_Shredder for leaving that $0$ in the answer. It shows you are just regurgitating symbols back at him!

 

[Office_Shredder]

Right, I think we coincidentally covered the only valid representation of this answer here

https://www.physicsforums.com/threa...-compared-to-my-solution.1046454/post-6812896

Nice work @ergospherical. I actually computed this via a pretty different looking starting integral if anyone else is still poking around at other solutions.

 

[pbuk]

Nice work @ergospherical. I actually computed this via a pretty different looking starting integral if anyone else is still poking around at other solutions.

I wonder what? This:

I put Z = max(X,Y) then considered P(Z < z) = P(X < z)P(Y < z), which is \begin{align*}
P(Z<z) = \frac{1}{2\pi} \left( \int_{-\infty}^{z} e^{-\frac{1}{2} x^2} dx \right)^2
\end{align*}

is the only way I can see?

 

[Office_Shredder]

I wonder what? This:

is the only way I can see?

You can write the expected value as a two dimensional integral by definition, and then solve from there.

 

[pbuk]

You can write the expected value as a two dimensional integral by definition, and then solve from there.

Isn't that what @ergospherical has done?

 

[Office_Shredder]

Isn't that what @ergospherical has done?

I don't see any integrals over 2 dimensional space there.

 

[PeroK]

Isn't that what @ergospherical has done?

An alternative is to note that it's equally likely that the first variable, $x$, is the maximum, so we have:
$$E(max(X, Y)) = E(X: X = max(X,Y)) = \frac 1 {2\pi}\int_{-\infty}^{\infty} xe^{-\frac 1 2 x^2} \int_{-\infty}^{x} e^{-\frac 1 2 y^2}\ dy \ dx$$$$= \frac 1 {2\pi} \int_0^{\infty}\int_{-\frac 3 4 \pi}^{\frac \pi 4} r^2e^{-\frac 1 2 r^2} \cos \theta \ d\theta \ dr = \frac 1 {\sqrt 2 \pi}\int_0^{\infty} r^2e^{-\frac 1 2 r^2} \ dr$$$$=\frac 1 {\sqrt 2 \pi}\sqrt{2\pi} = \frac 1 {\sqrt \pi}$$