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Let Z = min{|X|,|Y|}, with $X , Y \sim Normal(0,1)$.

Then, show that $E(Z) = \frac{2(\sqrt{2} - 1)}{\sqrt{\pi}}$

So far I got

$F_{Z}(z) = 1-[1 - F_{|X|}(z)][1-F_{|Y|}(z)]$

$F_{Z}(z) = 1-4F_{X}(-z)F_{Y}(-z)$

$F_{Z}(z) = 1-\frac{4}{2\pi}\int_{-\infty}^{-z}\int_{-\infty}^{-z}e^{-\frac{x^{2}+y^{2}}{2}}dxdy$

But I do not know where to go from here.

Harry
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    You may want to review this question to see if it gives you ideas on how to proceed: https://math.stackexchange.com/questions/1030946/what-is-the-expected-value-of-min-x-y-max-x-y-assuming-x-and/1030973#1030973 – heropup Nov 30 '17 at 06:07
  • Awesome, I got it. Thank you so much! – Harry Nov 30 '17 at 06:23
  • If you could answer your own question, that would be nice. I tried working on it myself but got a weird answer. – Remy Nov 30 '17 at 06:26
  • Thank you, I think I just went the wrong route. But I worked it out, finally. – Harry Nov 30 '17 at 06:56
  • @harry, could there be a type on your problem such as $E(z) = \frac{2\sqrt{2}-1}{\sqrt{\pi}}$. Could you let me know the source of the problem. I think I reworked and I am getting a slightly different result. – Satish Ramanathan Dec 01 '17 at 07:51
  • @SatishRamanathan It's from Probability and Random Variables - a beginner's guide by David Stirzaker, Section 6.15, #15. – Harry Dec 04 '17 at 23:15

2 Answers2

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We know that $F_{|X|}(x)=Pr(-x\leq{X}\leq{x})=F_{X}(x)-F_{X}(-x)$

Hence, $f_{|X|}(x)=f_{X}(x)+f_{X}(-x)=2f_{X}(x)$

Therefore, $f_{|X|}(x)=\sqrt{\frac{2}{\pi}}e^{\frac{-x^{2}}{2}}$ and similarly, $f_{|Y|}(y)=\sqrt{\frac{2}{\pi}}e^{\frac{-y^{2}}{2}}$

Then, by symmetry of Normal curve, we can simplify to:

$E[min(|X|,|Y|)] = \frac{2}{\pi}(\int_{x=0}^{\infty}\int_{y=0}^{x}ye^{\frac{-x^{2}}{2}}e^{\frac{-y^{2}}{2}}dydx + \int_{y=0}^{\infty}\int_{x=0}^{y}xe^{\frac{-x^{2}}{2}}e^{\frac{-y^{2}}{2}}dxdy)$

$E[min(|X|,|Y|)] = \frac{4}{\pi}\int_{x=0}^{\infty}\int_{y=0}^{x}ye^{\frac{-x^{2}}{2}}e^{\frac{-y^{2}}{2}}dydx$

$E[min(|X|,|Y|)] = \frac{4}{\pi}\int_{x=0}^{\infty}e^{\frac{-x^{2}}{2}}(1-e^{\frac{-x^{2}}{2}})dx$

$E[min(|X|,|Y|)] = \frac{4}{\pi}\int_{x=0}^{\infty}(e^{\frac{-x^{2}}{2}}-e^{-x^{2}})dx$

$E[min(|X|,|Y|)] = \frac{4}{\pi}(\frac{1}{2})\int_{0}^{\infty}(\sqrt{2\pi}\frac{1}{\sqrt{2\pi}}e^{\frac{-x^{2}}{2}}-\sqrt{\pi}\frac{1}{\sqrt{\pi}}e^{-x^{2}})dx$

Since, we have 2 pdf's:

$Normal(0,1) : \frac{1}{\sqrt{2\pi}}e^{\frac{-x^{2}}{2}}$

$Normal(0,\frac{1}{2}) : \frac{1}{\sqrt{\pi}}e^{-x^{2}}$

Thus,

$E[min(|X|,|Y|)] = \frac{2}{\pi}(\sqrt{2\pi}-\sqrt{\pi})$

$E[min(|X|,|Y|)] = \frac{2({\sqrt{2}-1})}{\sqrt{\pi}}$

Hence, proved.

Satish Ramanathan
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Harry
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1

I have not checked your solution. Your initial approach is not wrong either. I have completed your first approach and came to the same result. Attached you will findenter image description here

Satish Ramanathan
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