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I have 2 random variables $X,Y$ with mean 0 and variance 1, their correlation is $\rho$. I need to prove this inequality

$$\mathrm E\left({\max}\left\{X^2 , Y^2\right\} \right) \leq 1 + \sqrt{1 - \rho^2}$$
I need some pointers as to how to solve this problem. Thanks!

Ali Caglayan
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SphericalCow
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  • If you know that their correlation $p$ is equal to $1$, what is $p$ doing in the equation below? Swap it for a $1$, and you'll have a much simpler right-hand side. – Arthur Aug 12 '15 at 10:40
  • Sorry that was a typo I changed it now – SphericalCow Aug 12 '15 at 10:46
  • The $1$ on the right makes me want to write $\max(X^2,Y^2) = X^2 \phi + Y^2(1-\phi) = Y^2 + (X^2-Y^2)\phi$ with $\phi = \Bbb{1}_{X^2 > Y^2}$. Not sure if it helps. – Alexandre Halm Aug 12 '15 at 12:06
  • https://math.stackexchange.com/questions/1395820/prove-that-exyx-y-leq-2-sqrt1-rho%C2%B2-where-rho-is-correlation?rq=1. – StubbornAtom Dec 18 '18 at 19:55

1 Answers1

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Here is an idea: (I dont know if it will work out but it may be worth exploring)

You could use the fact that $max(a,b)=\frac{a+b+|a-b|}{2}$. Then use the linearity of E, that gives you $E(max(X²,Y²)) \leq 1+\frac{E(|(X+Y)(X-Y)|)}{2}$ Now we just need to prove that $E(|(X+Y)(X-Y)|) \leq 2\sqrt{1-p²}$
GBQT
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