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I have $\xi$ and $\eta$ with following properties: $E\xi = E\eta = 0$, $D\xi = D\eta = 1$. And the correlation coefficient: $\rho = \rho (\xi, \eta)$.

I want to prove the following inequality:

$$ E \max (\xi^2, \eta^2) \leq 1 + \sqrt{1 - \rho^2}.$$

I don't know how to start as r.v.'s are not independent and since I can't use standard approach:

$$ P( \max (\xi^2, \eta^2) \leq x) = P( \xi^2 \leq x, \ \eta^2 \leq x ) \neq P( \xi^2 \leq x)P( \eta^2 \leq x).$$

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  • $\begingroup$ Have you showed the result when $\xi$ and $\eta$ are Gaussian? $\endgroup$ – Davide Giraudo Nov 19 '12 at 12:52
  • $\begingroup$ @DavideGiraudo: no I haven't. Actually this problem is from very beginning of my probability theory book even before continuous random variables but I think it holds for any distribution. $\endgroup$ – grozhd Nov 19 '12 at 13:05
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As $\max\{a,b\}=\frac 12(a+b+|a-b|)$, we just have to show that $$E|X^2-Y^2|\leqslant 2\sqrt{1-\rho^2}.$$ We have by Cauchy-Schwarz inequality that \begin{align} E|X^2-Y^2|&=E|X-Y|\cdot |X+Y|\\ &\leqslant\sqrt{E(X-Y)^2}\sqrt{E(X+Y)^2}\\ &=\sqrt{2-2\rho}\sqrt{2+2\rho}\\ &=2\sqrt{1-\rho^2}. \end{align}

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