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I don't really know how to start proving this question.

Let $\xi$ and $\eta$ be random variables with $E(\xi)=E(\eta)=0$, $V(\xi)=V(\eta)=1$, and correlation coefficient $r$. Show that $$ E(\max(\xi^2,\eta^2))\leq1+\sqrt{1-r^2}. $$ Does anyone here have any idea for starting this question please?

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Any luck with the answer below? – Did May 13 '12 at 18:53

Here is an answer to the first version of the post, now modified: $$\max(\xi^2,\eta^2)\leqslant\xi^2+\eta^2\implies\mathrm E(\max(\xi^2,\eta^2))\leqslant\mathrm E(\xi^2)+\mathrm E(\eta^2)=2\leqslant1+\sqrt{1+r^2}$$

Here are some detailed hints leading to a solution of the revised version.

First, $\max(\xi^2,\eta^2)=\frac12(\xi^2+\eta^2)+\frac12|\eta^2-\xi^2|$ hence, using the notation $s=\sqrt{1-r^2}$, one sees that the desired inequality is equivalent to $\mathrm E|\eta^2-\xi^2|\leqslant2s$. Second, there exists a random variable $\zeta$ such that $\mathrm E(\zeta)=\mathrm E(\zeta\xi)=0$, $\mathrm E(\zeta^2)=1$ and $\eta=r\xi+s\zeta$. Hence, one wants to show that $\mathrm E|\kappa|\leqslant2$, where $$ \kappa=s\zeta^2+2r\xi\zeta-s\xi^2. $$ If, by chance, the pointwise inequality
$$ -\xi^2-\zeta^2\leqslant\kappa\leqslant\xi^2+\zeta^2\tag{$\ast$} $$ holds, the proof is over. Now, remember that $s=\sqrt{1-r^2}$ and show that $(\ast)$ holds.

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Makes sense, but how did you come up with it? – Andrew MacFie Mar 24 '15 at 16:19

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