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If $x$ and $y$ are independent standardized random variables with zero mean and unit variance, what are the upper and lower bounds of $$E[\max(x,y)]$$

Can I get a useful reference to this?

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1 Answer 1

Note that $2\mathbb E(\max(x,y))=\mathbb E(|x-y|)$. Since $\mathbb E(|x-y|)^2\leqslant\mathbb E((x-y)^2)=2$, one knows that $\mathbb E(\max(x,y))\leqslant\sqrt{2}/2$. Standard normal distributions yield $\mathbb E(\max(x,y))=1/\sqrt{\pi}$.

On the other hand, if $x$ and $y$ are i.i.d. with $\mathbb P(x=n)=\mathbb P(x=-n)=1/(2n^2)$ and $\mathbb P(x=0)=1-1/n^2$, then $\mathbb E(\max(x,y))=1/n$, hence no universal positive lower bound for $\mathbb E(\max(x,y))$ is valid for every distributions of $x$ and $y$.

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What about negative lower bounds? –  Eckhard Jan 11 '13 at 14:24
    
@Eckhard Sorry? $\max(x,y)\geqslant x$ hence $E(\max(x,y))\geqslant E(x)=0$. –  Did Jan 11 '13 at 17:23
    
Obviously. Sorry for the stupid question. –  Eckhard Jan 11 '13 at 17:39

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