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If I have two zero mean random variables say $x$ and $y$, and there is a function $f$ such that $$\mathrm{Var}(x+f(x)) \approx \mathrm{Var}(x)$$ and $$\mathrm{Var}(y+f(y)) \approx \mathrm{Var}(y)\; .$$

If we constrain $f$ such that $E{f(x)}=E{f(y)}=0$ and $x > f(x)$, for all $x$, Can I say that

$E[(x+f(x))\cdot(y+f(y))] \approx E(xy) \; ?$

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    Dear user12268, I can't agree with you. Compose a function by $\mathbb{E}$ is like integrate a such function. In our case, X>f(X) "everywhere", so \mathbb{E}[X] > \mathbb{E}[f(X)].2011-06-18

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