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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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If $X$ is a zero mean random variable and $X > f(X)$, why we still have $\mathbb{E}[f(X)] = 0$? (It is supposed to be a comment, but I do not have enough reputation.) –  Hsueh-Yung Lin Jun 17 '11 at 17:05
    
Unless you provide the precise and mathematical meaning of the symbol $\approx$, it seems difficult to understand your question, let alone to answer it. –  Did Jun 17 '11 at 18:36
    
@Hsueh-Yung Lin Even if $x$ is a zero mean random variable, and $x>f(x)$, both can be zero mean. Consider f() as a ladder function (generally used in quantization of a scalar variable). –  user12268 Jun 18 '11 at 11:36
    
No. $ $ $ $ $ $ –  Did Jun 18 '11 at 11:55
    
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 $\mathbb{E}[X] > \mathbb{E}[f(X)]$. –  Hsueh-Yung Lin Jun 18 '11 at 16:56

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