Consider a random variable $X$, with $E(x)=\mu_x$ and $V(x)=\sigma^2_x$.
It seems intuitive to me that, for two functions $f(x)$ and $g(x)$, where $f(x) \times g(x)=x$, it must be true that
$$ E\left[f(x)g(x)\right]= E(x) = \mu_x $$
However, expanding this expected value leads to:
$$ E\left[f(x)g(x)\right] = E\left[f(x)\right]E\left[g(x)\right] +Cov\left[f(x),g(x)\right]$$
But I fail to see how the right-hand side in the above expression can be shown to be always equal to $\mu_x$, except in the trivial case where e.g. $f(x)=cx$ and $g(x)=\frac{1}{c}$.
What about a less trivial case but still "simple", like $f(x)=x^2$ and $g(x) = \frac{1}{x}$?