If $X$ and $Y$ are independent, is it true that $$ E[\frac{X}{Y}] = E[X\cdot\frac{1}{Y}]=\text{cov}(X,\frac{1}{Y})+E[X]E[\frac{1}{Y}]$$ and therefore $$E[\frac{X}{Y}] =E[X]E[\frac{1}{Y}].$$ Any intuitive explanation of why the covariance is always zero?
What about $E[\frac{XY}{ZZ}]$ when $X$, $Y$, $Z$ are independent?
If $X$ and $Y$ are independent, the so are $X$ and $Z = f(Y) = \frac{1}{Y}$. Therefore, the $cov(X, \frac{1}{Y}) = cov(X,Z)$ is zero due to the independence between $X$ and $Z$. Also, recall that independence implies uncorrelatedness,