Expectation of X divided by Y Say we have two continuous independent random variables $X$ and $Y$. I understand that 
$E[XY] = E[X]E[Y]$
but what about $E[X/Y]$. Can I say:
$E[Xg(Y)]$ where $g(x) = \dfrac{1}{x}$
$E[X]E[g(Y)]$ by linearity
Then solve for each? If not then how does one go about solving this type of problem?
 A: Let us analyze this through the lens of the product distribution. First let there be a random variable $Z = {1 \over Y}$. Since $Y$ is independent of $X$, $Z$ is also independent of $X$.
For two independent random variables, the expectation of their product is the product of their expectations. This can be proved through the Law of Total Expectation. 
$$\mathbb E(XZ) = \mathbb E_Z\mathbb E_{XZ | Z}(XZ | Z))$$
$$\text{and given } Z \text{ is a constant in } XZ \text{ of inner expression...}$$
$$=\mathbb E_Z(Z \cdot \mathbb E_{X | Z}(X))$$
$$\text{and due to independence...}$$
$$=\mathbb E_X(X) \cdot \mathbb E_Z(Z)$$
To come back to your original problem...
$$\mathbb E\left({X \over Y}\right)$$
$$= \mathbb E\left(X \cdot {1 \over Y}\right)$$
$$=\mathbb E(X \cdot Z)$$
$$=\mathbb E(X) \cdot \mathbb E(Z)$$
$$=\mathbb E(X) \cdot \mathbb E\left({1 \over Y}\right)$$
However, it is very rarely the case that $\mathbb E\left({1 \over Y}\right) = {1 \over \mathbb E(Y)}$
Jensen's inequality states that if $U$ is a random variable and $\varphi$ is a convex function, then $\varphi(\mathbb {E}(U)) \leq \mathbb E(\varphi(U))$. 
If $U$ is strictly positive, then $1 \over U$ is convex. Then ${1 \over \mathbb {E}(U)} = \mathbb E\left({1 \over U}\right)$ only if $U$ has zero variance.
