What is the expectation of one random variable divided over another (both independent)? Suppose I have $X,Y$, which are independent random variables. 
Why is it that $E(\frac{X}{Y}) = E(X)E(\frac{1}{Y})$? 
Also, why is it that $E(X^2Y^2)=E(X^2)E(Y^2)$? How is it that the square of an independent random variable is also independent in relation to $Y$ or $Y^2$? Thanks! 
 A: Basically, if $X$ and $Y$ are independent, then also $f(X)$ and $g(Y)$ are independent if $f$ and $g$ are measurable functions:
$$\eqalign{
P(f(X) \in A,\ g(Y) \in B) &= P\left(X \in f^{-1}(A),\ Y \in g^{-1}(B)\right) \\
& =  P\left(X \in f^{-1}(A)\right) \ P\left(Y \in g^{-1}(B)\right) \\
& = P\left(f(X) \in A\right) \ P\left(g(Y) \in B\right).
}$$
In particular all continuous functions (like the $f(x)=1/x$ and $f(x)=x^2$ in your examples) are Borel-measurable, and hence also $X$ and $1/Y$ as well as $X^2$ and $Y^2$ are independent.
A: let $Z=\frac{1}{Y}$, then we have:
\begin{equation}
E(XZ) = \int \int XZ p(X,Z) \mathrm{d}x \mathrm{d}z
\end{equation}
but $X$ and $Z$ are independent, so $p(X,Z) = p(X)p(Z)$ we have
\begin{equation}
E(XZ) = \int \int x .z . p(X=x).p(Z=z) \mathrm{d}x \mathrm{d}z
\end{equation}
which can be arranged as:
\begin{equation}
E(XZ) = \int x . p(X=x) \mathrm{d}x \int z.p(Z=z) \mathrm{d}z = E(X)E(Z)
\end{equation}
using the same argument you can show that $E(X^2Y^2) = E(X^2).E(Y^2)$
does that help?
