Expected value of the product of functions of two independent random variables If $X$ and $Y$ are independent random variables, are the statements below true
$$E(e^{X+Y} ) = E(e^X)\times E(e^Y)$$
and 
$$E(X^2\times Y^2) = E(X^2)\times E(Y^2),$$
where $E(\cdot)$ = expectation?
 A: Yes, both statements are correct.
In general if $X$ and $Y$ are independent:
$$E \left[ u(X)v(Y) \right]=E \left[u(X)\right]\times E\left[ v(Y)\right]$$
provided the expectations exist.
A: I'll just consider the case of continuous random variables with densities, which should give you enough to go on for the general case. If $X$ and $Y$ are independent random variables with densities $f_1$ and $f_2$, then their joint density can be written as a product:
$f_{X,Y}(x,y) = f_1(x)\cdot f_2(y)$
In that case,
$E[X^2\cdot Y^2] = \int\int x^2y^2f_{X,Y}(x,y)\hspace{1pt}dx\hspace{1pt}dy = \int x^2y^2f_1(x)f_2(y)\hspace{1pt}dx\hspace{1pt}dy$.
Do you see where to go from here?
A: Yes, because $E(PQ)=E(P)E(Q)$ when the random variables $P$ and $Q$ are independent. In each case you can simply define new random variables that are functions of the first.


*

*$P=e^X, Q=e^Y \implies PQ=e^X e^Y= e^{X+Y}$

*$P=X^2, Q=Y^2$

If $X$ and $Y$ are independent , how do you knew that $X^2$ and $Y^2$ are independent?

This can be looked at in two ways.
First of all, and this is what I was relying on above, you can appeal to our everyday understanding of how the world works. E.g.  let $X$ and $Y$ be the results of rolling two dice. $X$ and $Y$ are independent. Now, let's say we square each result. We clearly haven't introduced any dependency by doing this, so $P=X^2$ and $Q=Y^2$ are independent.
Secondly, you can look more deeply at the underlying mathematics. The general result is JohnK's answer and a specific instance of that is justified in korrok's answer.
Expectation of two random variables $X$, $Y$ is defined as the sum of the products of the values of those random variables times their joint probabilities. For continuous random variables this is
$$\mathrm{E}(XY)=\int\int xy \; f_{XY}(x,y) \;\mathrm{d}x\mathrm{d}y $$
where the integrals are over the range that $X$ and $Y$ can take, and $f_{XY}$ is the joint probability density of $X$ and $Y$.
In the more general case of $\alpha(x)$ as some function of $x$ and $\beta(y)$ as some function of $y$ (you had $\alpha = \beta = x\mapsto x^2$), the expectation of their product is defined similarly.
$$\mathrm{E}(\alpha(X)\beta(Y))=\int\int \alpha(x)\beta(y) \; f_{XY}(x,y) \;\mathrm{d}x\mathrm{d}y $$
Because $X$ and $Y$ are independent, you can factorize the probability density  $f_{XY}$ into the product of the probability density $f_X(x)$ for $X$ and the probability density $f_Y(y)$ for $Y$, i.e.: $f_{XY} = f_X(x)f_Y(y)$. So
$$\mathrm{E}(\alpha(X)\beta(Y))=\int\int \alpha(x)\beta(y) \; f_X(x)f_Y(y)\;\mathrm{d}x\mathrm{d}y $$
Rearranging the integrand we see that the integrand is the product of terms that only depend on $x$ and terms that only depend on $y$ so the integral itself can be split into two. Each of those two integrals is the definition of an expectation.
$$\begin{align}
\mathrm{E}(\alpha(X)\beta(Y))
&= \int\int \alpha(x)f_X(x) \; \beta(y)f_Y(y)\;\mathrm{d}x\mathrm{d}y \\
&= \int \alpha(x)f_X(x) \;\mathrm{d}x \int\beta(y)f_Y(y)\;\mathrm{d}y \\
&= \mathrm{E}( \alpha(X))\mathrm{E}(\beta(Y))
\end{align}$$
