Joint probability distribution of sum and product of two random variables Let $X$ and $Y$ be two discrete random variables. I know the joint probability distribution of the vector $(X,Y)$, namely $P(X = x, Y = y)$ for all $x$ and $y$ in the sample spaces $\Omega_X$ and $\Omega_Y$, respectively. Using the joint distribution, I discovered that $X$ and $Y$ are conditionally dependent. 
Now let $U := X + Y$ and $V := X \cdot Y$ be two random variables. I need to calculate the distribution of both $U$ and $V$. Is it correct to calculate the distribution of $U$ as
\begin{align}
  P(U = u) = \sum_{x \in \Omega_X, y \in \Omega_Y \text{ such that } x \cdot y = u} P(X = x, Y = y),
\end{align}
even though $X$ and $Y$ are conditionally dependent?
Furthermore I need to calculate the joint distribution of the probabilty vector $(U,V)$. I know that the distribution can be calculated as
\begin{align}
  P(U = u, V = v) = P(U = u) \cdot P(V = v),
\end{align}
if $U$ and $V$ are conditionally independent. However, to show that $U$ and $V$ are conditionally independent, I would look at the joint distribution, which I have to calculate. How does one calculate the joint distribution of $(U,V)$, if the variables are conditionally dependent? Furthermore, is there any way to find out whether $U$ and $V$ are conditionally dependent or not, without using the joint probability distribution? Since $U$ and $V$ are defined as sum and product of two random variables (which are conditionally dependent), I have the feeling that $U$ and $V$ are conditionally dependent.
 A: Yes, if you know the joint probability distribution of $X,Y$ then the distributions of $U$ and $V$ are, respectively:
$$\begin{align}
\mathsf P(U=u) & = \sum_\overbrace{x\in\Omega_X, u-x\in\Omega_Y} \mathsf P(X=x, Y=u-x)
\\
\mathsf P(V=v) & = \sum_\overbrace{x\in\Omega_X, v/x\in\Omega_Y} \mathsf P(X=x, Y=v/x)
\end{align}$$

The joint distribution on $U,V$, would similarly be $$\mathsf P(U=u, V=v) = \sum_\overbrace{x\in\Omega_X, u-x\in\Omega_Y, v/x\in\Omega_Y} \mathsf P(X=x, Y=u-x, Y=v/x)$$
However, for any given pair of $U=u,V=v$ there is in fact only two corresponding pair of $X=\Box,Y=\Box$.
$\begin{align}
(U=X+Y) \wedge (V= XY) & \iff  
\left(X= \dfrac{U\pm\sqrt{U^2-4V}}{2}\right)
\wedge \left(Y=\dfrac{U\mp\sqrt{U^2-4V}}{2}\right)
\\[4ex]
\therefore \mathsf P(U=u, V=v)
 & = \mathsf P\left(X= \dfrac{u\pm\sqrt{u^2-4v}}{2}, Y=\dfrac{u\mp\sqrt{u^2-4v}}{2}\right) 
\\[1ex]
 & = {\mathsf P\left(X= \tfrac{u+\sqrt{u^2-4v}}{2}, Y=\tfrac{u-\sqrt{u^2-4v}}{2}\right)
 + \mathsf P\left(X= \tfrac{u-\sqrt{u^2-4v}}{2}, Y=\tfrac{u+\sqrt{u^2-4v}}{2}\right)}
\end{align}$
