# 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.

• Thanks for asking. I was wondering this myself! – amcalde Dec 10 '14 at 23:13
• Implicit in Graham Kemp's answer is that it doesn't matter whether $X$ and $Y$ are dependent. (You speak of them being "conditionally dependent" without specifying the event with respect to which they are conditionally dependent.) – joriki Jul 1 '18 at 7:11

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}