random variablesstochastic-processes
Given two random uniform variables, $U$ and $V$, that are uniformly distributed over [0,1], how do you calculate the joint pdf of $X$, $Y$ where $X = F(U,V)$ and $Y = G(U,V)$ and where is the joint pdf zero.
Specifically, where $$X = + \sqrt(U) \\ Y = UV$$
I understand this is similar to Joint PDF of two random variables and their sum but I am not interested in the marginal distribution and looking for more of an explanation.
I will try to address the question you posed in the comments, namely:
Given 3 independent random variables $U$, $V$ and $W$ uniformly distributed on $(0,1)$, find the joint probability distribution function of $X=U+V$ and $Y=U+W$.
Gives $0<x<2$ and $0<y<2$, we first compute $F_{X,Y}(x,y)$:
$$\begin{eqnarray}
F_{X,Y}(x,y) &=& \mathbb{P}\left(X \leqslant x, Y \leqslant y\right) = \mathbb{P}\left(U+V \leqslant x, U+W \leqslant y\right) = \mathbb{E}\left(\mathbb{P}\left(U+V \leqslant x, U+W \leqslant y|U\right)\right)\\
&=& \mathbb{E}\left(F_V(x-U)F_W(y-U)\right) = \mathbb{E}\left(\min(1, \max(x-U,0))\min(1, \max(y-U,0))\right)
\end{eqnarray}
$$
The pdf, being a derivative of the cdf, will then read:
$$
f_{X,Y}(x,y) = \frac{\partial}{\partial x}
\frac{\partial}{\partial y} F_{X,Y}(x,y) = \mathbb{E}\left( [1+U>x>U] [1+U > y>U] \right) = \mathbb{P}\left(U < x < 1+U \land U < y<1+U\right)
$$
The evaluation of the latter expectation is straightforward but tedious, so I asked Mathematica for help:
No. If one of the variables is discrete and the other continuous, they can't have a common density [neither with respect to the Lebesgue-measure, nor the counting measure].
Best Answer
For any $x, y$ in $[0, 1]$, \begin{align} \Pr(X \le x, Y \le y) & = \Pr(\sqrt U \le x, UV \le y) \\ & = \Pr\left(\sqrt U \le x, U \le \frac yV\right) \\ & = \int_0^1 \Pr\left(U \le x^2, U \le \frac yV \mid V = v\right) dv \\ & = \int_0^1 \min\left\{x^2, \frac yv\right\} dv \\ & = \begin{cases} x^2 & ; x^2 \le y \\ \int_0^{y/x^2} x^2 dv + \int_{y/x^2}^1 \frac yv dv & ; x^2 \ge y \\ \end{cases}\\ & = \begin{cases} x^2 & ; x^2 \le y \\ y - y \log y + 2y\log x & ; x^2 \ge y \\ \end{cases} \end{align} To obtain the probability density function, you take the derivative with respect to $x$ and $y$: $$ f_{X,Y}(x,y) = \begin{cases} 0 & ; x^2 \le y \\ 2/x & ; x^2 \ge y \\ \end{cases} $$