probabilityrandom variables
If two random variables are uniformly distributed over a region, how do you in general find the joint PDF of those random variables?
For example, if $(X,Y)$ is distributed uniformly over the region $-2\leq x\leq 2$, $0\leq y\leq 1-x^2$, how could you derive the joint density function of $X$ and $Y$?
I believe you have to integrate, but with what integrand? Thank you.
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:
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 $y$: $$ f_{X,Y}(x,y) = \begin{cases} 0 & ; x^2 \le y \\ 2/x & ; x^2 \ge y \\ \end{cases} $$
Best Answer
the easiest way is to calculate the Area of the domain region and, the joint pdf is its reciprocal
In the example you posted, the domain area is the following
$$\int_{-1}^{1}[1-x^2]dx=4/3$$
thus
$$f_{XY}(x,y)=\frac{3}{4}\mathbb{1}_{[-1;1]}(x)\cdot\mathbb{1}_{[0;1-x^2]}(y)$$
you can set also $x \in \mathbb{R}$ but thea area does not changes due to the fact that $y \ge 0$
observing the drawing of the joint domain, you get that
$$C\int_{-1}^{1}\left[ \int_0^{1-x^2}dy \right]dx=1$$
that is $C=3/4$
Now observe that the above double integral is, geometrically, the volume of a solid figure with base the green region and constant height C that is your uniform density.
thus
$$V=1=C\cdot A$$
that means
$$C=\frac{1}{A}$$