Find distribution of random vector

Question

Probability density function of random vector $(X,Y)$ is given by formula $f(x,y)=2\cdot\mathbb{1}_D(x,y)$, where $D=\left\{(x,y)\in\mathbb{R}^2:x\in[0,1],y\in[0,1],y<-x+1\right\}$. Find distribution of vector $(X,Y)$.

Overall this is pretty simple exercise, but I am having a problem with specifying proper intervals.

My attempt:

$\begin{cases} 0 &\text{ for } x<0 \vee y<0 \\ \star &\text{ for } x\in[0,1), y\in[0;-x+1] \\ \star\star &\text{ for } x\in[0,1],y\ge1\\\star\star\star &\text{ for } x\in[0;-y+1],y\in[0,1) \\ \star\star\star\star &\text{ for } x\ge1,y\in[0,1) \\ 1 &\text{ for } x\ge1,y\ge1\end{cases}$

I have a doubt about $\star$ and $\star\star\star$, or more precisely - aren't these two basically the same?

Answer 1

Examine the shape of the support.

$$D=\{(x,y)\in\Bbb R^2: 0\leq x, 0\leq y, x+y<1\}$$

$D$ is the isosceles right triangle, $\triangle(0,0)(0,1)(1,0)$, which has two sides on the $xy$-axii, the hypotenuse along the line $x+y=1$.

The regions you need to observe for the CDF are $$\begin{cases} \; 0 &:& (x< 0)~\vee~(y< 0) &\text{below or left from the axii} \\[1ex] \star\,,\,\star\star\star &:& (0\leq x < 1) \wedge (0\leq y< 1)\wedge (x+y<1) & \text{inside the triangle body} \\[1ex] &:&(0\leq x< 1)\wedge(0\leq y< 1)\wedge (1\leq x+y) & \text{the other half of the square} \\[1ex] \star\star&:& (0\leq x< 1)\wedge (1-x\leq y) & \text{above the square} \\[1ex] \star\star\star\star &:& (1-y\leq x)\wedge (0\leq y < 1) & \text{right of the square } \\[1ex] \;1 &:& (1\leq x)\wedge (1\leq y) & \text{above and right of the unit square}\end{cases}$$