How do i determining boundaries of Probability Density Function integral

Question

Question: We draw a random vector $(X, Y )$ uniformly from the diamond $(-1,0)- (0,1)-(1,0)-(0,-1)$ without its lower-right corner.(see figure)

Determine the joint pdf of X and Y

I know that

$f_{X,Y}(x,y) = 2/3$

But I don't know the boundaries of $x$ and $y$ when $f_{X,Y} = 2/3$

How do I determine the boundaries in this case? Thank you.

Answer 1

Indeed $f_{X,Y}(x,y)=\begin{cases}2/3&:& (x,y)\in \mathcal S\\0 &:& (x,y)\notin \mathcal S\end{cases}$

The support you seek, $\mathcal S$, is the union of two triangles.

When $X$ is negative, what are the bounds for $Y$?   When $X$ is positive, what are the bounds for $Y$?

$${\mathcal S}={{\{(x,y): (-1\leq x\leq 0) \wedge (\Box\leq y\leq \Box)\}}\cup{\{(x,y):(0<x\leq 1)\wedge(\Box\leq y\leq\Box)\}}}$$

Fill in the boxes with the appropriate variables, then verify that $$\int_{-1}^1\int_{\Box}^{\Box}\frac 23~\mathrm d y~\mathrm d x+\int_0^1\int_\Box^\Box \frac 23~\mathrm d y~\mathrm d x =1$$