Can't understand a specific step in finding $\operatorname{Cov}(X,Y)$

Question

Given the following joint PDF: $$f_{X,Y}(t,s)=\begin{cases}\frac{1}{4}&,(t,s)\in D \\ 0 &,\text{ else }\end{cases}$$

where $D=\{(t,s):|t|+s\le 2\,,\,s\ge 0\}$. I need to find $\operatorname{Cov}(X,Y)$.

For doing that, I started by finding the marginal density functions, so I would be able to calculate $E[X],E[Y]$.

According to the solution:

\begin{equation} f_{Y}(s)= \begin{cases} \frac{1}{2}(2-s) & , 0\leq s \leq 1 \\ 0 & ,\text{ else } \end{cases} \end{equation}

I got to the same one, except the region. I have no clue why $0\leq s \leq 1 $ is the correct region? This is the plot:

Could anyone please explain me how this region was picked?

Answer 1

$f_Y(s)=\int_{-(2-s)}^{2-s} \frac 1 4 \, dt= \frac 12 (2-s)$ for $0\leq s \leq 2$.

Answer 2

The marginal density of $Y$ is given by

\begin{equation} f_{Y}(s)= \begin{cases} \frac{1}{2}(2-s) & 0\leq s \leq 2 \\ 0 & \text {elsewhere} \end{cases} \end{equation}

Observe that if instead

\begin{equation} f_{Y}(s)= \begin{cases} \frac{1}{2}(2-s) & 0\leq s \leq 1 \\ 0 & \text {elsewhere} \end{cases} \end{equation} then $$\int_{0}^{1} f_Y (s)\ \text {ds} = \frac 3 4 \neq 1.$$

and you may find that $\operatorname {Cov} (X,Y) = 0$ i.e. $X$ and $Y$ are uncorrelated.