I'm a little confused about how to approach the below question:
Q: Let $\,W=Y-X$ and determine the CDF of $W$.
I've read the solution, here, but this was not very clear and I was wanting an approach from calculus if possible (unless this is too messy).
I understand that $w=-1$ is the point that splits the region but am not sure how to set the integrals up - any help appreciated.
The question is to find the cdf of $W$ where $W=Y-X$ in order to do that we know that the joint density of $W$ is given by
$$ F_{W}(w) = \iint_{D} f_{X,Y}(x,y) dx dy \tag{1}$$
Where the joint density is split into two uniform densities
$$ f_{X,Y}(x,y) =\begin{align}\begin{cases} \frac{1}{2} & 0 \leq x \leq 1 , 0\leq y\leq 1 \\ \\ \frac{3}{2} & 1 \leq x \leq 2 , 0 \leq y \leq 1 \end{cases} \end{align} \tag{2}$$
It notes the following
$$ P(W \leq w) = P(Y-X \leq w ) =P(Y \leq X +w) \tag{3}$$
Which gives us this picture
While you could use calculus and I actually spent a while preparing an answer with it. There is no point. They are made of triangles it notes that in the answer.
Now instead if you actually look at the picture. I am pasting it
The area of a triangle is $A = \frac{1}{2}b \cdot h $ $$ I_{1} = f_{X,Y}(x,y) \frac{1}{2} \frac{(2+w)}{2} (2+w) \tag{4} $$
$$ I_{1} = \frac{3}{2} \frac{1}{2} \frac{(2+w)}{2} (2+w) \tag{5} $$ $$ I_{1} = \frac{3}{2} \frac{1}{2} \frac{(2+w)^{2}}{2} \tag{6} $$
The other region is $1+w$ which would be only $\frac{1}{2}$ for the joint density. It is the triangular region in the middle.
$$ I_{2} = f_{X,Y}(x,y) \frac{1}{2} (1+w)(1+w) \tag{7} $$ $$ I_{2} = \frac{1}{2} \frac{1}{2} (1+w)^{2} \tag{8} $$
Look at the top now. See the triangle. The base is $-w$ and the height is $\frac{-w}{2}$
$$ I_{3} = \frac{1}{2}\frac{-w}{2} \cdot -w = \frac{w^{2}}{4}\tag{9} $$
We want the region for $w \in (-1,0)$ so now the area of the whole right side is
$$ I_{4} = \frac{1}{2} 1 \cdot 1 = \frac{1}{2} \tag{10}$$
If we subtract off the triangle before we get this area
So we let
$$ I_{5} = f_{X,Y}(x,y)(I_{4} -I_{3}) \tag{11} $$ $$ I_{5} = \frac{3}{2}(\frac{1}{2} -\frac{w^{2}}{4}) \tag{12} $$
Note this a cumulative density. We just added the region for $I_{1}$ now let's add $I_{2}$ call it $I_{6}$ and make our $F_{W}(w)$
$$ I_{6} = I_{2} + I_{5} \tag{13}$$
$$ I_{6} = \frac{1}{2} \frac{1}{2} (1+w)^{2} + \frac{3}{2}(\frac{1}{2} -\frac{w^{2}}{4}) \tag{14}$$
$$ f_{W}(w) =\begin{align}\begin{cases} 0 & \textrm{ if } w < -2 \\ \\ \frac{3}{2} \frac{1}{2} \frac{(2+w)^{2}}{2} & \textrm{ if } -2 \leq w \leq -1 \\ \\ \frac{1}{2} \frac{1}{2} (1+w)^{2} + \frac{3}{2}(\frac{1}{2} -\frac{w^{2}}{4}) & \textrm{ if } -1 < w \leq 0 \\ \\ 1 & \textrm{ if } w > 0 \end{cases} \end{align} \tag{15}$$
fixing this like they did
$$ f_{W}(w) =\begin{align}\begin{cases} 0 & \textrm{ if } w < -2 \\ \\ \frac{3}{8} (2+w)^{2} & \textrm{ if } -2 \leq w \leq -1 \\ \\ \frac{1}{8}\big(-w^{2} +4w +8 \big) & \textrm{ if } -1 < w \leq 0 \\ \\ 1 & \textrm{ if } w > 0 \end{cases} \end{align} \tag{16}$$