Let $$\displaystyle{ f_{X,Y}(x,y)=\left\{ \begin{array}{cc} \frac18 & |x|+|y|\leq 2 \\ 0 & |x|+|y|>2 \end{array}\right. }$$ be density function of continuous random variable $X,Y$. for every $|y|<2$ find $f_{X|Y}(x|y)$
So my idea here was to find $f_y$ and than use the formula of $$f_{X|Y}(x|y)=\frac{f_{X,Y}(x,y)}{f_Y(y)}$$
so the because $|x| \leq 2-|y|$ I can say that $-2 \leq x \leq 2$ and than $$f_Y = \int_{-2}^{2}f_{X,Y}dx$$.
Why is this wrong?
Because $(X,Y)$ support is a square with vertices $(0,2);(2,0);(0,-2);(-2,0)$
thus integrating you get
$$f_Y(y)=\frac{1}{8}\int_{-2-y}^{y+2}dx=\frac{y+2}{4}$$
for $-2<y\leq 0$
and
$$f_Y(y)=\frac{1}{8}\int_{y-2}^{2-y}dx=\frac{2-y}{4}$$
for $0<y<2$
Concluding:
$$f_Y(y)=\frac{y+2}{4}\cdot\mathbb{1}_{(-2;0)}(y)+\frac{2-y}{4}\cdot\mathbb{1}_{(0;2)}(y)$$
... a triangle distribution