Suppose x and y are uniformly jointly distributed over the trapezoid $(A)$ with vertices at $(-2,0)$, $(0,2)$, $(1,2)$ and $(1,0)$. What is the density of y given $x \leq 0$ or in other words what is $p(y|x \leq 0)$?
Here's the progress (?) I made,
$p_{XY}(x,y) = 1/4 \quad \forall x,y \in A$ $P(Y\leq y|X\leq x) = \dfrac{P(X\leq x,Y\leq y)}{P(X\leq x)}$
I am not sure if my approach is correct or not. Thanks for the help, in advance!
It really helps to draw a picture of the trapezoid and mark the region $\{(x,y):x\leq 0, y \leq t\}$. For t between 0 and 2 the latter is the trapezoid with vertices $(-2,0),(t-2,t),(0,t),(0,0)$. The area of a trapeziod can be calculated by splitting it into a triangle and a rectangle. The result is $P\{Y\leq t,X\leq 0\}=2t-\frac {t^{2}} 2$. Also $P\{X\leq 0\}=2$ and the ratio is $t-\frac {t^{2}} 4$. The density is $1-t/2$ for $0<t<2$.