You choose a random point inside the triangle with vertices $A=(0,0),B=(1,3),C=(2,0)$.
Let $(X,Y)$ be the random double variable that indicates the point chosen.
Find the density of $(X,Y)$
Ok, we have a an isosceles triangle inside the rectangle defined by variables $X\sim U(0,2)\Rightarrow f_X(x)=\frac{1}{2}\mathbb{I}_{[0<x<2]}(x)$ and $Y\sim U(0,3)\Rightarrow f_Y(y)=\frac{1}{3}\mathbb{I}_{[0<y<3]}(y)$.
Would somebody please explain me why the solution is:
$$(X,Y) \sim U(T)$$
$$f_{X,Y}(x,y)=\dfrac{1}{area(T)}=\dfrac13$$
Thanks in advance for any help!
Best Answer
The triangle is defined by $T = \left\{(x,y): y \in [0,3], \; |x -1| \leq 1-\dfrac{y}{3} \right\}$, and has area 3. So the pdf is $ f(x,y) = \dfrac{1}{3} \mathbf{1}_{(x,y) \in T}$.