Density of a point

Question

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 $(X,Y)$

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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!

Answer 1

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}$.

Answer 2

Firstly, I would like to thank all those who answered. Ok, that's what I understood.

I draw the bisector to separate the triangle in two sections: one for $y<x$ and the other for $y>x$. Further, I use the mean of the triangle to separate the section where $x>1$ from the section $x<1$. Now I observe that:

• if $(y<x \cup x>1)\Rightarrow x-1<1-\frac{y}{3}$, in the sense that the length of the segment $x-1$ is, for each points of this area, smaller than the length of the segment $1-\frac{y}{3}$. By analogy,
• if $(y<x \cup x<1)\Rightarrow 1-x<1-\frac{y}{3}$.

It follows $|x-1|<1-\frac{y}{3}$, and like extremes of integration for $\int\int f(x,y)dxdy$ we have $0<y<3$ and $\frac{y}{3}<x<2-\frac{y}{3}$. Furthermore (probably?), we can apply the same reasoning in the area with $y>x$. However, I continue to not understand why:

1) this fact can help me to find the joint density $f(x,y)$

2) why $f(x,y)=\frac{1}{area(Triangle)}$, with $area(Triangle)=\frac{base\times heigth}{2}=\frac{2\times 3}{2}=\frac{1}{3}$.