TITLE: Joint PDF of two random variables in a triangle
QUESTION [8 upvotes]: Let the random variables $X$ and $Y$ have a joint PDF which is uniform
  over the triangle with vertices at $(0, 0), (0, 1 )$ and $(1, 0)$.

Find the joint PDF of $X$ and $Y$.


So apparently the answer is $2$. And it's related with the area of this triangle that easily we can see is $1/2$. How do we get that answer? I guess I don't understand the meaning of joint PDF. Could someone explain that to me too?

REPLY [11 votes]: The random variable $(X,Y)$ is uniformly distributed over the region, lets call it $R$, i.e. the pdf 
$f_{X,Y}(x,y) = k$ for some constant on the region. Now lets integrate over the region
$$\int\int_R f_{X,Y}(x,y)dxdy = \int\int_R k \:dxdy$$
$$\int\int_R f_{X,Y}(x,y)dxdy = k * area(R)$$
$$\int\int_R f_{X,Y}(x,y)dxdy = k * \frac{1}{2}$$
We also know that 
$$\int\int_R f_{X,Y}(x,y)dxdy = 1$$
so
$$k * \frac{1}{2} = 1 \Rightarrow k = 2 $$
And we end up with this pdf:
$$f_{X,Y}(x,y) =
\begin{cases}
2,  & (x,y) \in R \\
0, & \text{otherwise}
\end{cases}
$$