Finding an unknown joint PDF?

Question

Knowing that $X$ and $Y$ are random variables, $f_{X,Y}(x,y)$ is their joint PDF, and \begin{align} f_{X,Y}(x,y)=k, \qquad(0<x<2; 0<y<x) \end{align} How can I compute $k$?

Answer 1

\begin{align} & \int_{x = 0}^{2}\int_{y = 0}^{x}f_{X, Y}(x, y)\, dy\, dx = 1 \\ \implies & \int_{x = 0}^{2}\int_{y = 0}^{x} k \, dy\, dx = 1 \\ \implies & k \int_{x = 0}^{2} x \, dx = 1 \\ \implies & k \dfrac{x^{2}}{2}\Bigg\vert_{0}^{2} = 1 \\ \implies & 2k = 1 \\ \implies & k = 0.5 \end{align}

Answer 2

We know that $$\int_0^2\int_{0}^{x} f_{X,Y}(x,y) \ dy \ dx =1$$ since this is a density.

So $$\int_0^2\int_{0}^{x} k \ dy \ dx =1$$

$$\int_0^2\big[ky\big]_0^x \ dx =1$$

$$\int_0^2kx \ dx =1$$

$$\left[\frac12kx^2\right]_0^2=1$$ $$2k=1$$ $$k=\frac12$$

Answer 3

Geometrically, $0\lt x \lt 2, 0\lt y \lt x$ is a triangle with vertices in $O(0,0), A(2,0), B(2,2)$ and its area is $2$. Thus, the integral of the (constant) PDF over that triangle is $k\cdot \operatorname{Area}(\triangle OAB)$, i.e. $1=2k$, which gives $k=\frac{1}{2}$.