I have a triangle bounded by $0 \leq x, y \leq 1$ and $x + y \geq 1.5$. I'm told that points are uniformly distributed within this triangle.
I am wondering how I can find the pdf? Is it simply solving for the following?
$$ \int_{0.5}^1 \int_{0.5}^1 f(x,y) dx dy = 1 $$
$f(x,y) = 4$ satisfies this equation, but is it unique?
By definition uniform density in a region has a constant value equal to the reciprocal Of the area of the region (and $0$ outside the region) . In this case the region is the triangle formed by the points $(1,1), (1,0.5)$ and $(0.5,1)$. So the density is $\frac 1 A$ for all points in this triangle where $A$ is the area. I will let you calculate the area.
Note that the limits for the integral are not the ones you have written. You should get $\int_{0.5}^{1}\int_{1.5-y}^{1} f(x,y) dxdy$.