Let $(X,Y)$ be the coordinates of a point uniformly chosen from a
quadrilateral with vertices $(0,0)$, $(1,0)$, $(1,1)$, $(0,2)$. Find
the marginal probability density functions of $X$ and $Y$.
Try
Well, these points are vertices of a quadriteral whose area is $\frac{2}{3}$. Since it is uniformly chosen the points, then
$$ f_{X,Y}(x,y) = \frac{1}{\text{Area(S)}} = \frac{1}{2/3} = \frac{3}{2} $$
Where $(x,y) \in S $, the quadrilateral. Hence,
$$ f_X(x) = \int\limits_0^2 \frac{2}{3} dy = \frac{4}{3} $$
and
$$ f_Y(y) = \int\limits_0^1 \frac{2}{3} = \frac{2}{3} $$
Is this correct?
Best Answer
This is the region:
The area should be $\frac32$. For $x \in (0,1)$, $$f_X(x) = \int_0^{2-x} \frac23 \, dy$$ For $y \in (0,2)$, $$f_Y(y) = \int_0^{\min(1,2-y)} \frac23 \, dx$$