Calculating marginal probability density when multivariate pdf's support is $0Suppose that multivariate pdf $f(x,y)$'s support is in $0<y<2$ and $y<x<3$. I now want to calculate marginal probability density function $f_X(x)$ and $f_Y(y)$. But arranging terms only get me to $0<y<x<2$.
How do I apply calculus here? 
 A: Let $f_{X,Y}(x,y)$ be the joint probability density function.
The cumulative density function for $Y$ is:
$$\begin{align}
F_Y(c)&=\int\int_\mathbb Rf_{X,Y}(x,y)dxdy\\
&=\int_{-\infty}^\infty\left(\int_{-\infty}^\infty f_{X,Y}(x,y)dx\right)dy\\
&=\int_{0}^c\left(\int_{y}^3 f_{X,Y}(x,y)dx\right)dy\\
\end{align}$$
And the probability density function fo $Y$ is:
$$f_Y(y)=F'(y)=\int_{y}^3 f_{X,Y}(x,y)dx$$
Similarly for $f_X(x)$.
A: The first thing to do to solve this kind of question is to vizualize the domain of $(X,Y)$. Here this is the interior of the polygon with vertices $(0,0)$, $(3,0)$, $(3,2)$ and $(2,2)$:

$\qquad\qquad\qquad\qquad$

Then intersecting the domain with lines $y=$constant or $x=$constant allows to guess the formulas.
The density $f_Y$ is zero except on $(0,2)$. For every $y$ in $(0,2)$, $$f_Y(y)=\int_y^3f(x,y)\mathrm dx.$$ The density $f_X$ is zero except on $(0,3)$. For every $x$ in $(0,2)$, $$f_X(x)=\int_0^yf(x,y)\mathrm dy.$$ Finally, for every $x$ in $(2,3)$, $$f_X(x)=\int_0^2f(x,y)\mathrm dy.$$
