Random vector with density on triangle and trapezoid I have a random vector $(X,Y)$ with the density $f_{XY}(x,y)=\tfrac{1}{5}$ on the trapezoid $T_1$ with vertex $(0,0),(0,1),(2,1),(3,0)$ and $f_{XY}(x,y)=\tfrac{3}{2}x$ on the triangle $T_2$ with vertex $(0,4),(1,4),(1,3)$.
I need to find:


*

*support and probability function for marginal $X$;

*support and probability function for $(Y|X=x), 0\le{x}\le1$;

*support and distribution for $Z=X+Y$


I started with support for $T_1$ and $T_2$:
$T_1$ ${(x,y):0\le{x}\le{-y+3}, 0\le{y}\le{1}}$
$T_2$ ${(x,y):0\le{x}\le{1}, 3\le{y}\le{-x+4}}$
Now I need to check that 
$\iint_{T_1}f_{XY}(x,y)\mathrm{d}x\mathrm{d}y + \iint_{T_2}f_{XY}(x,y)\mathrm{d}x\mathrm{d}y = 1$, so:
$\frac{1}{5}\int_0^{1}(\int_0^{-y+3}{}\mathrm{d}x)\mathrm{d}y + \frac{3}{2}\int_0^1(\int_3^{-x+4}\mathrm{d}y)x\mathrm{d}x = 1$
For $T_1$ I got $\frac{1}{2}$ and for $T_2$  $\frac{1}{4}$, but the sum is not 1. Why?
I need to find the error before proceed with other questions :-)

First edit and first attempt:


*

*support and probability function for marginal $X$:


$f_x(x)\begin{cases}\frac{3}{2}x^2+\frac{1}{5} & 0\leq{x}\leq{1}\\ \frac{1}{5} & 1\leq{x}\leq{2}\\ \frac{-x+3}{5} & 2\leq{x}\leq{3} \end{cases}$


*

*distribution function for marginal $X$:


$F_x(x)\begin{cases}\frac{x^3}{2}+\frac{x}{5} & 0\leq{x}\leq{1}\\ \frac{7}{10} + \frac{x}{5} & 1\leq{x}\leq{2}\\ \frac{7}{10} + \frac{1}{5} -\frac{x^2}{10} + 3x & 2\leq{x}\leq{3} \end{cases}$

*

*I think your marginal pdf for $X$ is right.

*I think your cdf for it is wrong, although the question doesn't require it. It should be:
$$ F_X(x) =
\begin{cases}
\int_{0}^{x} (3u^2/2+ 1/5)\;du &= \dfrac{x^3}{2} + \dfrac{x}{5},  & \text{if $0\leq x\leq 1$} \\
\\
F(1) + \int_{1}^{x} (1/5)\;du &= \dfrac{x}{5} + \dfrac{1}{2},  & \text{if $1\leq x\leq 2$} \\
\\
F(2) + \int_{2}^{x} (\frac{-u+3}{5})\;du &= \dfrac{-x^2}{10} + \dfrac{3x}{5} + \dfrac{1}{10},  & \text{if $2\leq x\leq 3.$} \\
\end{cases}$$
$$\\$$


*Conditional pdf of $Y$ given $X$ with $0\leq X\leq 1$:


In this range for $X$ we have two ranges for $Y$ to handle.
For $0\leq y\leq 1$:
\begin{align}
f_{Y|X}(y|x) &= \dfrac{f_{X,Y}(x,y)}{f_X(x)} = \dfrac{1/5}{1/5 + 3x^2/2} = \dfrac{2}{2 + 15x^2}.
\end{align}
For $4-x\leq y\leq 4$:
\begin{align}
f_{Y|X}(y|x) &= \dfrac{f_{X,Y}(x,y)}{f_X(x)} = \dfrac{3x/2}{1/5 + 3x^2/2} = \dfrac{15x}{2 + 15x^2}.
\end{align}
$$\\$$


*Support and pdf of $Z=X+Y$:


Assuming you've already drawn the regions above on the $xy-$plane, now imagine a line $x+y=c$ moving across. Values of $c$ are in the support of $Z$ if the line crosses a region of support for $(X,Y)$. So the support for $Z$ is $0\leq Z\leq 3$ and $4\leq Z\leq 5$.
It also shows the cases we should consider:
For $0\leq z\leq 1$:
$$P(Z=z) = \int_{x=0}^{z} P(X=x\cap Y=z-x)\;dx = \int_{0}^{z}\dfrac{1}{5}\;dx = \dfrac{z}{5}.$$
For $1\leq z\leq 3$:
$$P(Z=z) = \int_{x=z-1}^{z} P(X=x\cap Y=z-x)\;dx = \int_{z-1}^{z}\dfrac{1}{5}\;dx = \dfrac{1}{5}.$$
For $4\leq z\leq 5$:
$$P(Z=z) = \int_{x=z-4}^{1} P(X=x\cap Y=z-x)\;dx = \int_{z-4}^{1}\dfrac{3x}{2}\;dx = \dfrac{3}{4}\left(-z^2+8z-15\right).$$
