Random vector $(X,Y)$ has uniform distribution at $D=\left\{(x,y)\in \mathbb{R}^2: |x|+|y|<1 \right\}$. Find conditional distribution of $Z=X+Y$ with the condition $\left\{X>0\right\}$.
My attempt:
$F_z(t)=\mathbb{P}\left( z\le t \ | \ x>0\right)=\mathbb{P}\left( x+y\le t \ | \ x>0\right)=\frac{\mathbb{P}\left( z\le t, \ x>0\right)}{\mathbb{P}\left( x>0\right)}= \begin{cases} 0, &\text{ for } t<-1 \\ \star &\text{ for } -1\le t<1 \\ 1, &\text{ for } t\ge 1 \end{cases}$
$\mathbb{P}\left( z\le t, \ x>0\right)=\int_0^{\frac{t+1}{2}} \int_{x-1}^{t-x} \text{d}y\text{d}x=\dots=\frac{\left( t+1\right)^2 }{4}$
$\mathbb{P}\left( x>0\right)=\frac{1}{2}$
So the answer seems to be
$\star= \frac{\frac{\left( t+1\right)^2 }{4}}{\frac{1}{2}}=\frac{\left( t+1\right)^2 }{2}$
But this is incorrect, since for $t=1^-$ this equals $2$. I must be doing some silly mistake here, caused by my lack of knowledge of the basic staff. Maybe I should multiply this by $\frac{1}{\text{area of rhombus}}$? I do not really understand what $|A|$ stands for in the formula for uniform distribution, $f(x,y)=\frac{1}{|A|}\cdot1{\hskip -2.5 pt}\hbox{l}_A(x,y)$
And the next task is to calculate the probability $\mathbb{P}\left( X+Y>0 \ | \ X>0\right)$
So I guess it is just $\mathbb{P}\left( X+Y>0 \ | \ X>0\right)=\frac{\mathbb{P}\left( X+Y>0, X>0\right)}{\mathbb{P}\left(X>0\right)}$
The area of $X+Y>0 \text{ and } X>0$ is $\frac{1}{2}+\frac{1}{4}=\frac{3}{4}$
So the answer would be
$\mathbb{P}\left( X+Y>0 \ | \ X>0\right)=\frac{\frac{\frac{3}{4}}{2}}{\frac{1}{2}}=\frac{3}{4}$
?