Joint distribution between sum and a component of its sum Let $X$ and $N$ be independent uniform random variables on $[0,1]$. Define,
\begin{equation}
Y=X+N
\end{equation}
I am interested in computing the joint distribution $P_{XY}$.
I have the following tried from my side. 
\begin{equation}
P_{XY}(x,y)=P_{X}P_{Y|X}(x,y)=P_{Y|X}(x,y)=P_{N}(y-x)
\end{equation}
Then, $P_{XY}(x,y)=1$ for $(x,y)\in S:=\{ (x,y): y\ge x, y \le 1+x \}$, defines the distribution. But I see that, 
\begin{equation}
\iint_S P_{XY}= \frac{1}{2} \neq 1 
\end{equation}
Where am I wrong?
Thank you
 A: Assume now that $x\in [0,1]$ and $y\in [0,2]$.
$$P(X\le x,Y \le y) = P(X\le x,X+N \le y)=\int_0^1 P(X\le x,X\le y -n\mid N=n) \, dn.$$
By independence, the conditional probability on RHS is equal to
$$P(X\le x,X \le y-n) = P(X\le \min (x,y-n)) = \max(\min (x,y-n),0).$$
Therefore
$$P(X\le x,Y \le y) = \int_0^1 \max (\min (x,y-n),0)) d n.$$
Then $\min (x,y-n)>0$ if and only if $y-n>0$, that is if and only if $n<y$. Therefore we have
$$(*)\quad P(X\le x,Y\le y) = \int_0^y \min(x,y-n) \, dn.$$
If $y<x$, the integral is $\int_0^y (y-n) \, dn = y^2/2$, while if $y>x$, then the integral is equal to
$$\int_0^{y-x} x \, dn +\int_{y-x}^y (y-n) \, dn=(y-x)x + x^2/2=x (y-x/2).$$
Therefore joint CDF is
$$ P(X\le x,Y\le y) = \min(x,y)(y- \min(x,y)/2).$$
Edit (08/24/20).  There's an error in $(*)$. It should be
$$ P(X\le x,Y\le y) = \int_0^{\min(1,y)} \min(x,y-n) dn.$$
A: Quick question...
Why can't we just say that $Y|X=x\sim U[x,x+1]$ for $x\in[0,1]$ fixed and immediately conclude $$f_{XY}(x,y)=f_{Y|X=x}(y|x)f_X(x)=1$$ whenever $0\leq x\leq y\leq x+1 \leq 2$ and $f_{XY}(x,y)=0$ elsewhere, just as was suggested in the original post?
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{1}\int_{0}^{1}\bracks{y = x + n}\dd x\,\dd n} =
\int_{0}^{1}\int_{0}^{1}\bracks{x = y - n}\dd x\,\dd n
\\[5mm] = &\
\int_{0}^{1}\bracks{0 \leq y - n \leq 1}\dd n =
\int_{0}^{1}\bracks{y - 1 \leq n \leq y}\dd n
\\[5mm] = &\
\bracks{0 \leq y \leq 1}\int_{0}^{y}\dd n +
\bracks{0 < y - 1 \leq 1}\int_{y - 1}^{1}\dd n
\\[5mm] = &\
\bbx{\bracks{0 \leq y \leq 1}\color{red}{y} +
\bracks{1 < y \leq 2}\color{red}{\pars{2 - y}}} \\ &\
\end{align}

