Ordered statistics of sum of uniform variates Let $X,Y,Z \sim \mathcal{U}[0,1]$. What will be the distribution of $W = \min(X+Y,Y+Z)$? 
I think that first of all I should find out the distributions followed by $X+Y$ and $Y+Z$, and then find out the ordered statistic, but I'm still getting a wrong answer.
Any help will be appreciated.
 A: It is evident that $W$ takes values in $[0,2]$ so it is enough to find $P(W\leq w)$ or equivalently $P(W>w)$ for $w\in[0,2]$:
$\begin{aligned}P\left(W>w\right) & =\int_{0}^{1}P\left(X+Y>w,Y+Z>w\mid Y=y\right)dy\\
 & =\int_{0}^{1}P\left(X>w-y,Z>w-y\mid Y=y\right)dy\\
 & =\int_{0}^{1}P\left(X>w-y,Z>w-y\right)dy\\
 & =\int_{0}^{1}P\left(X>w-y\right)P\left(Z>w-y\right)dy\\
 & =\int_{0}^{1}P\left(X>w-y\right)^{2}dy
\end{aligned}
$
The third equality is based on independence of $Y$ wrt $X$ and $Z$.
The fourth equality is based on independence of $X$ and $Z$.
The fifth equality is based on the fact that $X$ and $Z$ have the same distribution.
Now it is time to discern cases.
If $w\in\left[0,1\right]$ then we proceed with:
$\begin{aligned}P\left(W>w\right) & =\int_{0}^{w}P\left(X>w-y\right)^{2}dy+\int_{w}^{1}P\left(X>w-y\right)^{2}dy\\
 & =\int_{0}^{w}\left(1-w+y\right)^{2}dy+\int_{w}^{1}1dy\\
 & =\left[\frac{1}{3}\left(1-w+y\right)^{3}\right]_{0}^{w}+1-w\\
 & =\frac{1}{3}-\frac{1}{3}\left(1-w\right)^{3}+1-w\\
 & =1-w^{2}+\frac{1}{3}w^{3}
\end{aligned}
$
If $w\in\left[1,2\right]$ then we proceed with:
$\begin{aligned}P\left(W>w\right) & =\int_{0}^{w-1}P\left(X>w-y\right)^{2}dy+\int_{w-1}^{1}P\left(X>w-y\right)^{2}dy\\
 & =\int_{0}^{w-1}0dy+\int_{w-1}^{1}\left(1-w+y\right)^{2}dy\\
 & =0+\left[\frac{1}{3}\left(1-w+y\right)^{3}\right]_{u-1}^{1}\\
 & =\frac{1}{3}\left(2-w\right)^{2}\\
 & =\frac{4}{3}-\frac{4}{3}w+\frac{1}{3}w^{2}
\end{aligned}
$
