Find the distribution of $|X-Y|$ if $X$ and $Y$ are i.i.d. uniform on $[0,1]$ $X$ and $Y$ are independent random variables uniformly distributed over $[0,1]$. I want to find the CDF of $|X-Y|$. I could use convolution but I wan't to calculate this more "directly". Here is my set-up so far:
Let $Z = |X-Y|$. Then, $F_Z(z) = P(Z \leq z) = P(|X-Y| \leq z)$.
We can split up this last inequality into two cases because of the absolute value:
$P(|X-Y| \leq z) = P(X-Y \leq z, X \geq Y) + P(Y-X \leq z, Y > X)$
We now compute the two terms separately with integration. Let's start with the first term:
To integrate this we need to find the correct limits of integration. The constraints we must satisfy are:


*

*$0 \leq x,y \leq 1$

*$x \geq y$

*$x - z \leq y$


All of this implies $\max\{x-z,0\} \leq y \leq \min\{1,x\}$. But, $x =1$ at most so we can replace $\min\{1,x\}$ with just $x$:


*

*$\implies \max\{x-z, 0\} \leq y \leq x$


We use these boundaries to set up the integral of the first term:
$P(X-Y \leq z , X \geq Y) = \int_0^1{\mathrm{d}x \int_{\max\{x-z,0\}}^x{\mathrm{d}y}}$
Should I continue proceeding this way? Is my set up correct? Thank you!
 A: Let $Z=|X-Y|$. To determine the density function of the random variable $Z $ we see that
$$
\begin{aligned}
\mathbb{P}(Z> z)&=\mathbb{P}(|X-Y|> z)\\
&=\mathbb{P}(X-Y>z)+\mathbb{P}(X-Y<-z)\\
&=\mathbb{P}(X>z+Y)+\mathbb{P}(Y>z+X)\\
&\text{then apply the law of total probability}\\
&=\int_0^{1-z}\mathbb{P}(X> z+y)f_Y(y)dy+\int_0^{1-z}\mathbb{P}(Y>z+x)f_X(x)dx\\
&=\int_0^{1-z}[1-(z+y)]dy+\int_0^{1-z}[1-(z+x)]dx\\
&=(1-z)^2-\left.\frac{y^2}{2}\right|_{y=0}^{1-z}+(1-z)^2-\left.\frac{x^2}{2}\right|_{x=0}^{1-z}\\
&=2\left[(1-z)^2-\frac{(1-z)^2}{2}\right]\\
&=(1-z)^2
\end{aligned}
$$
Thus 
$$
F_Z(z)=1-\mathbb{P}(Z>z)=1-(1-z)^2=1-1+2z-z^2=2z-z^2  
$$
and finally
$$
F_Z(z)=\left\{
\begin{aligned}
0&,\quad z<0\\
2z-z^2&, \quad 0\leq z<1\\
1&,\quad z\geq 1
\end{aligned}
\right.
$$
A: Since $X$ and $Y$ are iid Uniform Random variables, the joint density of $(X, Y)$ is    $$f(x,y)=1, \qquad 0<x<1, \qquad 0<y<1 $$
Given that $Z=|X-Y|.$ Assume another variable $U=Y.$
The above set of transformation from $$S_{x,y}=\{(x,y): 0<x<1, 0<y<1 \}$$ to $$S_{Z,U}=\{ (z,u): 0<z<1,0<u<1, z+u\leq 1  \}  $$
is not one-to-one.
Consider two sets of transformations:
\begin{eqnarray*}
Z_1&=& X-Y, \textrm{ if } X>Y\\
U&=&Y
\end{eqnarray*} 
and
\begin{eqnarray*}
Z_2&=& Y-X, \textrm{ if }X<Y\\
U&=&Y
\end{eqnarray*} 
These are one-to-one.
The solutions to the set of  linear equations $z_1=x-y,u=y $  and $z_2=x-y,u=y $ are 
\begin{eqnarray*}
x^{(1)}&=&z_1+u\\
y^{(1)}&=&u
\end{eqnarray*}
and
\begin{eqnarray*}
x^{(1)}&=&u-z_2\\
y^{(1)}&=&u
\end{eqnarray*}
The Jagobians of the first transformation is $1$ and that of second one is $-1.$
The joint density of $(z,u)$ is 
\begin{eqnarray*}
h(z,u)&=&f(x^{(1)},y^{(1)})|J_1|+f(x^{(2)},y^{(2)})|J_2|\\
&=&1+1\\
&=&2, \qquad 0<u<1,\qquad 0<x<1,  \mbox{ such that }u+z\leq 1
\end{eqnarray*}
The marginal density of $Z$ is 
$$ h(z)=\int_0^{1-z}h(z,u)du=\int_0^{1-z}2du=2(1-z)$$
The distribution function can be obtained
$ F_Z(z)=\int_0^z h(w)dw $ which is 
\begin{eqnarray*}
F_Z(z)&=&0,\qquad  z\leq 0\\
&=& \int_0^z 2(1-w)dw=2z-z^2,\qquad 0\leq w< 1\\
&=& 1 \qquad 1\leq w
\end{eqnarray*}
