CDF of absolute value of difference in random variables Let $X$ and $Y$ be independent random variables, uniformly distributed in the interval $[0,1]$. Find the CDF and the PDF of $|X - Y|$?
Attempt
Let $Z = |X - Y|$, so for $z \geq 0$, the CDF $F_{Z}(z) = \mathbf{P}(Z \leq z) = \mathbf{P}(|X - Y| \leq z) = \mathbf{P}(-z \leq X - Y \leq z)$, which is where the algebra becomes confusing. Since they are independent, the joint pdf of $X$ & $Y$ is simply 1, as long as $(X,Y)$ belong to the unit square.
The solution suggests a plot the event of interest as a subset of the unit square and find its area. Any hints?
 A: The graphical approach is often a good one when dealing with uniform distributions, because we can interpret probabilities as areas.
Draw the unit square, and the line $y=x$ within it. Now let $z\in [0,1]$. What you have to find is the area of the polygon:
$$[0,1]^2\cap \{(x,y)\in\mathbb{R}^2:|x-y|\leq z\}$$
The set $\{(x,y)\in\mathbb{R}^2:|x-y|\leq z\}$ is just like a band centered around the first diagonal of the plane, can you see this?
More precisly, the polygon you're looking for is delimited by the points $(0,0), (0,z), (1-z,1),(1,1),(1,1-z),(z,0)$. 
With a little bit of geometry, it's easy to determine its area. Now that you have the cdf, compute its derivative to get the pdf.
A: For fixed $z\in\mathbb{R}$ we have:$$F_Z(z)=P\left(\left|X-Y\right|\leq z\right)=\int\int1_{\left(-\infty,z\right]}\left(\left|x-y\right|\right)f_{X,Y}(x,y)dxdy$$
where $f_{X,Y}(x,y)$ denotes the density of $\langle X,Y\rangle$.
Substituting this density  we arrive at:
$$F_Z(z)=P\left(\left|X-Y\right|\leq z\right)=\int_{0}^{1}\int_{0}^{1}1_{\left(-\infty,z\right]}\left(\left|x-y\right|\right)dxdy$$
Can you work this out?
If the CDF is found then the PDF can be found by differentiating.
A: The area of the square that is between the two lines is the probability of the event $\{|X-Y|\le t\}$ for $0\le t\le 1$ ($t=0.5$ in the graph).

The area of the two triangles is the same and the area of one triangle is $(1-t)^2/2$. Hence, the area of the square that is between the two lines is given by $1-2\cdot (1-t)^2/2=1-(1-t)^2$ and the probability
$$
\Pr\{|X-Y|\le t\}=1-(1-t)^2
$$
for $0\le t\le 1$.
