Convolution formula for $Z = |Y-X|$, where $Y$ and $X$ are independent uniform random variables on $[0,L]$

Question

Convolution formula for $Z = |Y-X|$, where $Y$ and $X$ are independent uniform random variables on $[0,L]$.

I know how to get it by getting the CDF first then differentiating. but I also want to know how to get the PDF directly from the convolution formula. Thanks in advance!

Answer 1

Note that it suffices to find the distribution of $Y-X$ via your favorite method (say, convolutions), and then the absolute value changes the density via $(f(x)+f(-x))/2$.

Answer 2

You have two uniform random variables $X \sim \mathcal{U}(0,L)$ and $-Y\sim \mathcal{U}(0,-L)$. The pdf of their sum is convolution of the two rectangular pdf (with equal support $L$), which is triangular with a support of $2L$. So the pdf of $D=X-Y$ becomes $$f_D(d)=\begin{cases}(\frac1L)(d+L), & -L\le d \le0\\(\frac1L)(L-d), & 0\le d\le L\\0, & \text{otherwise}\end{cases}$$ Due to symmetry, the pdf of the absolute value becomes $$f_Z(z)=\begin{cases}(\frac2L)(L-z), & 0\le z\le L\\0, & \text{otherwise}\end{cases}$$