Find the density function of the random variable $Z=X+Y$ Let $X$ and $Y$ be independent and uniformly distributed random variable on the intervals $[0,3]$ and $[0,1]$, respectively. Find the density function of the random variable $Z=X+Y$.
I find $f(x,y)=1/3$ and the range of $Z$ to be $[0,4]$, but I cannot find the density function of $Z$. 

In general, to find the pdf, you should find the cdf first, and then differentiate the function.
$F[z] = P(Z \leq z)  = \begin{cases} \frac {1}{6} z^2 & 0\leq z \leq 1 \\ 
\frac {1}{6} + \frac {1}{3} (x-1) & 1 \leq z \leq 3 \\
1- \frac {1}{6} (4-z)^2 & 3 \leq z \leq 4 \\ \end{cases}$
This gives that 
$f(z) = \begin{cases} \frac {1}{3} z & 0 \leq z \leq 1 \\ \frac {1}{3}  & 1 \leq z \leq 3 \\ \frac {1}{3} (4-z) & 3 \leq z \leq 4 \\\ \end{cases}$
