Density of sum of two independent uniform random variables on $[0,1]$ I am trying to understand an example from my textbook.
Let's say $Z = X + Y$, where $X$ and $Y$ are independent uniform random variables with range $[0,1]$. Then the PDF
is
$$f(z) = \begin{cases}
z & \text{for $0 < z < 1$} \\
2-z & \text{for $1 \le z < 2$} \\
0 & \text{otherwise.}
\end{cases}$$
How was this PDF obtained? 
Thanks 

If we want to use a convolution, let $f_X$ be the full  density function of$X$, and let $f_Y$ be the full density function of $Y$. Let $Z=X+Y$. Then 
$$f_Z(z)=\int_{-\infty}^\infty f_X(x)f_Y(z-x)\,dx.$$
Now let us apply this general formula to our particular case. We will have $f_Z(z)=0$ for $z\lt 0$, and also for $z\ge 2$. Now we deal with the interval from $0$ to $2$. It is useful to break this down into two cases (i) $0\lt z\le 1$ and (ii) $1\lt z\lt 2$.
(i) The product $f_X(x)f_Y(z-x)$ is $1$ in some places, and $0$ elsewhere. We want to make sure we avoid calling it $1$ when it is $0$. In order to have $f_Y(z-x)=1$, we need $z-x\ge 0$, that is, $x\le z$. So for (i), we will be integrating from $x=0$ to $x=z$. And easily
$$\int_0^z 1\,dx=z.$$
Thus $f_Z(z)=z$ for $0\lt z\le 1$.
(ii) Suppose that $1\lt z\lt 2$. In order to have $f_Y(z-x)$ to be $1$, we need $z-x\le 1$, that is, we need $x\ge z-1$. So for (ii) we integrate from $z-1$ to $1$. And easily
$$\int_{z-1}^1 1\,dx=2-z.$$
Thus $f_Z(z)=2-z$ for $1\lt z\lt 2$. 
Another way: (Sketch) We can go after the cdf $F_Z(z)$ of $Z$, and then differentiate. So we need to find $\Pr(Z\le z)$.
For a few fixed $z$ values, draw the lines with equation $x+y=z$ on an x-y axis plot. Draw the square $S$ with corners $(0,0)$, $(1,0)$, $(1,1)$, and $(0,1)$. 
Then $\Pr(Z\le z)$ is the area of the part $S$ that is "below" the line $x+y=z$. That area can be calculated using basic geometry. For example, when z is 2, the whole square area is under the line so Pr=1. There is a switch in basic shape at $z=1$. 
