Two independent random variable X,Y which is U(0,1)
what is the pdf of Z=X+Y?
$F_Z(z)$ = \begin{cases} 0, & \ z\le0 \\ \int_0^z \int_0^{z-y} 1 \,dxdy=z^2/2, & \ 0\lt z\le1 \\ 1-\int_\bbox[yellow]{z-1}^1 \int_{z-y}^1 1 \,dxdy=1-(2-z)^2/2,& \ 1\lt z\le2 \\ 1 & \ z\ge2 \end{cases}
I understand I have to differential $F_z(z)$
The thing I can't understand is the highlight part.
Why does it starts with z-1?
Convolution of probability densities
Let $X$ and $Y$ be any two independent (real-valued) random variables with densities $f_X$ and $f_Y$. Define $Z \equiv X + Y$. Note that $$ F_Z(z) = \mathbb{P}(Z \leq z) = \mathbb{P}(X + Y \leq z) = \int_{-\infty}^{\infty} f_X(x) \mathbb{P}(x + Y \leq z) dx = \int_{-\infty}^{\infty} f_X(x) F_Y(z - x) dx. $$ Differentiating, $$ f_Z(z) = \int_{-\infty}^{\infty} f_X(x) f_Y(z - x) dx. $$
Convolution of i.i.d. uniform distributions
If $X$ and $Y$ are i.i.d. $U(0,1)$, then $f_x = f_Y = \boldsymbol{1}_{(0,1)}$ where $\boldsymbol{1}$ is the indicator function. Plugging this into the integral of the previous section, $$ f_Z(z) = \int_0^1 \boldsymbol{1}_{(0,1)}(z - x) dx. $$ Proceeding by cases, $$ f_Z(z) = \begin{cases} \int_0^z dx = z & \text{if } 0 \leq z \leq 1 \\ \int_\bbox[yellow]{z-1}^1 dx = 2 - z & \text{if } 1 \leq z \leq 2. \\ \end{cases} $$