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Find the density of the sum $Z = X+Y$ when $X$ and $Y$ are independent, standard uniform random variables.

$$f_X(x) = 1\quad\mathrm{if}\quad 0\le x \le 1$$ $$f_Y(y) = 1\quad\mathrm{if}\quad 0\le y\le 1$$ $$\begin{align} f_Z(z) & = \int_{-\infty}^\infty f_X(z-y)f_Y(y) dy \\ & = \int_0^1 f_X(z-y) dy \\ \end{align}$$ I don't understand the following part of the solution, specifically how the range for $z$ gets separated and the respective bounds for the integrals are determined.

The integrand is 0 unless $0\le z-y\le 1$ or $z-1\le y\le z$. So:

if $0\le z\le 1, f_Z(z) = \int_0^z dy = z$

if $1<z \le 2, f_Z(z) = \int_{z-1}^1 dy = 2-z$

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$f_X(z-y)f_Y(y)$ is either $=1$ or $0$. More specifically, it is $=1$ iff $$\tag10\le z-y\le1\text{ and }0\le y\le 1$$ or equivalently $$\tag2 \max\{0,z-1\}\le y\le \min\{1,z\}.$$ We can compare the two bounds and find $$\tag3 \max\{0,z-1\}\le \min\{1,z\}\quad \text{iff }0\le z \le 2.$$ If the upper bound is at less than the lower bound, $(2)$ cannot be fulfilled, that is we have $$f_Z(z)=\int_{-\infty}^\infty 0\, dy\quad \text{if }z<0\text{ or }z>2.$$ On the other hand, for $0\le z\le 2$ we find $$f_Z(z)=\int_{-\infty}^\infty f_X(z-y)f_Y(y)\,dy=\int_{\max\{0,z-1\}}^{\min\{1,z\}} 1\,dy=\min\{1,z\}-\max\{0,z-1\}$$ ans observe that all this boils down to $$f_Z(z)=\begin{cases}0&\text{if }z\le 0\text{ or }z\ge2\\z&\text{if }0\le z\le 1\\2-z&\text{if }1\le z\le 2.\\\end{cases}$$

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$$f_Z(z)=\int_0^1\mathbf 1_{z-1\leqslant y\leqslant z}\,\mathrm dy=\mathrm{Leb}(|0,1]\cap[z-1,z])=\ldots$$

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