Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
up vote 1 down vote accepted

$f_X(z-y)f_Y(y)$ 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 $$f_Z(z)=\begin{cases}0&\text{if }z\le 0\text{ or }z\ge2\\z&\text{if }0\le z\le 1\\2-z&\text{if }1\le z\le 2.\\\end{cases}$$

share|cite|improve this answer

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.