Find the density of $Z=U_{1}+U_{2}$

Question

Let $U_{1}$ and $U_{2}$ be independent, and uniform on $[0,1]$. Find the density of $Z=U_{1}+U_{2}$.

Here is what I have:

The joint p.d.f of $U_{1}$, $U_{2}$ is:

$f(U_{1},U_{2}) = 1 $ when $U_{1},U_{2} \in [0,1]$;

$f(U_{1},U_{2}) = 0 $ otherwise.

Then I am missing on here, could anyone give me hints to answer this question?

Hannah, draw U1 and U2 as x,y axes - what corresponds to Z=z on the graph? (ie what combination of U1 and U2 give you a given value z of U1+U2? — seanv507, Oct 14 '16 at 16:52
I added the `self-study` tag as it is appropriate. You have to give us more indications of what you tried than the joint density of $(U_,U_2)$. — Xi'an, Oct 14 '16 at 17:11
Four different methods to solve this problem are presented at http://stats.stackexchange.com/questions/41467/consider-the-sum-of-n-uniform-distributions-on-0-1-or-z-n-why-does-the/43075?s=1|0.0000#43075. Additional answers appear at http://stats.stackexchange.com/questions/195433. I found these posts by searching on [uniform sum density](http://stats.stackexchange.com/search?tab=votes&q=uniform%20sum%20density). — whuber, Oct 14 '16 at 17:47

score 1 · Answer 1 · answered Oct 14 '16 at 16:44

1

The convolution of the densities is the density of the sum of independent realisations.

answered Oct 14 '16 at 16:44

Horst Grünbusch

5

This seems better posed as a comment than an answer to the question... – Math1000 Oct 14 '16 at 16:57

score 1 · Answer 2 · edited Apr 13 '17 at 12:44

1

A similar question has been address here

If $U_1$ and $U_2$ are independent random variables the PDF of $Z=U_1+U_2$ is given by

$f_{z}(z)=\int_{-\infty}^\infty f_{U_1}(z-u_1)\, f_{U_2}(u_2)\, du_2,$

A key difference here is you need the marginal distribution given by:

$f_{U_1}(u_1)=\int_{0}^1 f_{U_1,U_2}(u_1,u_2)du_2 $

$f_{U_2}(u_2)=\int_{0}^1f_{U_1,U_2}(u_1,u_2)du_1 $

edited Apr 13 '17 at 12:44

Community

answered Oct 14 '16 at 17:33

2 Answers2