# Sum of Two Independent Random Variables (uniform)

• February 3rd 2011, 06:12 PM
oisduda
Sum of Two Independent Random Variables (uniform)
Suppose X and Y are Uniform(-1, 1) such that X and Y are independent and identically distributed. What is the density of Z = X + Y?

Here is what I have done so far (I am new to this forum, so, my formatting is very bad). I know that
fX(x) = fY(x) = 1/2 if -1<x<1 and 0 otherwise

The density of Z will be given by
fZ(z)= integral fX(z-y)fY(y)dy bounds of integration from -infinity to infinity

fY(y) = integral 1/2 if -1<y<1 and 0 otherwise
So,

fZ(z) = (1/2)fX(z-y)dy (bounds of integration -1 to 1)

The integrand = 1/2 if -1<z-y<1 or z-1<y<z+1
or 0 otherwise

This is where I get stuck and not sure how to proceed. Thanks
• February 3rd 2011, 07:38 PM
chisigma
If x has PDF f(x) and y has PDF g(y), the z=x+y has PDF...

$\displaystyle \varphi(z)= \int_{- \infty}^{+ \infty} f(\xi)\ g(z-\xi)\ d \xi$ (1)

If f(*) and g(*) are 'rectangular' then $\varphi(*)$ is 'triangular'...

Kind regards

$\chi$ $\sigma$
• February 3rd 2011, 09:37 PM
matheagle
This is just like the sum of two U(0,1).
It's easy to solve it via geometry.
Draw a square (-1,1) by (-1,1)
Obtain $F_Z(z)=P(X+Y\le z)$ by figuring the area "under" x+y<z for -2<z<2.
The third dimension is just 1/4, since the density is constant over that square.
• February 4th 2011, 01:58 AM
mr fantastic
Quote:

Originally Posted by matheagle
This is just like the sum of two U(0,1).
It's easy to solve it via geometry.
Draw a square (-1,1) by (-1,1)
Obtain $F_Z(z)=P(X+Y\le z)$ by figuring the area "under" x+y<z for -2<z<2.
The third demsion is just 1/4, since the density is constant over that square.

Indeed. @OP: http://www.dartmouth.edu/~chance/tea...k/Chapter7.pdf page 8.