# Pdf of the sum of two distributions

1. Mar 21, 2006

### shan

I'm not too sure where to post this so feel free to move it :)

Anyway I'm hoping someone could explain the answer of this problem to me (I would ask my lecturer but he's conveniently away for the week for a meeting).

Suppose X and Y are iid continuous random variables with density f. Then X+Y has density:

f_X+Y(z) = integral from infinity to neg infinity of f(x)f(z-x)dx

which is the convolution of the densities of X and Y.

Use this to determine the pdf (probability density function) and distribution function of X+Y when X and Y are iid U(0,1). (independently and identically distributed, Uniform(0,1)).

The answer starts by showing the region where f(x)f(z-x) is non-zero ie the area between z-x=0 and z-x=1 (and I understand this part, if X and Y are uniform, the sum will be 0 outside (0,2)). It then says that the density function is:

For 0<z<1
f_X+Y(z) = (integral from z to 0 of 1dx) = (x evaluated at z and 0) = z.

For 1<z<2
f_X+Y(z) = (integral from 1 to z-1 of 1dx) = (x evaluated at 1 and z-1) = 2-z.

I don't understand why for 1<z<2, the integral is from 1 to z-1? Where did these boundaries come from?

The other part is the distribution function which is supposedly:

For 0<z<1
F_X+Y(z) = (integral from z to 0 of wdw) = (w^2/2 evaluated at z and 0) = z^2/2

For 1<z<2
F_X+Y(z) = (integral from z to 0 of f_X+Y(w)dw) = (integral from 1 to 0 of f_X+Y(w)dw + integral from z to 0 of f_X+Y(w)dw)
= (integral from 1 to 0 of wdw + integral from z to 1 of (2-w)dw) = 1^2/2 + (2w-w^2/2) evaluated at z and 1)
= 1/2 + (2z-z^2/2)-(2-1/2) = 2z-z^2/2-1

Again, I'm having problems with where the numbers in the integral came from when 1<z<2.

Sorry for the hard read >_< I can't seem to preview the latex images and since I don't know how to use them very well I thought I could do without...