8. Suppose that X and Y are independent continuous random variables, and each is uniformly distributed on the interval [0,1] (thus the pdfs for X and Y are zero outside of this interval and equal to one on [0,1]).
(a) Find the mean and variance for X+Y.
(b) Calculate and graph the pdf for X+Y (using the convolution formula, and be careful to remember that the pdf of X vanishes outside of [0,1], etc).
fX+Y = fX * fY
X + Y = Z
The Attempt at a Solution
I know that the density functions for X and Y are both 1 since the continuous distribution is 1/(b-a). To get the density function for Z, I calculated [tex]\int[/tex] fX(x) * fY(z-x) dx from 0 to z. Since this simplifies to [tex]\int[/tex] 1 dx from 0 to z. I get fZ(z) = z.
Using the density function of Z, I calculate the mean, which is E(Z = z) = [tex]\int[/tex] z2 dz from 0 to 1. I get 1/3 for my mean when I evaluate the integral.
To get the second moment, i.e. E(Z2) : [tex]\int[/tex] z3 dz from 0 to 1. When I evaluate this I get 1/4. To get the variance I then get E(Z2) - E(Z)2 which is 1/4 - 1/9 = 5/36.
I don't know if this is right though.