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Pdf of the sum of two distributions

  1. Mar 21, 2006 #1
    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...
  2. jcsd
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