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Homework Help: Determining the distribution function

  1. Apr 7, 2006 #1
    I've gotten a weird answer after doing the problem but I'm stuck as to where I messed up.

    The density function is this:
    [tex]f_{X} (x) = \frac{1}{6}x[/tex] for [tex]0<x\leq2[/tex]
    [tex] = \frac{1}{3}(2x-3)[/tex] for 2<x<3
    and 0 otherwise

    And the question is to find the distribution function.

    So integrating for the first part from 0 to x:
    [tex]\int \frac{1}{6}u du = \frac {1}{6} \frac{x^2}{2} = \frac{x^2}{16}[/tex]
    for 0<x<=2

    I have a big problem with the second, part. This is what I did (integrating from 2 to x):
    [tex]\int \frac{1}{3} (2u-3) du = \frac{1}{3} [u^2-3u] = \frac{1}{3} (x^2-3x - (2^2-6)) = \frac{x^2}{3} - x + \frac{2}{3}[/tex]
    for 2<x<3

    I know my answer for the second part is wrong as when x=2, the distribution function = 0 and when x=3, the distribution function = 2/3. But the distribution shouldn't be broken up like that at x=2 and supposedly at x=3, it should = 1. So did I forget to do something to the end points or did I not integrate properly?
    Last edited: Apr 7, 2006
  2. jcsd
  3. Apr 7, 2006 #2


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    The distribution function is given by:

    [tex]F_X(x) := \int_{-\infty}^x f_X(x) \, dx[/tex]

    So when you computed [itex]\int_2^x f_X(x) \, dx[/itex], you computed the wrong thing, and there's no reason you should have gotten the right answer.

    P.S. 6*2 is not 16.
  4. Apr 7, 2006 #3
    whoops sorry, typo :blushing: that first part is
    [tex]\int_0^x \frac{1}{6}u du = \frac {1}{6} \frac{x^2}{2} = \frac{x^2}{12}[/tex]

    which shows I don't really understand what I'm doing but now that you mentioned it...

    is the second part then given by [tex]\int_0^2 \frac{1}{6}u du + \int_2^x \frac{1}{3} (2u-3) du[/tex]? ie I forgot to add the first part of f(x)?
  5. Apr 7, 2006 #4


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    Right. Of course, there should also be a [itex]\int_{-\infty}^0 f_X(u) \, du[/itex] component as well. (But you know it's zero, so I suppose that's why you left it out)
  6. Apr 7, 2006 #5
    Thank you very much for your help :)
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