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Stats/prob: finding cumulative distribution function

  1. Feb 27, 2013 #1
    1. The problem statement, all variables and given/known data

    given pdf:

    f(x) = 2/3x for 0<=x<=1
    f(x) = 2/3 for 1<x<=2
    f(x) = 0 elsewhere

    Find the CDF.


    2. Relevant equations



    3. The attempt at a solution

    I've found:

    F(x) = 0 for x<= 0
    F(x) = 1 for x>=2
    F(x) = (1/3)x2 for 0<=x<=1

    and I found:

    F(x) = (2/3)x for 1<x<=2

    However, the last bit is incorrect. It should be F(x) = (2/3)x -(1/3)

    I'm unclear as to why. I think it has something to do with solving for the constant of integration, but I'm not sure exactly.
     
  2. jcsd
  3. Feb 27, 2013 #2

    Mute

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    Homework Helper

    Yes, it has to do with the integration constant. Your CDF has to be continuous, so you need to fix that constant value so that

    $$\lim_{\epsilon \rightarrow 0} F(1 - \epsilon) = \lim_{\epsilon \rightarrow 0} F(1 + \epsilon).$$

    That is, the limit of the CDF on either side of x = 1 have to be the same, which wouldn't be true if you didn't fix the constant of integration to be -1/3 just above x=1.
     
  4. Feb 27, 2013 #3
    Could you clarify.. what is epsilon?
     
  5. Feb 27, 2013 #4

    Ray Vickson

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    Science Advisor
    Homework Helper

    You have F correct for x ≤ 1 and for x ≥ 2. Since the random variable has a finite density function, its F(x) must be a continuous function, and since F'(x) = 2/3 on [1,2], F must increase linearly with slope 2/3, starting from F(1) = 1/3 and ending at F(2) = 1. You can figure out what the formula must be for F(x) in the region 1 ≤ x ≤ 2.

    Basically, you need to use
    [tex] F(x) = F(1) + \int_{1}^{x} f(t) \, dt, \: 1 \leq x \leq 2.[/tex]
    Note that in this calculation there is NO constant of integration!
     
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