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CDF of a variable with a negative exponent in its PDF

  1. May 3, 2012 #1
    Ok, this one's got me stumped!

    Let's take as an example the probability density function for a random variable X so that:

    f(x) = [itex]\frac{4}{3x^{3}}[/itex] 1≤x<2
    f(x) = [itex]\frac{x}{12}[/itex] 2≤x≤4
    f(x) = 0

    So the CDF for this variable comes out as:

    F(x) = [itex]\frac{-2}{3x^{2}}[/itex] 1≤x<2
    F(x) = [itex]\frac{x^{2}}{24}[/itex] 2≤x≤4

    So how can the CDF be negative for 1≤x<2, the CDF is P(X≤x), so to my mind that makes no sense. And secondly I have never seen a CDF with a discontinuity like in this one. At x=2, it jumps from -1/6 to 1/6

    This made me think that I should ignore the negative sign in the CDF for 1≤x<2, but then for 1≤x<2 F(x) is a decreasing function, how can that make sense?

    Someone enlighten me... please!
  2. jcsd
  3. May 3, 2012 #2
    Ok I've figured out why the minus sign shouldn't have been there, I forgot that what you actually do to find the CDF is:

    [itex]\int_{-∞}^{x} p(t)dt[/itex]

    Which sorts out the minus sign, other problem still remains?

    EDIT: I'm talking rubbish this hasn't fixed a thing!

    EDIT2: Turns out it fixed both problems. CDF is now a lovely increasing function that starts at 0 and ends at 1. Turns out I'd been doing an incorrect short cut on CDFs and never noticed because it has never not worked until tonight!
    Last edited: May 3, 2012
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