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Homework Help: Determining the distribution function F(x): Statistics/Probability

  1. Feb 28, 2010 #1
    1. The problem statement, all variables and given/known data

    For the random variable X with probability density function determine the distribution function F(x)

    http://img20.imageshack.us/img20/3314/questiontsy.jpg [Broken]

    2. Relevant equations

    Cumulative Distribution Function

    3. The attempt at a solution

    Integrate f(x) to get the distribution function:
    F(X) =
    • 0, x ≤ -1
    • x[tex]^{2}[/tex]/2 + x, -1 < x ≤ 0
    • x - x[tex]^{2}[/tex]/2, 0 < x ≤ 1
    • 0, x > 1

    The actual answer is
    http://img297.imageshack.us/img297/426/answer.jpg [Broken]

    I understand that the equation must be integrated to get the distribution function, but where do the constants (1/2) come from and how are they calculated?
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 28, 2010 #2
    You can't just take a piecewise density function and find the indefinite integral of each piece separately. Notice your F(x) isn't continuous (any integrable density function must have a continuous cumulative distribution function), so it has to be incorrect.

    Remember the cumulative distribution function is [itex]\int_{-\infty}^{x}f(t)\,dt,[/itex] where f is your density function (t is just a dummy variable). Now obviously the values of f for x less than or equal to 1 contributes nothing to the cumulative distribution function, so to find the values of F for [itex]-1<x\leq0,[/itex] you need to compute
    [tex]\int_{-\infty}^{x}f(t)\,dt = \int_{-\infty}^{x} t + 1\,dt = \int_{-1}^{x} t + 1\,dt. [/tex]
  4. Feb 28, 2010 #3
    Thanks, that helps clarify a bit. I solved the integral you posted and got the correct answer, but tried to solve the other 2 and got an answer different than what the professor gave.

    To get the F for [itex]0<x\leq-1,[/itex] I used this integral:
    [tex]\int_{-\infty}^{x}f(t)\,dt = \int_{-\infty}^{x} 1 - t\,dt = \int_{0}^{x} 1 - t\,dt. [/tex]

    but since F(0) = 0, the answer I got isn't what the professor posted. Does the 1/2 carry over from the F for [itex]-1<x\leq0,[/itex] carry over to this one?

    And for the [itex]x>1,[/itex] obviously the F is just 0 + Constant, is it just known that the value of this is 1 since it must add up to one?
    Last edited: Feb 28, 2010
  5. Feb 28, 2010 #4
    Of course! I mean what you first calculated was the cumulative distribution function for x between -1 and 0, so F(0) = 1/2, not 0. I think you should try drawing the density out to get a geometric feel for the problem for now; this isn't hard once you do a few more of these.

    For the last "piece" F(x) should just be 1, but remember you should be able to get this value by computing F(1). The cumulative density function should be continuous, and you should check this at the endpoints of each piece.
  6. Feb 28, 2010 #5
    I actually do have a graph of the density function.
    http://img64.imageshack.us/img64/8127/graphbo.png [Broken]

    But if F(0) = 1/2, when solving the integral above and doing F(x) - F(0), wouldn't that make it (x^2)/2 + x - (1/2)? What I meant was F(0) = 0 in context of the integral above, then adding the actual value of F(0) from the -1 < x ≤ 0 piece.

    I am realizing these seem relatively easy as I work through it, but there's just a few small things I'm unsure of and can't find examples of in my textbook or lecture notes.
    Last edited by a moderator: May 4, 2017
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