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Problem with density function

  1. May 1, 2013 #1
    Hello. I have problem with one task of probability theory. Hope that someone will be able to help me!

    The task:

    The rod is dividing in two parts accidentally. (Division is evenly distributed, dont know if its good saying in english). Find the density function of random variable X which is equal to longer part of rod divided by smaller part of rod.

    So X can get infinity values i think. The smaller is 1, when rod is divided 0.5/0.5=1. And coming up to infinity when smaller part of rod coming up to 0. for example 0.99/0.01=99... and so on.

    I dont know but i think that density function of this X does not exist. Or im wrong?
     
  2. jcsd
  3. May 1, 2013 #2

    HallsofIvy

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    If the shorter length is y, the longer length is 1- y so the function is X= y/(1- y). Further, you seem to be saying that "y" is equally likely to be any number between 0 and 1/2.
     
  4. May 1, 2013 #3

    haruspex

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    It's often clearer conceptually to deal with the CDF and deduce the pdf later. HoI has given you the formula for X, so you want F(x) = P[X < x] = P[Y/(1-Y) < x]. Can you proceed from there?
     
  5. May 1, 2013 #4
    I think the task ask to do not the same as you say. I think we need 1-y/y... where 1-y is equal to longer part and y shorter...If i need dividing shorter from longer it much easier. or i dont understand you?
    Maybe my opinion that this function does not exist is true?
     
  6. May 1, 2013 #5

    Ray Vickson

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    The parts (u,1-u) have zero probability for u = 0 or u = 1 because the endpoints of the interval [0,1] have zero probability for a continuous distribution. Therefore, we are (with probability 1) never dividing by zero, so a value X = ∞ has probability 0. We can even change to the open interval u ε (0,1), and so division by zero never happens---not even with probability zero. So, X does have a well-defined density function. Having X unbounded above is not a problem---lots of random variables are unbounded above, below, or both, and we deal with them with no problem. Of course, there is the issue of whether or not X has finite mean and/or variance, etc, but that is a separate question from the existence of a density for X.
     
    Last edited: May 1, 2013
  7. May 1, 2013 #6
    ok understand.
     
  8. May 1, 2013 #7
    But cant invent what this function shuold be. Maybe some hints? I thought more than an hour and have no idea.
     
  9. May 1, 2013 #8

    haruspex

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    Did you understand what I suggested? Let Y be the r.v. representing the distance from the nearer end. Y is uniformly distributed on [0, 0.5]. The CDF of X is F(x) = P[X < x] = P[(1-Y)/Y < x]. Can you evaluate that?
     
  10. May 2, 2013 #9
    thank you haruspex. your hint was very helpfull. :smile:
     
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