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Probability density function of a random number

  1. Mar 12, 2009 #1
    1. The problem statement, all variables and given/known data
    Let X be a random number from (0,1). Find the probability density function of Y = 1/X.


    2. Relevant equations



    3. The attempt at a solution
    I keep thinking this is easier than I am making it out to be, but the only places I find anything similar searching is on two exams at two different schools, so it must be more interesting than I thought.

    I am attempting to first find the distribution function and then differentiate to get the density function (not sure that is the right way to approach)

    for any x between 0 and 1, the probability distribution would be
    0 if x < 0
    x if 0 [tex]\leq[/tex] x < 1
    1 if x [tex]\geq[/tex] 1

    for y = 1/x
    x = 1/y
    so F(y) = 1/y and F'(y) = f(y) = -1/[tex]y^{2}[/tex]

    I know the answer is 1/[tex]y^{2}[/tex] and not negative, but I have the suspicion that I am doing it wrong anyway and my answer is just coincidentally close to correct. Please help
     
  2. jcsd
  3. Mar 12, 2009 #2
    You can tell F(y)=1/y is wrong by taking y=1/2. Consider [itex]\mathbb{P}(Y=k)[/itex] for some k. Now try to get the expression in terms of X.

    Good luck.
     
  4. Mar 13, 2009 #3

    statdad

    User Avatar
    Homework Helper

    Two approaches.
    First, if [tex] Y = 1/X[/tex], then you are correct that [tex] 1 \le Y < \infty [/tex].
    Now

    [tex]
    P(Y \le y) = P(1/X \le y) = P(X \ge \frac 1 y)
    [/tex]

    You forgot to reverse the inequality. Write out a form for [tex] P(X \ge 1/y) [/tex] and differentiate that to get the density.

    Second: if you want to do a change of variables, you need to multiply be the absolute value of the derivative.
     
  5. Mar 13, 2009 #4

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    Just picking a nit here, but that should be U(0,1), not just (0,1).

    That is not correct. This fails two key tests of whether a function is a CDF. This function is monotonically decreasing for y>1 (CDFs must be monotonically increasing functions) and F(y)→0 as y→∞ (CDFs must have F(y)→1 as y→∞).

    You are on the right track, however.
     
  6. May 31, 2010 #5
    PF Mentor,
    Can you post the complete answer?
     
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