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The inverse of uniform random variable

  1. Nov 4, 2011 #1
    Hi all
    I'm looking for solving this problem to find the closed form solution if it is possible:

    [itex]Y=\frac{1}{X}[/itex]

    Where X is uniform random variable > 0
    I know the expected value for X which is [itex]\overline{X}[/itex]

    is there a method to find the expected value of Y which is [itex]\overline{Y}[/itex] in term of [itex]\overline{X}[/itex] as closed form solution?

    I know how to calculate it easily using numerical solution, but I need it for modeling problem and I need the analytical solution.

    Thanks
     
  2. jcsd
  3. Nov 4, 2011 #2

    chiro

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    Hey giglamesh and welcome to the forums.

    What is your statistics and probability background like?

    A standard intro year course in university will give you the tools to solve this problem. Do you know about transformation theorems in statistics?

    http://www.ncur20.ws/presentations/2/216/presentation.pdf
     
  4. Nov 4, 2011 #3
    hello chiro
    Thanks for replying

    I have a background with Random Variables and stochastic processes
    I've read about MLE once but never use it in my applications, I remember that it is used to estimate the random variable from sample data vectors.

    which is not what I'm looking for.

    Maybe I didn't explain my problem well:

    X is random variable I know only it's expected variable
    Y=1/X is a random variable I need to know it's expected value using only E[X]

    I'll check the transformation methods you mentioned, I know there are Laplace and Z-transform, I've used Z-transform but it didn't give the required result.

    I'll try to search more.
    Thanks
     
  5. Nov 4, 2011 #4

    micromass

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    Aah, that changes it. Given a continuous random variable X with pdf [itex]f_X[/itex] and a function g, we can always calculate

    [tex]E[g(X)]=\int_{-\infty}^{+\infty}{g(x)f_X(x)dx}[/tex]

    So in your case, you need to calculate

    [tex]E[1/X]=\int_{-\infty}^{+\infty}{\frac{f_X(x)}{x}dx}[/tex]

    So if X is uniform(1,2) for example, then

    [tex]E[1/X]=\int_1^2{\frac{1}{x}dx}=\log{2}[/tex]
     
  6. Nov 4, 2011 #5
    hi micromass
    Actually X is discrete I need to say: X is not uniform but Y=1/X is constructed as a uniform distribution from X, that means gives that X=3 then Y=1/3
    P(y)=E[1/X]
    I know only X then I need to get E[1/X] using only E[X] which is known but the distribution of X is not known.

    Thanks for replying
     
  7. Nov 4, 2011 #6
    I think what chiro said makes sense for me right now

    X 0 1 2 3 ..........H
    P(Y|X=i)=1/i 1 0.5 1/3 .......1/H

    From the second line I'll try to estimate the PMF of Y using MLE, I'll try it
     
  8. Nov 4, 2011 #7
    I just closed this thread, I will open new one and try to make it more clear.
     
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