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How to find the expected value of a continuous variable with pdf fy= y^(-2)?

  1. Oct 11, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the expected value of a continuous variable y with pdf fy= alpha*y^-2, 0<y<infinity.

    I know it is the integral from zero to infinity of y*fy, but I don't know where to go from there. I'm then supposed to use the expected value to find the method of moments estimator.

    Does this pdf fall under a certain type of distribution?
    Any help is appreciated.
     
  2. jcsd
  3. Oct 11, 2012 #2

    Simon Bridge

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    What is f(y) if y≤0?

    The next step is to write out the integral and carry it out.
    But I'm thinking there is information not included?
     
  4. Oct 11, 2012 #3
    @Simon Bridge.

    No. That's all I'm given.

    f(y;alpha)= alpha*y^(-2) 0<alpha<= y< infinity.

    It says to determine E(Y) and then find the method of moments estimator of alpha.


    The integral of y*fy gives me [alpha* lny] evaluated from zero to infinity, but isn't it infinity?
     
  5. Oct 11, 2012 #4

    Simon Bridge

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    $$f(y;\alpha) = \frac{\alpha}{y^2}: 0 < \alpha \leq y < \infty$$
    Well then it cannot be done because the function in not determined for part of the range... though the above is already different from what you gave in post #1. Perhaps the extra information is implicit in the context - in your course notes for eg?

    Perhaps you are supposed to assume the pdf is zero outside the given range?
    But maybe you are supposed to evaluate from ##\alpha## to ##\infty##?
    Still not much help -

    And what about normalization? Isn't $$\int_{-\infty}^\infty f(y)dy=1$$ ... unless ##\alpha## is a function of y?

    I'd go ask the person who set the problem. (After checking with other students.)

    http://en.wikipedia.org/wiki/Method_of_moments_(statistics [Broken])
     
    Last edited by a moderator: May 6, 2017
  6. Oct 11, 2012 #5

    HallsofIvy

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    It is standard to define a probability density on a restricted range- it does not have to be from negative infinity to infinity. However, the first given problem, from 0 to infinity has that the problem that the integral goes to infinity. The revised problem, from [itex]\alpha[/itex] to infinity, is good:
    [tex]\int_\alpha^\infty \frac{\alpha}{y^2}dy= \alpha\int_\alpha^\infty y^{-2}dy= \alpha\left[-\frac{1}{y}\right]_\alpha^\infty= 1[/tex]

    And the expected value is just [itex]\int yf(y)dy[/itex] which, for this problem, is
    [tex]\alpha\int_{\alpha}^\infty \frac{1}{y}dy[/tex]
    Now that has a problem at infinity!
     
  7. Oct 11, 2012 #6

    Ray Vickson

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    There IS NO SUCH probability density function! What you have integrates to +∞ for any α > 0, so cannot integrate to 1. Are you really sure you have written the question correctly?

    RGV
     
  8. Oct 11, 2012 #7

    HallsofIvy

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    No, it doesn't. The integral of [itex]\alpha y^{-2}[itex], from [itex]\alpha[/itex] to infinity is 1 so it is a pdf. It just does not have finite expected value.
     
  9. Oct 11, 2012 #8

    Ray Vickson

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    OK, I missed the revised statement that had y ε [α,∞) instead of [0,∞).

    However, there is still a problem with EY, so I still wonder if the question has been stated correctly.

    RGV
     
  10. Oct 11, 2012 #9

    Simon Bridge

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    ... are the other moments finite?
     
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