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Math Help - PGF of Discrete Random Variable

  1. #1
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    PGF of Discrete Random Variable

    Hi Everyone,

    I am trying to show that the discrete random variable....

    P(R=j) = (j+1) p^j {(1-p)}^2, j = 0,1,2,.... 0<p<1

    Has a P.G.F.....

    GR(s) = ( \frac{1-p}{1-ps})^2

    Above is meant to be the whole fraction to the power 2, but i was having some Latex Syntax problems showing that

    I have tried changing s^j and p^j to ps^jand taking it outside the summation when beginning to try and find the PGF but i didnt get anywhere, help needed please!

    Also maybe \Sigma (j+1) a^j {(1-a)}^2 = 1 for any a where \mida \mid < 1 can be used? i cant see how though!
    Last edited by sirellwood; March 12th 2010 at 01:59 PM.
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  2. #2
    MHF Contributor matheagle's Avatar
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    you placed the math OUTSIDE the [tex] tex [ /math]

    Quote Originally Posted by sirellwood View Post
    Hi Everyone,

    I am trying to show that the discrete random variable....

    P(R=j) = (j+1) p^j(1-p)^2, j = 0,1,2,.... 0<p<1

    Has a P.G.F.....

    GR(s) = \left(\frac{1-p}{1-ps}\right)^2

    Above is meant to be the whole fraction to the power 2, but i was having some Latex Syntax problems showing that

    I have tried changing s^j and p^j to ps^jand taking it outside the summation when beginning to try and find the PGF but i didnt get anywhere, help needed please!

    Also maybe \Sigma (j+1) a^j {(1-a)}^2 = 1 for any a where \mida \mid < 1 can be used? i cant see how though!

    The trick in these is to differentiate the geometric, that brings downs that power of j.

    Inside the interval of convergence you can..........

    {d\over dx} \sum_{n=0}^{\infty}x^n=\sum_{n=1}^{\infty}nx^{n-1}
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  3. #3
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    Thanks matheagle! any ideas on how to solve the question? :-)
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  4. #4
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    Sorry matheagle, first time id loaded your post, the extra bit at the bottom hadnt loaded! :-)
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