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Math Help - Second moment for geometric distribution

  1. #1
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    Second moment for geometric distribution

    Hello,

    I am stuck trying to calculate the second moment of the geometric distribution.


    E(x^2) = Sum (x^2 * f(x))
    = Sum (x^2 * p * q^(x-1))

    I am not sure how to progress further from here - do you have any pointers? Thanks
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by sharpe View Post
    Hello,

    I am stuck trying to calculate the second moment of the geometric distribution.


    E(x^2) = Sum (x^2 * f(x))
    = Sum (x^2 * p * q^(x-1))

    I am not sure how to progress further from here - do you have any pointers? Thanks
    What about
    \displaystyle E(x^2)=\sum_{x=1}^\infty x^2 p q^{x-1}=\sum_{x=1}^\infty x p \frac{d}{dq} q^x=\frac{d}{dq}\sum_{x=1}^\infty x p q^x}=\frac{d}{dq} \Big(q E(x)\Big)=\ldots
    Since you know the value of the first moment, E(x), you are essentially done.
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  3. #3
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    Quote Originally Posted by Failure View Post
    What about
    \displaystyle E(x^2)=\sum_{x=1}^\infty x^2 p q^{x-1}=\sum_{x=1}^\infty x p \frac{d}{dq} q^x=\frac{d}{dq}\sum_{x=1}^\infty x p q^x}=\frac{d}{dq} \Big(q E(x)\Big)=\ldots
    Since you know the value of the first moment, E(x), you are essentially done.


    fantastic thanks
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  4. #4
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    sorry, if I can just check - does that give the right answer?

    E(x) is 1 / (1-q). Which gives E(x^2) = d/dq (q/(1-q)

    = 1 / (1-q)^2 = 1/ p^2

    I know the answer is (1+q)/p^2; so this looks wrong?
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  5. #5
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    Quote Originally Posted by sharpe View Post
    sorry, if I can just check - does that give the right answer?

    E(x) is 1 / (1-q). Which gives E(x^2) = d/dq (q/(1-q)

    = 1 / (1-q)^2 = 1/ p^2

    I know the answer is (1+q)/p^2; so this looks wrong?

    In case anyone is interested, I found this by finding E(X(X-1)) first which is markedly easier. You then take the approach suggested, except using a second derivative.
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