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    • Thread Starter

    1. The problem statement, all variables and given/known data

    Hi, I have the probabilty density:

    p_{n}=(1-p)^{n}p , n=0,1,2...

    and I am asked to find the characteristic function: p(k)= <e^{ikn}>

    and then use this to determine the mean and variance of the distribution.

    2. Relevant equations

    I have the general expression for the characteristic function :

    \sum\limits^{\infty}_{n=0} \frac{(-ik)^m}{m!} <x^{m}> * ,
    from which can equate coefficients of k to find the moments.

    3. The attempt at a solution

    So I have  <e^{-ikn}>=\sum\limits^{\infty}_{n-0} (1-p)^{n}p e^{-ikn}

    I understand the solution given in my notes which is that this is equal to, after some rearranging etc, expanding out using taylor :

     1 + \frac{(1-p)}{ p} (-k + 1/2 (-ik)^{2} + O(k^3) ) + \frac{(1-p)^{2}} { p^2 } ( (-ik)^{2} + O(k^3))

    and then equating coefficients according to * However my method was to do the following , and I'm unsure why it is wrong:

     <e^{-ikn}>=\sum\limits^{\infty}_{n=0} (1-p)^{n} p e^{-ikn} =  \sum\limits^{\infty}_{n=0} (1-p)^{n}p \frac{(-ik)^n}{n!}

    And so comparing to *  \implies  \sum\limits^{\infty}_{n=0} (1-p)^{n}p = \sum\limits^{\infty}_{n=0} <x^{n}>

    Anyone tell me what I've done wrong? thank you, greatly appreciated.

    (Original post by xfootiecrazeesarax)

     <e^{-ikn}>=\sum\limits^{\infty}_{n=0} (1-p)^{n} p e^{-ikn} =  \sum\limits^{\infty}_{n=0} (1-p)^{n}p \frac{(-ik)^n}{n!}
    I don't understand why the 2nd equality holds here (I am doubtful that it does...)
    • Thread Starter

    (Original post by DFranklin)
    I don't understand why the 2nd equality holds here (I am doubtful that it does...)
    expanded my exponential incorrectly, falling alseep, thank you !
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