Geometric distribution , cumulative from characteristic function

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.

Reply 1

DFranklin

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...)

Reply 2

xfootiecrazeesarax

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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Geometric distribution , cumulative from characteristic function

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