- #1

binbagsss

- 1,269

- 11

## Homework Statement

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.

## Homework Equations

[/B]

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.

## 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.