# Moments from characteristic function geometric distribution

1. Jan 6, 2017

### binbagsss

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.

2. Jan 6, 2017

### Ray Vickson

Simplify:
$$p (1-p)^n e^{-ikn} = p x^n, \; \text{where} \; x = (1-p)e^{-ik}$$
Now just sum the geometric series $\sum x^n$.

3. Jan 8, 2017

### binbagsss

Yeah that's fine, and in the end you get the 'solution given' which I said I understand. That's what's been done before expanding out.

My problem was wondering what is wrong with the method I post after that...

Last edited: Jan 8, 2017
4. Jan 8, 2017

### Ray Vickson

You seem to be saying that $p(1-p)^n e^{-ikn}$ is the same as $p(1-p)^n (-ik)^n/n!$, but this is obviously false.

Last edited: Jan 8, 2017