Proof for Mean of Geometric Distribution

Question

I am studying the proof for the mean of the Geometric Distribution

http://www.randomservices.org/random/bernoulli/Geometric.html (The first arrow on Point No. 8 on the first page).

It seems to be an arithco geometric series (which I was able to sum using

http://en.wikipedia.org/wiki/Arithmetico-geometric_sequence#Sum_to_infinite_terms) However, I have not able to find any site which uses this simple property above.

Instead, they differentiate.

The way the differentiation works is: 1. You have n*x^n-1, so you integrate that to get x^n, and add the differentiation to "balance". 2. You interchange the differentiation and summation (slightly complicated topic). 3. Complete the summation (geometric series). 4. Complete the differentiation. 5. Get your answer.

Questions:

Is there anything wrong in arriving at the formula the way I have done. Isn't it better to use the arithco-geometric formula then go through all that calculus just to convert an arithco-geometric series into a geometric one.

Answer 1

Here is one approach: using $\sum_{j=0}^{\infty}(1-p)^j=\frac{1}{p}$ only. We have $\sum_{n=1}^{\infty}np(1-p)^{n-1}=p\sum_{n=1}^{\infty}n(1-p)^{n-1}$ hence let's focus on $\sum_{n=1}^{\infty}n(1-p)^{n-1}$:

\begin{align} \sum_{n=1}^{\infty}n(1-p)^{n-1}&=\sum_{n=1}^{\infty}n(1-p)^{n-1}\\ &=(1-p)^0\\ &+ (1-p)^1+(1-p)^1\\ &+ (1-p)^2+(1-p)^2+(1-p)^2\\ &+ ...\\ &=\sum_{j=0}^{\infty}\sum_{n=0}^{\infty}(1-p)^{j+n} (\mbox{ first sum the "columns"})\\ &=\sum_{j=0}^{\infty}\frac{(1-p)^j}{p}\\ &=\frac{1}{p^2} \end{align}