Expectation – Proof for Mean of Geometric Distribution

expectation

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

http://www.math.uah.edu/stat/bernoulli/Geometric.html
(The first arrow on Point No. 8 on the first page).

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

http://en.wikipedia.org/wiki/Arithmetico-geometric_sequence#Sum_to_infinite_terms)

Here are the steps I took to arrive at the result:

Mean of Geometric Distribution: $E(N)
= \sum_{n=1}^{\infty} [n p (1 – p)^{n-1}]=S_n$

An arithmetico-geometric series is $a + (a + d)r + (a + 2d)r^2+\cdots$

$E(n)$ is then an arithmetico-geometric series with

  • first term: $a= p$
  • common difference $d= p$
  • common ratio $r= (1 – p)$
  • $1 – r = 1 – (1 – p) = 1 – 1 + p = p$

The formula for the sum to infinity of an arithmetico-geometric series is (from the link above):

$$ \lim_{n\to\infty} S_{n}= \frac{a}{(1-r)} + \frac{rd}{(1 – r^2)} = \frac{p}{p} + \frac{(1-p)p}{p^2} = \frac{p^2 + p – p^2}{p^2} = \frac{p}{p^2} = \frac{1}{p}$$

Note: I have not checked the proof / correctness of the formula given on the wikipedia page.

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 $nx^{n-1}$, 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.

Best Answer

There are actually three different proofs offered at the link there so your question "why do you differentiate" doesn't really make sense*, since it's clear from the very place you link to that there are multiple methods. You may have located a fourth way.

* The answer is "I don't!"; I've never used that way (though I've seen it done).

The way I've seen it done probably most often is to compute $(1-p)S$, which has the same terms as $S$ but shifted by one. We then obtain $pS$ by term-by-term subtraction, which cancels to a simpler case on the RHS.

[This looks to me like it would be a special case of your approach.]

$E(X^2)$ can be done the same way (though again there are multiple approaches that work).

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