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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 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.

Starlight
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  • How did you reformulate the density function into an arithmetico-geometric series? Otherwise, the result for the sum of geometric series is very well known and it is not unusual to transform a mathematical expression into another, familiar one via simple operations to make the solution apparent. It doesn't mean your approach is wrong, but you might have needed to check Wikipedia to get the result. – Fato39 May 19 '15 at 12:37
  • Your reference gives three distinct derivations. The other two seem to answer your question. – whuber May 19 '15 at 12:46
  • @whuber I am not sure I understand your comment. How, specifically do the other two derivations helps. – Starlight May 19 '15 at 13:16
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    @Fato39 The density function does not need to be re-formulated. Rather, the formula for expectation requires us to sum n * p * (1-p)^(n-1) to infinity. This is an arithco-geometric series with a (first term) = p, d (common difference) = p, and r(common ratio) = (1 - p). After looking at other derivations, I get the feeling that this differentiation trick is required in other derivations (like that of the variance of the same distribution). Hence, that is why it is used. – Starlight May 19 '15 at 13:16
  • I see. I can find no objection to your approach then, apart from that the arithmetico-geometric result is not widely known (the Wikipedia talk page also mentions that). – Fato39 May 19 '15 at 13:22
  • I dont know calculus too much - so I try to find other proofs. Also, that proof requires you to interchange summation and differentiation which requires some maths that I have not even heard of. – Starlight May 19 '15 at 13:29
  • @Fato39 But perhaps the fact that there is a banner at the top of the Wikipedia page in question stating that there are some unresolved mathematical issues that still need to be considered might give the OP pause... – Dilip Sarwate May 19 '15 at 13:44
  • @DilipSarwate The banner says that "the article does not summarize the key points". It does not question the formula itself. Nowwhere does it say that there are "mathematical issues". – Starlight May 19 '15 at 14:18
  • @Starlight Oh, cut the crap! The top item in the banner (the one immediately above the one you give above) says "This article needs attention from an expert on the subject. The specific problem is: Mathematics." To quote your words back to you; could you be a little more factual? – Dilip Sarwate May 19 '15 at 14:25
  • @DilipSarwate What about these links: http://www.artofproblemsolving.com/wiki/index.php/Arithmetico-geometric_series and http://gradestack.com/General-Aptitude/Sequence-Series-Number/Arithmetic-Geometric/15300-3083-6564-study-wtw. Also, as I said, it would be much more useful if you told me my approach was right or wrong instead of critiquing it. I don't think the derivation of the formula is too complicated to check - it is there on the same wikipedia page. Did you have a look at the derivation? – Starlight May 19 '15 at 14:27
  • @Fato39 The math formatting really improved the readability. Do you do that manually, or is there an editor where you can enter it WYSIWYG-style and then copy-paste it into the stackexchange editor. – Starlight May 20 '15 at 11:50
  • I did it by hand. If you don't know them, many LaTeX editors (IDEs) have buttons for common math operations that you can click and they provide you with the LaTeX commands as well as instructions on how to input your arguments. – Fato39 May 20 '15 at 12:00

2 Answers2

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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).

Glen_b
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  • @Glen_b The OP has not actually found a fourth way of summing the series (other than by blindly applying a formula that he found on a Wikipedia page which itself admits, in a banner at the very top, that there are some unresolved mathematical issues in the contents of that page). – Dilip Sarwate May 19 '15 at 13:50
  • @DilipSarwate Could you be a little more factual. A. The series is an arithco-geometric series. I have checked that myself. B. The forumula for the sum on an infinite arithco-geometric series can also be found http://www.artofproblemsolving.com/wiki/index.php/Arithmetico-geometric_series and http://gradestack.com/General-Aptitude/Sequence-Series-Number/Arithmetic-Geometric/15300-3083-6564-study-wtw. Note that I am not a mathematician, and applied math usually consists of applying formula that others have found. – Starlight May 19 '15 at 14:14
  • @DilipSarwate Also it would be much more helpful if you could check the math behind what i just did (and tell me if something is wrong), rather than questioning the formula. Note that i still had to recognise that the series in question was an arithoc-geometric one. Also note that none of the issues raised in the banner pertain to the correctness of the formulae on that page. – Starlight May 19 '15 at 14:15
  • Starlight If you want anyone to "check what you have done" you must present what you have done, exactly, in the context of the mean of the geometric -- a sequence of steps that can be examined. Which is to say, give us something specific to check. – Glen_b May 19 '15 at 22:57
  • @Glen_b You are right. I added the steps stating the series, and summing it. This assumes that the formula for the sum of the series is right (will see if I check that - not sure I understand that derivation too much). – Starlight May 20 '15 at 09:48
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    Okay, your problem there is, you don't know if that derivation you rely on is correct, so you have a gap in your proof. So why not take the approach I suggested in my answer (though previously I had $p$ where you have $1-p$): write out $S=p+2p(1-p)+3p(1-p)^2+...$ and then under it write $(1-p)S$ (with the terms shifted right one) and then subtract term by term (you need all three series to be absolutely convergent) to get $(1-p)S$. You then have a much simpler problem. If you follow that approach, you may see how to prove the original result via induction -- but you won't need it. – Glen_b May 20 '15 at 10:23
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Khan Academy used the approach I thought of in the question above. Though they have done it in a much more intuitive way.

The most important thing is that it shows that the approach was valid.

Starlight
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