The expected value of a Geometric Series

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The discussion focuses on proving the expected value of a geometric distribution, expressed as μ = 1/p, without using moment generating functions. The user begins with the definition of expected value and manipulates the summation but encounters difficulty in evaluating it. Participants suggest differentiating the summation formula for y^r to derive a solution. Clarifications are made regarding the proper handling of factors in the summation process. The conversation emphasizes the importance of correctly applying differentiation to evaluate the series involved in the expected value calculation.
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I'm supposed to prove that in a geometric distribution, the expected value,

<br /> \mu = \frac{1}{p}<br />

without the use of moment generating functions (whatever that is)

I start off with the very definition of the expected value.

<br /> \mu_x = E(x) = \sum x \cdot p \cdot (1-p)^{x-1}<br />

<br /> \mu_x = p \sum x \cdot (1-p)^{x-1}<br />

<br /> \mu_x = p \sum x \cdot (1-p)^x \cdot (1-p)^{-1}<br />

<br /> \mu_x = \frac{p}{1-p} \sum x \cdot (1-p)^x<br />

Now I get stuck because I don't know how to evaluate the summation. Can anyone help me out?

btw, x starts from 1 to n
 
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can you sum y^r as r goes from 1 to n. what if you differentiate both sides?
 
I am assuming that y^r is (1-p)^x. If I would convert the summation into its series and differentiate both sides, what would be the derivative of \mu_x?
 
erm, what? i indicated to you how to sum a certain kind of series, the series you wanted to sum. I'm not doing anything with differentiating mu_x.
 
err... sorry, my bad. So, when you said differentiate both sides I thought both sides of the equation. What you really mean is that in order to evaluate the summation you need to differentiate, am I understanding it right?
 
you know a formula :

S(n) = sum 1 to n of y^r

that is anequation in y, diff wrt to y and you'll find a formula for a sum that looks a lot like the one you want to sum in your problem. you've pulled that factor of 1/(1-p) out when you shouldn't have: it'll make it more transparent when you put it back in.
 
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