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The expected value of a Geometric Series

  1. Sep 29, 2004 #1
    I'm supposed to prove that in a geometric distribution, the expected value,

    [tex]
    \mu = \frac{1}{p}
    [/tex]

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

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

    [tex]
    \mu_x = E(x) = \sum x \cdot p \cdot (1-p)^{x-1}
    [/tex]

    [tex]
    \mu_x = p \sum x \cdot (1-p)^{x-1}
    [/tex]

    [tex]
    \mu_x = p \sum x \cdot (1-p)^x \cdot (1-p)^{-1}
    [/tex]

    [tex]
    \mu_x = \frac{p}{1-p} \sum x \cdot (1-p)^x
    [/tex]

    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
     
  2. jcsd
  3. Sep 29, 2004 #2

    matt grime

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    can you sum y^r as r goes from 1 to n. what if you differentiate both sides?
     
  4. Sep 29, 2004 #3
    I am assuming that [tex]y^r[/tex] is [tex](1-p)^x[/tex]. If I would convert the summation into its series and differentiate both sides, what would be the derivative of [tex]\mu_x[/tex]?
     
  5. Sep 29, 2004 #4

    matt grime

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    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.
     
  6. Sep 29, 2004 #5
    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?
     
  7. Sep 30, 2004 #6

    matt grime

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    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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