I'm supposed to prove that in a geometric distribution, the expected value,(adsbygoogle = window.adsbygoogle || []).push({});

[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

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# The expected value of a Geometric Series

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