Expectation of Geometric Distribution/Formulation 2/Proof 2

Theorem

$\map X \Omega = \set {0, 1, 2, \ldots} = \N$

$\map \Pr {X = k} = p \paren {1 - p}^k$

Then the expectation of $X$ is given by:

$\map E X = \dfrac {1-p} p$

Proof

By Moment Generating Function of Geometric Distribution, the moment generating function of $X$ is given by:

$\map {M_X} t = \dfrac p {1 - \paren {1 - p} e^t}$

for $t < -\map \ln {1 - p}$, and is undefined otherwise.

From Moment in terms of Moment Generating Function:

$\expect X = \map { {M_X}'} 0$

From Moment Generating Function of Geometric Distribution: First Moment:

$\map { {M_X}'} t = \dfrac {p \paren {1 - p} e^t } {\paren {1 - \paren {1 - p} e^t}^2 }$

Hence setting $t = 0$:

\(\ds \map { {M_X}'} 0\)

\(=\)

\(\ds \dfrac {p \paren {1 - p} } {\paren {1 - \paren {1 - p} }^2 }\)

\(\ds \)

\(=\)

\(\ds \dfrac {p \paren {1 - p} } {p^2 }\)

\(\ds \)

\(=\)

\(\ds \dfrac {1 - p} p\)

$\blacksquare$