Expectation of Geometric Distribution/Formulation 1/Proof 2

Theorem

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

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

Then the expectation of $X$ is given by:

$\expect X = \dfrac p {1 - p}$

Proof

From the Probability Generating Function of Geometric Distribution:

$\map {\Pi_X} s = \dfrac q {1 - p s}$

where $q = 1 - p$.

From Expectation of Discrete Random Variable from PGF:

$\expect X = \map {\Pi'_X} 1$

We have:

\(\ds \map {\Pi'_X} s\)

\(=\)

\(\ds \map {\frac \d {\d s} } {\frac q {1 - p s} }\)

\(\ds \)

\(=\)

\(\ds \frac {q p} {\paren {1 - p s}^2}\)

Derivatives of PGF of Geometric Distribution

Plugging in $s = 1$:

\(\ds \map {\Pi'_X} 1\)

\(=\)

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

\(\ds \)

\(=\)

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

as $q = 1 - p$

Hence the result.

$\blacksquare$