Expectation of Geometric Distribution/Proof 2

Theorem

Let $X$ be a discrete random variable with the geometric distribution with parameter $p$ for some $0 < p < 1$.

Formulation 1

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

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

Then the expectation of $X$ is given by:

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

Formulation 2

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

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

Then the expectation of $X$ is given by:

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

Proof

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.

$\blacksquare$