Variance of Geometric Distribution/Proof 2

Theorem

Let $p \in \R$ be a real number such that $0 < p < 1$.

Let $X$ be a discrete random variable with the geometric distribution with parameter $p$.

Formulation 1

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

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

Then the variance of $X$ is given by:

$\var X = \dfrac p {\paren {1-p}^2}$

Formulation 2

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

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

Then the variance of $X$ is given by:

$\var X = \dfrac {1 - p} {p^2}$

Proof

From Variance of Discrete Random Variable from PGF:

$\var X = \map {\Pi_X} 1 + \mu - \mu^2$

where $\mu = \expect X$ is the expectation of $X$.

From the Probability Generating Function of Geometric Distribution:

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

where $q = 1 - p$.

From Expectation of Geometric Distribution:

$\mu = \dfrac p q$

From Derivatives of PGF of Geometric Distribution:

$\map {\Pi_X} s = \dfrac {2 q p^2} {\left({1 - ps}\right)^3}$

Putting $s = 1$ using the formula $\map {\Pi_X} 1 + \mu - \mu^2$:

$\var X = \dfrac {2 q p^2} {\paren {1 - p}^3} + \dfrac p q - \paren {\dfrac p q}^2$

and hence the result, after some algebra.

$\blacksquare$