Variance of Shifted Geometric Distribution/Proof 2

Theorem

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

Then the variance of $X$ is given by:

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

Proof

From Variance of Discrete Random Variable from PGF, we have:

$\var X = \map {\Pi''_X} 1 + \mu - \mu^2$

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

From the Probability Generating Function of Shifted Geometric Distribution, we have:

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

where $q = 1 - p$.

From Expectation of Shifted Geometric Distribution, we have:

$\mu = \dfrac 1 p$

From Derivatives of PGF of Shifted Geometric Distribution, we have:

$\map {\Pi''_X} s = \dfrac {p q} {\paren {1 - q s}^3}$

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

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

and hence the result, after some algebra.

$\blacksquare$