# Variance of Shifted Geometric Distribution/Proof 1

## 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 the definition of Variance as Expectation of Square minus Square of Expectation:

$\var X = \expect {X^2} - \paren {\expect X}^2$
$\ds \expect {X^2} = \sum_{x \mathop \in \Img X} x^2 \map \Pr {X = x}$

To simplify the algebra a bit, let $q = 1 - p$, so $p + q = 1$.

Thus:

 $\ds \expect {X^2}$ $=$ $\ds \sum_{k \mathop \ge 0} k^2 p q^{k - 1}$ Definition of Shifted Geometric Distribution, with $p + q = 1$ $\ds$ $=$ $\ds \sum_{k \mathop \ge 1} k^2 p q^{k - 1}$ The term in $k=0$ is zero, so we change the limits $\ds$ $=$ $\ds \sum_{k \mathop \ge 1} k \paren {k + 1} p q^{k - 1} - \sum_{k \mathop \ge 1} k p q^{k - 1}$ splitting sum up into two $\ds$ $=$ $\ds \sum_{k \mathop \ge 1} k \paren {k + 1} p q^{k - 1} - \frac 1 p$ Second term is Expectation of Shifted Geometric Distribution $\ds$ $=$ $\ds p \frac 2 {\paren {1 - q}^3} - \frac 1 p$ Derivative of Geometric Sequence: Corollary $\ds$ $=$ $\ds \frac 2 {p^2} - \frac 1 p$ putting $p = 1 - q$ back in and simplifying

Then:

 $\ds \var X$ $=$ $\ds \expect {X^2} - \paren {\expect X}^2$ $\ds$ $=$ $\ds \frac 2 {p^2} - \frac 1 p - \frac 1 {p^2}$ Expectation of Shifted Geometric Distribution: $\expect X = \dfrac 1 p$ $\ds$ $=$ $\ds \frac 1 {p^2} - \frac 1 p$ $\ds$ $=$ $\ds \frac {1 - p} {p^2}$

$\blacksquare$