# Expectation of Shifted Geometric Distribution

## Theorem

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

Then the expectation of $X$ is given by:

$\expect X = \dfrac 1 p$

## Proof 1

From the definition of expectation:

$\expect X = \ds \sum_{x \mathop \in \Omega_X} x \map \Pr {X = x}$

By definition of shifted geometric distribution:

$\expect X = \ds \sum_{k \mathop \in \Omega_X} k p \paren {1 - p}^{k - 1}$

Let $q = 1 - p$:

 $\ds \expect X$ $=$ $\ds p \sum_{k \mathop \ge 0} k q^{k - 1}$ as $\Omega_X = \N$ $\ds$ $=$ $\ds p \sum_{k \mathop \ge 1} k q^{k - 1}$ The term in $k = 0$ vanishes $\ds$ $=$ $\ds p \frac 1 {\paren {1 - q}^2}$ Derivative of Geometric Sequence $\ds$ $=$ $\ds \frac p {p^2}$ as $q = 1 - p$ $\ds$ $=$ $\ds \frac 1 p$

$\blacksquare$

## Proof 2

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 Discrete Random Variable from PGF, we have:

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

We have:

 $\ds \map {\Pi'_X} s$ $=$ $\ds \map {\frac \d {\d s} } {\frac {p s} {1 - q s} }$ $\ds$ $=$ $\ds \frac p {\paren {1 - q s}^2}$ Derivatives of PGF of Shifted Geometric Distribution

Plugging in $s = 1$:

 $\ds \map {\Pi'_X} 1$ $=$ $\ds \frac p {\paren {1 - q}^2}$ $\ds$ $=$ $\ds \frac p {p^2}$ as $q = 1 - p$ $\ds$ $=$ $\ds \frac 1 p$

Hence the result.

$\blacksquare$