# Expectation of Geometric Distribution

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## Theorem

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

Then the expectation of $X$ is given by:

$E \left({X}\right) = \dfrac p {1-p}$

## Proof 1

From the definition of expectation:

$\displaystyle E \left({X}\right) = \sum_{x \mathop \in \Omega_X} x \Pr \left({X = x}\right)$

By definition of geometric distribution:

$\displaystyle E \left({X}\right) = \sum_{k \mathop \in \Omega_X} k p^k \left({1 - p}\right)$

Let $q = 1 - p$:

 $\displaystyle E \left({X}\right)$ $=$ $\displaystyle q \sum_{k \mathop \ge 0} k p^k$ $\quad$ as $\Omega_X = \N$ $\quad$ $\displaystyle$ $=$ $\displaystyle q \sum_{k \mathop \ge 1} k p^k$ $\quad$ The term for $k=0$ is zero $\quad$ $\displaystyle$ $=$ $\displaystyle q p \sum_{k \mathop \ge 1} k p^{k-1}$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle q p \frac 1 {\left({1 - p}\right)^2}$ $\quad$ Derivative of Geometric Progression $\quad$ $\displaystyle$ $=$ $\displaystyle \frac p {1 - p}$ $\quad$ as $q = 1 - p$ $\quad$

$\blacksquare$

## Proof 2

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

$\Pi_X \left({s}\right) = \dfrac q {1 - ps}$

where $q = 1 - p$.

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

$E \left({X}\right) = \Pi'_X \left({1}\right)$

We have:

 $\displaystyle \Pi'_X \left({s}\right)$ $=$ $\displaystyle \frac {\mathrm d} {\mathrm d s} \left({\frac q {1 - p s} }\right)$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \frac {q p} {\left({1 - p s}\right)^2}$ $\quad$ Derivatives of PGF of Geometric Distribution $\quad$

Plugging in $s = 1$:

 $\displaystyle \Pi'_X \left({1}\right)$ $=$ $\displaystyle \frac {q p} {\left({1 - p}\right)^2}$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \frac p {1 - p}$ $\quad$ as $q = 1 - p$ $\quad$

Hence the result.

$\blacksquare$