Expectation of Geometric Distribution

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:

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

By definition of geometric distribution:

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

Let $q = 1 - p$:

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

$\blacksquare$

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 P.G.F., we have:

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

We have:

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

Plugging in $s = 1$:

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

Hence the result.

$\blacksquare$