Expectation of Geometric Distribution

From ProofWiki
Jump to: navigation, search

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.

$\blacksquare$