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:

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

By definition of geometric distribution:

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

Let $q = 1 - p$:

\(\ds \expect X\)

\(=\)

\(\ds q \sum_{k \mathop \ge 0} k p^k\)

as $\Omega_X = \N$

\(\ds \)

\(=\)

\(\ds q \sum_{k \mathop \ge 1} k p^k\)

as the $k = 0$ term is zero

\(\ds \)

\(=\)

\(\ds q p \sum_{k \mathop \ge 1} k p^{k - 1}\)

\(\ds \)

\(=\)

\(\ds q p \frac 1 {\paren {1 - p}^2}\)

Derivative of Geometric Sequence

\(\ds \)

\(=\)

\(\ds \frac p {1 - p}\)

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 PGF, we have:

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

We have:

\(\ds \Pi'_X \left({s}\right)\)

\(=\)

\(\ds \frac {\mathrm d} {\mathrm d s} \left({\frac q {1 - p s} }\right)\)

\(\ds \)

\(=\)

\(\ds \frac {q p} {\left({1 - p s}\right)^2}\)

Derivatives of PGF of Geometric Distribution

Plugging in $s = 1$:

\(\ds \Pi'_X \left({1}\right)\)

\(=\)

\(\ds \frac {q p} {\left({1 - p}\right)^2}\)

\(\ds \)

\(=\)

\(\ds \frac p {1 - p}\)

as $q = 1 - p$

Hence the result.

$\blacksquare$