Expectation of Geometric Distribution/Formulation 1

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Theorem

Let $X$ be a discrete random variable with the geometric distribution with parameter $p$ for some $0 < p < 1$.

$\map X \Omega = \set {0, 1, 2, \ldots} = \N$
$\map \Pr {X = k} = \paren {1 - p} p^k$


Then the expectation of $X$ is given by:

$\map E X = \dfrac p {1-p}$


Proof 1

From the definition of expectation:

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

By definition of geometric distribution:

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

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:

$\map {\Pi_X} s = \dfrac q {1 - ps}$

where $q = 1 - p$.


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

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


We have:

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


Plugging in $s = 1$:

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

Hence the result.

$\blacksquare$