Expectation of Geometric Distribution/Formulation 2/Proof 1

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Theorem

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


Then the expectation of $X$ is given by:

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


Proof

From the definition of expectation:

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

By definition of geometric distribution:

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

Let $q = 1 - p$:

\(\ds \expect X\) \(=\) \(\ds p \sum_{k \mathop \ge 0} k q^k\) as $\Omega_X = \N$
\(\ds \) \(=\) \(\ds p \sum_{k \mathop \ge 1} k q^k\) as the $k = 0$ term is zero
\(\ds \) \(=\) \(\ds pq \sum_{k \mathop \ge 1} k q^{k - 1}\)
\(\ds \) \(=\) \(\ds p q \frac 1 {\paren {1 - q}^2}\) Derivative of Geometric Sequence
\(\ds \) \(=\) \(\ds \frac {p q} {p^2}\) as $q = 1 - p$
\(\ds \) \(=\) \(\ds \frac {1 - p} p\)

$\blacksquare$