Expectation of Geometric Distribution/Proof 1

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Theorem

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


Formulation 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:

$\expect X = \dfrac p {1 - p}$


Formulation 2

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


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:

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

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 $q = 1 - p$

$\blacksquare$