Expectation of Geometric Distribution/Formulation 2

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Theorem

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

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


Then the expectation of $X$ is given by:

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


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:

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

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 $q = 1 - p$
\(\ds \) \(=\) \(\ds \frac {1 - p} p\)

$\blacksquare$


Proof 2

By Moment Generating Function of Geometric Distribution, the moment generating function of $X$ is given by:

$\map {M_X} t = \dfrac p {1 - \paren {1 - p} e^t}$

for $t < -\map \ln {1 - p}$, and is undefined otherwise.

From Moment in terms of Moment Generating Function:

$\expect X = \map { {M_X}'} 0$

From Moment Generating Function of Geometric Distribution: First Moment:

$\map { {M_X}'} t = \dfrac {p \paren {1 - p} e^t } {\paren {1 - \paren {1 - p} e^t}^2 }$

Hence setting $t = 0$:

\(\ds \map { {M_X}'} 0\) \(=\) \(\ds \dfrac {p \paren {1 - p} } {\paren {1 - \paren {1 - p} }^2 }\)
\(\ds \) \(=\) \(\ds \dfrac {p \paren {1 - p} } {p^2 }\)
\(\ds \) \(=\) \(\ds \dfrac {1 - p} p\)

$\blacksquare$


Also presented as

The Expectation of Geometric Distribution is also presented in the form:

$\var X = \dfrac q p$

where $q$ has been defined conventionally as $q = 1 - p$.


Sources

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