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:
- $\expect X = \dfrac p {1 - p}$
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:
- $\map {\Pi_X} s = \dfrac q {1 - p s}$
where $q = 1 - p$.
From Expectation of Discrete Random Variable from PGF:
- $\expect X = \map {\Pi'_X} 1$
We have:
| \(\ds \map {\Pi'_X} s\) | \(=\) | \(\ds \map {\frac \d {\d s} } {\frac q {1 - p s} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {q p} {\paren {1 - p s}^2}\) | 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.
$\blacksquare$
Proof 3
From the definition of expectation:
- $\ds \expect X = \sum_{x \mathop \in \Omega_X} x \map \Pr {X = x}$
Then
| \(\ds \expect X\) | \(=\) | \(\ds \sum_{k \mathop \in \N} k p^k \paren {1 - p}\) | Definition of Geometric Distribution | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{k \mathop \ge 1} k p^k \paren {1 - p}\) | as the $k = 0$ term is zero | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{k \mathop \ge 1} k p^k - k p^{k+1}\) | Real Multiplication Distributes over Addition |
By the Ratio Test, both $\sum_{k \mathop \ge 1} k p^k$ and $\sum_{k \mathop \ge 1} k p^{k+1}$ converge absolutely.
From Absolutely Convergent Real Series is Convergent, both series converge.
| \(\ds \sum_{k \mathop \ge 1} k p^k - k p^{k+1}\) | \(=\) | \(\ds \sum_{k \mathop \ge 1} k p^k - \sum_{k \mathop \ge 1} k p^{k+1}\) | Convergent Series can be Added Term by Term | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{k \mathop \ge 1} k p^k - \sum_{k \mathop \ge 2} \paren {k - 1} p^k\) | Translation of Index Variable of Summation | |||||||||||
| \(\ds \) | \(=\) | \(\ds p + \sum_{k \mathop \ge 2} k p^k - \sum_{k \mathop \ge 2} \paren {k - 1} p^k\) | Moving out the first term | |||||||||||
| \(\ds \) | \(=\) | \(\ds p + \sum_{k \mathop \ge 2} k p^k - \paren {k - 1} p^k\) | Recombining the two convergent series | |||||||||||
| \(\ds \) | \(=\) | \(\ds p + \sum_{k \mathop \ge 2} p^k\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds p + \frac {p^2}{1-p}\) | Sum of Infinite Geometric Sequence | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {p \paren {1-p} }{1-p} + \frac {p^2} {1-p}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac p {1-p}\) |
$\blacksquare$