Expectation of Geometric Distribution/Formulation 1/Proof 3
Jump to navigation
Jump to search
Theorem
- $\map X \Omega = \set {0, 1, 2, \ldots} = \N$
- $\map \Pr {X = k} = \paren {1 - p} p^k$
Then the expectation of $X$ is given by:
- $\expect X = \dfrac p {1 - p}$
Proof
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 | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {p \paren {1-p} }{1-p} + \frac {p^2} {1-p}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac p {1-p}\) |
$\blacksquare$