Expectation of Geometric Distribution/Formulation 2/Proof 1
Jump to navigation
Jump to search
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 $q = 1 - p$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {1 - p} p\) |
$\blacksquare$