Expectation of Geometric Distribution/Formulation 1/Proof 1
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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:
- $\ds \expect X = \sum_{x \mathop \in \Omega_X} x \map \Pr {X = x}$
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 $q = 1 - p$ |
$\blacksquare$