Expectation of Geometric Distribution/Proof 2
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Theorem
Let $X$ be a discrete random variable with the geometric distribution with parameter $p$ for some $0 < p < 1$.
Formulation 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:
- $\expect X = \dfrac p {1 - p}$
Formulation 2
- $\map X \Omega = \set {0, 1, 2, \ldots} = \N$
- $\map \Pr {X = k} = p \paren {1 - p}^k$
Then the expectation of $X$ is given by:
- $\map E X = \dfrac {1-p} p$
Proof
From the 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.
$\blacksquare$