Variance of Geometric Distribution/Formulation 1/Proof 2
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 variance of $X$ is given by:
- $\var X = \dfrac p {\paren {1-p}^2}$
Proof
From Variance of Discrete Random Variable from PGF, we have:
- $\var X = \map {\Pi''_X} 1 + \mu - \mu^2$
where $\mu = \map E x$ is the expectation of $X$.
From the Probability Generating Function of Geometric Distribution, we have:
- $\map {\Pi_X} s = \dfrac q {1 - ps}$
where $q = 1 - p$.
From Expectation of Geometric Distribution, we have:
- $\mu = \dfrac p q$
From Derivatives of PGF of Geometric Distribution, we have:
- $\map {\Pi''_X} s = \dfrac {2 q p^2} {\paren {1 - ps}^3}$
Putting $s = 1$ using the formula $\map {\Pi''_X} 1 + \mu - \mu^2$:
\(\ds \var X\) | \(=\) | \(\ds \dfrac {2 q p^2} {\paren {1 - p}^3} + \dfrac p q - \paren {\dfrac p q}^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {2 \paren {1 - p} p^2} {\paren {1 - p}^3} + \dfrac p {1 - p} \paren {\dfrac {\paren {1 - p}^2 } {\paren {1 - p}^2} } - \dfrac {p^2} {\paren {1 - p}^2 } \paren {\dfrac {1 - p} {1 - p} }\) | multiplying by $1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {2 p^2 - 2 p^3 + p - 2 p^2 + p^3 - p^2 + p^3} {\paren {1 - p}^3}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {p - p^2} {\paren {1 - p}^3}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac p {\paren {1 - p}^2}\) |
$\blacksquare$