Variance of Geometric Distribution/Formulation 2/Proof 2
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Theorem
- $\map X \Omega = \set {0, 1, 2, \ldots} = \N$
- $\map \Pr {X = k} = p \paren {1 - p}^k$
Then the variance of $X$ is given by:
- $\var X = \dfrac {1 - p} {p^2}$
Proof
By Moment Generating Function of Geometric Distribution, the moment generating function of $X$ is given by:
- $\map {M_X} t = \dfrac p {1 - \paren {1 - p} e^t}$
for $t < -\map \ln {1 - p}$, and is undefined otherwise.
From Variance as Expectation of Square minus Square of Expectation:
- $\ds \var X = \expect {X^2} - \paren {\expect X}^2$
From Expectation of Geometric Distribution/Formulation 2, we have:
- $\expect X = \dfrac {1 - p} p$
From Moment Generating Function of Geometric Distribution: Second Moment:
- $\map { {M_X}''} t = \dfrac {p \paren {1 - p} e^t + p \paren {1 - p}^2 e^{2t} } {\paren {1 - \paren {1 - p} e^t}^3 }$
From Moment in terms of Moment Generating Function, we also have:
- $\expect {X^2} = \map { {M_X}''} 0$
Setting $t = 0$, we obtain the second moment:
\(\ds \map {M''_X} 0\) | \(=\) | \(\ds \expect {X^2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {p \paren {1 - p} + p \paren {1 - p}^2 } {\paren {1 - \paren {1 - p} }^3 }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {p \paren {1 - p} + p \paren {1 - p}^2 } {p^3 }\) |
So:
\(\ds \var X\) | \(=\) | \(\ds \dfrac {p \paren {1 - p} + p \paren {1 - p}^2 } {p^3 } - \dfrac {\paren {1 - p}^2} {p^2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {p \paren {1 - p} + p \paren {1 - p}^2 } {p^3 } - \dfrac {\paren {1 - p}^2} {p^2} \paren {\dfrac p p}\) | multiplying by $1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {1 - p} {p^2}\) |
$\blacksquare$