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 $1$
\(\ds \) \(=\) \(\ds \dfrac {1 - p} {p^2}\)

$\blacksquare$