Variance of Geometric Distribution/Formulation 1

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Theorem

Let $X$ be a discrete random variable with the geometric distribution with parameter $p$ for some $0 < p < 1$.

$\map X \Omega = \set {0, 1, 2, \ldots} = \N$
$\map \Pr {X = k} = \paren {1 - p} p^k$


Then the variance of $X$ is given by:

$\var X = \dfrac p {\paren {1-p}^2}$


Proof 1

From the definition of Variance as Expectation of Square minus Square of Expectation:

$\var X = \expect {X^2} - \paren {\expect X}^2$

From Expectation of Function of Discrete Random Variable:

$\ds \expect {X^2} = \sum_{x \mathop \in \Omega_X} x^2 \map \Pr {X = x}$


Let $q = 1 - p$


Thus:

\(\ds \expect {X^2}\) \(=\) \(\ds \sum_{k \mathop \ge 1} k^2 q p^k\) Geometric Distribution: Formulation 1, with $p + q = 1$
\(\ds \) \(=\) \(\ds p \sum_{k \mathop \ge 1} k^2 q p^{k-1}\)
\(\ds \) \(=\) \(\ds p \paren {\frac 2 {q^2} - \frac 1 q}\) Variance of Shifted Geometric Distribution: Proof 1


Then:

\(\ds \var X\) \(=\) \(\ds \expect {X^2} - \paren {\expect X}^2\)
\(\ds \) \(=\) \(\ds p \paren {\frac 2 {\paren {1 - p}^2} - \frac 1 {1 - p} } - \frac {p^2} {\paren {1 - p}^2}\) Expectation of Geometric Distribution: Formulation 1: $\expect X = \dfrac p {1 - p}$
\(\ds \) \(=\) \(\ds \frac {2 p} {\paren {1 - p}^2} - \frac p {1 - p} \paren {\frac {1 - p} {1 - p} } - \frac {p^2} {\paren {1 - p}^2}\) multiplying by $1$
\(\ds \) \(=\) \(\ds \frac {2p - p + p^2 - p^2} {\paren {1 - p}^2}\)
\(\ds \) \(=\) \(\ds \frac p {\paren {1 - p}^2}\)

$\blacksquare$


Proof 2

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 $1$
\(\ds \) \(=\) \(\ds \dfrac {2 p^2 - 2 p^3 + p - 2 p^2 + p^3 - p^2 + p^3} {\paren {1 - p}^3}\)
\(\ds \) \(=\) \(\ds \dfrac {p - p^2} {\paren {1 - p}^3}\)
\(\ds \) \(=\) \(\ds \dfrac p {\paren {1 - p}^2}\)

$\blacksquare$