Variance of Geometric Distribution/Proof 1

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Theorem

Let $p \in \R$ be a real number such that $0 < p < 1$.

Let $X$ be a discrete random variable with the geometric distribution with parameter $p$.


Formulation 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:

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


Formulation 2

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


Then the variance of $X$ is given by:

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


Proof

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:

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


To simplify the algebra a bit, let $q = 1 - p$, so $p + q = 1$.


Thus:

\(\ds \expect {X^2}\) \(=\) \(\ds \sum_{k \mathop \ge 1} k^2 q p^k\) Definition of Geometric Distribution, 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: $\expect X = \dfrac p {1 - p}$
\(\ds \) \(=\) \(\ds \frac p {\paren {1 - p}^2}\) after some algebra

$\blacksquare$