Variance of Geometric Distribution/Formulation 2/Proof 1

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

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:

\(\ds \expect {X^2}\)

\(=\)

\(\ds \sum_{k \mathop \ge 1} k^2 p q^k\)

Geometric Distribution: Formulation 2, with $p + q = 1$

\(\ds \)

\(=\)

\(\ds \sum_{k \mathop \ge 1} \paren {k^2 - k + k} p q^k\)

adding $0$

\(\ds \)

\(=\)

\(\ds \sum_{k \mathop \ge 1} \paren {k^2 - k} p q^k + \sum_{k \mathop \ge 1} k p q^k\)

splitting sum

\(\ds \)

\(=\)

\(\ds p q^2 \sum_{k \mathop \ge 1} k \paren {k - 1} q^{k-2} + p q \sum_{k \mathop \ge 1} k q^{k-1}\)

\(\ds \)

\(=\)

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

Derivative of Geometric Sequence/Corollary and Derivative of Geometric Sequence

\(\ds \)

\(=\)

\(\ds p q^2 \dfrac 2 {p^3 } + p q \dfrac 1 {p^2 }\)

\(\ds \)

\(=\)

\(\ds \dfrac {2 q^2} {p^2 } + \dfrac q p\)

Then:

\(\ds \var X\)

\(=\)

\(\ds \expect {X^2} - \paren {\expect X}^2\)

\(\ds \)

\(=\)

\(\ds \dfrac {2 q^2} {p^2 } + \dfrac q p - \frac {q^2} {p^2}\)

Expectation of Geometric Distribution: Formulation 2: $\expect X = \dfrac {1 - p} p = \dfrac q p$

\(\ds \)

\(=\)

\(\ds \dfrac {q^2} {p^2 } + \dfrac q p \paren {\dfrac p p}\)

multiplying by $1$

\(\ds \)

\(=\)

\(\ds \frac {q \paren {q + p} } {p^2}\)

\(\ds \)

\(=\)

\(\ds \frac {1 - p} {p^2}\)

$p + q = 1$

$\blacksquare$