Variance of Geometric Distribution/Formulation 1

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 $1$

\(\ds \)

\(=\)

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

\(\ds \)

\(=\)

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

\(\ds \)

\(=\)

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

$\blacksquare$