Variance of Geometric Distribution/Formulation 2
Theorem
Let $X$ be a discrete random variable with the geometric distribution with parameter $p$ for some $0 < p < 1$.
- $\map X \Omega = \set {0, 1, 2, \ldots} = \N$
- $\map \Pr {X = k} = p \paren {1 - p}^k$
Then the variance of $X$ is given by:
- $\var X = \dfrac {1 - p} {p^2}$
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:
\(\ds \expect {X^2}\) | \(=\) | \(\ds \sum_{k \mathop \ge 1} k^2 p q^k\) | 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 $1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {q \paren {q + p} } {p^2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {1 - p} {p^2}\) | $p + q = 1$ |
$\blacksquare$
Proof 2
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 $1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {1 - p} {p^2}\) |
$\blacksquare$
Also presented as
The Variance of Geometric Distribution is also presented in the form:
- $\var X = \dfrac q {p^2}$
where $q$ has been defined conventionally as $q = 1 - p$.
Sources
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): geometric distribution