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 |
$\blacksquare$