Variance of Shifted Geometric Distribution
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Theorem
Let $X$ be a discrete random variable with the shifted geometric distribution with parameter $p$.
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:
- $\ds \expect {X^2} = \sum_{x \mathop \in \Img X} x^2 \map \Pr {X = x}$
To simplify the algebra a bit, let $q = 1 - p$, so $p + q = 1$.
Thus:
\(\ds \expect {X^2}\) | \(=\) | \(\ds \sum_{k \mathop \ge 0} k^2 p q^{k - 1}\) | Definition of Shifted Geometric Distribution, with $p + q = 1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop \ge 1} k^2 p q^{k - 1}\) | The term in $k=0$ is zero, so we change the limits | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop \ge 1} k \paren {k + 1} p q^{k - 1} - \sum_{k \mathop \ge 1} k p q^{k - 1}\) | splitting sum up into two | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop \ge 1} k \paren {k + 1} p q^{k - 1} - \frac 1 p\) | Second term is Expectation of Shifted Geometric Distribution | |||||||||||
\(\ds \) | \(=\) | \(\ds p \frac 2 {\paren {1 - q}^3} - \frac 1 p\) | Derivative of Geometric Sequence: Corollary | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 2 {p^2} - \frac 1 p\) | putting $p = 1 - q$ back in and simplifying |
Then:
\(\ds \var X\) | \(=\) | \(\ds \expect {X^2} - \paren {\expect X}^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 2 {p^2} - \frac 1 p - \frac 1 {p^2}\) | Expectation of Shifted Geometric Distribution: $\expect X = \dfrac 1 p$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {p^2} - \frac 1 p\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {1 - p} {p^2}\) |
$\blacksquare$
Proof 2
From Variance of Discrete Random Variable from PGF, we have:
- $\var X = \map {\Pi_X} 1 + \mu - \mu^2$
where $\mu = \expect X$ is the expectation of $X$.
From the Probability Generating Function of Shifted Geometric Distribution, we have:
- $\map {\Pi_X} s = \dfrac {p s} {1 - q s}$
where $q = 1 - p$.
From Expectation of Shifted Geometric Distribution, we have:
- $\mu = \dfrac 1 p$
From Derivatives of PGF of Shifted Geometric Distribution, we have:
- $\map {\Pi_X} s = \dfrac {p q} {\paren {1 - q s}^3}$
Putting $s = 1$ using the formula $\map {\Pi_X} 1 + \mu - \mu^2$:
- $\var X = \dfrac {p q} {\paren {1 - q}^3} + \dfrac 1 p - \paren {\dfrac 1 p}^2$
and hence the result, after some algebra.
$\blacksquare$
Also presented as
The Variance of Shifted 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
- 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): Appendix $13$: Probability distributions
- 2021: Richard Earl and James Nicholson: The Concise Oxford Dictionary of Mathematics (6th ed.) ... (previous) ... (next): Appendix $15$: Probability distributions