Variance of Shifted Geometric Distribution/Proof 2
Jump to navigation
Jump to search
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
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.
$\blacksquare$
Sources
- 1986: Geoffrey Grimmett and Dominic Welsh: Probability: An Introduction ... (previous) ... (next): $\S 4.3$: Moments: Example $21$