Variance of Shifted Geometric Distribution/Proof 1
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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
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: $\expect X = \dfrac 1 p$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {p^2} - \frac 1 p\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {1 - p} {p^2}\) |
$\blacksquare$
Sources
- 1986: Geoffrey Grimmett and Dominic Welsh: Probability: An Introduction ... (previous) ... (next): $\S 2.4$: Expectation: Example $24$