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

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