Variance of Geometric Distribution

Theorem
Let $X$ be a discrete random variable with the geometric distribution with parameter $p$.

Then the variance of $X$ is given by:
 * $\operatorname{var} \left({X}\right) = \dfrac p {\left({1 - p}\right)^2}$

Proof 1
From the definition of Variance as Expectation of Square minus Square of Expectation:
 * $\operatorname{var} \left({X}\right) = E \left({X^2}\right) - \left({E \left({X}\right)}\right)^2$

From Expectation of Function of Discrete Random Variable:
 * $\displaystyle E \left({X^2}\right) = \sum_{x \in \Omega_X} x^2 \Pr \left({X = x}\right)$

To simplify the algebra a bit, let $q = 1 - p$, so $p+q = 1$.

Thus:

Then:

Proof 2
From Variance of Discrete Random Variable from P.G.F., we have:
 * $\operatorname{var} \left({X}\right) = \Pi''_X \left({1}\right) + \mu - \mu^2$

where $\mu = E \left({x}\right)$ is the expectation of $X$.

From the Probability Generating Function of Geometric Distribution, we have:
 * $\Pi_X \left({s}\right) = \dfrac q {1 - ps}$

where $q = 1 - p$.

From Expectation of Geometric Distribution, we have:
 * $\mu = \dfrac p q$

From Derivatives of PGF of Geometric Distribution, we have:
 * $\Pi''_X \left({s}\right) = \dfrac {2 q p^2} {\left({1 - ps}\right)^3}$

Putting $s = 1$ using the formula $\Pi''_X \left({1}\right) + \mu - \mu^2$:
 * $\displaystyle \operatorname{var} \left({X}\right) = \frac {2 q p^2} {\left({1-p}\right)^3} + \frac p q - \left({\frac p q}\right)^2$

and hence the result, after some algebra.