Variance of Shifted Geometric Distribution

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

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

Proof
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:
 * $$E \left({X^2}\right) = \sum_{x \in \operatorname{Im} \left({X}\right)} x^2 \Pr \left({X = x}\right)$$

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 Shifted Geometric Distribution, we have:
 * $$\Pi_X \left({s}\right) = \frac {ps} {1 - qs}$$

where $$q = 1 - p$$.

From Expectation of Shifted Geometric Distribution, we have:
 * $$\mu = \frac 1 p$$

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

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

and hence the result, after some algebra.