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:

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Then:

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