Expectation of Shifted Geometric Distribution

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

Then the expectation of $$X$$ is given by:
 * $$E \left({X}\right) = \frac 1 p$$

Proof
From the definition of expectation:
 * $$E \left({X}\right) = \sum_{x \in \operatorname{Im} \left({X}\right)} x \Pr \left({X = x}\right)$$

By definition of shifted geometric distribution:
 * $$E \left({X}\right) = \sum_{k \in \operatorname{Im} \left({X}\right)} k p \left({1 - p}\right)^{k-1}$$

Let $$q = 1 - p$$:

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