Expectation of Shifted Geometric Distribution/Proof 1

Proof
From the definition of expectation:


 * $\expect X = \displaystyle \sum_{x \mathop \in \Omega_X} x \map \Pr {X = x}$

By definition of shifted geometric distribution:
 * $\expect X = \displaystyle \sum_{k \mathop \in \Omega_X} k p \paren {1 - p}^{k - 1}$

Let $q = 1 - p$: