Expectation of Shifted Geometric Distribution/Proof 1

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

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

Proof
From the definition of expectation:


 * $\displaystyle E \left({X}\right) = \sum_{x \mathop \in \Omega_X} x \Pr \left({X = x}\right)$

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

Let $q = 1 - p$: