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:
 * $E \left({X}\right) = \dfrac 1 p$

Proof 1
From the definition of expectation:
 * $\displaystyle E \left({X}\right) = \sum_{x \in \Omega_X} x \Pr \left({X = x}\right)$

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

Let $q = 1 - p$:

Proof 2
From the Probability Generating Function of Shifted Geometric Distribution, we have:
 * $\displaystyle \Pi_X \left({s}\right) = \frac {ps} {1 - qs}$

where $q = 1 - p$.

From Expectation of Discrete Random Variable from P.G.F., we have:
 * $E \left({X}\right) = \Pi'_X \left({1}\right)$

We have:

Plugging in $s = 1$:

Hence the result.