Expectation of Shifted Geometric Distribution/Proof 2

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 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 PGF, we have:


 * $E \left({X}\right) = \Pi'_X \left({1}\right)$

We have:

Plugging in $s = 1$:

Hence the result.