Expectation of Geometric Distribution

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

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

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

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

Let $$q = 1 - p$$:

$$ $$ $$ $$ $$

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

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.