Geometric Distribution Gives Rise to Probability Mass Function

Theorem
Let $X$ be a discrete random variable on a probability space $\struct {\Omega, \Sigma, \Pr}$.

Let $X$ have the geometric distribution with parameter $p$ (where $0 < p < 1$).

Then $X$ gives rise to a probability mass function.

Shifted Geometric Distribution
The same result applies to the shifted geometric distribution.

Proof
By definition:


 * $\map \Omega X = \N = \set {0, 1, 2, \ldots}$


 * $\map \Pr {X = k} = p^k \paren {1 - p}$

Then:

The above result is valid, because $0 < p < 1$.

So $X$ satisfies $\map \Pr \Omega = 1$, and hence the result.