Geometric Distribution Gives Rise to Probability Mass Function/Shifted

Theorem
Let $Y$ be a discrete random variable on a probability space $\left({\Omega, \Sigma, \Pr}\right)$.

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

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

Proof
By definition:


 * $\Omega \left({Y}\right) = \N_{>0} = \left\{{1, 2, 3, \ldots}\right\}$


 * $\Pr \left({Y = k}\right) = p \left({1 - p}\right)^{k-1}$

Then:

So $Y$ satisfies $\Pr \left({\Omega}\right) = 1$, and hence the result.