Geometric Distribution Gives Rise to Probability Mass Function

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

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.

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:


 * $\Omega \left({X}\right) = \N = \left\{{0, 1, 2, \ldots}\right\}$


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

Then:

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

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

Proof for Shifted Geometric Distribution
By definition:


 * $\Omega \left({Y}\right) = \N^* = \left\{{1, 2, 3, \ldots}\right\}$


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

Then:

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