# 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:

 $\displaystyle \Pr \left({\Omega}\right)$ $=$ $\displaystyle \sum_{k \mathop \ge 0} p^k \left({1 - p}\right)$ by definition of geometric distribution $\displaystyle$ $=$ $\displaystyle \left({1 - p}\right) \sum_{k \mathop \ge 0} p^k$ $\displaystyle$ $=$ $\displaystyle \left({1 - p}\right) \frac 1 {1 - p}$ Sum of Infinite Geometric Progression $\displaystyle$ $=$ $\displaystyle 1$

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

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

$\blacksquare$