# 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.

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:

$\map \Omega X = \N = \set {0, 1, 2, \ldots}$
$\map \Pr {X = k} = p^k \paren {1 - p}$

Then:

 $\ds \map \Pr \Omega$ $=$ $\ds \sum_{k \mathop \ge 0} p^k \paren {1 - p}$ Definition of Geometric Distribution $\ds$ $=$ $\ds \paren {1 - p} \sum_{k \mathop \ge 0} p^k$ $\ds$ $=$ $\ds \paren {1 - p} \frac 1 {1 - p}$ Sum of Infinite Geometric Sequence $\ds$ $=$ $\ds 1$

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

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

$\blacksquare$