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.

$\blacksquare$