Geometric Distribution Gives Rise to Probability Mass Function

From ProofWiki
Jump to navigation Jump to search

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.

$\blacksquare$