Geometric Distribution Gives Rise to Probability Mass Function
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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.
$\blacksquare$