Geometric Distribution Gives Rise to Probability Mass Function/Shifted
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Theorem
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:
- $\Omega \left({Y}\right) = \N_{>0} = \left\{{1, 2, 3, \ldots}\right\}$
- $\Pr \left({Y = k}\right) = p \left({1 - p}\right)^{k-1}$
Then:
\(\ds \Pr \left({\Omega}\right)\) | \(=\) | \(\ds \sum_{k \mathop \ge 1} p \left({1 - p}\right)^{k-1}\) | by definition of shifted geometric distribution | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{j \mathop \ge 0} p \left({1 - p}\right)^j\) | putting $j = k-1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 1\) | Geometric Distribution Gives Rise to Probability Mass Function |
So $Y$ satisfies $\Pr \left({\Omega}\right) = 1$, and hence the result.
$\blacksquare$
Sources
- 1986: Geoffrey Grimmett and Dominic Welsh: Probability: An Introduction ... (previous) ... (next): $\S 2.2$: Examples