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.

$\blacksquare$


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