Geometric Distribution Gives Rise to Probability Mass Function

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:


 * $$\operatorname{Im} \left({X}\right) = \N = \left\{{0, 1, 2, \ldots}\right\}$$


 * $$\Pr \left({X = k}\right) = p \left({1 - p}\right)^k$$

Then:

$$ $$ $$ $$

The above result is valid, because $$0 < p < 1$$ and so $$0 < 1 - p < 1$$.

So $$X$$ satisfies $$\Pr \left({\Omega}\right) = 1$$, and hence the result.

Proof for Shifted Geometric Distribution
By definition:


 * $$\operatorname{Im} \left({Y}\right) = \N^* = \left\{{1, 2, 3, \ldots}\right\}$$


 * $$\Pr \left({Y = k}\right) = p \left({1 - p}\right)^{k-1}$$

Then:

$$ $$ $$

So $$Y$$ satisfies $$\Pr \left({\Omega}\right) = 1$$, and hence the result.