ThmDex – An index of mathematical definitions, results, and conjectures.
Limit of distribution function of geometric random positive integer scaled by reciprocal of index
Formulation 0
Let $G_n \in \text{Geometric}(\theta_n)$ be a D4001: Geometric random positive integer for each $n \in \mathbb{N}$ such that
(i) $\lambda \in (0, \infty)$ is a D5407: Positive real number
(ii) \begin{equation} \lim_{n \to \infty} \theta_n n = \lambda \end{equation}
Let $a \in (0, \infty)$ be a D5407: Positive real number.
Then \begin{equation} \lim_{n \to \infty} \mathbb{P} \left( \frac{G_n}{n} \leq a \right) = 1 - e^{- \lambda a} \end{equation}
Subresults
R4996
Proofs
Proof 0
Let $G_n \in \text{Geometric}(\theta_n)$ be a D4001: Geometric random positive integer for each $n \in \mathbb{N}$ such that
(i) $\lambda \in (0, \infty)$ is a D5407: Positive real number
(ii) \begin{equation} \lim_{n \to \infty} \theta_n n = \lambda \end{equation}
Let $a \in (0, \infty)$ be a D5407: Positive real number.
Since $\lim_{n \to \infty} \theta_n n = \lambda$, then $\lim_{n \to \infty} - \theta_n = 0$ and $\lim_{n \to \infty} - a \theta_n n = - a \lambda$. Therefore, using results
(i) R4805: Dual probability distribution function for geometric random positive integer
(ii) R3304: Approximating sequence for the natural exponential function

we have \begin{equation} \mathbb{P} \left( \frac{G_n}{n} > a \right) = \mathbb{P} ( G_n > n a ) = (1 - \theta_n)^{n a} = e^{- \lambda a} \end{equation} and thus \begin{equation} \mathbb{P} \left( \frac{G_n}{n} \leq a \right) = 1 - e^{- \lambda a} \end{equation} This is what was required to be shown. $\square$