How do I evaluate the limit $\lim_{n \to +\infty}(1-(1-p/n)^{[nx]})=1-e^{-px}$?

Question

Let $(X_n)_{n\ge 1}$ be a sequence of random variables , $\forall n\ge 1, X_n \sim Geo(p/n)$ with $p>0$ fixed. Study the convergence of $Y_n=X_n/n$ in distribution. (without using charateristic functions)

Solution

Convergence in distribution is equivalence to convergence of the cumulative functions so

$F_{Y_n}(x)=\mathbb{P}(Y_n\le x)=\mathbb{P}(X_n\le xn)=F_{X_n}(nx)$

Considering the cumulative function for the geometric distribution with parameter $p/n$ is $F_{X_n}(x)=1-(1-p/n)^{[x]}, $where $[x]$ is the integer part (floor function), I have that $\lim_{n \to +\infty} F_{X_n}(nx)=\lim_{n \to +\infty}(1-(1-p/n)^{[nx]}) $

If it weren't for that [.] , i could just say that $\lim_{n \to +\infty}(1-(1-p/n)^{nx})=1-e^{-px}$ and conclude the limit is an exponential distribution.

a)How do I deal with that [.]? I feel I can't just suppose x is an integer since the cumulative function is not only defined for integer numbers.

Answer 1

Note that $\lfloor u\rfloor=u+O(1)$ for all $u>0$, so we have \begin{aligned} \log\left(1-\frac pn\right)^{\lfloor xn\rfloor} &=\{xn+O(1)\}\left\{-\frac pn+O\left(1\over n^2\right)\right\} \\ &=-px+O\left(\frac1n\right) \end{aligned}

Consequently, we have $$ \left(1-\frac pn\right)^{\lfloor nx\rfloor}=e^{-px}+O\left(\frac1n\right) $$ given that $p$ and $x$ are fixed.