Geometric distributions converging to exponential distribution This is a pretty common question in probability and there are already a few answers on the site. HOWEVER, my Professor has put a little twist on it and I can't piece it together anymore.
QUESTION:
Let $(X_n)$ be a sequence of geometric random variables, each with parameter $p_n$ respectively. Suppose $p_n \rightarrow 0$. Let $(\theta_n)$ be a sequence of positive real numbers with $\frac{p_n}{\theta_n} \rightarrow \lambda$ for some $0<\lambda < \infty$. Let $Y_n = \theta_n X_n$. Show that $Y_n$ converges in distribution to an exponential distribution with parameter $\lambda$.
MY ATTEMPT:
Fix $x \in \mathbb{R}^+$. Let $Z$ be an exponential random variable with parameter $\lambda$. Then we have
\begin{align*}
P(Y_n \leq x) &=P\bigg(X_n \leq \frac{x}{\theta_n}\bigg)\\
&= 1-(1-p_n)^{\big\lfloor \frac{x}{\theta_n}\big\rfloor} \\
& \approx 1-e^{-\frac{p_n}{\theta_n} x} \tag{large n}\\
& \rightarrow 1- e^{-\lambda x} \\
&= P(Z \leq x),
\end{align*}
as desired.
PROBLEMS WITH MY SOLUTION
In my third line, I am just using my intuition. I am unable to prove this, nor can I find any theorems which might help.
Any input is appreciated! Thanks :)
 A: First write $$\lim_{n\to \infty} \log\left((1-p_n)^{\lfloor x/\theta_n\rfloor}\right) = \lim_{n\to\infty} \lfloor x/\theta_n\rfloor \log(1-p_n).$$
Since $p_n\to 0$ as $n\to\infty,$ we can write the above as
$$\lim_{n\to\infty} -\lfloor x/\theta_n\rfloor p_n.$$
Now $x/\theta_n -1 \le \lfloor x/\theta_n\rfloor \le x/\theta_n$ and $p_n\to 0.$ So Sandwich applies and tells us that the above limit equals $-\lambda x.$ Therefore we conclude that $$\lim_{n\to \infty}(1-p_n)^{\lfloor x/\theta_n\rfloor} =e^{-\lambda x}.$$
A: Let's try to give a clear meaning to your "$\approx$" with linear approximations. By definition of the floor function
$$\frac{x}{\theta_n}-1<\left\lfloor \frac{x}{\theta_n}\right\rfloor\leq \frac{x}{\theta_n}$$
Then write the Taylor expansion
$$\ln \left(1-p_n\right)=-p_n+o(p_n)$$
For $n$ large enough, this expression is negative, therefore
$$-\left(\frac{x}{\theta_n}-1\right)p_n+o(1)>\left\lfloor \frac{x}{\theta_n}\right\rfloor\ln\left(1-p_n\right)\geq -\frac{x}{\theta_n}p_n+o(1)$$
where I simplified the $o(p_n)$ using $o(xp_n/\theta_n)=o(x\lambda)=o(1)$ for $x>0$ and $o(p_n)=o(1)$. Finally, applying $\exp$ yields
$$e^{-\frac{x}{\theta_n}p_n}e^{o(1)+p_n}>e^{\left\lfloor \frac{x}{\theta_n}\right\rfloor\ln\left(1-p_n\right)}\geq e^{-\frac{x}{\theta_n}p_n}e^{o(1)}$$
Taking the limit proves your result.
