Question
Suppose the following holds $$ n(\theta - X_{(n)}) \xrightarrow{L} E(0, \theta). $$ How can this be transformed as follows? $$ \frac{n}{\theta}(\theta - X_{(n)}) \xrightarrow{L} E(0, 1). $$
What I know
$E(0, \theta)$ denotes the exponential distribution with density $\frac{1}{\theta}e^{-x/\theta}$ when $x > 0$.
from $n(\theta - X_{(n)}) \xrightarrow{L} E(0, \theta)$, it can be transformed as follows. $$ n(\theta - X_{(n)}) \xrightarrow{L} 1 - e^{-x/\theta}\\ \therefore \frac{n}{\theta}(\theta - X_{(n)}) \xrightarrow{L} \frac{1}{\theta}\left(1 - e^{-x/\theta} \right), $$
$$ CDF: \frac{1}{\theta}\left(1 - e^{-x/\theta} \right) \implies PDF: \frac{1}{\theta^2}e^{-x/\theta}.\ ?? $$
From the definition of convergence in law, we have
$$ \lim_{n\rightarrow \infty} \mathbf P[n(\theta - X_n)\leq x] = 1 - \exp(-x/\theta). $$
Now consider $\frac{n}{\theta}(\theta - X_n)$, then
$$ \begin{align} \lim_{n\rightarrow \infty} \mathbf P\left[\frac{n}{\theta}(\theta - X_n) \leq x\right] &= \lim_{n\rightarrow \infty} \mathbf P\left[n(\theta - X_n) \leq \theta x\right] \\ & = 1 - \exp(-x) \end{align} $$
Which is to say that this later sequence of variables converges in law to $E(0,1)$.