I am working on the following problem.
Let $X_1, X_2, ... ,X_n$ be iid Uniform$(0,\theta)$. Show that if $Z_n=n(\theta-X_{(n)})$, $Z_n$ converges in distribution to exponential with mean $\theta$.
My work so far is that through transformation $f(z)=\frac{1}{\theta} \left(1-\frac{z/\theta}{n}\right)^{n-1} $
so $$\lim_{n\to \infty} \frac{1}{\theta} \left(1-\frac{z/\theta}{n}\right)^{n-1}=\frac{1}{\theta}e^{-z/\theta}$$
therefore it is true.
I have some iffy parts that I would like to confirm.
1), My understanding is that the cumulative density function has to approach to the converging function, i.e. $$\lim_{n \to \infty} F_n(x) = F(x)$$ by definition, so I am not sure if what I did really works or not.
If not, I will try to derive the cdf for $f(z)$ and try from there.
2), I remember that $$\lim_{n \to \infty} \left( 1+\frac{x}{n} \right)^n = e^x$$ but what if the exponent was $n-1$ instead? Intuitively I think that it should go to the same limit, but how can I show work appropriately?
I would go for CDF's at once (BTW it stands for "cumulative distribution function")
First of all let us do it for iid $Y_1,Y_2,\dots$ that have standard uniform distribution (i.e. $\theta=1$ in this context). This to avoid the in my view annoying parameters.
Then for $x>0$ and $n$ large enough: $$F_n(x)=P\left(n\left(1-Y_{(n)}\right)\leq x\right)=1-P\left(Y_{(n)}<1-\frac{x}{n}\right)=1-\left(1-\frac{x}n\right)^n$$
This reveals that $\lim_{n\to\infty}F_n(x)=F(x)$ for every $x\in\mathbb R$ where $F$ denotes the CDF of standard exponential distribution.
This justifies the statement $$n\left(1-Y_{(n)}\right)\stackrel{d}{\to}Z$$where $Z$ has standard exponential distribution.
Now take $X_i:=\theta Y_i$ for $i=1,2,\dots$ and it follows immediately that $$n\left(\theta-X_{(n)}\right)\stackrel{d}{\to}\theta Z$$
Here $\theta Z$ has exponential distribution with mean $\theta$.