Assume $X_i\sim \mathrm{Uniform}(0,1)$ and $X_{(1)}<\dots<X_{(n)}$ the order statistics. For any fixed k, find the asymptotic distribution of $nX_{(k)}$ as $n \rightarrow\infty$.
What we learned in class is that if $k=np$, then $\sqrt{n}(X_{(k)}-p)\rightarrow \mathcal{N}(0,p(1-p))$, but I don't know how to proceed with $nX_{(k)}$. Maybe it has something to do with $k$ being fixed?
Any help would be appreciated.
First of all note that $X_{(k)}$ has a $\mathsf{Beta}(k,n-k+1)$ distribution (discussed here for example).
It helps to recall that distribution functions of Beta distribution and Binomial distribution are directly related, as is the case with those of Gamma distribution and Poisson distribution.
In particular, we have the following identities:
For $a,b>0$ and $0<x<1$,
$$\int_0^x \frac{u^{a-1}(1-u)^{b-1}}{B(a,b)}\,du=1-\sum_{j=0}^{\lfloor{a-1}\rfloor}\binom{a+b-1}{j}x^j(1-x)^{a+b-1-j} \tag{1}$$
And for $x,p>0$,
$$\int_0^x \frac{e^{-u}u^{p-1}}{\Gamma(p)}\,du=1-\sum_{j=0}^{\lfloor{p-1}\rfloor}\frac{e^{-x}x^j}{j!} \tag{2}$$
Let us take some particular values of $k$.
For $k=1$, we directly have
\begin{align} P(nX_{(1)}\le t)&=1-P\left(X_{(1)}>\frac tn\right) \\&=1-\left\{P\left(X_1>\frac tn\right)\right\}^n \\&=\begin{cases}0 &,\text{ if }t<0 \\1-\left(1-\frac tn\right)^n &,\text{ if }0\le t<n \\1 &,\text{ if }t\ge n \end{cases} \\&\stackrel{n\to \infty}\longrightarrow \begin{cases} 1-e^{-t}&,\text{ if }t\ge 0 \\ 0&,\text{ if }t<0 \end{cases} \\&=P(Y_1\le t)\,, \end{align}
where $Y_1\sim \mathsf{Exp}(1)$, i.e. an exponential distribution with mean $1$.
For $k=2$, we use relation $(1)$ to get
\begin{align} P(nX_{(2)}\le t)&=P\left(X_{(2)}\le \frac tn\right) \\&=\begin{cases} 0&,\text{ if }t<0 \\ 1-\left(1-\frac tn\right)^n -\binom{n}{1}\frac tn\left(1-\frac tn\right)^{n-1} &,\text{ if }0\le t<n \\ 1&,\text{ if }t\ge n \end{cases} \\&\stackrel{n\to \infty}\longrightarrow \begin{cases} 1-e^{-t}-te^{-t} &,\text{ if }t\ge 0 \\ 0&,\text{ if }t<0 \end{cases}\quad, \\&=P(Y_1+Y_2\le t)\,, \end{align}
where $Y_1,Y_2$ are i.i.d $\mathsf{Exp}(1)$. In other words, the limiting distribution is a Gamma distribution.
Asymptotic distribution of $nX_{(k)}$ for a general $k$ can be derived in a similar fashion.