Determine the limiting distribution of Uniform Order Statistic I have a random sample of size $n$ from a uniform distribution 
$$U(0, \theta)$$
And I've proven that the pdf of $Y_n$, the n-th order statistic of the sample is:
$$
f_{Y_n}(y) = \frac{n}{\theta^n}  y^{n-1}  \quad \quad, 0<y<\theta , 
$$$$
f_{Y_n}(y) = 0  \quad \quad \quad \quad ,\text{elsewhere}
$$ 
Now, what I'm trying to do next is calculating the limiting distribution of $Y_n$, and I'm not sure how to do that.
Am I supposed to calculate the limit of the pdf as $ n \rightarrow \infty $ ? or the cdf?
Any help regarding the steps I need to do is appreciated!
 A: If you look at the cdf of $Y_n$, $$F_n(\delta)=\mathbb{P}(Y_n\le\delta)=(\delta/\theta)^n\qquad0\le\delta\le\theta\,,$$ you get that $F_n(\delta)$ converges to zero when $0\le\delta<\theta$, which means that $Y_n$ converges in probability to $\theta$:
$$Y_n\stackrel{\text{prob}}{\longrightarrow} \theta\,.$$ This implies that $Y_n$ also converges in distribution to the constant value random variable $\theta$:
$$Y_n\stackrel{\text{dist}}{\longrightarrow} \theta\,.$$ To get a more precise description of the asymptotic behaviour of $Y_n$, you need to zoom around $\theta$, i.e., to consider $(\theta-Y_n)$ scaled by a power of $n$, $n^\alpha$, so that, while $(\theta-Y_n)$ converges to zero in distribution and $n^\alpha$ goes to infinity, the product $$n^\alpha(\theta-Y_n)$$ converges to a standard distribution (in distribution). 
This is e.g. the case for the Central Limit theorem: if the mean of $X$ is well-defined, $\bar{X}_n-\mathbb{E}[X]$ converges to zero in distribution, while $$\sqrt{n}(\bar{X}_n-\mathbb{E}[X])=n^{1/2}(\bar{X}_n-\mathbb{E}[X])$$ converges to a Normal distribution (in distribution).
To answer your question you thus have to find the right scale $n^\alpha$ (there is only one!) and then deduce the associated limiting distribution. Hint: Remember that 
$$\lim_{n\to\infty} \left\{1-\frac{\beta}{n} \right\}^n = \exp\{-\beta\}\,.$$
A: As suggested by Xi'an, work with the CDF
$$
  F_n(t) = \Pr(Y_n\leq t)=\bigcap_{i=1}^n \Pr(X_i\leq t) = \prod_{i=1}^n \Pr(X_1\leq t) \, .
$$
Hence,
$$
  F_n(t) = \begin{cases} 
0 &,& t<0\,; \\
\left(t/\theta\right)^n &,& 0\leq t<\theta\,; \\
1 &,& t\geq\theta \, .
\end{cases}
$$
Plot a graph of this $F_n$.
For $t<0$ and $t\geq\theta$ you're done. Now, what is $\lim_{n\to\infty}F_n(t)$ for $0\leq t<\theta$?
Put everything together to find $F(t)$ such that $F_n(t)\to F(t)$ for every $t$ when $n\to\infty$.
Graph this $F$. What is the interpretation of $F$?
Is this result intuitive? Imagine yourself drawing huge samples from a $U[0,\theta]$ distribution and computing the sample maximum every time.
Can you code a short simulation in R which confirms your results?
