Limiting Distribution of $Z_n$ Let $Y_n$ denote the $n^{th}$ order statistic of a random sample from the uniform distribution, $U(0,\theta)$ and let $Z_n=n(\theta-Y_n)$. The probability density function of $Z_n$ is $$h_n(z)=\frac{(\theta-z/n)^{n-1}}{\theta^n},0<z<n\theta$$ and  0 elsewhere. 
Find the limiting distribution of $Z_n$. 
Is there any hint to this question? I do not know where to start? 
Should I find the cdf first and then prove it by taking $n$ to infinity. 
 A: Your idea with finding the cdf and then letting $n\to\infty$ is a good idea. After all $Z_n\to Z$ in distribution means that $F_{Z_n}(z)\to F_{Z}(z)$ for all points in the continuity set of $F_Z$ where $F$ denotes cdf. You can use Scheffes theorem and work with densities but this requires a theorem and more machinery.
For notation let $Y_n$ be the maximum of $n$ i.i.d uniform random variables $X_1,\dotsc, X_n$ on $[0,\theta]$.
We claim that $Z_n$ converges in distribution to an exponential distribution with expected value $\theta$. To see this note that for $z>0$ we have that
$$
P(Z_n\leq z)=P\left(Y_n\ge \theta-\frac{z}{n}\right)=1-P(Y_n<\theta-\frac{z}{n})=1-\left(\theta-\frac{z}{n}\right)^n\frac{1}{\theta^n}\tag{0}
$$
(for $z\leq 0$ obviously $P(Z_n\leq z)=0$) where in the last equality in $(0)$ we used the fact that
$$
P(Y_n<y)=P(Y_n\leq y)=P(X_1\leq y)^n=\frac{y^n}{\theta^n}
$$
Hence
$$
P(Z_n\leq z)=1-\left
(1-\frac{z/\theta}{n}
\right)^n\to1-e^{-z/\theta}
$$
as $n\to\infty$ which is the cdf of an exponential distribution with expected value $\theta$.
