Weakly converges to exponential distribution with parameter 1 Let $\{x_n\}_{n=1}^{+\infty}$ independent identically distributed uniformly on $[a,b]$ random variables, $f_n=\max\{x_1, x_2,..x_n\}$, $g_n=\min\{x_1, x_2,..x_n\}$. Prove that for $n\frac{b-f_n}{b-a}$ weakly converges to exponential distribution with parameter $1$.
 A: Set $Y_n := \frac{n(b-f_n)}{b-a}$ and let $c < d$ be given. Then we have
\begin{align*}
  P(c \le Y_n < d) &= P\bigl((b-a)c \le n(b-f_n) < (b-a)d\bigr) \\
   &= P\left( b-\frac{(b-a)d}n < f_n \le b-\frac{(b-a)c}n\right)\\
   &= P\left(f_n \le b-\frac{(b-a)c}n\right) - P\left(f_n \le b-\frac{(b-a)d}n\right)\\
   &= \prod_{i=1}^n P\left(X_n \le b-\frac{(b-a)c}n\right) - \prod_{i=1}^n P\left(X_n \le b-\frac{(b-a)d}n\right)\\
\end{align*}
If $c \ge 0$, the first product equals finally (i. e. for large enough $n$)
\[
\left[\frac 1{b-a} \left(b-\frac{(b-a)c}n - a\right) \right]^n
   = \left(1 - \frac cn\right)^n
\]
and, as this implies $d > 0$, the second one equals in this case $(1 - \frac dn)^n$ So, for $[c,d) \subseteq [0,\infty)$, we have 
\[ P(c \le Y_n < d) \to \exp(-c) - \exp(-d) = \int_c^d \!\exp(-x)\, dx\]
If now $d \le 0$ the second product and the are equal 1, hence 
\[ P(c \le Y_n < d) \to 0 \]
in this case. If finally, $c < 0 < d$, then 
\[
P(c \le Y_n < d) \to 1 - \exp(-d) = \int_0^d \!\exp(-x)\, dx \]
so, in each case
\[
  P(c \le Y_n < d) \to \int_c^d \chi_{[0,\infty)}(x)\exp(-x)\, dx \]
that is $Y_n \to \operatorname{Exp}(1)$ weakly.
