# 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 $1$.

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see if you can show it for the interval $[0,1]$ first. Also what's the use of defining $g_n$? – tom Oct 16 '12 at 10:02

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.