1
$\begingroup$

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 $1$.

$\endgroup$
  • 1
    $\begingroup$ see if you can show it for the interval $[0,1]$ first. Also what's the use of defining $g_n$? $\endgroup$ – tom Oct 16 '12 at 10:02
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.