Convergence in distribution of order statistics range (from uniform distributed sample) I've been working on this assignment:
Let $\{X_n\}_{n \ge 1} \stackrel{i.i.d}\sim Uniform(0,1)$. Define $U_n = min\{X_1, \dots, X_n\}$ and $V_n = max\{X_1, \dots, X_n\}$.
(a) Prove that $Z_n = V_n - U_n$ converges almost surely to 1.
(b) Find a sequence $\{a_n\}_{n \ge 1}$ such that $a_n(1 - Z_n)$ converges in distribution and find its limit.
Part (a) seems to be intuitive because you are uniformly sampling from $[0,1]$, so eventually you'll cover the whole interval, making the range $Z_n$ to be the interval length.
But for part (b) I found that the distribution for $Z_n$ is $F(x) = {x^{n-2} \over {n+1}}$. Defining $Y_n = a_n(1-Z_n)$ you have:
$F_{Y_n}(x) = P(Y_n \leq x) = P(a_n(1-Z_n) \leq x) = P(Z_n \geq \left(1 - {x \over a_n}\right))=1-P(Z_n \leq \left(1 - {x \over a_n}\right))$
$=1-{1 \over n+1}\left(1 - {x \over a_n}\right)^{n-2}$
There you can spot that defining $a_n = n-2$, you could get a kind of exponential distribution. But I haven't figured out any way to by pass the term $1 \over n+1$ which makes the whole thing to converge to 1.
Any help would be appreciated!
 A: I don't think that the distribution of $Z_n$ that you derived is correct.
$$f_{Z_n}(z) = n(n-1)z^{n-2}(1-z)$$
Then, $Y_n = a_n(1-Z_n)$ has the distribution,
$$f_{Y_n}(y) = \frac{n(n-1)y}{a_n^2}\left(1-\frac{y}{a_n}\right)^{n-2}$$
The limiting distribution of $Y_n$ with $a_n = n-2$  is $gamma(y|\alpha = 2, \beta = 1)$.
Edit:
Here is the derivation for the distribution of $Z_n$,
$$\begin{align}f_{X_{(1)},X_{(n)}}(x_1, x_n) &= \frac{n!f_{U}(x_1)f_U(x_n)F_{U}(x_1)^{(1-1)}(F_{U}(x_n)-F_{U}(x_1))^{n-1-1}F_{U}(x_n)^{n-n}}{(1-1)!(n-1-1)!(n-n)!}
\\\\ &= n(n-1)(x_n-x_1)^{n-2}, \,\,\, 0 < x_1 < x_n < 1\end{align}$$
Now, let $V = X_{(1)}, R = X_{(n)}-X_{(1)} \implies |detJ| = 1$
$$f_{V,R}(v,r) = f_{X_{(1)}, X_{(n)}}(v, r+v) \cdot 1 = n(n-1)r^{n-2}, \,\,\,\, 0 < v < 1-r, 0 < r < 1$$
$$\implies f_{R}(r) = \int_{0}^{1-r}f_{V,R}(v,r) dv = \int_{0}^{1-r}n(n-1)r^{n-2}dv = n(n-1)r^{n-2}(1-r)$$
Here $R = Z_n$. Note that I used the joint pdf of two order statistics $(X_{(i)}, X_{(j)})$ where $i=1, j=n$ which is given by,
$$\begin{align}f_{X_{(i)},X_{(j)}}(x_1, x_n) &= \frac{n!f_{X}(x_i)f_X(x_j)F_{X}(x_i)^{(i-1)}(F_{X}(x_j)-F_{X}(x_i))^{j-i-1}F_{X}(x_j)^{n-j}}{(i-1)!(j-i-1)!(n-j)!}\end{align}$$
