Let $X_i ∼ U(0, 1), i = 1, . . . , 20, iid$. IQR = $F^{−1}(.75)−F^{−1}(.25)$ = $X_{(15)}−X_{(5)}$ in this example as n = 20.
a. Find the distribution of the random variable W = IQR.
b. Devise a function of this random variable such that it has expectation of the length of the support of distribution. Do the same thing for the range
c. Consider both of these quantities from (b) as ways to guess (estimate) the length of the support. Compute the relative efficiency (the ratio of mean square errors, it doesn’t matter which is in the numerator) of these two quantities.
So far, I've looked at an example discussing the range of a random sample from a uniform distribution, but I'm having trouble applying that example to the IQR.
Information I know for this problem includes:
$F(x)=x$ for $0<x<1$.
For order statistics: $f_k(x)= \frac{n!}{((k-1)!(n-k)!)}*{f(x)F(x)^{k-1}(1-F(x))^{n-k}}$
Bivariate joint PDF for $Y_1$ (the min) and $Y_n$ (the max) is $g(y_1,y_n)=n(n-1)[F(y_n)-F(y_1)]^{n-2} f(y_1)f(y_n)$
I've tried plugging in definitions and going from there, but I get stuck at a really difficult integral that even Wolfram can't compute, so I think I may be doing something wrong. Any help is appreciated!
From this answer we see that $W\sim\mathsf{Beta}(15-5,20-15+5+1)$ with density $$ f_W(t) = \frac{(20)!}{(10)!9!}t^9(1-t)^{10}\cdot\mathsf 1_{(0,1)}(t). $$ Since $$ \mathbb E[W] = \int_0^1 \frac{(20)!}{(10)!9!}t^{10}(1-t)^{10}\ \mathsf dt =\frac{10}{21}, $$ letting $g(w) = \frac{21}{10}w$ yields $$ \mathbb E[g(W)] = \frac{21}{10}\mathbb E[W] = 1. $$