shortest length

Given two i.i.d uniform random points x and y on the interval [0,1] , what
is the average size of the smallest of the three resulting intervals?

A:
Think of minimum as a function on triangle (0,0), (0,1) and (1,1). The
"height" is 1/3, and the "base" is 1/2. Then the volume is 1/18. There
is another triangle (0,0), (1,0) and (1,1). So the total volume is 1/9.
It is original integration.
2/18 for the smallest one, 5/18 for the middle one, 11/18 for the largest
one.
Min for n+1 partition (n variables) is 1/(n+1)^2

Max for n+1 partition is int_0_inf{1-[1-exp(-(n+1)x)]^(n+1)}dx

using the fact that uniformly sampling a simplex is the same as drawing i.i.
d. exp(-x) random variables and then normalizing them, the result is a 
direct application of order statistics of i.i.d. r.v.s. for example the 
expectation of the longest interval is just

int_0^inf (1-exp(-x))^10 exp(-x) x dx

the expectations of the rest can be easily obtained similarly.

假设 x1 <= x2 <= ... <= xn,  xi is the size of ordered intervals.

考虑 以下 n 个变量:
y1 = (n) * (x1);
y2 = (n-1) * (x2 - x1);
...
yn = (1) * (xn - x{n-1});

y1,...,yn 为随机分布且满足：
1 = \sum_{i=1}^{n} yi
易有 E[yi] = 1/n;

所以 E[x1] = 1/n * E[y1] = 1/n^2
E[x2] = E[x1] + E[x2-x1] = E[x1] + 1/(n-1) * E[y2] = 1/n^2 + 1/n(n-1)
...
E[xn] = 1/n*n + 1/n(n-1) + ... +1/n*1

Given G(n)x_i=e_i in terms of identical dist., where e_i is exponentially 
distributed, and G(n)is independent of x_i and Gamma(n,1) distributed with
expectation sum(e_i),
it can be shown that G(n)y_i=G(n)(n+1-i)(x_(i)-x_(i-1))=(n+1-i)(e_(i)-e_(i-1))=
e_i in terms of identical dist., based on the following facts:
1)joint density of e_(i) is n!exp(-sum_i=1^n(a_i)) where 0=2)sum_i=1^n(a_i)=sum_i=1^n((n+1-i)(a_i-a_i-1))
3)Jocabian of the transformation matrix
e_(1)=Y_1/n
e_(2)=Y_1/n+Y_2/(n-1)
...
e_(n)=Y_1/n+Y_2/(n-1)+ ...Y_i/(n+1-i)...+Y_n
is 1/n!
4)from 1),2),and 3) we have joint density of Y_i=(n+1-i)(e_(i)-e_(i-1)) is 
exp(-sum_i=1^n(b_i)) where b_i>=0.

Therefore G(n)y_i=Y_i in terms of identical dist., E(y_i)=E(e_i)/(nE(e_i))
=1/n