问题变成求该单纯形中点的最大坐标与最小坐标的平均值。由于各坐标的对称性，只须求出x1在小单纯形 x1>=x2>=x3>=...>=xn 和 x1<=x2<=x3<=...<=xn （即图中红绿部分）中的平均值，即单纯形重心的x1坐标。这两个小单纯形的顶点分别为
(1,0,...0)
(1/2,1/2,0,...0)
(1/3,1/3,1/3,...,0)
......
(1/n,1/n,...,1/n)
和
(0,...0,1)
(0,...,0,1/2,1/2)
......
(1/n,1/n,...,1/n)
它们重心的x1坐标分别为 (1+1/2+1/3+...+1/n)/n, 及1/n^2。
This method is enticing indeed. However, if the original sample points (not arc lengths) are uniform on the interval [0 1], by known results of order statistics, the arec lengths {X_i} do NOT follow uniform distribution any more. What you calculated are not expectation/mean. Of course you can choose the joint distribution to be uniform (per unit area) on the N-D surface patch. Noticing that individual X_i will not follow the uniform distribution; it'd be an equally daunting problem (when N is large) to find the distribution and realize it. By the way, N=2 is the only time when the sample sites and arc lengths both follow uniform distribution. |
The arc lengths are not uniformly distributed, they have a density function (n-1)*(1-x)^(n-2). This can be calculated directly from definition or from my model. But the joint ditribution of all n arc lengths does follow a uniform distribution over the simplex. However the original problem was not clearly defined. We may need to wait until February to know what IBM thinks. |
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