IBM 一月解答（图）

constant

一个周长为1的圆圈，随机分成N段。问最长一段和最短一段长度的期望值。

 

先定义一个名词。一个n维单纯形n维空间中最简单的多面体。平面上是三角形，三维空间中是四面体。

 

这个问题的样本空间是n-1维单纯形 x1+x2+...+xn = 1。n=3时如下图。

 

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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。www.ddhw.com

 

野 菜 花

  你自己第一名的答案总是标准答案了

constant

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. www.ddhw.com

Archi

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. www.ddhw.com