预期的成对方形欧几里德点之间的距离 -- probability-theory 领域 和 randomized-algorithms 领域 和 euclidean-distance 领域 cs 相关 的问题

expected pairwise square euclidean distance between points

3

问题

$$theta（d）$$表示最大可能的距离是d。

im跟随这条路径： https://stats.stackexchange.com/questions/22488/probability-that-uniformly-random-points-in-a-rectangle-have-euclidean-distance

How can I show that the expected pairwise square euclidean distance between points in $$X$$ is $$xcex98(d)$$?

Where $$X$$ is a $$(x_1,...x_n)$$ of points generated uniformly at random in the unit, d is d-dimensional cube , $$x=(x(1),...x(d))$$ the generic point has its -th component $$x(i)$$ chosen uniformly at random in$$[0,1]$$independently of other components and points.

$$\Theta(d)$$ represent the largest possible distance is d.

I try to reconduct this problem to Bertrand Paradox but i dont think is right. Maybe I that show that $$E(||xxe2x88x92y||2)=xcex98(d)$$ , because is a hint but i dont know how.

i m following this path: https://stats.stackexchange.com/questions/22488/probability-that-uniformly-random-points-in-a-rectangle-have-euclidean-distance

but is different to my point.

Thanks.

回答列表

1

let $$vec {x}， vec {y}$$是两个随机$$d$$$$[0,1] ^ d$$均匀地选择的等级矢量。也就是说，$$x_1， ldots，x_d，y_1， ldots，y_d$$都是均匀分布的均匀随机样本，而不是$$[0,1]$$。然后 $$mathbb {e} [ | vec {x} - vec {y} | ^ 2] = mathbb {e} left [ sum_ {i = 1} ^ d（x_i-y_i）^ 2 light] = sum_ {i = 1} ^ d mathbb {e} [（x_i-y_i）] ^ 2 = d operatorname * { mathbb {e}} _ {x，y sim [0,1]} [（x-y）^ 2]。$$$$c = mathbb {e} [（x-y）^ 2]$$。然后在$$[0,1] ^ d$$$$cd$$中的预期平方距离。

Let $$\vec{x},\vec{y}$$ be two random $$d$$-dimensional vectors chosen uniformly and independently from $$[0,1]^d$$. That is, $$x_1,\ldots,x_d,y_1,\ldots,y_d$$ are all uniform random samples of the uniform distribution over $$[0,1]$$. Then $$\mathbb{E}[\|\vec{x}-\vec{y}\|^2] = \mathbb{E}\left[\sum_{i=1}^d (x_i-y_i)^2\right] = \sum_{i=1}^d \mathbb{E}[(x_i-y_i)]^2 = d \operatorname*{\mathbb{E}}_{x,y \sim [0,1]} [(x-y)^2].$$ Let $$C = \mathbb{E}[(x-y)^2]$$. Then the expected squared distance of two points in $$[0,1]^d$$ is $$Cd$$.

It is not hard to calculate $$C$$ explicitly: $$C = \mathbb{E}[((x-1/2) - (y-1/2))^2] = \mathbb{E}[(x-1/2)^2] + 2\mathbb{E}[(x-1/2)(y-1/2)] + \mathbb{E}[(y-1/2)^2] = \\ 2\mathbb{E}[(x-1/2)^2] + 2\mathbb{E}[x-1/2] \mathbb{E}[y-1/2] = 2\mathbb{E}[(x-1/2)^2] = \\ 2\int_0^1 (x-1/2)^2 \, dx = 2\int_0^1 x^2-x+\frac{1}{4} \, dx = 2\left(\frac{1}{3} - \frac{1}{2} + \frac{1}{4}\right) = \frac{1}{6}.$$

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