o***n 发帖数: 921 | 1 有一个半径为1的单位圆,在圆内任取两点,问两点间距的分布?
假设点是均匀分布的,即点出现在(r,theta)的概率密度为 P(r,theta) = r/Pi. | p*****k 发帖数: 318 | 2 the pdf is: 2x/pi*[2*arccos(x/2)-x*sqrt(4-x^2)/2]*dx, 0<=x<=2.
(see wilmott.com/messageview.cfm?catid=26&threadid=64937)
one interesting observation is that the expression in the parentheses is the area of the intersection of two unit circles with their centers separated by x, and i always wonder whether there is an easy proof based on that, but i never figured out how... | h*****0 发帖数: 4889 | 3 可以考虑两个点AB距离为x时,原来的单位圆圆心必须位于单位圆A和单位圆B的交集中
。再加上一个旋转自由度就出来了。
the area of the intersection of two unit circles with their centers
separated by x, and i always wonder whether there is an easy proof based on
that, but i never figured out how
【在 p*****k 的大作中提到】![](/moin_static193/solenoid/img/up.png) : the pdf is: 2x/pi*[2*arccos(x/2)-x*sqrt(4-x^2)/2]*dx, 0<=x<=2. : (see wilmott.com/messageview.cfm?catid=26&threadid=64937) : one interesting observation is that the expression in the parentheses is the area of the intersection of two unit circles with their centers separated by x, and i always wonder whether there is an easy proof based on that, but i never figured out how...
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