s***t
发帖数: 69 | 1 想出来的,不知道怎么算,一条线长度为1,在上面任意切两点(每点都是U[0,1])
分成三段,求最长的那段期望值E[L_max],怎么算?? simulation的话是0.68左右
另外一题,x,y,z都是U[0,1], 求 E[max{x,y,z}]这题简单点,怎么算,答案应该是3/4
谢谢! | y**t
发帖数: 50 | 2 The answer to the problem is 99/162 and can be obtained by dividing into cases
let x and y are random numbers between 0,1.
One case is y>x and this case can be further divided into 0
and 1/2
The other case gives the same contribution.
The approach to second problem is you may assume that x>y>z and do an integral
.
The result is 6 times the integral. |
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