t*******y
发帖数: 637 | 1 就是那个[0,1]上任取两点, uniform distribution
求三个segment最小的那个的期望 | t******t
发帖数: 40 | 2 这个怎么做啊？没思路呀。模拟的结果是0.2左右。同问。 | d***y
发帖数: 65 | 3 A brute force solution. Let M be the length of minimum segment,
X be the first point, Y be the second.
Then M=min{X,Y,|Y-X|,1-X,1-Y}.
We can find the cdf and pdf of M by computing
P(M>t)=P(X>t,Y>t,|Y-X|>t,1-X>t,1-Y>t)
=1-6t+9t^2 (This can be obtained by drawing the lines of the constraints on
the unit sqaure)
and the pdf of M is simply 6-18t.
Taking expectation by integration, we get E(M)=1/9. | p*****k
发帖数: 318 | 4 an equivalent way to find P(M>=t) in douby's post:
since M>=t, so every segment>=t, hence one could cut off
3*t from the total length, and the two points are now chosen
on (1-3*t) without constraint. this gives the prob:
(1-3*t)^2 | L******2
发帖数: 274 | 5 I think the final result of douby is correct, except that the expression P(M
>t) should be
P(M>t)=P(Y-X>t,X>t,1-Y>t)+P(X-Y>=t,Y>t,1-X>t)
=2P(Y-X>t,X>t,1-Y>t) where 0<=t<=1/3. |
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