IBM一月

constant · 发表于 2006-1-3 20:55:43

还没时间想，难度也不知道。

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。

constant · 发表于 2006-1-4 00:04:13

好象不是很难，但不太容易说清楚，算++.5 吧。

Archi · 发表于 2006-1-4 21:40:35

Use the following property:

Let y be the minimum of set {x_i}, z the maximum, i=1,2,...N. Then the distribution functions:

F_Y(y) = P(Y less than y) = 1-P(Y greater than y) = 1-P(all x_i greater than y), and
F_Z(z) = P(Z less than z) = P(all x_i less than z).

Next the probability can be calculated using conditional probabilities, noticing that the random variables x_i are not independent.

The expectations can be obtained once the distribution functions are available.

But the process is not trivial. I used symbolic toolbox in MATLAB and got the minimum length expectation to be 0.011. The maximum length seems more nasty.

Any smart way of doing this?

constant · 发表于 2006-1-4 22:46:35

The answer depends on N. Obviously the minimum < 1/N, so your answer would be wrong for N > 90.

Archi · 发表于 2006-1-4 23:44:04

1) The numerical result (0.011) follows from N=10.

2) Let the arc lengths be {u_i}, i=1,2,...,10. Apparently all u_i have identical distribution, which can be deduced from my previous argument: F_U(u)=1-(1-u)^9. We can check it by examing the end conditions and the mean arc length (0.1 of course). Notice that u_i are not independent as they sum up to 1.

By the way, if we unwrap the circle and make it a [0 1] line segment, starting and ending at any of of the 10 sample points, then the positions of the remaining 9 points are iid r.v. with uniform distribution in [0 1]. After sorting, the position expectations are 1/10, 2/10, etc. This is related to the proposed problem but different.

constant · 发表于 2006-1-7 00:17:43

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IBM一月

好象不是很难，但不太容易说清楚，算++.5 吧。

回复：Tedious

回复：回复：Tedious

Forgot to mention...

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