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