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Consider a loop of string with unit length. Take n cuts independently and randomly along the string, what is the expected length of the smallest and the largest piece?

1 Ansicht (letzte 30 Tage)
This is what I did.
The probability is (1+(1-n)x)^n
So, expected value of x is it integral for x varies from 0 to 1/n which evaluates to 1/n^2
If this is right how should I write the code?
  7 Kommentare
Guillaume
Guillaume am 9 Feb. 2016
Bearbeitet: Guillaume am 9 Feb. 2016
I think Jan's point was that a loop, having no start and end point, can't really have a length. Perimeter might be a more appropriate term.
Vin Sen  Lee
Vin Sen Lee am 10 Feb. 2016
So, i just got clarification on this problem. My task is to write a function whose input is a number of cuts and output is the 1x2 array of the form [xmin xmax] where xmin is your experimental expected value of the smallest cut and xmax is your experimental expected value of the largest cut.

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Antworten (1)

Are Mjaavatten
Are Mjaavatten am 8 Feb. 2016
Your question is not very clear. The code below is an answer to: How can I code a test of this result?
N=100000; % Number of samples
n=8; % Number of cuts
d = zeros(N,n); % Allocate space for results
for i = 1:N
a = sort(rand(1,n)); % Draw random cut poins and distribute them along the string
b = [a(end)-1,a]; % Join ends
d(i,:) = sort(diff(b)); % Sort the pieces by length
end
mean_lengths = mean(d); % mean_lengths(i) is the mean length of the i'th shortest piece
disp(mean_lengths);
  2 Kommentare
Walter Roberson
Walter Roberson am 9 Feb. 2016
Bearbeitet: Walter Roberson am 9 Feb. 2016
mean_lengths(end) is the mean of the longest.
The shortest out of all of the runs is min(d(:)) and the longest out of all of the runs is max(d(:)) (those might occur on different runs.)

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