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?
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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
Walter Roberson
am 8 Feb. 2016
The probability of what?
Jan
am 8 Feb. 2016
Does a loop have a length? Are the randomly distributed cuts set according to a equasl, normal Gauss , Lorenz, etc distribution?
Walter Roberson
am 8 Feb. 2016
"unit" length according to the title.
Vin Sen Lee
am 9 Feb. 2016
Guillaume
am 9 Feb. 2016
'randomly distributed' does not mean anything. You can have 'uniform' distribution, 'normal' distribution, 'log normal' distribution, and many more.
Vin Sen Lee
am 10 Feb. 2016
Antworten (1)
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
Vin Sen Lee
am 9 Feb. 2016
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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