Simple Matlab Random Number Generation
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I have to get 5 random numbers a1, a2, a3, a4, a5 where each a1, a2, a3, a4, a5 should be between [-0.5, 0.5] and sum i.e. a1 + a2 + a3 + a4 + a5 = 1.
How should I do it?
4 个评论
bym
2011-2-27
I don't think the problem statement is consistent. There is some probability that you could draw [.5 .5 .5 .5 .5]
Paulo Silva
2011-2-27
Hi Sam, why "Simple Matlab Random Number Generation"? it's not that simple.
Sam Da
2011-2-27
Paulo Silva
2011-2-28
I deleted my answer (the one that was accepted but it wasn't the best one) and voted on Bruno's and Matt's answers.
Please reselect (Sam or someone who can (admins?!)) the best answer, thank you.
采纳的回答
chris hinkle
2011-2-27
0 个投票
Create 5 arrays that have all possible combinations of these numbers then generate a random number that is between 1 and length of array and then use that value as the index for the array and viola, there's your number. The size of these arrays can be controlled by the resolution you go to.
更多回答（2 个）
Bruno Luong
2011-2-27
To generate true uniform distribution, the correct method is not quite straightforward. I strongly recommend Roger Stafford's FEX,
http://www.mathworks.com/matlabcentral/fileexchange/9700-random-vectors-with-fixed-sum
3 个评论
Jan
2011-2-27
This is defintely the best answer.
the cyclist
2011-2-27
Agreed that this is the definitive answer. Specifically for Sam's solution:
X = randfixedsum(5,10000,1,-0.5,0.5);
Matt Tearle
2011-2-27
Very nice!
Matt Tearle
2011-2-27
How about a brute-force approach?
ntot = 0;
n = 10000;
x = zeros(n,5);
while ntot<n
r = rand(100,4)-0.5;
r5 = 1 - sum(r,2);
idx = (r5>-0.5) & (r5<0.5);
tmp = [r(idx,:),r5(idx)];
nidx = min(size(tmp,1),n-ntot);
x(ntot+1:ntot+nidx,:) = tmp(1:nidx,:);
ntot = ntot + nidx;
end
1 个评论
the cyclist
2011-2-28
My first reaction to this solution was that, as a rejection method (with a loop, no less!), it would be much slower than Roger's method. The reality is that is does comparably well, speed-wise. I haven't done a full-blown comparison, but I think the reason is two-fold. First, you "semi-vectorized" by pulling chunks of random numbers at a time. Second, and I think more importantly, the accept/reject fraction is pretty good. (It might not be so favorable otherwise, like if the marginals were on [0,1] and still had to sum to 1.)
This solution is highly intuitive, and I believe leads to marginal distributions and correlations between summands that are identical to Roger's solution.
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