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generate random number with total sum is 10

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Khairul nur
Khairul nur on 13 Jan 2021
Commented: Khairul nur on 4 Feb 2021
How can i generate a table of 6x2 with sum of the row is 10 for example:
0.6 0.4
0.5 0.5
0.7 0.3
i try this way:
r = rand(1, 3); % Start with 3 random numbers that don't sum to 1.
r = r / sum(r) % Normalize so the sum is 1.
however the sum is vertically. How i want to make the sum horizontally?
theSum = sum(r) % Check to make sure. Should be 1
tqvm

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Accepted Answer

Bruno Luong
Bruno Luong on 13 Jan 2021
Edited: Bruno Luong on 13 Jan 2021
For the case of more generic sum of n variables Xi is equal to 1.
If the apriori distribution is supposed to be uniform density
Xi ~ U(0,1)
If the conditioning is
sum(Xi) = 1
we can show that the Xi conditional probability distribution has density of
pdf(Xi | sum Xi = 1) = (n-1)*(1-x)^(n-2).
Note that for n=2, the conditioning distribution happens to be a uniform distribution. This is not true of n > 2.
One neatly way to generate a proper conditioning distribution is using n independant exponential distribution then normalize by the sum
n = 4;
Y = -log(rand(100000,n));
X = Y ./ sum(Y,2); % Each row is a realization of the conditining
We can check it gives the expected conditioning distribution
x=linspace(0,1);
pdf=@(x)(n-1)*(1-x).^(n-2);
histogram(X(:),'Normalization','pdf');
hold on
plot(x,pdf(x),'MarkerSize',3);
title('exponential normamlization');
If you want further to constraint the Xi with in a lower and upper bounds, Roger Stafford's randfixedsum function does the job.

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More Answers (2)

James Tursa
James Tursa on 13 Jan 2021
Edited: James Tursa on 13 Jan 2021
Not sure from your wording if you want the row sums to be 1 or 10. If it is 1, then
r = rand(6,1);
r(:,2) = 1 - r(:,1);
Modifying your code you could also do this:
r = rand(6,2);
r = r ./ sum(r,2);

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Bruno Luong
Bruno Luong on 13 Jan 2021
Careful in selecting the methods if you need a correct conditioning distribution.
James's second method won't give a proper "uniform" conditioning distribution
r=rand(100000,2); r=r./sum(r,2);
histogram(r(:))
The first method gives a more "healty" distribution
r = rand(100000,1); r(:,2) = 1 - r(:,1);
histogram(r(:))
PS: This method won't extend if you need more than 2 that sum to 1. Need other methods for that case.
James Tursa
James Tursa on 13 Jan 2021
@Bruno: Yes, the 2nd method posted was just to show OP how to modify his version of the code to get the row sums = 1. But if uniform distribution is desired then I would typically just point to Roger Stafford's randfixedsum solution as you have done. OP needs to pick a method for the distribution he wants.

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Khairul nur
Khairul nur on 3 Feb 2021
The first one works better, but if i change the matrix column into 3..the 2nd col become null or 0..how to fix this?

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Khairul nur
Khairul nur on 3 Feb 2021
opps..my bad.. the first one refer to as in above comments:
r = rand(6,1);
r(:,2) = 1 - r(:,1);
Bruno Luong
Bruno Luong on 3 Feb 2021
Well you have been warned, I wrote in my comment;
"PS: This method won't extend if you need more than 2 that sum to 1. Need other methods for that case."
And I already give alternative method that can be extended in my answer using exponential pdf.
Khairul nur
Khairul nur on 4 Feb 2021
i already try it out..its working..tqvm

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