generate random number with total sum is 10

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

Answer 1

[Bruno Luong]

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.

Answer 2

[James Tursa]

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);

[Khairul nur]

many thanks...its work!

[Bruno Luong]

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]

@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.

Answer 3

[Khairul nur]

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?

[Bruno Luong]

You post in the wrong place (answer and not comment) so no one know what you refer to in "The first one".

[Khairul nur]

opps..my bad.. the first one refer to as in above comments:

r = rand(6,1);
r(:,2) = 1 - r(:,1);

[Bruno Luong]

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]

i already try it out..its working..tqvm