rand matrix for FM

Hi all,I want to generate a random matrix so that the sum of all the elements is zero. I mean the random numbers are so selected that the total sum of the matrix goes to zero.Regards. Early reply will be highly appreciated.

for example

a = rand(3)

a =

    0.3000    0.5000    0.8000
   -0.1000   -0.4000   -0.6000
   -0.4000    0.8000   -0.9000

>> sum(sum(a))

ans =

     0

[Stephen23]

Integer values only, or with decimal fraction?

[Offroad Jeep]

Both for integer and decimals.............

Answer 1

[Roger Stafford]

If your x values are subject to common upper and lower bounds, you can use my 'randfixedsum' function in the File Exchange, located at:

http://www.mathworks.com/matlabcentral/fileexchange/9700-random-vectors-with-fixed-sum

It is designed to give a uniform distribution on the hyperplane of values satisfying the condition of a predetermined sum - in your case a sum of zero.

[John D'Errico]

And of course, this is the best answer.

Answer 2

[Zoltán Csáti]

I recommend you to generate the matrix of the required size and then modify one element of it so that the sum holds. E.g.

 A = rand(3);
 totalSum = sum(sum(A));
 A(end,end) = A(end,end) - totalSum;

Then the sum will give you zero, aside from the round-off error.

[John D'Errico]

It depends on how you want the elements themselves to be distributed. See that if they should originally be bounded in the interval [-1,1], then by the final shift, they often will no long be so bounded.

For example...

A = rand(3) *2 - 1;
A = A - sum(A(:))/numel(A)
A =
    0.53571     -0.78404      0.51328
   -0.81271     -0.68212     -0.24218
   -0.32253      0.77053       1.0241
sum(A(:))
ans =
  -2.2204e-16

See that while the sum is now zero, that now one of the elements actually exceeded 1, even though the original elements fell inside [-1,1].

[Zoltán Csáti]

Yes, you are right I didn't think of this aspect.

Answer 3

[Brendan Hamm]

How about you create a random matrix and then subtract from each element the sum(matrix(:))/numel(matrix).

n = 4;
A = rand(4);
s = sum(A(:))/numel(A);
A = A - s;
sum(A(:))

[John D'Errico]

It depends on how you want the elements themselves to be distributed. See that if they should originally be bounded in the interval [-1,1], then by the final shift, they often will no long be so bounded.

For example...

A = rand(3) *2 - 1;
A = A - sum(A(:))/numel(A)
A =
    0.53571     -0.78404      0.51328
   -0.81271     -0.68212     -0.24218
   -0.32253      0.77053       1.0241
sum(A(:))
ans =
  -2.2204e-16

See that while the sum is now zero, that now one of the elements actually exceeded 1, even though the original elements fell inside [-1,1].

[Brendan Hamm]

Naturally. But the question is vague and it is never specified that the values should be U([-1 1]) or originate from any other distribution? This is just an easy way to get a matrix which is random and has a sum of elements equal to zero, as per request.

[Offroad Jeep]

Brendan,,,,,,,,, can you please tell how I can put the condition............

[Roger Stafford]

The vagueness has to do with how you expect the set of vectors for which the sum has a given value, to be statistically distributed. Consider each vector as a point in n-dimensional vector space. What kind of statistical density function do you desire for the hyperplane therein where the vector sums are a given value? Is it to be uniform, gaussian, etc., and is it bounded?

[Brendan Hamm]

They are currently U([-c c]) where c i s the sum of the originally sampled elements divided by the number of elements. You want them to be U([-1 1]), then just divide by the magnitude of the largest resulting element now:

A = A / max(abs(A(:)));