Main diagonal operations problem

Hi guys, I need your help.

I want to create a matrix(4,4) in which the main diagonal have values between 0.3 and 1 and the other cells assume values ​​such as to have a horizontal sum equal to 1.

By now I'm using the following code but the only result is to have a main diagonal composed by the same numbers:

x = eye(4)
x(1,1) = 1+(0.3-1)*rand(1,1)
x(2,2) = x(1,1)
x(3,3) = x(1,1)
x(4,4) = x(1,1)

Any suggestion?

PS : I've tried even with diag

[Stephen23]

"...the other cells assume values ​​such as to have a horizontal sum equal to 1"

Are there any other requirements on the other elements? Positive, negative, fractional values, integer, >1, >100, >1e100 ... what values are allowed?

What is the "horizontal sum": do you mean to sum along the 2nd dimension (i.e. along each row) ?

[Giuseppe Pintori]

the values must be positive and by horizontal sum i mean that the sum of each row must be equal to 1

Answer 1

[Stephen23]

Using Roger Stafford's FEX submission randfixedsum (must be downloaded first):

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

>> N = 4; % matrix size
>> M = nan(N,N); % preallocate
>> V = 0.3+(1-0.3)*rand(1,1) % diagonal value
V =
      0.47505
>> M(~eye(N)) = randfixedsum(N-1,N,1-V,0,1); % other values
>> M = M.'; % transpose
>> M(1:N+1:end) = V % assign diagonal value
M =
      0.47505      0.40657    0.0087969      0.10958
      0.12917      0.47505      0.21287       0.1829
      0.35794      0.15519      0.47505     0.011825
      0.41335     0.032696     0.078907      0.47505

Checking the sum of each row:

>> sum(M,2)
ans =
     1
     1
     1
     1

and diagonal:

>> diag(M)
ans =
      0.47505
      0.47505
      0.47505
      0.47505

[John D'Errico]

Interesting. I wonder if your method, using a rejection scheme, and mine, which conditions on the diagonal element, will produce the same distribution in the end? I'm not sure it terribly matters, but it is just a technical question for future reference to me. I'm not sure at the moment, since these things can get tricky.

[Giuseppe Pintori]

Thanks, you're phenomenal! Your help is really precious.

[John D'Errico]

And the answer after some thought is, yes, the result will be technically different distributions of elements in the array. But I have no clue as to which is more appropriate given the question. Your solution will generate a sample that is uniform within each row in the sample space, so yours may be more appropriate.

[Bruno Luong]

@John, I have some formula on the distribution for you (you might know it already).

for X, m vectors of R^n as returned by

X = randfixedsum(n,m,1,0,1)

X(:,i) = { x uniform in [0,1]^n such that sum(x) = 1 }, i=1...,m.

the projection of any component #k, x(k,:) is on [0,1] has the pdf that is a polynomial of order (n-2):

P(x) := (n-1)(1-x).^(n-2) = sum_ {k=0,1..., n-2} p_{n,k} x^k;

with p_{n,k} = (n-1)*(-1)^k* nchoosek(n-2,k);

Example:

n=4;
m = 1e6;
X = randfixedsum(n,m,1,0,1);

P=arrayfun(@(k) (n-1)*(-1)^k*nchoosek(n-2,k),n-2:-1:0);

figure()

ax1 = subplot(2,1,1);
ezplot(ax1,@(x) (n-1)*(1-x).^(n-2),[0 1]);
legend(ax1,"PDF n="+n)

ax2 = subplot(2,1,2);
histogram(X(:),100); % by symmetry all components have the same PDF

linkaxes([ax1 ax2],'x');
xlim(ax1,[0 1]);
ylim(ax1,[0 n-1]);

The brutefoce rejection method from Stephen's (before-edited) code implies the diagonal elements follow the PDF of polynomial form of 2nd order 3*(1-x)^2, truncated bellow 0.3.

In your case as you generate the diagonal as uniform, therefore it's certainly different than the distribution from Stephen's method (polynomial).

The PDF formula can be showed recursively on n, without much difficulty technically.

[John D'Errico]

Yes. I had come to that conclusion after some thought too. But there still seems to be the question of what is the true goal, since only Giuseppe can know that. I think Stephen's solution comes closer than mine.

On the other hand, HAD I generated the diagonal elements using a better distribution than uniform, then my solution would be an avenue to not needing to use a rejection while loop at all.

[Bruno Luong]

Oh I see your point. I think you are right by forcing the right PDF to generate D, it's equivalent to rejection method. This I'm pretty sure it's right because of the special property of simplex and barycentric coordinates.

So the right way (I simplify the procedure to a single row without loosing the generality) is

N = 4;
dmin = 0.3;
dmax = 1;

d = dmax-(dmax-dmin)*rand().^(1/(N-1)); % rather than dmin+(dmax-dmin)*rand;
v = (1-d) * randfixedsum(N-1,1,1,0,1);

then insert d to v....

Answer 2

[John D'Errico]

Easy enough, it seems. First, determine the diagonal elements.

x = diag(rand(1,4)*.7 + .3);

Next, you need to choose the other row elements randomly so the sum will be 1. But that sum will now depend on the diagonal element you just chose. Stilll simple, as long as you use randfixedsum, by Roger Stafford, found on the file exchange.

for i = 1:4
  x(i,setdiff(1:4,i)) = randfixedsum(3,1,1 - x(i,i),0,1)';
end

Did it work? Of course.

x
x =
      0.83586     0.075979     0.057706     0.030454
     0.012356      0.85664      0.11425     0.016757
      0.13748      0.21163      0.43081      0.22009
      0.15838     0.037488      0.16129      0.64284
>> sum(x,2)
ans =
            1
            1
            1
            1

Find randfixedsum here:

https://www.mathworks.com/matlabcentral/fileexchange/9700-random-vectors-with-fixed-sum?s_tid=prof_contriblnk

Answer 3

[Matt J]

x=eye(4);
x(1:5:end)=0.7*rand(4,1)+0.3

[Giuseppe Pintori]

sorry, I forgot to write that the diagonal must be composed of identical values

Answer 4

[Bruno Luong]

Here is a method that has two advantages:

• without the need of Roger's FEX randfixedsum
• Produce matrix with rigourous uniform conditional probability

N = 4; % matrix size

% diagonal lo/up bounds
dmin = 0.3;
dmax = 1;
% random (common) diagonal value
d = dmax-(dmax-dmin)*rand().^(1/(N-1)); % Edit see comment above, equiv to rejection method
% d = dmin+(dmax-dmin)*rand;

% Generate N random vectors of length N-1 required sum == (1-d)
V = -log(rand(N-1,N)); % Marsaglia's [1961] method 
V = V .* ((1-d)./sum(V,1));

% Arrange in the final matrix
A = zeros(N);
isdiag = sparse(1:N,1:N,true);
A(isdiag) = d;
A(~isdiag) = V(:);
A = A.';

% Check result
disp(A)
sum(A,2)

EDIT change after this disccusion with John