How to implement the Gaussian radial basis function in MATLAB ?

Hello all, I am dealing with the following optimization problem

where is a column vector of dimension , is also a column vector of dimension , is matrix of dimension , is row vector of and is the Gaussian radial basis function, where is the variance.

My query is how to code this Gaussian radial basis function (RBF) ?

Any help in this regard will be highly appreciated.

Answer 1

[Torsten]

Before you start the optimization, check whether K in your case is positive-definite.

Lt = 10;
p = rand(Lt,4);
sigma = 1.0;
for l = 1:Lt
    for j = 1:Lt
        K(l,j) = exp((norm(p(l,:)-p(j,:)))^2/(2*sigma^2));
    end
end
K

K = 10×10

1.0000 1.0558 1.2332 1.1752 1.1144 1.7121 1.4345 1.4970 1.3174 1.4356 1.0558 1.0000 1.1059 1.1852 1.0453 1.4602 1.3978 1.2603 1.1884 1.3441 1.2332 1.1059 1.0000 1.4993 1.1769 1.7198 1.2359 1.2245 1.2960 1.8679 1.1752 1.1852 1.4993 1.0000 1.3279 1.1864 1.5427 1.3336 1.2247 1.0896 1.1144 1.0453 1.1769 1.3279 1.0000 1.6242 1.4504 1.5273 1.1608 1.4210 1.7121 1.4602 1.7198 1.1864 1.6242 1.0000 1.7290 1.2760 1.2118 1.1176 1.4345 1.3978 1.2359 1.5427 1.4504 1.7290 1.0000 1.5769 1.2315 1.9993 1.4970 1.2603 1.2245 1.3336 1.5273 1.2760 1.5769 1.0000 1.4254 1.5342 1.3174 1.1884 1.2960 1.2247 1.1608 1.2118 1.2315 1.4254 1.0000 1.2412 1.4356 1.3441 1.8679 1.0896 1.4210 1.1176 1.9993 1.5342 1.2412 1.0000

eig(K)

ans = 10×1

-1.7051 -0.8915 -0.6203 -0.3001 0.0081 0.0171 0.0346 0.0637 0.1039 13.2897

[charu shree]

Thank you so much sir for your quick response.....

It means K will be a matrix of dimension .

Sir, Could you please tell how did you get by looking at function that it should be a matrix and also why we have to check whether K is positive definite?

[Torsten]

Sir, Could you please tell how did you get by looking at function that it should be a matrix and also why we have to check whether K is positive definite?

https://en.wikipedia.org/wiki/Radial_basis_function

Since you have a constraint quadratic optimization problem, forget about the positive-definiteness of K.

[charu shree]

I had just checked, the eigen values are positive, for total 96 eigen values (as L_t in my case is 96), 15 eigen values are zero. So is it correct ?

[Matt J]

Torsten's calculation of K gives a matrix that isn't positive definite only because of a missing minus sign,

Lt = 10;
p = rand(Lt,4);
sigma = 1.0;
clear K
for l = 1:Lt
    for j = 1:Lt
        K(l,j) = exp( -(norm(p(l,:)-p(j,:)) )^2/(2*sigma^2));
    end
end

eig(K)

ans = 10×1

0.0021 0.0122 0.0286 0.0498 0.0724 0.1386 0.4154 0.6368 0.9973 7.6469

[charu shree]

Ok....Thank you sir...

[Torsten]

@Matt J

Thank you for the correction.

Is it somehow obvious that the matrix will be positive-definite ?

Answer 2

[Matt J]

If you have the Statistics Toolbox, you can avoid a loop by using pdist2

Lt = 10;
p = rand(Lt,4);
sigma = 1.0;

K=exp(-pdist2(p,p).^2 / 2/sigma^2);

[charu shree]

Oh... Thats very effective ...