How we can fix the fmincon?

I have a GARCH with an exogenous variable with a delta coefficient. I analysed the maximum likelihood estimation to find the four parameters.

There is somethign wrong.

omega = theta(1,1);
alpha = theta(2,1);
beta = theta(3,1);
delta = theta(4,1);
lb = [0.0000001;
0;
0;
0.0000001];

ub = [1000000;
    0.999999;
    0.999999;
    100000];

 A = [0 1 1 0;1 0 0 1];
b = [0.9999999; 0]; 

%Aeq =[1 0 0 0];
%beq =[0];

What I should do for input = omega + delta >0.

I think this is the problem of the output.

[x,fval,exitflag,output,lambda,grad,hessian] = fmincon('log_lk_GARCHVIX',theta0,A,b,[],[],lb,ub)

The output give me most of parameters equatl to zero and 1 equal to 1.

[Karim Hasouna]

  x=  0.0003
    0.1585
    0.1579
    0.0010

Now give me this output with the exitflag=1. What I want ot understand is why the last value delta = 0.0010 follows the lower value that I put in the lb[ ..., ..., ..., 0.0010]. It means that is useless teh last parameter in my analysis?

[Matt J]

Well, a minimum will follow lb if the objective is "positively sloped" in the neighborhood of lb. For example, consider the simpler, one-variable problem

min. x
s.t. x>=lb

Clearly the minimum will always be at x=lb.

Answer 1

[Matt J]

 A = [0 1 1 0;-1 0 0 -1];
 b = [0.9999999; 0]; 

[Karim Hasouna]

Local minimum possible. Constraints satisfied.

fmincon stopped because the size of the current step is less than
the value of the step size tolerance and constraints are 
satisfied to within the value of the constraint tolerance.

<stopping criteria details>
x = 4×1    
    0.0001
    0.3536
    0.6464
    0.0000

fval = -4.1948e+03
exitflag = 2
output = 
         iterations: 55
          funcCount: 358
    constrviolation: 0
           stepsize: 1.3190e-10
          algorithm: 'interior-point'
      firstorderopt: 25.1935
       cgiterations: 101
            message: '↵Local minimum possible. Constraints satisfied.↵↵fmincon stopped because the size of the current step is less than↵the value of the step size tolerance and constraints are ↵satisfied to within the value of the constraint tolerance.↵↵<stopping criteria details>↵↵Optimization stopped because the relative changes in all elements of x are↵less than options.StepTolerance = 1.000000e-10, and the relative maximum constraint↵violation, 0.000000e+00, is less than options.ConstraintTolerance = 1.000000e-06.↵↵'

lambda = 
         eqlin: [0×1 double]
      eqnonlin: [0×1 double]
       ineqlin: [2×1 double]
         lower: [4×1 double]
         upper: [4×1 double]
    ineqnonlin: [0×1 double]

grad = 4×1    
   25.2244
   -1.2960
   -1.2880
    1.7870

hessian = 4×4    
1.0e+10 *

    5.8135    0.0001   -0.0003    0.1405
    0.0001    0.0000   -0.0000   -0.0000
   -0.0003   -0.0000    0.0000   -0.0000
    0.1405   -0.0000   -0.0000    0.0354

t_omega = 28.9752
t_alpha = 5.1986
t_beta = 9.5768
t_delta = 0.3876

sigma2_hat_T1 = 6.8021e-04

[Matt J]

Glad you got it working, but please Accept-click the answer to indicate that your question is resolved.

[Karim Hasouna]

Is it working? It should be minimum satisfied. exitflag equal to 1. I think there is a mistake. Also the omega and the delta are 0 and is weird.

[Matt J]

exitflag=1 is a more confident termination flag than exitflag=2, but exitflag=2 is still pretty good. There is no obvious mistake in the use of fmincon that I can see. That's not to say there couldn't be a mistake in your problem formulation...

[Walter Roberson]

You are being stopped by a constraint. exitflag 1 can only occur if you are not being stopped by a constraint and the jacobian is all nondecreasing (indicating a local minimum).

[Matt J]

Hmmm. I don't think that can be true, based on the following example. Clearly the optimization would keep going if it were not for the lower bound lb=0. Yet, we get an exitflag of one.

>> [x,fval,exitflag]=fmincon(@(x)x , 2, [],[],[],[],0,inf)

Local minimum found that satisfies the constraints.

x =

   2.0000e-08


fval =

   2.0000e-08


exitflag =

     1