How we can fix the fmincon?

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Karim Hasouna 2020-8-15

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编辑： Matt J 2020-8-16

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.

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Karim Hasouna 2020-8-15

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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 2020-8-15

编辑：Matt J 2020-8-15

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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.

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Answer 1

Matt J 2020-8-15

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

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Karim Hasouna 2020-8-15

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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

Walter Roberson 2020-8-16

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 2020-8-16

编辑：Matt J 2020-8-16

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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