Please help with fmincon function

Question

dav 2013-11-9

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/105520-please-help-with-fmincon-function

评论： dav 2013-11-10

采纳的回答： Walter Roberson

在 MATLAB Online 中打开

Hi,

I am trying to estimate parameters of an arch model using MLE.

If my code works fine I should get estimates 0.1 and 0.4 ( which I used to create the data)

However, I am having some issues and the estimates I get are wrong.

Can someone help me with this please. Given below are the codes I used.

% data generation
clc;
clear;
p=1;
T = 3000;
ra = zeros(T+2000,1);
seed=123;
rng(seed);
ra = randn(T+2000,1); 
epsi=zeros(T+2000,1);
simsig=zeros(T+2000,1);
a0=0.1; a1=0.4;
unvar = a0/(1-a1);
for i = 1:T+2000
                 if (i==1) 
                     simsig(i) = a0+a1*((a0)/(1-a1));
                     s=(simsig(i))^0.5;
                     epsi(i) = ra(i) * s;
                   else                     
                       simsig(i) = a0+ a1*(epsi(i-1))^2;
                       s=(simsig(i))^0.5;
                       epsi(i) = ra(i)* s;
                   end
  end
yt = epsi(2001:T+2000);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%ESTIMATION %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
theta0 = [1;1];
A=[0 1];
b=0.99999;
[theta, opt] = fmincon(@(theta)...
               lach(theta,yt),theta0,A,b,[],[],0,1);
% function lach
function L = lach(theta,y)
w     =  theta(1); 
alpha =  theta(2);
y2   = y.^2;
[T1,K] = size(y2); 
ht    = zeros(T1,1);
ht(1) = sum(y2)/T1;
for i=2:T1
    ht(i)=w + alpha*y2(i-1);
end
sqrtht  = sqrt(ht);
x       = y./sqrtht;
l = -0.5*log(2*pi) - log(sqrtht) - 0.5*(x.^2);
l=-l;
L = sum(l);

Also, if you can kindly explain the difference between the following two ways of calling fmincon, that will be highly appreciated.

[theta, opt] = fmincon(@(theta)...
               lach(theta,yt),theta0,A,b,[],[],0,1);
[theta, opt] = fmincon(@lach,theta0,A,b,[],[],0,1);

thanks

When writing these codes I got some help from codes I found online.

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dav 2013-11-10

thanks

Matt J 2013-11-10

编辑：Matt J 2013-11-10

在 MATLAB Online 中打开

If my code works fine I should get estimates 0.1 and 0.4 ( which I used to create the data)

What happens when you run fmincon with 0.1 and 0.4 as your initial guess? What is the final value for opt in that case and what was it for

theta =
      0.0002
    -13.4557

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Answer 1

Walter Roberson 2013-11-10

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/105520-please-help-with-fmincon-function#answer_114702

在 MATLAB Online 中打开

[theta, opt] = fmincon(@lach,theta0,A,b,[],[],0,1);

will not have any way of getting y into the routine lach. What you have now,

[theta, opt] = fmincon(@(theta) lach(theta,yt),theta0,A,b,[],[],0,1);

is fine.

The message "Local minimum possible. Constraints satisfied." is a message you would get if it was successful in finding a minimum. fmincon cannot tell whether the minimum it finds is a local minimum or the global minimum so it is being conservative in warning you that what it has find might be a local minima.

The message "fmincon stopped because the size of the current search direction is less than" and so on, is less common. You are getting it because you are hitting against the minimum for the first pararameter. The algorithm has detected that it cannot take two more steps towards the minimum, using the minimum step size that it has been configured to use. You could pass an options structure to specify a smaller minimum step size, but the effect of that would just be to get you ever closer to the boundary 0.

I suspect that you are running into a local minimum. That happens with fmincon(). You can try experimenting with different theta0 (different starting points.) If you had the global optimization toolbox, you could use a routine that would run fmincon() with a variety of different starting points, automatically, looking for the best result out of the set. The global optimization toolbox also has other minimization routines that can be used.