least absolute deviation matlab

3 次查看（过去 30 天）

显示更早的评论

dav 2013-7-2

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/80847-least-absolute-deviation-matlab

采纳的回答： Matt J

Hi,

Is the a way to do parameter estimation using Least Absolute deviation with constraints in matlab.

If so, please let me know of a place to read about it.

thanks.

采纳的回答

Matt J 2013-7-2

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/80847-least-absolute-deviation-matlab#answer_90595

在 MATLAB Online 中打开

If you are talking about the problem

min_x sum( abs(r(x)))

it is equivalent to

min_{x,d} sum(d)
s.t.
     r(x) <=d
    -r(x)<=d

The above formulation is differentiable if r(x) is differentiable, so FMINCON can handle it. You can obviously add any additional constraints you wish, so long as they too are differentiable.

14 个评论
显示 12更早的评论隐藏 12更早的评论

dav 2013-7-18

编辑：dav 2013-7-18

在 MATLAB Online 中打开

thank you.

I used your code to test the data set I have. BOTH parameter estimates should be 0.1. However, the estimates I get using your code are very different. is there a way to fix it. From a standard theory I know that when you regress the y vector I have mentioned on the x vector I should get parameters, both equal to 0.1

clc;
clear;
p=1;
T = 300;
a0 = 0.1; a1 = 0.1; 
seed=123;
ra = randn(T+2000,1); 
epsi=zeros(T+2000,1);
simsig=zeros(T+2000,1);
unvar = a0/(1-a1);
for i = 1:T+2000
                 if (i==1) 
                     simsig(i) = unvar;
                     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
epsi2 = epsi.^2;
y = epsi2(2001:T+2000); % THIS IS THE DATA SET I WANT TO TEST. 
len = length(y);
x = zeros(len,p);
for i = 1:p
       x(1+i:len,i) = y(1:len-i,1);  % THIS IS THE x VECTOR
end
       N=length(x);
       e=ones(N,1);
       f=[0,0,e.'];
       A=[x(:),e,-speye(N);-x(:), -e, -speye(N)];
       b=[y(:);-y(:)];
       lb=zeros(N+2,1);
       ub=inf(N+2,1); ub(1)=1;
       %%Perform fit and test
       p=linprog(f,A,b,[],[],lb,ub);
       slope=p(1),
       intercept=p(2),