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least absolute deviation matlab
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Accepted Answer
Matt J
on 2 Jul 2013
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 Comments
dav
on 2 Jul 2013
Edited: dav
on 2 Jul 2013
Hi,
Thanks for your answer.
I have a dataset yt (all positive - size (100,1)). I need to fit a straight line to this data with the constraints that intercept parameter is greater than zero and slope parameter is between 0 1nd 1.
Could you please help me with this code? It is a little difficult to figure it out using your answer.
Thanks much!
Matt J
on 2 Jul 2013
Edited: Matt J
on 2 Jul 2013
You cannot impose strict inequality constraints. The minimum can be ill-defined, for example in the minimization of f(a)=a^2 over a>0. If a=0 is not allowed, then what do you consider the minimizing point?
dav
on 2 Jul 2013
The data set is yt.
clc;
clear;
T = 300;
a0 = 0.1; a1 = 0.4;
ra = zeros(T+2000,1);
seed=123;
rng(seed);
ra = trnd(5,T+2000,1);
ytn = [];
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;
yt = epsi2(2001:T+2000);
Matt J
on 3 Jul 2013
Edited: Matt J
on 15 Jul 2013
I don't understand your code, but here's an example for you to study
%%Fake data
xi=0:10;
yi=2*xi+bt+randn(size(xi))/2;
%%Linprog parameters
N=length(xi);
e=ones(N,1);
f=[0,0,e.'];
A=[xi(:),e,-speye(N);-xi(:), -e, -speye(N)];
b=[yi(:);-yi(:)];
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),
yf=slope*xi+intercept;
plot(xi,yi,xi,yf)
Matt J
on 15 Jul 2013
Edited: Matt J
on 15 Jul 2013
Recall that the reformulation of the problem requires additional constraints
r(x) <=d
-r(x)<=d
where r(x) is your residual. The A,b pair that you've cited is the implementation of these additional constraints. bt is the true intercept of the simulated line data, yi. You can choose any value for it that you want, for simulation purposes, and see if the p(2) returned by linprog ends up being a good estimate of it.
dav
on 18 Jul 2013
Edited: dav
on 18 Jul 2013
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),
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