Quantile regression and linprog

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Patrick
Patrick 2014-2-9
评论： Patrick 2014-9-24
I'm trying to implement the linear program at page 2 of this link http://faculty.arts.ubc.ca/pschrimpf/628/hw5-quantile.pdf . I can't seem to get it to work though. As far as I have understood the estimate should be retrievable as the last element of the solution, but this is very small (<10e-10) and it should be 1.0675 according to another quantile regression algorithm (basically the true parameter is set to 1). Any thoughts?
% set seed
rng(1);
% set parameters
n=30;
tau=0.5;
% create regressor and regressand
x=rand(n,1);
y=x+rand(n,1)/10;
% number of regressors (1)
m=size(x,2);
% vektors and matrices for linprog
f=[tau*ones(n,1);(1-tau)*ones(n,1);zeros(m,1)];
A=[eye(n),-eye(n),x;
-eye(n),eye(n),-x;
-eye(n),zeros(n),zeros(n,m);
zeros(n),-eye(n),zeros(n,m)];
b=[y;
y
zeros(n,1);
zeros(n,1)];
% get solution bhat=[u,v,beta] and exitflag (1=succes)
[bhat,~,exflag]=linprog(f',A,b);

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回答（1 个）

Gittetier
Gittetier 2014-9-24
Hello Patrick, probably this no longer of any interest to you. But I think there is a mistake in the proposition you cite. When you split up the equality condition into two inequalities one of the Y should turn negative. So there is a mistake in your matrix b. In front of the second Y there should be a minus sign. Best Gittetier
  1 个评论
Patrick
Patrick 2014-9-24
Hello Gittetier, well you are quite right. I had actually forgotten this question. I solved it not long after, but had already forgotten about this.
I have the corrected code on my computer at home, and should probably post the corrections. Thanks for reminding me. Patrick

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