Constraints on Parameter Estimation

I am trying to fit linear regression model and predict parameters without intercept. I have written my code as under;

tbl=table(yobs,x1,x2,x3);

mdl = fitlm(tbl,'yobs ~ x1 + x2 + x3 - 1')

but I am getting the estimates which are negative but in my model all parameters should be positive. LB>=0 and UB=inf. How to set these constraints while doing the prediction.

Answer 1

[Torsten]

Use lsqlin instead of fitlm.

[Faizan Lali]

I am using lsqlin with lb=[0.5 0 0] and ub=[inf inf inf], but it is not changing the 1st parameter. I mean whatever I give the value as lb it it spitting out same for Beta1, but changing the values of other two betas.

[Torsten]

C = [x1 x2 x3];
d = yobs;
lb = [0.5 0 0];
ub = [inf inf inf];
sol = lsqlin(C,d,[],[],[],[],lb,ub)

where x1, x2, x3 and yobs are column vectors of the same length doesn't work ?

[Matt J]

Perhaps the columns of C need to be normalized.

C = [x1 x2 x3];
  c=vecnorm(C,2,1);

d = yobs+1;

lb = [0.5 0 0];
ub = [inf inf inf];

sol = lsqlin(C./c,d,[],[],[],[],lb,ub);

sol=sol./c(:)

[Faizan Lali]

Thank you, it worked. but it is not giving me the estimates with low SE. I mean the parameter estimates are way different from global values and it is not giving the best fit.

[Torsten]

This is the best fit you can get without intercept and the constraints you want to impose on the parameters.

[Torsten]

@Matt J

According to the documentation,

yobs ~ x1 + x2 + x3 - 1

means a three-variable linear model without intercept.

Thus the "-1" just means: no constant term, not

yobs = p1*x1 + p2*x2 + p3*x3 - 1

Very confusing.