4
$\begingroup$

For solving an unconstrained LS regression $$\hat{y}=w_1.x_1+w_2.x_2+w_3.x_3+w_4.x_4 + \epsilon$$ I use the following normal equation: $$W^*=(X^{\top}X)^{-1}X^{\top}Y $$

If I want to introduce a constraint on all of the parameters, e.g. the ridge regression $w_i^2<c$, for $\forall i$, I can formulate the Lagrangian:

$$\mathcal{L(W,\lambda)} = (Y - XW)^2 - \lambda (W)^2.$$

and also obtain a matrix form solution, given by: $$W^*=(X^{\top}X-\lambda I)^{-1}X^{\top}Y $$

The question is: how may I formulate the above equation if I only want to impose constraints on some of the coefficients, e.g. $w_1=w_2$ with $w=[w_1, w_2, ..., w_d]$?

I can get to the Lagrangian, which would be: $$ \mathcal{L(W,\lambda)} = (Y - XW)^2 - \lambda (w_1-w_2)$$ but I can't get to the matrix solution for $W^*$.

I'm searching for a manual solution (i.e. with no python or R code). Thanks in advance and sorry for the not-rigorous notation.

$\endgroup$
5
  • 1
    $\begingroup$ With $w_1=w_2$, just look at the linear predictor $w_0+w_1 x_1 + w_2 x_2 + \dotsm = w_0 + w_1 (x_1+x_2) + \dotsm $! So jus remove one of them, and add the corresponding predictors. Here is a similar but more complicated example: stats.stackexchange.com/questions/248779/… $\endgroup$ Commented Jun 27, 2018 at 9:42
  • 1
    $\begingroup$ What could I do to impose a constraint that w1 should be "around" 1000 and w3 "around" 50? $\endgroup$
    – MrT77
    Commented Jan 30, 2019 at 5:59
  • $\begingroup$ Maybe a Bayes solution, with a prior distribution with mean of $W1$ 1000 and of $w3$ 50? and some prior variance expressing how certain you are about those restrictions. Or, not going that route, using regularization but with offsets in addition, that is , $w1$ is represented by w1+offset(Id(50*w1)) in R. $\endgroup$ Commented Jan 30, 2019 at 7:29
  • $\begingroup$ Given your comment "What could I do to impose a constraint that w1 should be "around" 1000 and w3 "around" 50" - would you like to shrink your coefficients towards particular prior centers? If so look at the answer here stats.stackexchange.com/questions/76925/… $\endgroup$ Commented Apr 12 at 13:15
  • $\begingroup$ If you are looking for least squares with box constraints on individual coefficients you can look at my answer here stats.stackexchange.com/questions/136563/… $\endgroup$ Commented Apr 12 at 13:16

1 Answer 1

0
$\begingroup$

Ridge is the exception, not the norm, in this case. Applying a quadratic term in the Lagrangian still allows for a closed solution of the regression, and is one of the reasons why Ridge is interesting. Using a different function to regularize will most likely not produce a closed form solution. As an example, L1-regularization, the LASSO, has no closed form solution for $W^*$.

While I could not show that your regularization has no closed form, I am persuaded that is does not; because this coupling of coordinates usually yields nasty results in matrix theory (see the uses of the Vandermonde determinant in Gaussian Random Matrices and you will see what I mean by nasty). This does not mean it is useless, just like LASSO is not useless, it just means that you should look for a numerical, iterative and algorithmic solution rather than a pretty closed form.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.