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-   -   Linear regression with constraint on the hypothesis set (http://book.caltech.edu/bookforum/showthread.php?t=4256)

barbacot 04-30-2013 01:25 PM

Linear regression with constraint on the hypothesis set
 
When applying the one-step equation for linear regression, the vector of weights is obtained directly with all its components.

What if we impose from the beginning a restriction on the form of the hypothesis, say h(x)=3+w1*x1, instead of the full linear form h(x)=w0+w1*x1? In other words, we want w0 to be 3, no matter what.

Is the one-step equation still applicable somehow?

To compare, if we were to apply the gradient descent with the same constraint, we could do it very easy, just by keeping w0 fixed at its initial value (3).

yaser 04-30-2013 01:35 PM

Re: Linear regression with constraint on the hypothesis set
 
Quote:

Originally Posted by barbacot (Post 10674)
When applying the one-step equation for linear regression, the vector of weights is obtained directly with all its components.

What if we impose from the beginning a restriction on the form of the hypothesis, say h(x)=3+w1*x1, instead of the full linear form h(x)=w0+w1*x1? In other words, we want w0 to be 3, no matter what.

Is the one-step equation still applicable somehow?

To compare, if we were to apply the gradient descent with the same constraint, we could do it very easy, just by keeping w0 fixed at its initial value (3).

You can transfer the fixed part to the other side of the equation, then solve for the remaining parameters only. In your example, it would be w_1x_1=y-3 solving for w_1 only (the matrix {\rm X} would be a column vector of x_1's).

barbacot 05-04-2013 04:22 AM

Re: Linear regression with constraint on the hypothesis set
 
Thank you, sir, I think I finally got it. So if I want to pin down a number of M weights, I just move the constant M terms to the y vector (subtracting from it), and I am left with a matrix X with d+1-M columns (the column of ones may also be gone). The result will be a vector of d+1-M weights.


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