Homework#6 Q3

invis · #1 08-15-2012, 01:36 PM

I have some problems with understanding the question #3.

Ok, we have a formula for linear regression with regularization:
$w_{reg} = (Z^T Z- \lambda I)^{-1} Z^T y$

But I cant catch this:
"add the term $\lambda /N \sum_{i=0}^7 w_i^2$ to the squared in sample error"

What

I suppose to use in this formula ? The

without regularization ? And why adding to squared error ?

yaser · #2 08-15-2012, 03:30 PM

Quote:

Originally Posted by invis

What

I suppose to use in this formula ? The

without regularization ? And why adding to squared error ?

The solution $w_{\rm reg} = (Z^T Z+ \lambda I)^{-1} Z^T y$ (notice the plus sign) comes from solving the case of augmented error where the regularization term is added to the in-sample mean-squared error.

invis · #3 08-16-2012, 04:22 AM

Just one more question

... at this time

Why when we computing $w_{lin}$ there is no eye matrix ? We do the same divide to w by

.

But eye matrix appears when we divide $(Z^T Z w + \lambda w)$ by w. That confusing me.

yaser · #4 08-16-2012, 05:34 AM

Quote:

Originally Posted by invis

Just one more question

... at this time

Why when we computing $w_{lin}$ there is no eye matrix ? We do the same divide to w by

.

But eye matrix appears when we divide $(Z^T Z w + \lambda w)$ by w. That confusing me.

The "divide by" is in fact multiplying by a matrix inverse. In the case of augmented error, in order to put things in terms of a matrix, we write $Z^T Z w + \lambda w$ as $Z^T Z w + \lambda I w$ , then factor out $Z^T Z + \lambda I$ which becomes the matrix to be inverted. In the linear regression case, there is no

term (or you can think of it as $\lambda=0$ killing that term).

ESRogs · #5 08-20-2012, 09:32 AM

In the lecture, the formula for $w_{reg}$ was derived for the case where the weights were being multiplied by Legendre polynomials. Is it still valid when the set of transformations is different, such as those specified in the problem?

yaser · #6 08-20-2012, 02:01 PM

Quote:

Originally Posted by ESRogs

In the lecture, the formula for $w_{reg}$ was derived for the case where the weights were being multiplied by Legendre polynomials. Is it still valid when the set of transformations is different, such as those specified in the problem?

The formula is valid for any nonlinear transformation into a ${\cal Z}$ .

melipone · #7 02-14-2013, 07:16 PM

Quote:

Originally Posted by yaser

The solution $w_{\rm reg} = (Z^T Z+ \lambda I)^{-1} Z^T y$ (notice the plus sign) comes from solving the case of augmented error where the regularization term is added to the in-sample mean-squared error.

I am confused. Are we supposed to rederive the result of Slide 11?

yaser · #8 02-14-2013, 07:38 PM

Quote:

Originally Posted by melipone

I am confused. Are we supposed to rederive the result of Slide 11?

No rederivation needed. You can always use the results given in the lectures.

Kekeli · #9 05-13-2013, 05:07 AM

My answers for any value of K are order(s) of magnitude higher that the options.

w_reg = (Z^\top Z + λI)^-1 Z^T y

As a Python noob, perhaps someone can confirm that I need the linalg.inv of the first term in parentheses, because we cannot use the pinv method when the weight term is present.
thanks.

jlaurentum · #10 05-13-2013, 07:48 PM

Kekeli:

I don't know about Python, but in R I was using the "chol2inv" function of the "Matrix" library to find the matrix inverse. It turns out this wasn't the right tool for the job. I ended up using "solve" from the base package to find the inverse. So in R, I used the following functions:

t(M) for the transpose of an M matrix
%*% for matrix-matrix or matrix-vector multiplication (or inner product)
diag(...) to generate a diagonal matrix such as needed for the $\lambda I$ part.
solve(M) to find the inverse of an M matrix.

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