LFD Book Forum
Page 1 of 2 1 2
Show 40 post(s) from this thread on one page

LFD Book Forum (http://book.caltech.edu/bookforum/index.php)
-   Homework 6 (http://book.caltech.edu/bookforum/forumdisplay.php?f=135)
-   -   Homework#6 Q3 (http://book.caltech.edu/bookforum/showthread.php?t=1012)

invis 08-15-2012 01:36 PM
Homework#6 Q3
 
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 w I suppose to use in this formula ? The w without regularization ? And why adding to squared error ? :confused:

yaser 08-15-2012 03:30 PM
Re: Homework#6 Q3
 
Quote:
Originally Posted by invis (Post 4095)
What w I suppose to use in this formula ? The w without regularization ? And why adding to squared error ? :confused:
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 08-16-2012 04:22 AM
Re: Homework#6 Q3
 
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 Z^T Z w.

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

yaser 08-16-2012 05:34 AM
Re: Homework#6 Q3
 
Quote:
Originally Posted by invis (Post 4102)
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 Z^T Z w.

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

ESRogs 08-20-2012 09:32 AM
Re: Homework#6 Q3
 
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 08-20-2012 02:01 PM
Re: Homework#6 Q3
 
Quote:
Originally Posted by ESRogs (Post 4182)
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 02-14-2013 07:16 PM
Re: Homework#6 Q3
 
Quote:
Originally Posted by yaser (Post 4097)
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 02-14-2013 07:38 PM
Re: Homework#6 Q3
 
Quote:
Originally Posted by melipone (Post 9393)
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 05-13-2013 05:07 AM
Re: Homework#6 Q3
 
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 05-13-2013 07:48 PM
Re: Homework#6 Q3
 
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:
  1. t(M) for the transpose of an M matrix
  2. %*% for matrix-matrix or matrix-vector multiplication (or inner product)
  3. diag(...) to generate a diagonal matrix such as needed for the \lambda I part.
  4. solve(M) to find the inverse of an M matrix.


All times are GMT -7. The time now is 07:00 AM.
Page 1 of 2 1 2
Show 40 post(s) from this thread on one page

Powered by vBulletin® Version 3.8.3
Copyright ©2000 - 2021, Jelsoft Enterprises Ltd.
The contents of this forum are to be used ONLY by readers of the Learning From Data book by Yaser S. Abu-Mostafa, Malik Magdon-Ismail, and Hsuan-Tien Lin, and participants in the Learning From Data MOOC by Yaser S. Abu-Mostafa. No part of these contents is to be communicated or made accessible to ANY other person or entity.