LFD Book Forum  

Go Back   LFD Book Forum > Book Feedback - Learning From Data > Chapter 1 - The Learning Problem

Reply
 
Thread Tools Display Modes
  #11  
Old 06-23-2017, 04:42 PM
RicLouRiv RicLouRiv is offline
Junior Member
 
Join Date: Jun 2017
Posts: 7
Default Re: Problem 1.9

Quick question on Part D...

I agree that s = ln(a / (1-a)) is the value of s that minimizes e^(-sa)U(s), where I'm using "a" to represent "alpha."

I think that, if you plug this value of s in, you get 2^(-b), where I'm using "b" to represent "beta."

So, 2^(-b) < e^(-sa)U(s), by definition of minimization.

Then, raising to the power of N, we get:

2^(-bN) < (e^(-sa)U(s))^N, but this inequality is the wrong way.

Any tips?
Reply With Quote
  #12  
Old 06-24-2017, 09:36 AM
RicLouRiv RicLouRiv is offline
Junior Member
 
Join Date: Jun 2017
Posts: 7
Default Re: Problem 1.9

Ah, maybe it's because the inequality holds for any s, it must hold for the min.
Reply With Quote
  #13  
Old 06-29-2017, 10:00 PM
htlin's Avatar
htlin htlin is offline
NTU
 
Join Date: Aug 2009
Location: Taipei, Taiwan
Posts: 562
Default Re: Problem 1.9

Quote:
Originally Posted by RicLouRiv View Post
Ah, maybe it's because the inequality holds for any s, it must hold for the min.
You got it. :-)
__________________
When one teaches, two learn.
Reply With Quote
  #14  
Old 10-22-2017, 04:19 AM
mygame182 mygame182 is offline
Junior Member
 
Join Date: Oct 2017
Posts: 2
Default Re: Problem 1.9

Quote:
Originally Posted by RicLouRiv View Post
Ah, maybe it's because the inequality holds for any s, it must hold for the min.
Excuse me, I can understand your last reply, but confuse in this one.
More tips please.

When 2^(-b) is the minimize of e^(-sa)U(s), i find that 1-a = 1/2 + e (e means epsilon), so P[u>=a] = P[u>= 1/2 - e] , but not P[u>=1/2+e].
Reply With Quote
  #15  
Old 10-23-2017, 08:23 AM
mygame182 mygame182 is offline
Junior Member
 
Join Date: Oct 2017
Posts: 2
Default Re: Problem 1.9

When 2^(-b) equal to the minimize of e^(-sa)U(s), i try to assume that 1-a=1/2-e, so that a = 1/2 + e, P[u>=a]=P[u>=1/2+e]

According to (b), P[u>=1/2+e] = P[u>=a] <= (e^(-sa)U(s))^N for any s , even if the minimize of e^(-sa)U(s) when s = ln(a / (1-a)).

Hence P[u>=1/2+e] <= 2^(-bN)
Reply With Quote
  #16  
Old 11-14-2017, 03:14 PM
don slowik don slowik is offline
Junior Member
 
Join Date: Nov 2017
Posts: 4
Default Re: Problem 1.9

That's correct. But it doesn't get us anywhere.
Reply With Quote
  #17  
Old 11-14-2017, 03:18 PM
don slowik don slowik is offline
Junior Member
 
Join Date: Nov 2017
Posts: 4
Default Re: Problem 1.9

Ahhhm, That was a reply to kongweihan's OP.
It is correct as shown by the first line: Since...
But ends up barking up the wrong tree.
There is a nice description on wikipedia of Chernoff bound.
It is similar to maciekleks path.
Reply With Quote
Reply

Thread Tools
Display Modes

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off

Forum Jump


All times are GMT -7. The time now is 04:23 AM.


Powered by vBulletin® Version 3.8.3
Copyright ©2000 - 2017, 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.