LFD Book Forum  

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

Reply
 
Thread Tools Display Modes
  #1  
Old 09-14-2015, 06:36 PM
castit castit is offline
Junior Member
 
Join Date: Sep 2015
Posts: 1
Default Exercise 1.10 part c

For part c where we plot estimates of the probability, are we supposed to be basing our estimates on all coins we flipped or only the three coins we made histograms for: c1, crand and cmin? And if only those three coins, are we making a different graph for each one or somehow combine the three coins into a single probability estimate?
Reply With Quote
  #2  
Old 01-12-2016, 02:31 AM
MaciekLeks MaciekLeks is offline
Member
 
Join Date: Jan 2016
Location: Katowice, Upper Silesia, Poland
Posts: 17
Default Re: Exercise 1.10 part c

I got lost with that part of the exercise, but I would like to cope with that. Can someone please explain this as a software engineer but not a statistician?
How to "plot estimates for P[|v-u|>epsilon] as a function of epsilon" based on data from the simulation?

P.S.1. At least the plot image would be helpful to imagine what the author(of the exercise) had in mind.

P.S.2. I read all the posts related to this exercise and I see, that more people have a problem with this point.
Reply With Quote
  #3  
Old 01-13-2016, 07:49 AM
magdon's Avatar
magdon magdon is offline
RPI
 
Join Date: Aug 2009
Location: Troy, NY, USA.
Posts: 595
Default Re: Exercise 1.10 part c

Fix \epsilon to say 0.1.

Now run the experiment, and compute |\mu-\nu|. Repeat. Some of the time, |\mu-\nu|>\epsilon. Compute the fraction of the time that |\mu-\nu|>\epsilon. You now have a pair:

(\epsilon=0.1, fraction of time |\mu-\nu|>\epsilon)

Repeat the whole process for \epsilon=0.2,0.3,\ldots and plot the fraction versus \epsilon.

Quote:
Originally Posted by MaciekLeks View Post
I got lost with that part of the exercise, but I would like to cope with that. Can someone please explain this as a software engineer but not a statistician?
How to "plot estimates for P[|v-u|>epsilon] as a function of epsilon" based on data from the simulation?

P.S.1. At least the plot image would be helpful to imagine what the author(of the exercise) had in mind.

P.S.2. I read all the posts related to this exercise and I see, that more people have a problem with this point.
__________________
Have faith in probability
Reply With Quote
  #4  
Old 01-14-2016, 10:01 AM
MaciekLeks MaciekLeks is offline
Member
 
Join Date: Jan 2016
Location: Katowice, Upper Silesia, Poland
Posts: 17
Default Re: Exercise 1.10 part c

Thank you, Professor.

I've allowed myself to post my plot. If I should not put it here, please let me know. Is this what we should see?



I understand that vmin result is the explanation of this sentence from the book: "h is fixed before you generate the data set" for the truth of the Hoeffding Inequality. Am I right? I also understand that in spite of the fact that crand is a random coin it holds the Hoeffding Inequality due to the randomness according to the Binomial distribution over every run of the experiment.

Last edited by MaciekLeks; 01-14-2016 at 10:56 AM. Reason: changed the source url of the plot to be more readable
Reply With Quote
  #5  
Old 02-09-2016, 06:29 AM
ntvy95 ntvy95 is offline
Member
 
Join Date: Jan 2016
Posts: 37
Default Re: Exercise 1.10 part c

I am wondering about the point of this exercise. Here is my guess:



hence:



It means that c_min always stands a higher chance that | v_min - u | > e compared to the other coins, so we will doubt if c_min obeys the Hoeffding bound? c_rand and c_1 however do not have this garuantee so they will obey the Hoeffding bound?
Reply With Quote
  #6  
Old 03-23-2016, 05:19 AM
ntvy95 ntvy95 is offline
Member
 
Join Date: Jan 2016
Posts: 37
Default Re: Exercise 1.10 part c

I am still trying to understand this exercise. I have seen the error in my above post but I cannot edit it so I will post a new one here:

Here is my argument guess after reading the dicussion here: Because 1,000 coins all share the same \mu and any two flips of one coin are independent from each other, and any two flips of two coins are also independent from each other, and the event a coin is randomly selected is independently from its flip result, so c_{rand} can be treated as a specific coin. Hence the distribution of \nu_{1} and \nu_{rand} is the same and it is binomial distribution. However, \nu_{min} has the different distribution and it is not binomial.

For example:

(*)

while:



Is my argument and calculation correct? I am still confused about the random coin (I think my above argument about random coin is still rather naive). Is there anything that distinguishes c_rand from c_1?

I see that the result (*): 1 - (1 - 0.5^(10))^(1000) = 0.62357620194... is very close to giridhar1202's experiemental result, and I also see that (*) is analogous to the coin example that you mentioned in Lecture 02's video. Is my view right?

Thank you very much in advance.
Attached Thumbnails
gif (5).gif   gif.gif  
Reply With Quote
  #7  
Old 07-17-2019, 10:14 PM
lfdreader lfdreader is offline
Junior Member
 
Join Date: Jul 2019
Posts: 1
Default Re: Exercise 1.10 part c

Can't see the plot. Any alternate links? Thank you!

Quote:
Originally Posted by MaciekLeks View Post
Thank you, Professor.

I've allowed myself to post my plot. If I should not put it here, please let me know. Is this what we should see?



I understand that vmin result is the explanation of this sentence from the book: "h is fixed before you generate the data set" for the truth of the Hoeffding Inequality. Am I right? I also understand that in spite of the fact that crand is a random coin it holds the Hoeffding Inequality due to the randomness according to the Binomial distribution over every run of the experiment.
Reply With Quote
  #8  
Old 07-26-2019, 04:17 AM
AlexS AlexS is offline
Junior Member
 
Join Date: Sep 2018
Posts: 4
Default Re: Exercise 1.10 part c

Is there correct answer?

Quote:
Originally Posted by ntvy95 View Post
I am still trying to understand this exercise. I have seen the error in my above post but I cannot edit it so I will post a new one here:

Here is my argument guess after reading the dicussion here: Because 1,000 coins all share the same \mu and any two flips of one coin are independent from each other, and any two flips of two coins are also independent from each other, and the event a coin is randomly selected is independently from its flip result, so c_{rand} can be treated as a specific coin. Hence the distribution of \nu_{1} and \nu_{rand} is the same and it is binomial distribution. However, \nu_{min} has the different distribution and it is not binomial.

For example:

(*)

while:



Is my argument and calculation correct? I am still confused about the random coin (I think my above argument about random coin is still rather naive). Is there anything that distinguishes c_rand from c_1?

I see that the result (*): 1 - (1 - 0.5^(10))^(1000) = 0.62357620194... is very close to giridhar1202's experiemental result, and I also see that (*) is analogous to the coin example that you mentioned in Lecture 02's video. Is my view right?

Thank you very much in advance.
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 09:15 AM.


Powered by vBulletin® Version 3.8.3
Copyright ©2000 - 2019, 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.