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-   -   Chapter 1 - Exercise 1.9 (http://book.caltech.edu/bookforum/showthread.php?t=4412)

goldsj3 09-11-2013 10:31 AM

Chapter 1 - Exercise 1.9
 
Based on the marbles example, I understand that u is the probability of choosing a red marble from the bin of marbles and v is the fraction of red marbles in the sample of chosen marbles. And I think that the LHS of the Hoeffding Inequality represents the probability of v deviating from u being greater than the error bar epsilon. Now, I am not sure how I should choose an epsilon to evaluate this probability and where I should begin. Can someone point me in the right direction?

magdon 09-11-2013 12:31 PM

Re: Chapter 1 - Exercise 1.9
 
If \mu=0.9 and \nu\le0.1, it implies that |\nu-\mu|>0.8^- (any number slightly less than 0.8). By the implication bound,

P[\nu\le0.1] \le P[|\nu-\mu|>0.8^-]

By looking at the RHS, one can identify \epsilon for applying the Hoeffding bound.

eshmrt 09-15-2015 02:51 PM

Re: Chapter 1 - Exercise 1.9
 
Since we're only looking at epsilon applied in one direction, meaning that we're only interested in nu being mu - 0.8 and not mu + 0.8, should we drop the leading 2 from the bound calculation? I'm not sure if I'm interpreting this bound correctly or simply overthinking things.

magdon 09-21-2015 04:27 PM

Re: Chapter 1 - Exercise 1.9
 
You are right. If al you want is the "one sided deviation", you could drop the factor of 2. However keeping the factor of 2 is a worse but still valid bound.


Quote:

Originally Posted by eshmrt (Post 12036)
Since we're only looking at epsilon applied in one direction, meaning that we're only interested in nu being mu - 0.8 and not mu + 0.8, should we drop the leading 2 from the bound calculation? I'm not sure if I'm interpreting this bound correctly or simply overthinking things.


henry2015 11-05-2017 07:17 AM

Re: Chapter 1 - Exercise 1.9
 
Quote:

Originally Posted by magdon (Post 11480)
If \mu=0.9 and \nu\le0.1, it implies that |\nu-\mu|>0.8^- (any number slightly less than 0.8). By the implication bound,

P[\nu\le0.1] \le P[|\nu-\mu|>0.8^-]

By looking at the RHS, one can identify \epsilon for applying the Hoeffding bound.

Hi, I am a bit confused.

I thought when it says \nu\le0.1, it means 0\le\nu\le0.1 because smallest probably is 0; hence, 0.8\le|\nu-\mu|\le0.9....I am sure I miss something.

Any pointer?

Thanks!

mike7 09-11-2018 03:03 PM

Re: Chapter 1 - Exercise 1.9
 
Quote:

Originally Posted by magdon (Post 11480)
If \mu=0.9 and \nu\le0.1, it implies that |\nu-\mu|>0.8^- (any number slightly less than 0.8). By the implication bound,

P[\nu\le0.1] \le P[|\nu-\mu|>0.8^-]

By looking at the RHS, one can identify \epsilon for applying the Hoeffding bound.

If I am understanding the implications of this correctly, \epsilon = 0.8 is too large within the Hoeffding bound, correct? One would need to select some \epsilon < 0.8, perhaps arbitrarily?


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