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?

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[vu>epsilon] as a function of epsilon" based on data from the simulation?:bow: 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. 
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? https://scontentfra31.xx.fbcdn.net...54&oe=57408E4A 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.:clueless: 
Re: Exercise 1.10 part c
I am wondering about the point of this exercise. Here is my guess:
http://www.vynguyen.net/wpcontent/q...d396812_l3.svg hence: http://www.vynguyen.net/wpcontent/q...91a4ba4_l3.svg 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? 
Re: Exercise 1.10 part c
2 Attachment(s)
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: http://book.caltech.edu/bookforum/at...1&d=1458732539 (*) while: http://book.caltech.edu/bookforum/at...1&d=1458732513 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). :clueless: 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? :clueless: Thank you very much in advance. 
Re: Exercise 1.10 part c
Can't see the plot. Any alternate links? Thank you!
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Re: Exercise 1.10 part c
Is there correct answer?
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