#1




Problem 5.1b
I get as far as this
P ≥ 1  4mH(N)^2/e^(N/32) but I'm not certain how to go about showing that P ≥ 1  mH(N)/2^N Maybe I have totally the wrong approach: I started with Eout(h) ≤ Ein(h) + sqr (8/N ln(4mH(N)^2/δ)) with P ≥ 1δ So perhaps there's another formula that would work better. I feel like I'm making a silly mistake somewhere but I'm not sure what it could be. 
#2




Re: Problem 5.1b
If the data is generated from a random (arbitrary) target function, then every dichotomy is equally likely. Since you can implement at most m(N) of these dichotomies, the number of dichotomies you cannot implement is at least 2^Nm(N). Each of these dichotomies you cannot implement has probability 1/2^N
Quote:
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Have faith in probability 
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