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:
Originally Posted by Sandes
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.
