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a.sanyal902 04-29-2013 01:24 PM

Lecture 6- 2N sample
 
Hi all,
At the end of lecture 6, Prof. Yaser replaces one sample with 2 samples of size N each. We then replace m(N) with m(2N) in the inequality.

However, these are 2 distinct samples. So won't the maximum no. of hypothesis actually be 2m(N), which in general is more than m(2N)) ?
The constraints of 2N points would hold if it were a single sample set, but here more no. of hypothesis are possible as the N-sets are distinct.

Is this the reason for the additional factors in the inequality?

yaser 04-29-2013 02:45 PM

Re: Lecture 6- 2N sample
 
Quote:

Originally Posted by a.sanyal902 (Post 10654)
Hi all,
At the end of lecture 6, Prof. Yaser replaces one sample with 2 samples of size N each. We then replace m(N) with m(2N) in the inequality.

However, these are 2 distinct samples. So won't the maximum no. of hypothesis actually be 2m(N), which in general is more than m(2N))?

I am not sure I completely got the point that you are making, so let me just address the above statement. The growth function normally grows very fast, so it will more than double in value when its argument doubles. When we consider all dichotomies on 2N points, whether we view them as one sample or two samples, the number to be considered is m(2N).

a.sanyal902 04-29-2013 10:22 PM

Re: Lecture 6- 2N sample
 
Thank you Prof! I was a little confused earlier. Much clearer now!


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