#21




Re: Q10 higher bound
There are no rules constraining the specific hypotheses in a hypothesis set. It can be any subset of the power set of the underlying set of points.

#22




Re: Q10 higher bound
Quote:
for N=1, H={+,} and shatters this 1 point for N=2, H={+,+} and does not shatter these two points. for N=3, H={++, ++, ++, +, +, +} and does shatter two points. What is the VC dimension of this hypothesis set? From the N=2 case, I might conclude that it's 1 (it fails to shatter two points), but from the N=3 case I might conclude it's 2 (it shatters 2 of the points within this set of points). Where am I going wrong? 
#23




Re: Q10 higher bound
Since the homework set was now due, I was wondering if the course staff could weigh in on this question. Did our discussion adequately cover this question or were there any nuances that we missed?

#24




Re: Q10 higher bound
The discussion has been great. I won't talk about the answer which you all know by now as I want to keep this thread rather than archive it with *ANSWER* threads.
__________________
Where everyone thinks alike, no one thinks very much 
#25




Re: Q10 higher bound
Quote:

#26




Re: Q10 higher bound
There is a proviso to my glib claim above that absolutely any subset can be a hypothesis. While this is true when the underlying set is finite or countable, when it is uncountably infinite I believe it only makes sense to have hypotheses which are measurable sets, according to the probability distribution associated with samples.
[Almost anything that makes any sense is likely to satisfy this, but it is not difficult to construct pathological examples even when the sample space is as simple as the unit interval with the uniform distribution]. 
#27




Re: Q10 higher bound
Quote:
Quote:

#28




Re: Q10 higher bound
Quote:

#29




Re: Q10 higher bound

Thread Tools  
Display Modes  

