- #36
bobby2k
- 127
- 2
Stephen Tashi said:The limit in the theorem is a special kind of limit. That's why there is an "a.s" in the notation of the limit. This is not the kind of limit used in ordinary calculus. So you can't say an "a.s" limit implies the existence of the kind of limit used in ordinary calculus.
Ok, I see, good point. I guess what you are saying is very logical because if it was that existence of the limit was implied, then it would be proof that relative frequencies was probabilities.Thanks for disproving the set inclusion question. I wanted a way to visualize what it meant that something had probability 1. In the text they say "we are essentially certain that the relative frequency will converge to the probability". But how can they say this. Even though we have said that an event have probability 1, we haven't defined that probability 1 means that it must happen have we?
I mean, how are we supposed to interpret the strong law of large numbers/the absolute certain convergence of relative frequency? My attempt earler to show set inclusion failed as you showed. Is there something else "concrete" that can be done to back up the statement "essentially certain".
Is it maybe possible in this case that P(K) = 1 imlies K = S. Because in your case an unncountable number of possibilities, but in this case we can enumerate each case((n=1,nA=1),(n=1,nA=0),(n=2,nA=0). etc.), so get less cardinality than your example. (I am on very thin ice here, but I am just trying to find a way to interpret the result.)
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