[過去ﾛｸﾞ] 現代数学の系譜 工学物理雑談 古典ガロア理論も読む74 (1002ﾚｽ)

現代数学の系譜 工学物理雑談 古典ガロア理論も読む74 http://rio2016.5ch.net/test/read.cgi/math/1564659345/

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787: 現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE [] 2019/08/14(水) 16:37:06.91 ID:rg2Nhb+h おサルさん、（>>692より）【必死のパッチ】やなｗｗ（＾＾； おサルさん、墓穴だなｗ https://mathoverflow.net/questions/151286/probabilities-in-a-riddle-involving-axiom-of-choice Probabilities in a riddle involving axiom of choice Dec 9 '13 (抜粋) asked Dec 9 '13 at 16:16 Denis I think it is ok, because the only probability measure we need is uniform probability on {0,1,…,N-1}, but other people argue it's not ok, because we would need to define a measure on sequences, and moreover axiom of choice messes everything up. Alexander Pruss Let's go back to the riddle. Suppose u￣ is chosen randomly. The most natural option is that it is a nontrivial i.i.d. sequence (uk), independent of the random index i which is uniformly distributed over [100]={0,...,99}. In general, Mj will be nonmeasurable (one can prove this in at least some cases). We likewise have no reason to think that M is measurable. But without measurability, we can't make sense of talk of the probability that the guess will be correct. Denis Dec 17 '13 at 15:21 Our choice of index i is made randomly, but for this we only need the uniform distribution on {0,…,n}. It is made independently of the opponent's choice. Tony Huynh Dec 9 '13 at 17:37 In order for such a question to make sense, it is necessary to put a probability measure on the space of functions f:N→R. Note that to execute your proposed strategy, we only need a uniform measure on {1,…,N}, but to make sense of the phrase it fails with probability at most 1/N, we need a measure on the space of all outcomes. The answer will be different depending on what probability space is chosen of course. If it were somehow possible to put a 'uniform' measure on the space of all outcomes, then indeed one could guess correctly with arbitrarily high precision, but such a measure doesn't exist. (引用終り) http://rio2016.5ch.net/test/read.cgi/math/1564659345/787

788: １３２人目の素数さん [] 2019/08/14(水) 16:48:21.03 ID:c6g6R1pg >>787 お前はすでに死んでいるｗ 測度は必要ない　必要だと主張する奴は皆馬鹿ｗ ついでにいうとTony Huynhはスレ主並みの考えナシの馬鹿ｗｗｗ http://rio2016.5ch.net/test/read.cgi/math/1564659345/788

790: １３２人目の素数さん [] 2019/08/14(水) 16:59:51.73 ID:MPteNw3f >>787 確率論の専門家の誤解を理解していない時点で何言っても無駄 残念 http://rio2016.5ch.net/test/read.cgi/math/1564659345/790

791: 現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE [] 2019/08/14(水) 17:01:40.68 ID:rg2Nhb+h >>773 補足 （>>787より抜粋） おサルさん、（>>692より）【必死のパッチ】やなｗｗ（＾＾； おサルさん、墓穴だなｗ https://mathoverflow.net/questions/151286/probabilities-in-a-riddle-involving-axiom-of-choice Probabilities in a riddle involving axiom of choice Dec 9 '13 (抜粋) asked Dec 9 '13 at 16:16 Denis I think it is ok, because the only probability measure we need is uniform probability on {0,1,…,N-1}, but other people argue it's not ok, because we would need to define a measure on sequences, and moreover axiom of choice messes everything up. Alexander Pruss Let's go back to the riddle. Suppose u￣ is chosen randomly. The most natural option is that it is a nontrivial i.i.d. sequence (uk), independent of the random index i which is uniformly distributed over [100]={0,...,99}. In general, Mj will be nonmeasurable (one can prove this in at least some cases). We likewise have no reason to think that M is measurable. But without measurability, we can't make sense of talk of the probability that the guess will be correct. Denis Dec 17 '13 at 15:21 Our choice of index i is made randomly, but for this we only need the uniform distribution on {0,…,n}. It is made independently of the opponent's choice. Tony Huynh Dec 9 '13 at 17:37 In order for such a question to make sense, it is necessary to put a probability measure on the space of functions f:N→R. Note that to execute your proposed strategy, we only need a uniform measure on {1,…,N}, but to make sense of the phrase it fails with probability at most 1/N, we need a measure on the space of all outcomes. The answer will be different depending on what probability space is chosen of course. If it were somehow possible to put a 'uniform' measure on the space of all outcomes, then indeed one could guess correctly with arbitrarily high precision, but such a measure doesn't exist. (引用終り) http://rio2016.5ch.net/test/read.cgi/math/1564659345/791

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