現代数学の系譜工学物理雑談古典ガロア理論も読む74

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78: 現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE  [sage] 2019/08/03(土) 09:41:52.10 ID:ja6T8EuE

（>>68より）
Pruss氏の指摘（2013）とほぼ同じことを指摘している（下記）

（なぜ、mathoverflow>>465 の手法が成立たないのか？ ”CONGLOMERABILITY”が成立ってないというのが、数学DR Alexander Pruss氏の指摘（2013）で、それを2018年の著書で詳しく解説している）
スレ65 https://rio2016.5ch.net/test/read.cgi/math/1557142618/750-754
https://books.google.co.jp/books?id=RXBoDwAAQBAJ&pg=PA77&lpg=PA77&dq=%22conglomerability%22+assumption+math&source=bl&ots=8Ol1uFrjJQ&sig=ACfU3U1bAurNGJm5872wDblskzsSgsU0iA&hl=ja&sa=X&ved=2ahUKEwioiPyV_IPiAhXHxrwKHUeaArUQ6AEwCXoECEoQAQ#v=onepage&q=%22conglomerability%22%20assumption%20math&f=false
Infinity, Causation, and Paradox 著者: Alexander R. Pruss Oxford University Press, 2018
P75
(抜粋)
2.5.3 COUNTABLE ADDITITVITY AND CONGLOMERABILITY
(引用終り)

（mathoverflowの”conglomerability”関連箇所）
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)
（Alexander Pruss氏）
＜12＞
(抜粋)
The probabilistic reasoning depends on a conglomerability assumption・・
But we have no reason to think the event of guessing correctly is measurable with respect to the probability measure induced by the random choice of sequence and index i, and we have no reason to think that the conglomerability assumption is appropriate.
A quick way to see that the conglomerability assumption is going to be dubious is to consider the analogy of the Brown-Freiling argument against the Continuum Hypothesis (see here for a discussion).
http://www.mdpi.com/2073-8994/3/3/636

http://rio2016.5ch.net/test/read.cgi/math/1564659345/78

686: 現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE  [] 2019/08/13(火) 18:50:52.94 ID:6aC3VJqe

>>671
＞PrussはRiddleを否定していないｗ

ほいよ
（>>78より）
英語不得意科目だねｗｗ（＾＾；
”The probabilistic reasoning depends on a conglomerability assumption・・
But we have no reason to think the event of guessing correctly is measurable with respect to the probability measure induced by the random choice of sequence and index i, and we have no reason to think that the conglomerability assumption is appropriate.”
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)
（Alexander Pruss氏）
＜12＞
(抜粋)
The probabilistic reasoning depends on a conglomerability assumption・・
But we have no reason to think the event of guessing correctly is measurable with respect to the probability measure induced by the random choice of sequence and index i, and we have no reason to think that the conglomerability assumption is appropriate.
A quick way to see that the conglomerability assumption is going to be dubious is to consider the analogy of the Brown-Freiling argument against the Continuum Hypothesis (see here for a discussion).
http://www.mdpi.com/2073-8994/3/3/636

http://rio2016.5ch.net/test/read.cgi/math/1564659345/686

711: 現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE  [] 2019/08/13(火) 23:53:47.99 ID:KPMDnjHM

>>705
>The riddle には確率のかの字も出てきませんよ？（＾＾

（>>692）【必死のパッチ】やなｗｗ（＾＾；
まず、表題が”Probabilities in a riddle involving axiom of choice”ｗ
次に、下記The Modificationが、時枝記事に相当する部分で
”I think it is ok, because the only probability measure we need is uniform probability on {0,1,…,N?1}, ”が、「おサルの確率論」だけどｗ
数学DR Pruss氏に木っ端みじんに論破されましたとさｗｗ（＾＾；
（>>78ご参照）

https://mathoverflow.net/questions/151286/probabilities-in-a-riddle-involving-axiom-of-choice
Probabilities in a riddle involving axiom of choice
asked Dec 9 '13 at 16:16
Denis
(抜粋)
The Modification: I would find the riddle even more puzzling if instead of 100 mathematicians, there was just one, who has to open the boxes he wants and then guess the content of a closed box. He can choose randomly a number i between 0 and 99, and play the role of mathematician number i.
In fact, he can first choose any bound N instead of 100, and then play the game, with only probability 1/N to be wrong.
In this context, does it make sense to say "guess the content of a box with arbitrarily high probability"?
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.

http://rio2016.5ch.net/test/read.cgi/math/1564659345/711

773: 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) 13:39:58.89 ID:rg2Nhb+h

（>>692より）【必死のパッチ】やなｗｗ（＾＾；

＜Prussのmathoverflowでの発言について＞
（サルの>>718>>722より）
(引用開始)
スレ主がコピペしたPrussの発言に対しPrussとDenisが議論をしている。その中で、Prussは「勝率99/100以上」を認めてるよ（＾＾；
For each fixed opponent strategy, if i is chosen uniformly independently of that strategy (where the "independently" here isn't in the probabilistic sense), we win with probability at least (n－1)/n. That's right.

「固定された各対戦相手の戦略について、iがその戦略とは無関係に均一に選択された場合（ここで「独立して」は確率的な意味ではありません）、少なくとも（n-1）/ nの確率で勝ちます。そのとおり。」
Pruss　敗北！！！！！！！
(引用終り)

あほサルの勝手読み
Pruss氏の発言のその後があるだろう？
”That's right. But now the question is whether we can translate this to a statement without the conditional "For each fixed opponent strategy".
　 ? Alexander Pruss Dec 19 '13 at 15:05 ”と
要するに、力点は、But以下の文にあるってことと
前文の”if i is chosen uniformly independently of that strategy”の部分が未証明だってことよｗ

＞Denisがそれを論破したところで議論が終わっている（＾＾；

サル知恵か（＾＾
Denisが、確率論が分かってないし、本を書くネタをmathoverflowで書くには余白と時間が限られているってことよ
だからPruss氏は、本を一冊書いて2018年に出版した（下記）

（参考>>78より）
https://mathoverflow.net/questions/151286/probabilities-in-a-riddle-involving-axiom-of-choice
Probabilities in a riddle involving axiom of choice Dec 9 '13

つづく

http://rio2016.5ch.net/test/read.cgi/math/1564659345/773

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