[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76 (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76 http://rio2016.5ch.net/test/read.cgi/math/1566715025/
上
下
前次
1-
新
通常表示
512ﾊﾞｲﾄ分割
ﾚｽ栞
抽出解除
ﾚｽ栞
このｽﾚｯﾄﾞは過去ﾛｸﾞ倉庫に格納されています｡
次ｽﾚ検索
歴削→次ｽﾚ
栞削→次ｽﾚ
過去ﾛｸﾞﾒﾆｭｰ
ﾘﾛｰﾄﾞ規制
です｡10分ほどで解除するので､
他のﾌﾞﾗｳｻﾞ
へ避難してください｡
287: 現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE [sage] 2019/08/29(木) 06:43:21.30 ID:aQWHRZvT >>276 >時枝解法で必要なのはP(C)だｗ さんざんそう言ってきただろｗ >P(A)が必要と言ってるのは確率論の専門家だｗ ええ、下記Denis "I think it is ok, because the only probability measure we need is uniform probability on {0,1,…,N-1}" で、厳密な数学の証明がないと、Pruss氏、確率論の専門家さんと私ね（＾＾ (>>241) そこを（数学的に厳密でないと）批判しているのが、Alexander Pruss氏だよ 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 answered The probabilistic reasoning depends on a conglomerability assumption, namely that given a fixed sequence u￣ , the probability of guessing correctly is (n?1)/n, then for a randomly selected sequence, the probability of guessing correctly is (n?1)/n. 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/1566715025/287
290: 現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE [sage] 2019/08/29(木) 07:12:20.19 ID:aQWHRZvT >>287 蛇足 余談だが 私は、”確率論の専門家さん”と呼んでいるだけどね おサルにかかると、確率論の専門家になるみたい まあ、そう言い切ってもらう方が おサルのバカさが、ある意味でキワダツので、面白いけどねｗ（＾＾ http://rio2016.5ch.net/test/read.cgi/math/1566715025/290
317: １３２人目の素数さん [] 2019/08/29(木) 22:07:05.20 ID:F6jSJdzt >>287 >>時枝解法で必要なのはP(C)だｗ さんざんそう言ってきただろｗ >>P(A)が必要と言ってるのは確率論の専門家だｗ >ええ、下記Denis "I think it is ok, because the only probability measure we need is uniform probability on {0,1,…,N-1}" >で、厳密な数学の証明がないと、Pruss氏、確率論の専門家さんと私ね（＾＾ なにが「ええ」だｗ おまえ自分で言っててわかってないだろｗ P(A)が必要と言ってるのは確率論の専門家であり、それは誤解なんだから、厳密な証明が無いの指摘はまったくの的外れなんだよおバカさん だいいちおまえ確率論の専門家の発言内容がわかってないじゃん 訳も分からずに尻馬に乗っかってるだけ だからいつまで経っても理解できない 分からないんだったら分かりませんと云えよ 分からないのに分かってるふりするなバカ キチガイザル http://rio2016.5ch.net/test/read.cgi/math/1566715025/317
330: 現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE [sage] 2019/08/29(木) 23:18:31.15 ID:aQWHRZvT >>313>>315>>320 ID:BgUyythSさんの考えは、 下記Denis "I think it is ok, because the only probability measure we need is uniform probability on {0,1,…,N-1}" と同じでしょ？（＾＾（>>287ご参照） で、厳密な数学の証明がないというのが、Pruss氏、確率論の専門家さんと、私ね（＾＾ (>>241) そこを（数学的に厳密でないと）批判しているのが、Alexander Pruss氏だよ 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 answered The probabilistic reasoning depends on a conglomerability assumption, namely that given a fixed sequence u￣ , the probability of guessing correctly is (n?1)/n, then for a randomly selected sequence, the probability of guessing correctly is (n?1)/n. 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/1566715025/330
352: 現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE [sage] 2019/08/30(金) 07:26:14.21 ID:exryDrPV >>348-349 >君が「６コ中の最大値である確率は、1/6」と思ったんだよね 思ってないよ だって、「サイコロに勝手な自然数６コを記載する」（>>346） だから、n1,n2,n3,n4,n5,n6∈N（自然数）でしょ？ 確率空間書いて、積分してみなよ それって、N（自然数）全体で、各n1,n2,n3,n4,n5,n6達に、測度1を与える話でしょ？ 積分（実は和）は∞に発散するだろ？ｗ（＾＾ で、あなたの考えは 下記Denis "I think it is ok, because the only probability measure we need is uniform probability on {0,1,…,N-1}" と同じでしょ？（＾＾（>>287ご参照） で、厳密な数学の証明がないというのが、Pruss氏、確率論の専門家さんと、私ね（＾＾ (>>241) そこを（数学的に厳密でないと）批判しているのが、Alexander Pruss氏だよ 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 answered The probabilistic reasoning depends on a conglomerability assumption, namely that given a fixed sequence u￣ , the probability of guessing correctly is (n?1)/n, then for a randomly selected sequence, the probability of guessing correctly is (n?1)/n. 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/1566715025/352
443: 現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE [] 2019/08/31(土) 20:17:00.54 ID:PbGhNKv4 >>441 >代表元だって同じでしょ 「代表元による代表番号の確率計算」は、数学的な厳密な扱いができてないのですよ！！（＾＾ そのあなたの考えは（>>352より） 下記Denis "I think it is ok, because the only probability measure we need is uniform probability on {0,1,…,N-1}" と同じでしょ？（＾＾（>>287ご参照） で、厳密な数学の証明がないというのが、Pruss氏、確率論の専門家さんと、私ね（＾＾ （そもそも、Denis氏発言に対する批判” 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.”もあるよ） (>>241) そこを（数学的に厳密でないと）批判しているのが、Alexander Pruss氏だよ 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 answered The probabilistic reasoning depends on a conglomerability assumption, namely that given a fixed sequence u￣ , the probability of guessing correctly is (n?1)/n, then for a randomly selected sequence, the probability of guessing correctly is (n?1)/n. 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/1566715025/443
466: 現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE [] 2019/08/31(土) 21:45:16.39 ID:PbGhNKv4 >>462 (引用開始) >しかし、現代数学内のカンニング手段は、まだ、見つかっていませんね 時枝解法ｗ すなわち同値類の代表元をカンニングする解法 同値類が分かってないサルに理解できないだけの話ｗ (引用終り) つー、>>443-444 （＾＾； あなたの主張は下記 同値類→代表→代表と問題の数列を比較した決定番号→複数列の決定番号の大小から、カンニング正解率は100列で確率99/100だ！ その最後の確率99/100 下記Denis "I think it is ok, because the only probability measure we need is uniform probability on {0,1,…,N-1}" と同じでしょ？（＾＾（>>287ご参照） で、厳密な数学の証明がないというのが、Pruss氏、確率論の専門家さんと、私ね（＾＾ （そもそも、Denis氏発言に対する批判” 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.”もあるよ） （詳しくは、>>443-444 ） (>>241) そこを（数学的に厳密でないと）批判しているのが、Alexander Pruss氏だよ 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. http://rio2016.5ch.net/test/read.cgi/math/1566715025/466
ﾒﾓ帳
(0/65535文字)
上
下
前次
1-
新
書
関
写
板
覧
索
設
栞
歴
ｽﾚ情報
赤ﾚｽ抽出
画像ﾚｽ抽出
歴の未読ｽﾚ
AAｻﾑﾈｲﾙ
Google検索
Wikipedia
ぬこの手
ぬこTOP
0.052s