現代数学の系譜工学物理雑談古典ガロア理論も読む76

[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)
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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

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