[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75 (1002ﾚｽ)
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157(5): 現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE 2019/08/17(土)07:32 ID:sbItYGIt(2/35) AAS
>>154-155
おサルが二匹か
踊らないおサルは、使えねーな （by サル回しより）
スレ74 2chｽﾚ:math
関連
おサル
I think it is ok, because the only probability measure we need is uniform probability on {0,1,…,N-1}
おれ（＾＾
but other people argue it's not ok, because we would need to define a measure on sequences,
（参考）
外部ﾘﾝｸ:mathoverflow.net
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.
つづく
158(3): 現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE 2019/08/17(土)07:32 ID:sbItYGIt(3/35) AAS
>>157
つづき
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).
外部ﾘﾝｸ:www.mdpi.com
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.
Here's an amusing thing that may help see how measurability enters into these things. Consider a single sequence of infinitely many independent fair coin flips. Our state space is Ω={0,1}^N, corresponding to an infinite sequence (Xi)^∞ i=0 of i.i.d. r.v.s with P(Xi=1)=P(Xi=0)=1/2.
Can you guess the first coin flip on the basis of all the others? You might think: "Of course not! No matter what function from the values of flips X1,X2,... to {0,1} is chosen, the probability that the value of the function equals X0 is going to be 1/2."
つづく
172: 2019/08/17(土)09:34 ID:h0TAsPzg(4/38) AAS
>>157
>need to define a measure on sequences
何べんでも繰り返してやるけど
朝鮮人スレ主のここが間違い
数列の測度は必要ない
貴様は負け犬
さっさと北朝鮮に帰れｗ
192(1): 2019/08/17(土)10:52 ID:+5QXhyrz(2/38) AAS
>>157
おまえ（＾＾
but other people argue it's not ok, because we would need to define a measure on sequences,
Pruss（＾＾
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.
Prussは間違いを認めた。未だに認められないのはおまえ一人（＾＾；
215(4): 現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE 2019/08/17(土)13:11 ID:sbItYGIt(16/35) AAS
>>202 補足
＜追加説明＞
１）
時枝記事（>>179より）や、
mathoverflowの Denis氏は（>>157-159より）
ある1つの箱が、高確率 (n-1)/n で的中できるという
２）
しかし、もともと箱は可算無限個あるから、
１個や２個の箱が高確率 (n-1)/n としても
残り、可算無限個の箱は
そもそも、i.i.d. 独立同分布で
サイコロなら1/6
コインなら1/2
で、残り無限個の箱は、既存の確率論・確率過程論の通り
確率論・確率過程論に頼らざるを得ないのだったｗ（＾＾
３）
では、本当にある箱の確率が、
高確率 (n-1)/n になるのだろうか？
４）Pruss氏はいう（>>158）
”Can you guess the first coin flip on the basis of all the others? You might think: "Of course not! No matter what function from the values of flips X1,X2,... to {0,1} is chosen, the probability that the value of the function equals X0 is going to be 1/2."
この議論は、任意の有限i番目に拡張できる
即ち
”Can you guess the n-th coin flip on the basis of all the others? You might think: "Of course not! No matter what function from the values of flips X0,X1,X2,...,Xi-1,Xi+1,... to {0,1} is chosen, the probability that the value of the function equals Xi is going to be 1/2."
となる
以上
ちょっと、確率論と確率過程論の知識があれば
この通り、
時枝不成立がお分かりだろう
（サルには、確率は難しいから理解は無理だろうね。知能が幼稚園児並みだからねｗ（＾＾； ）
558(7): 現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE 2019/08/21(水)07:03 ID:6H2tIaYx(2/10) AAS
>>557
つづき
それで直観について補足すると下記
（PDFでP642）
Yet two puzzles remain. First: why do we not have Pr(p < q) = 1/2, given the symmetry of the
set-up? Despite the mathematical argument above, the intuition that p < q and q < p are equiprobable
remains strong. What should we say about this intuition? Freiling appears to think we should give such
intuitions priority. He writes (in a slightly different context) that his argument [1] depends upon
a principle
...not meant to be a mathematical statement of the Lebesgue measurability of a certain type of
set. Rather, it is an expression of an obvious, almost physical intuition concerning the
inherently nonmathematical notions of prediction, accuracy, and time independence.
Yet an appeal to symmetry, where we cannot produce a coherent mathematical model, is unreliable,
as we know from the many paradoxes associated with the Principle of Insufficient Reason.
ってところ
これが、きっと、Pruss氏が、この論文を引用した理由で
（>>157より）
Denis
I think it is ok, because the only probability measure we need is uniform probability on {0,1,…,N-1}
が、Pruss氏は「それDenisの直観でしかない」ということよ（＾＾；
つづく
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