現代数学の系譜工学物理雑談古典ガロア理論も読む75

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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

159(2): 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(4/35) AAS
>>158
つづき

Denis
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
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)
以上

168(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)09:31 ID:sbItYGIt(6/35) AAS
>>158 補足
>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."

前振りで、下記をば
これ、i.i.d. 独立同分布に尽きる気がします

（参考）
スレ74 2chｽﾚ:math
外部ﾘﾝｸ[pdf]:www.f.waseda.jp
「確率過程とその応用」逆瀬川浩孝
P3
1.2.1 時間パラメータ
時間パラメータ集合が離散の場合は T = {0, 1, 2, ...} とすることが多く、
その場合、確率過程は {Xn, n = 0, 1, 2, ...} と書かれることが多い。
一般には T = {t0, t1, t2, ...}、あるいは、すべての整数、とする場合もある。

「時間パラメータ」は「時刻」に限定されるものではない。
例えば、生産ラインから出荷される製品の故障をチェックするためのモデル化を考えるならば、
Xn は n 番目に出荷する製品の状態を表す確率変数（不良品ならば 1、正常ならば 0）と考えると都合がよいので、
その場合は nは製品番号を表すことになる。
光ファイバーの傷をチェックするための確率過程モデルとして、
X(t) をファイバーの終端から長さ t の点での状態を表す確率変数と考えると、t は長さを表す。

P11
2.3 確率変数 (random variable)
ランダム事象が起きた時に定まる数量を、そのランダム事象（根元事象、標本点）に対応させる関数を確率変数という。
P25
n 個の確率変数 X1, X2, ..., Xn が互いに独立で同分布に従う場合は i.i.d. (independent, identically distributed) と呼ばれ、良く使われる。

つづく

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（＾＾；）

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