https://en.wikipedia.org/wiki/White_noise White noise Depending on the context, one may also require that the samples be independent and have identical probability distribution (in other words independent and identically distributed random variables are the simplest representation of white noise).[3] (引用終り) 以上 0007現代数学の系譜 雑談 ◆yH25M02vWFhP 2021/04/05(月) 07:36:02.85ID:DsMvJGEN>>6 補足
ということで時枝戦略を否定したいなら時枝戦略について語って下さい。 時枝戦略とまったく無関係な戦略をいくら語ったところでナンセンス以外のなにものでもないです。 0019１３２人目の素数さん2021/04/13(火) 08:19:46.40ID:0m7k3PSf A 時枝戦略 B 箱の中身を確率変数とする戦略 C その他の戦略
Answers 12 answered Dec 11 '13 at 21:07 Alexander Pruss
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 Symmetry and the Brown-Freiling Refutation of the Continuum Hypothesis by Paul Bartha Symmetry 2011, 3(3), 636-652; ).
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. Start with P being the completion of the natural product measure on Ω.
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."
That's a fine argument assuming the function is measurable. But what if it's not? Here is a strategy: Check if X1,X2,... fit with the relevant representative. If so, then guess according to the representative. If not, then guess π. (Yes, I realize that π not∈{0,1}.) Intuitively this seems a really dumb strategy. After all, we're surely unlikely to luck out and get X1,X2,... to fit with the representative, and even if they do, the chance that X0 will match it, given the rest of the sequence, seems to be only 1/2.
What we have then is this: 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. 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
今度はあなたの番である.片端から箱を開けてゆき中の実数を覗いてよいが，一つの箱は開けずに閉じたまま残さねばならぬとしよう. どの箱を閉じたまま残すかはあなたが決めうる. 勝負のルールはこうだ. もし閉じた箱の中の実数をピタリと言い当てたら，あなたの勝ち. さもなくば負け. 勝つ戦略はあるでしょうか?」 ↑ 出題者が先に出題実数列を固定し、その後に回答者の数当てが開始される、という順序がしっかり明記されてます。 よって、Prussさんは箱入り無数目成立を完全に認めたことになります。 0053１３２人目の素数さん2021/04/17(土) 12:24:59.40ID:et8jrAa6 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さんは間違いを認めることができました。数学Drの彼にとってはさぞ不本意だったことでしょう。 大学1年4月の課程さえちんぷんかんぷんの誰かさんは間違いを認められないようですけど。 0054１３２人目の素数さん2021/04/17(土) 12:35:54.91ID:et8jrAa6https://mathoverflow.net/questions/151286/probabilities-in-a-riddle-involving-axiom-of-choice mathoverflow の The Riddle でも 「・・・Then all boxes are closed, and the next mathematician can play.・・・」 と、先に出題実数列を固定し、その後回答者の数当てが開始されるという順序が明記されてるんですけどねｗ つまりPrussさんの But now the question is whether we can translate this to a statement without the conditional "For each fixed opponent strategy". は後付けの言い訳でしかないんですけどねｗ 0055１３２人目の素数さん2021/04/17(土) 12:42:21.71ID:et8jrAa6>>51 パズルだから数学に非ずとでも言いたいのでしょうかね？ 数学パズルという数学の分野があることも知らない白痴ですか？
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.
（参考） http://www.sipta.org/isipta13/proceedings/papers/s029.pdf 8th International Symposium on Imprecise Probability: Theories and Applications, Compiegne, France, 2013 Two theories of conditional probability and non-conglomerability Teddy SeidenfeldMark J. SchervishJoseph B. KadaneCarnegie Mellon University
Abstract Conglomerability of conditional probabilities issuggested by some (e.g., Walley, 1991) as necessary forrational degrees of belief. Here we give sufficientconditions for non-conglomerability of conditionalprobabilities in the de Finetti/Dubins sense. Thesesufficient conditions cover familiar cases where P(?) is acontinuous, countably additive probability. In thisregard, we contrast the de Finetti/Dubins sense ofconditional probability with the more familiar account ofregular conditional distributions, in the fashion ofKolmogorov.
1 Introduction Consider a finitely, but not necessarily countablyadditive probability P(?) defined on a sigma-field of setsB, each set a subset of the sure-event Ω. In other terms,<Ω, B, P> is a (finitely additive) measure space.We begin by reviewing the theory of conditionalprobability that we associate with de Finetti (1974) andDubins (1975).
This account of conditional probability is not the usualtheory from contemporary Mathematical Probability,which we associate with Kolmogorov (1956). That theory, instead, defines conditional probability throughregular conditional distributions, as follows. 0065現代数学の系譜 雑談 ◆yH25M02vWFhP 2021/04/18(日) 08:06:02.76ID:0Dh4aVIp>>64 >正直、conglomerabilityは難しすぎ
P1より Player 1 chooses a countably infinite sequence x = (xn) n∈N of real numbers, and puts them in boxes labeled 1, 2, ...
P2より Remark. When the number of boxes is finite Player 1 can guarantee a win with probability 1 in game1, and with probability 9/10 in game2, by choosing the xi independently and uniformly on [0, 1] and {0, 1,..., 9}, respectively. (引用終り)
補足 つまり infinite sequence x = (xn) n∈N が、確率変数の加算無限族です Player 1が、出題者です。 ” puts them in boxes labeled 1, 2, ...”と記されている
the xi independently and uniformly は、IIDと同じ意味 ”When the number of boxes is finite”つまり、有限族の場合
主題者 Player 1 が、”uniformly on [0, 1]”つまり、区間[0, 1]からランダムに実数を入れると、 ”a win with probability 1 in game1”
”{0, 1,..., 9}” つまり、一桁の0〜9の整数を入れると、 ”with probability 9/10 in game2”だという
これが、落語（今回はパズルですが）の ”落ち”（オチ）です （普通の確率論＆確率過程論どおりｗ！）
”When the number of boxes is finite”つまり、有限族の場合と言っても、上限はないのです その極限では、加算無限（n→∞）ですからね 0076１３２人目の素数さん2021/04/18(日) 22:38:47.53ID:p6YzeXU0>>75 >つまり >infinite sequence x = (xn) n∈N が、確率変数の加算無限族です 「infinite sequence x = (xn) n∈N of real numbers」 とあるので実数列ですねー real numbers とは実数のことですよ？辞書引きましょうね
>”When the number of boxes is finite”つまり、有限族の場合と言っても、上限はないのです >その極限では、加算無限（n→∞）ですからね え？？？ 自然数に上限は無いですがどの自然数も有限値ですよ？∞は自然数ではありませんよ？ あなた有限と無限の区別もつかないんですか？こりゃ酷い。
そもそも、無限列で数当てできないことを直接的に示せるなら最初からそうすればいい。 それができないから有限列に逃げようとする。この時点で既に頭がオワッテイル。 バカに数学はできない。 0080１３２人目の素数さん2021/04/19(月) 02:41:58.43ID:LErD3ySh 瀬田くんの理屈によると満室の無限ホテルに新たな客は泊まれないことになりますねー ヒルベルト先生も思わず苦笑いするでしょうねー 0081１３２人目の素数さん2021/04/19(月) 10:58:53.76ID:LErD3ySh 無限は有限と同じと妄想する瀬田くんの >氏の結論部分は、はっきりと質問のstrategyを否決しています！（＾＾ も妄想なんでしょうねー Prussが否決してるという部分を一向に示さないしねー 0082１３２人目の素数さん2021/04/19(月) 11:06:32.01ID:LErD3ySh こちらは妄想症の瀬田くんと違いはっきりと示しますよ？ PrussはThe Modification（＝箱入り無数目）成立をはっきり認めてます。 ↓ 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. 0083１３２人目の素数さん2021/04/19(月) 11:12:59.35ID:LErD3ySh この For each fixed opponent strategy, if i is chosen uniformly independently of that strategy の部分は、The Modification（＝箱入り無数目）の条件と完全に符合します。 すなわちPrussはThe Modification（＝箱入り無数目）の成立を完全に認めました。 言い訳は一切通りません。 0084１３２人目の素数さん2021/04/19(月) 11:23:51.79ID:LErD3ySh>>45 >時枝先生は >Ｐ（ｎ1＞ｎ2）＝1/2 などと論じていない。おまえの妄想。 >何と論じているかはさんざんに教えただろ。どこに脳みそ落としたんだ？さっさと探してこい。
Consequences The statement given in the introduction follows immediately by taking M to be an infinite model of the theory. The proof of the upward part of the theorem also shows that a theory with arbitrarily large finite models must have an infinite model; sometimes this is considered to be part of the theorem.
Proof sketch Upward part First, one extends the signature by adding a new constant symbol for every element of M. The complete theory of M for the extended signature σ' is called the elementary diagram of M. In the next step one adds κ many new constant symbols to the signature and adds to the elementary diagram of M the sentences c ≠ c' for any two distinct new constant symbols c and c'. Using the compactness theorem, the resulting theory is easily seen to be consistent. Since its models must have cardinality at least κ, the downward part of this theorem guarantees the existence of a model N which has cardinality exactly κ. It contains an isomorphic copy of M as an elementary substructure.[3][4]:100?102 (引用終り) 以上 0088現代数学の系譜 雑談 ◆yH25M02vWFhP 2021/04/19(月) 23:54:30.84ID:fsi/ILI7>>86
