The reason we do not have countable additivity differs depending on whether the probability of. particular ticket winning is zero or infinitesimal. If the probability is exactly zero, then we lack countable additivity because 1 = P(E1 ∨ E2 ∨・・・) if En is the probability of ticket n being picked (it’s certain that some ticket or other is picked) whereas P(E1) + P(E2) +・・・ = 0 + 0 + ・・・ = 0.
If on the other hand, P( En ) = α for some (positive) infinitesimalsα, then things are more complicated. The standard systems for construction of infinitesimal do not in general define a countable in finite sum of infinitesimals, at least in our case where the summands are the same. Thus, the required equation P(E1 ∨ E2 ∨・・・ ) =P(E1 ) + P(E2)+・・・ does not hold, since although the left .hand side is defined, the right-hand side is not. In our infinite fair lottery case, we can intuitively see why we shouldn't be able to have a meaningful sum. For consider our infinite sum: α +α+α+α+ ・・・ = (α +α) +(α +α) +・・・ =2α+2α+ ・・・ =2(α+α+・・・ ). If the value of this sum is x, then x =2x, But if x is not zero, then we can divide both side, by x to yield 1 = 2, and so x must be zero. However, x cannot be zero since it must be at least as big as α, and hence a contradiction follows, from the assumption that the sum has a value. The lack of countable additivity in the case of an infinite lottery is responsible for a phenomenon known as non-conglomerability. A probability function P is conglomerable with respect to a partition E1,E2,・・・(a partition is a collection of pairwise disjoint event such that their disjunction is the whole space of possibilities ) provided there is no event A and real number such that for all I we have P(A｜Ei ) <= a and yet P(A) > a. Conglomerability is a very plausible properly.
Suppose you are certain that some event in the partition will occur. If you also know for sure that whatever event in that partition you learn occurs, your probability for A will be at most a, then how could your rational probability for A be more than a ? Conglomerability is closely related to van Fraassen’s very plausible Reflection Principle which says that if one is rationally certain that one will have a certain rational credence, one should already have that credence now (van Fraassen 1984). But typically, where there is no countable additivity, there is lack of conglomerability (Shervish, Seidenfeld, and Kadane 1984). In the case of the countably infinite fair lottery, we can see the lack of conglomerability directly. Let E be the event that the ticket picked will be even and O the event that it will be odd. By finite additivity, P(E) + P(O) = 1, so at least one of the two events must have probability at least 1/2, (Intuitively, they both have probability exactly 1/2, but I don't need that for the argument.) Suppose that P(E) >= 1/2 (the argument in the case where P(O)>= 1/2 will be very similar).
Then consider the partition provided by the following sets: E1= {2, 1, 3) E2= {4, 5, 7 } E3= {6, 9, 11 } E4. = {8, 13, 15 } Observe now that each En contains exactly one even number and two odd ones. Thus, by the fairness of the lottery, P( E| En) = 1/3. Thus, P(E|En) < 1/2 for all n, but by assumption P(E) >=1/2, and conglomerability is violated. Where conglomerability is absent, one gets strange results such as reasoning to a foregone conclusion and paying not to receive information (Kadane, Schervish,and Seidenfeld 1996), just as we saw in Section 2.5. And the symmetry puzzle in Section 2.4 is also a non-conglomerability puzzle. Taking the original two-ticket version, the probability that my ticket number is bigger than yours is initially within an infinitesimal of 1/2. But the conditional probability that my ticket number is bigger than yours given what my ticket number is - whatever that may be - is at most an infinitesimal, and so conglomerability is violated. One possible response to my preceding paradoxes is that non-conglomerability needs to be accepted when dealing with countably infinite fair lotteries, and non-conglomerability just happens to have a number of paradoxical consequences. But the cost of accepting non-conglomerability is high, namely many paradoxical consequences. It is better to take non-conglomerability in these lotteries to be both a paradox in its own right and the mathematical root of a number of other paradoxes. (引用終り) 0754現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE 2019/05/19(日) 21:38:36.07ID:hAkw8ZY7>>750 ＜conglomerabilityとは？＞
１）>>750 P(E1 ∨ E2 ∨・・・ ) =P(E1 ) + P(E2)+・・・ Classical mathematical probability theory assumes all probability functions to be countably additive. But in the countably infinite fair lottery, we do not have countable additivity. ２）>>751 The lack of countable additivity in the case of an infinite lottery is responsible for a phenomenon known as non-conglomerability. A probability function P is conglomerable with respect to a partition E1,E2,・・・(a partition is a collection of pairwise disjoint event such that their disjunction is the whole space of possibilities ) provided there is no event A and real number such that for all I we have P(A｜Ei ) <= a and yet P(A) > a. ３）Conglomerability is closely related to van Fraassen’s very plausible Reflection Principle which says that if one is rationally certain that one will have a certain rational credence, one should already have that credence now (van Fraassen 1984). But typically, where there is no countable additivity, there is lack of conglomerability (Shervish, Seidenfeld, and Kadane 1984). In the case of the countably infinite fair lottery, we can see the lack of conglomerability directly. ４）Thus, P(E|En) < 1/2 for all n, but by assumption P(E) >=1/2, and conglomerability is violated. Where conglomerability is absent, one gets strange results such as reasoning to a foregone conclusion and paying not to receive information (Kadane, Schervish,and Seidenfeld 1996), just as we saw in Section 2.5. And the symmetry puzzle in Section 2.4 is also a non-conglomerability puzzle. ５）ということで、以上の要点抜粋をまとめると、無限個の宝くじのように、可算無限の微小な和を加えると conglomerabilityが保証されないので、paradoxになると ６）こういう説明を、数学Dr Alexander Pruss氏は、例のmathoverflowでもしているんだな
哲学の先生であり、かつ数学Drで mathematicianです（下記） そして、証明書くには、余白が狭すぎると思ったんだ、2013年に https://en.wikipedia.org/wiki/Alexander_Pruss Alexander Pruss Alexander Robert Pruss (born January 5, 1973) is a Canadian mathematician, philosopher, Professor of Philosophy (引用終り)
>そうでなければ反論されたままでスレが終わることはないはずだ。
証明書くには、余白が狭すぎると思って、Pruss氏は本を書いた。出版されたのは2018年だ それが、>>750の”Infinity, Causation, and Paradox 著者: Alexander R. Pruss Oxford University Press, 2018”
A probability function P is conglomerable with respect to a partition E1,E2,・・・(a partition is a collection of pairwise disjoint event such that their disjunction is the whole space of possibilities ) provided there is no event A and real number such that for all I we have P(A｜Ei ) <= a and yet P(A) > a. 0760１３２人目の素数さん2019/05/20(月) 07:35:12.21ID:1R9eH7ri>>757 >数学Dr Alexander Pruss氏は、mathoverflowで、例の riddle も >conglomerabilityが保証されていないという指摘をしている
違うよ 同値類の index M が、無限宝くじでのｎ→∞と類似だと主張している 明らかに、index Mが、自然数全体を渡るからね 細かい証明は書いていないが（余白が狭い（＾＾ ）
しかし ”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 として、証明つき文献を挙げている それも含めて読め
これは擬似パラドックスと呼ばれ、前述した「真の」パラドックスとは別物である。 (引用終わり) 0778１３２人目の素数さん2019/05/22(水) 22:18:32.77ID:jEl/5QQ+ Paradox ということは一見当てられないように見えて実際は当てられるってことじゃんｗ つまり、時枝は正しいってことじゃんｗ 0779現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE 2019/05/22(水) 23:35:32.87ID:fJqlfrJg ご託は、本一冊読んでから言えｗ（＾＾ （>>772） 数学DRにして、Canadian mathematician, philosopher, Professor of Philosophy （>>45ご参照） の Alexander Pruss先生
2018に Oxford University Pressから、 （下記）「Infinity, Causation, and Paradox」を出版している 証明は、そこにあるよ。一冊買って読め！ｗ（＾＾；
https://books.google.co.jp/books?id=RXBoDwAAQBAJ&;pg=PA77&lpg=PA77&dq=%22conglomerability%22+assumption+math&source=bl&ots=8Ol1uFrjJQ&sig=ACfU3U1bAurNGJm5872wDblskzsSgsU0iA&hl=ja&sa=X&ved=2ahUKEwioiPyV_IPiAhXHxrwKHUeaArUQ6AEwCXoECEoQAQ#v=onepage&q=%22conglomerability%22%20assumption%20math&f=false Infinity, Causation, and Paradox 著者: Alexander R. Pruss Oxford University Press, 2018 0780現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE 2019/05/22(水) 23:36:23.39ID:fJqlfrJg 大学の学生なら、大学の図書にリクエストして買わせろ！ｗ（＾＾ 0781現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE 2019/05/22(水) 23:38:21.39ID:fJqlfrJg (>>753より) One possible response to my preceding paradoxes is that non-conglomerability needs to be accepted when dealing with countably infinite fair lotteries, and non-conglomerability just happens to have a number of paradoxical consequences. But the cost of accepting non-conglomerability is high, namely many paradoxical consequences. It is better to take non-conglomerability in these lotteries to be both a paradox in its own right and the mathematical root of a number of other paradoxes. (引用終り) 0782１３２人目の素数さん2019/05/23(木) 07:25:12.23ID:iS575wL5>>779
数学DRにして、Canadian mathematician, philosopher, Professor of Philosophy の Alexander Pruss先生（>>45ご参照） 著書2018に Oxford University Press 「Infinity, Causation, and Paradox」（>>779） P75”2.5.3 Countable Additivity and Conglomerability” & 77 ・One possible response to my preceding paradoxes is that non-conglomerability needs to be accepted when dealing with countably infinite fair lotteries, and non-conglomerability just happens to have a number of paradoxical consequences. ・But the cost of accepting non-conglomerability is high, namely many paradoxical consequences. ・It is better to take non-conglomerability in these lotteries to be both a paradox in its own right and the mathematical root of a number of other paradoxes. (引用終り)
＜意訳説明＞ ・先に述べたparadoxたちへの、一つの可能な答えは、non-conglomerabilityを受け入れることである、可算無限の宝くじの扱いで。そうすると、non-conglomerabilityは、また多くのパラドックスを導く（a number of paradoxical consequences） ・しかし、non-conglomerabilityを受け入れるコストは高い、多くのパラドックスを導く（many paradoxical consequences）から ・これらの（無限）宝くじの扱いでnon-conglomerabilityを採用することは、それ自身一つのparadoxであり、他の数学的な数多くのparadoxたちの根源になる （”It is better”は、反語だろう） ＜意訳説明終り＞
なお、数学DR Pruss先生がここでいう”paradox”は、>>776の擬似パラドックスではなく、数学的矛盾を含んだパラドックスであることは、明白 ∵その前段で（>>751より） In our infinite fair lottery case, we can intuitively see why we shouldn't be able to have a meaningful sum. For consider our infinite sum: α +α+α+α+ ・・・ = (α +α) +(α +α) +・・・ =2α+2α+ ・・・ =2(α+α+・・・ ). If the value of this sum is x, then x =2x, But if x is not zero, then we can divide both side, by x to yield 1 = 2, and so x must be zero. However, x cannot be zero since it must be at least as big as α, and hence a contradiction follows, from the assumption that the sum has a value. The lack of countable additivity in the case of an infinite lottery is responsible for a phenomenon known as non-conglomerability. (引用終わり) と書かれているのだから
ここでもベイズ推計には触れられています （>>44より） ”Second, the best probability models have properties analogous to non-conglomerability, motivating a proposed extension of that concept (and corresponding limits on Bayesian conditionalization).” ですね 0790現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE 2019/05/23(木) 16:44:16.72ID:QDC/QX0Z>>786 補足 >”non-conglomerability” >標本空間Ωが、無限の場合に、確率ゼロのような事象が出てくる（根元事象がその例） >確率ゼロの事象を、素朴に、時枝やmathoverflowのriddleのように直感で扱うと >ドツボにハマって、”paradox”になる
ここで １） 数学DR Pruss先生は、その回答で ”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)” (注：* u^(→)は、uの上に→をのせた記号（原文ご参照）) と書かれていて、「nontrivial i.i.d. sequence (uk)」とあるので、明らかに数列 ukを、確率変数として扱っている
２） 同じく数学DR Tony Huynh氏の回答で ”Suppose that for each index i we sample a real number Xi from the normal distribution so that the Xis are independent random variables. If there is only person, no matter which boxes they view, they gain no information about the un-opened boxes due to independence. ” だと、”the Xis are independent random variables”（確率変数）だとある
なお、いま気づいたが、数学DR Tony Huynh氏の ”no matter which boxes they view, they gain no information about the un-opened boxes due to independence. ”は 時枝記事の後半の 「私たちの戦略は頓挫してしまう. n番目の箱にXnのランダムな値を入れられて，ある箱の中身を当てようとしたって， その箱のX と他のX1，X2，X3，・・・がまるまる無限族として独立なら， 当てられっこないではないか－－他の箱から情報は一切もらえないのだから.」 と符合していますね
３） さらに、Hart氏は、PDFで（>>26より） ”P2 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.” とあって、boxes を”the xi independently and uniformly”として 確率変数で扱っている。（finiteもinfiniteも、同様に扱えることは、確率過程論を読めば分かる）
４） 時枝が、記事の中で、箱を確率変数の族として扱っていることは、 上記数学DR Tony Huynh氏の関連引用示した通りです
５） なお、mathoverflowでは、 数学DR Pruss先生と、数学DR Tony Huynh氏が、箱を確率変数として扱うことについて 異議を唱える者なし～！！（＾＾
Blavatnik School of Government 2019/05/15 にライブ配信 Public lecture by mathematician Dr Masaki Kashiwara, 2018 Kyoto Prize Laureate for Basic Sciences
Dr Kashiwara established the theory of D-modules, thereby playing a decisive role in the creation and development of algebraic analysis. His numerous achievements have exerted great influence on various fields of mathematics and contributed strongly to their development.
The Kyoto Prize is an international award to honour those who have contributed significantly to the scientific, cultural, and spiritual betterment of humankind. The awards are held annually in November, in Kyoto, Japan. The Laureates travel to Oxford in the following May where the Blavatnik School of Government at the University of Oxford is pleased to host them.
１）前振り （>>717より） >小数展開がすべて循環節となる唯一の代表元がとれる >という主張が間違ってるといいたいんだね？ （>>725より） 「唯一の代表元」がアウトだな （下記より） ”1つの同値類は、それに含まれている元のうちどれをとっても、それを代表元とする同値類はもとと同じ集合になる（代表元の取替えによって不変である）” https://ja.wikipedia.org/wiki/%E5%90%8C%E5%80%A4%E9%96%A2%E4%BF%82 同値関係 一つの同値類 X に対して、[x] = X となる S の元 x を1つ定めることを、X の代表元として x をとるという。 1つの同値類は、それに含まれている元のうちどれをとっても、それを代表元とする同値類はもとと同じ集合になる（代表元の取替えによって不変である） (引用終り)
２） 数学DR Tony Huynh氏も、mathoverflowの回答で ”Suppose that for each index i we sample a real number Xi from the normal distribution so that the Xis are independent random variables. If there is only person, no matter which boxes they view, they gain no information about the un-opened boxes due to independence. ” で、確率変数だと
３） Hart氏は、PDFで ”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.” と確率変数で扱っている。（finiteもinfiniteも、同様に扱えることは、確率過程論を読めば分かる）
４） 時枝氏が、記事の中で、箱を確率変数の族として扱っていることは、 すでに>>799の数学DR Tony Huynh氏の関連引用に示した通り
(mathoverflowより引用開始) 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.
Just let P′ be an extension on which the set of representatives has measure 1 and note that the dumb strategy works on the set of representatives. (引用終り)
The Cahn?Hilliard equation (after John W. Cahn and John E. Hilliard) is an equation of mathematical physics which describes the process of phase separation, by which the two components of a binary fluid spontaneously separate and form domains pure in each component. If {\displaystyle c} c is the concentration of the fluid, with style c=±1 indicating domains, then the equation is written as
The Cahn?Hilliard equations finds applications in diverse fields: in complex fluids and soft matter (interfacial fluid flow, polymer science and in industrial applications). The solution of the Cahn?Hilliard equation for a binary mixture demonstrated to coincide well with the solution of a Stefan problem and the model of Thomas and Windle.[1] Of interest to researchers at present is the coupling of the phase separation of the Cahn?Hilliard equation to the Navier?Stokes equations of fluid flow. 0848１３２人目の素数さん2019/05/25(土) 18:46:23.20ID:7RDo9CNI>>847 貴様に数学は理解できないから 数学板から失せろよ 馬鹿 0849◆QZaw55cn4c 2019/05/25(土) 18:49:50.18ID:xFGGCWUD>>848 そんな口調では相手にされませんよ 徹底的に純論理的に論破しないといけません、これは根気のいることですが、それ以外では排除できませんよ 0850現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE 2019/05/25(土) 18:52:07.47ID:TO6o1I3I>>842-846 突然だが 数学セミナー2015年11月号の記事『箱入り無数目』の最初が 2015/12/20 ガロアスレに紹介された
そうすると 2018に Oxford University から 本１冊出版している ことが分ったんだ https://books.google.co.jp/books?id=RXBoDwAAQBAJ&;pg=PA77&lpg=PA77&dq=%22conglomerability%22+assumption+math&source=bl&ots=8Ol1uFrjJQ&sig=ACfU3U1bAurNGJm5872wDblskzsSgsU0iA&hl=ja&sa=X&ved=2ahUKEwioiPyV_IPiAhXHxrwKHUeaArUQ6AEwCXoECEoQAQ#v=onepage&q=%22conglomerability%22%20assumption%20math&f=false Infinity, Causation, and Paradox 著者: Alexander R. Pruss Oxford University Press, 2018
で、Prussの経歴とか分るかもと検索すると、下記が分ってね。数学のプロだと （>>45） https://en.wikipedia.org/wiki/Alexander_Pruss Alexander Pruss Canadian mathematician, philosopher After earning a Ph.D. in Mathematics at the University of British Columbia in 1996 and publishing several papers in Proceedings of the American Mathematical Society and other mathematical journals,[4]
