Transitive Probability Constraints Take 2












0














We know that $P(B|A)$ and $P(C|B)$ don't generally constraint $P(C|A)$. But do $P(B|A)$, $P(neg A|neg B)$, $P(C|B)$, $P(neg B|neg C)$ jointly constrain $P(C|A)$ and $P(neg A|neg C)$? In particular, let $t > frac{1}{2}$ be such that



$t leq$ $P(B|A)$, $P(neg A|neg B)$, $P(C|B)$, $P(neg B|neg C)$.



What does this tell us about $P(C|A)$ and $P(neg A|neg C)$? In particular, does it imply $P(C|A)$, $P(neg A|neg C)$ $leq 1 - t$ can't both hold?






probability







share|cite|improve this question





























    0














    We know that $P(B|A)$ and $P(C|B)$ don't generally constraint $P(C|A)$. But do $P(B|A)$, $P(neg A|neg B)$, $P(C|B)$, $P(neg B|neg C)$ jointly constrain $P(C|A)$ and $P(neg A|neg C)$? In particular, let $t > frac{1}{2}$ be such that



    $t leq$ $P(B|A)$, $P(neg A|neg B)$, $P(C|B)$, $P(neg B|neg C)$.



    What does this tell us about $P(C|A)$ and $P(neg A|neg C)$? In particular, does it imply $P(C|A)$, $P(neg A|neg C)$ $leq 1 - t$ can't both hold?






    probability







    share|cite|improve this question



























      0












      0








      0







      We know that $P(B|A)$ and $P(C|B)$ don't generally constraint $P(C|A)$. But do $P(B|A)$, $P(neg A|neg B)$, $P(C|B)$, $P(neg B|neg C)$ jointly constrain $P(C|A)$ and $P(neg A|neg C)$? In particular, let $t > frac{1}{2}$ be such that



      $t leq$ $P(B|A)$, $P(neg A|neg B)$, $P(C|B)$, $P(neg B|neg C)$.



      What does this tell us about $P(C|A)$ and $P(neg A|neg C)$? In particular, does it imply $P(C|A)$, $P(neg A|neg C)$ $leq 1 - t$ can't both hold?






      probability







      share|cite|improve this question















      We know that $P(B|A)$ and $P(C|B)$ don't generally constraint $P(C|A)$. But do $P(B|A)$, $P(neg A|neg B)$, $P(C|B)$, $P(neg B|neg C)$ jointly constrain $P(C|A)$ and $P(neg A|neg C)$? In particular, let $t > frac{1}{2}$ be such that



      $t leq$ $P(B|A)$, $P(neg A|neg B)$, $P(C|B)$, $P(neg B|neg C)$.



      What does this tell us about $P(C|A)$ and $P(neg A|neg C)$? In particular, does it imply $P(C|A)$, $P(neg A|neg C)$ $leq 1 - t$ can't both hold?






      probability




      probability






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 30 at 15:52

























      asked Nov 30 at 14:31









      King Kong

      1507




      1507






















          1 Answer
          1






          active

          oldest

          votes


















          1














          (-----Answer to the old question-----)



          Consider the probability function $P$ that assigns the following probabilities to atomic events:
          $$P(neg A wedge B wedge C) = frac{2}{3}$$
          $$P(A wedge B wedge neg C) = frac{1}{12}$$
          $$P(neg A wedge neg B wedge C) = frac{1}{12}$$
          $$P(neg A wedge neg B wedge neg C) = frac{1}{6}$$
          The other atomic events get null probability by the function $P$.



          With such assignment we have that $P(B|A)=1$, $P(C|B)=frac{8}{9}$, $P(neg A|neg B)=1$ and $P(neg B|neg C)=frac{2}{3}$. Note that $P(C|A)=0$ under the distribution $P$. Therefore, the implication you suggest does not hold (take $t=frac{2}{3}$, for example).






          share|cite|improve this answer























          • fantastic, thanks a lot. I now realise that the property i'm after is actually a little different. I want to know whether $t leq P(C|B), P(neg B|neg C), P(B|A), P(neg A|neg B)$ guarantees that $P(C|A)$, $P(neg A|neg C) leq 1 - t$ can't both hold. I'm editing the question to reflect this. My apologies for the mistake.
            – King Kong
            Nov 30 at 15:52













          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3020156%2ftransitive-probability-constraints-take-2%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1














          (-----Answer to the old question-----)



          Consider the probability function $P$ that assigns the following probabilities to atomic events:
          $$P(neg A wedge B wedge C) = frac{2}{3}$$
          $$P(A wedge B wedge neg C) = frac{1}{12}$$
          $$P(neg A wedge neg B wedge C) = frac{1}{12}$$
          $$P(neg A wedge neg B wedge neg C) = frac{1}{6}$$
          The other atomic events get null probability by the function $P$.



          With such assignment we have that $P(B|A)=1$, $P(C|B)=frac{8}{9}$, $P(neg A|neg B)=1$ and $P(neg B|neg C)=frac{2}{3}$. Note that $P(C|A)=0$ under the distribution $P$. Therefore, the implication you suggest does not hold (take $t=frac{2}{3}$, for example).






          share|cite|improve this answer























          • fantastic, thanks a lot. I now realise that the property i'm after is actually a little different. I want to know whether $t leq P(C|B), P(neg B|neg C), P(B|A), P(neg A|neg B)$ guarantees that $P(C|A)$, $P(neg A|neg C) leq 1 - t$ can't both hold. I'm editing the question to reflect this. My apologies for the mistake.
            – King Kong
            Nov 30 at 15:52


















          1














          (-----Answer to the old question-----)



          Consider the probability function $P$ that assigns the following probabilities to atomic events:
          $$P(neg A wedge B wedge C) = frac{2}{3}$$
          $$P(A wedge B wedge neg C) = frac{1}{12}$$
          $$P(neg A wedge neg B wedge C) = frac{1}{12}$$
          $$P(neg A wedge neg B wedge neg C) = frac{1}{6}$$
          The other atomic events get null probability by the function $P$.



          With such assignment we have that $P(B|A)=1$, $P(C|B)=frac{8}{9}$, $P(neg A|neg B)=1$ and $P(neg B|neg C)=frac{2}{3}$. Note that $P(C|A)=0$ under the distribution $P$. Therefore, the implication you suggest does not hold (take $t=frac{2}{3}$, for example).






          share|cite|improve this answer























          • fantastic, thanks a lot. I now realise that the property i'm after is actually a little different. I want to know whether $t leq P(C|B), P(neg B|neg C), P(B|A), P(neg A|neg B)$ guarantees that $P(C|A)$, $P(neg A|neg C) leq 1 - t$ can't both hold. I'm editing the question to reflect this. My apologies for the mistake.
            – King Kong
            Nov 30 at 15:52
















          1












          1








          1






          (-----Answer to the old question-----)



          Consider the probability function $P$ that assigns the following probabilities to atomic events:
          $$P(neg A wedge B wedge C) = frac{2}{3}$$
          $$P(A wedge B wedge neg C) = frac{1}{12}$$
          $$P(neg A wedge neg B wedge C) = frac{1}{12}$$
          $$P(neg A wedge neg B wedge neg C) = frac{1}{6}$$
          The other atomic events get null probability by the function $P$.



          With such assignment we have that $P(B|A)=1$, $P(C|B)=frac{8}{9}$, $P(neg A|neg B)=1$ and $P(neg B|neg C)=frac{2}{3}$. Note that $P(C|A)=0$ under the distribution $P$. Therefore, the implication you suggest does not hold (take $t=frac{2}{3}$, for example).






          share|cite|improve this answer














          (-----Answer to the old question-----)



          Consider the probability function $P$ that assigns the following probabilities to atomic events:
          $$P(neg A wedge B wedge C) = frac{2}{3}$$
          $$P(A wedge B wedge neg C) = frac{1}{12}$$
          $$P(neg A wedge neg B wedge C) = frac{1}{12}$$
          $$P(neg A wedge neg B wedge neg C) = frac{1}{6}$$
          The other atomic events get null probability by the function $P$.



          With such assignment we have that $P(B|A)=1$, $P(C|B)=frac{8}{9}$, $P(neg A|neg B)=1$ and $P(neg B|neg C)=frac{2}{3}$. Note that $P(C|A)=0$ under the distribution $P$. Therefore, the implication you suggest does not hold (take $t=frac{2}{3}$, for example).







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 30 at 15:56

























          answered Nov 30 at 15:24









          DavidPM

          16618




          16618












          • fantastic, thanks a lot. I now realise that the property i'm after is actually a little different. I want to know whether $t leq P(C|B), P(neg B|neg C), P(B|A), P(neg A|neg B)$ guarantees that $P(C|A)$, $P(neg A|neg C) leq 1 - t$ can't both hold. I'm editing the question to reflect this. My apologies for the mistake.
            – King Kong
            Nov 30 at 15:52




















          • fantastic, thanks a lot. I now realise that the property i'm after is actually a little different. I want to know whether $t leq P(C|B), P(neg B|neg C), P(B|A), P(neg A|neg B)$ guarantees that $P(C|A)$, $P(neg A|neg C) leq 1 - t$ can't both hold. I'm editing the question to reflect this. My apologies for the mistake.
            – King Kong
            Nov 30 at 15:52


















          fantastic, thanks a lot. I now realise that the property i'm after is actually a little different. I want to know whether $t leq P(C|B), P(neg B|neg C), P(B|A), P(neg A|neg B)$ guarantees that $P(C|A)$, $P(neg A|neg C) leq 1 - t$ can't both hold. I'm editing the question to reflect this. My apologies for the mistake.
          – King Kong
          Nov 30 at 15:52






          fantastic, thanks a lot. I now realise that the property i'm after is actually a little different. I want to know whether $t leq P(C|B), P(neg B|neg C), P(B|A), P(neg A|neg B)$ guarantees that $P(C|A)$, $P(neg A|neg C) leq 1 - t$ can't both hold. I'm editing the question to reflect this. My apologies for the mistake.
          – King Kong
          Nov 30 at 15:52




















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Mathematics Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid …



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.





          Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


          Please pay close attention to the following guidance:


          • Please be sure to answer the question. Provide details and share your research!

          But avoid …



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3020156%2ftransitive-probability-constraints-take-2%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          -

          Popular posts from this blog

          Wiesbaden

          Image
          Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
          Read more

          Marschland

          Image
          Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
          Read more

          Dieringhausen

          Image
          Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
          Read more