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

          To store a contact into the json file from server.js file using a class in NodeJS

          Image
          0 I have created three files -contactPresentation.js, contactService.js, contacts.json. I will run the contactService.js file. And it will navigate to contactPresentation.js file that will only choose an option to either add or display a file. If add(), then add the data into add of service file.And the data will be added to the contacts.json file . And if display(), then display the display() of service file. And the display will take the data from contacts.json file. However, i know my code is flawed. Please correct it . I am also getting an error - this.contacts.push(new ContactService(userName, contactNumber, emailId)); ^ TypeError: Cannot read property 'push' of undefined at ContactService.add (e:NodeJSProjectcontactManager2contactService.js:57:23) at new C
          Read more

          Redirect URL with Chrome Remote Debugging Android Devices

          Image
          1 How can i redirect an URL with Chrome Remote Debugging Android Devices? like Requestly Chrome extension on desktop. google-chrome debugging url redirect testing share | improve this question edited Nov 24 '18 at 8:16 sachinjain024 14.7k 23 73 129 asked Nov 23 '18 at 16:09 Momen Momen 91 10
          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