Is it valid to draw event probability conclusions by comparison of calculations?












1












$begingroup$


I'm a mathematics/probability novice, but interested in the use and abuse of these things in other fields (historical, legal, etc.). Sorry about the lack of technicality here.



Here's a stupid hypothetical to explain the question.



You are friends with Vicky. Vicky is married to David. You wonder whether David has had an affair, but Vicky has not said anything to you.



Part 1: Probability of Vicky remaining silent



Vicky has not said anything. If Vicky knows of an affair, there is a 60% chance she would remain silent. If Vicky is ignorant of any affair, there is (obviously) a 100% chance she would remain silent.



P(silent|know) = 0.6. 
P(silent|ignorant) = 1.


In either case ("know" or "ignorant"), it is more probable that Vicky would remain silent.



Is it valid to compare these two results to consider the relative probability of "know" and "silent"?



I.e.:



P(silent|know) < P(silent|ignorant)


therefore it is more probable that Vicky's silence comes from "ignorant", and less probable that it comes from "know".



I have seen people do this comparison, and conclude that Vicky more probably does not know of any affair. Is that a valid inference?



Part 2: Probability that Vicky is ignorant



Assuming that the inference of Part 1 is correct, we may run the same kind of problem again:



Vicky is ignorant of any affair. If David had an affair, there is a 30% chance Vicky would remain ignorant. If David did not have an affair, there is (obviously) a 100% chance Vicky would remain ignorant.



P(ignorant|affair) = 0.3
P(ignorant|no-affair) = 1


Is it valid to then compare these to determine the relative probability of "affair" and "no-affair"?



P(ignorant|affair) < P(ignorant|no-affair) 


Thus, it is more probable that Vicky remains ignorant on "no-affair" than on "affair" - so "no-affair" is more probable.



Again - is this a valid inference?



What's the point?



Leaving aside all the other theoretical/methodological questions, someone could take this as a probabilistic argument to show David probably did not have an affair:



"silent" is more probably derived from "ignorant" than "know".
"ignorant" is more probably derived from "no-affair" than "affair"
therefore, the fact of "silence" allows the conclusion "no affair" is more probable than "affair".


In other words, the fact that Vicky hasn't said anything means that on the balance of probability (excluding all other factors), David has probably not had an affair.



So - is all that valid, or is there some sort of illegitimate transfer or comparison happening here?










share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    I'm a mathematics/probability novice, but interested in the use and abuse of these things in other fields (historical, legal, etc.). Sorry about the lack of technicality here.



    Here's a stupid hypothetical to explain the question.



    You are friends with Vicky. Vicky is married to David. You wonder whether David has had an affair, but Vicky has not said anything to you.



    Part 1: Probability of Vicky remaining silent



    Vicky has not said anything. If Vicky knows of an affair, there is a 60% chance she would remain silent. If Vicky is ignorant of any affair, there is (obviously) a 100% chance she would remain silent.



    P(silent|know) = 0.6. 
    P(silent|ignorant) = 1.


    In either case ("know" or "ignorant"), it is more probable that Vicky would remain silent.



    Is it valid to compare these two results to consider the relative probability of "know" and "silent"?



    I.e.:



    P(silent|know) < P(silent|ignorant)


    therefore it is more probable that Vicky's silence comes from "ignorant", and less probable that it comes from "know".



    I have seen people do this comparison, and conclude that Vicky more probably does not know of any affair. Is that a valid inference?



    Part 2: Probability that Vicky is ignorant



    Assuming that the inference of Part 1 is correct, we may run the same kind of problem again:



    Vicky is ignorant of any affair. If David had an affair, there is a 30% chance Vicky would remain ignorant. If David did not have an affair, there is (obviously) a 100% chance Vicky would remain ignorant.



    P(ignorant|affair) = 0.3
    P(ignorant|no-affair) = 1


    Is it valid to then compare these to determine the relative probability of "affair" and "no-affair"?



    P(ignorant|affair) < P(ignorant|no-affair) 


    Thus, it is more probable that Vicky remains ignorant on "no-affair" than on "affair" - so "no-affair" is more probable.



    Again - is this a valid inference?



    What's the point?



    Leaving aside all the other theoretical/methodological questions, someone could take this as a probabilistic argument to show David probably did not have an affair:



    "silent" is more probably derived from "ignorant" than "know".
    "ignorant" is more probably derived from "no-affair" than "affair"
    therefore, the fact of "silence" allows the conclusion "no affair" is more probable than "affair".


    In other words, the fact that Vicky hasn't said anything means that on the balance of probability (excluding all other factors), David has probably not had an affair.



    So - is all that valid, or is there some sort of illegitimate transfer or comparison happening here?










    share|cite|improve this question









    $endgroup$















      1












      1








      1


      1



      $begingroup$


      I'm a mathematics/probability novice, but interested in the use and abuse of these things in other fields (historical, legal, etc.). Sorry about the lack of technicality here.



      Here's a stupid hypothetical to explain the question.



      You are friends with Vicky. Vicky is married to David. You wonder whether David has had an affair, but Vicky has not said anything to you.



      Part 1: Probability of Vicky remaining silent



      Vicky has not said anything. If Vicky knows of an affair, there is a 60% chance she would remain silent. If Vicky is ignorant of any affair, there is (obviously) a 100% chance she would remain silent.



      P(silent|know) = 0.6. 
      P(silent|ignorant) = 1.


      In either case ("know" or "ignorant"), it is more probable that Vicky would remain silent.



      Is it valid to compare these two results to consider the relative probability of "know" and "silent"?



      I.e.:



      P(silent|know) < P(silent|ignorant)


      therefore it is more probable that Vicky's silence comes from "ignorant", and less probable that it comes from "know".



      I have seen people do this comparison, and conclude that Vicky more probably does not know of any affair. Is that a valid inference?



      Part 2: Probability that Vicky is ignorant



      Assuming that the inference of Part 1 is correct, we may run the same kind of problem again:



      Vicky is ignorant of any affair. If David had an affair, there is a 30% chance Vicky would remain ignorant. If David did not have an affair, there is (obviously) a 100% chance Vicky would remain ignorant.



      P(ignorant|affair) = 0.3
      P(ignorant|no-affair) = 1


      Is it valid to then compare these to determine the relative probability of "affair" and "no-affair"?



      P(ignorant|affair) < P(ignorant|no-affair) 


      Thus, it is more probable that Vicky remains ignorant on "no-affair" than on "affair" - so "no-affair" is more probable.



      Again - is this a valid inference?



      What's the point?



      Leaving aside all the other theoretical/methodological questions, someone could take this as a probabilistic argument to show David probably did not have an affair:



      "silent" is more probably derived from "ignorant" than "know".
      "ignorant" is more probably derived from "no-affair" than "affair"
      therefore, the fact of "silence" allows the conclusion "no affair" is more probable than "affair".


      In other words, the fact that Vicky hasn't said anything means that on the balance of probability (excluding all other factors), David has probably not had an affair.



      So - is all that valid, or is there some sort of illegitimate transfer or comparison happening here?










      share|cite|improve this question









      $endgroup$




      I'm a mathematics/probability novice, but interested in the use and abuse of these things in other fields (historical, legal, etc.). Sorry about the lack of technicality here.



      Here's a stupid hypothetical to explain the question.



      You are friends with Vicky. Vicky is married to David. You wonder whether David has had an affair, but Vicky has not said anything to you.



      Part 1: Probability of Vicky remaining silent



      Vicky has not said anything. If Vicky knows of an affair, there is a 60% chance she would remain silent. If Vicky is ignorant of any affair, there is (obviously) a 100% chance she would remain silent.



      P(silent|know) = 0.6. 
      P(silent|ignorant) = 1.


      In either case ("know" or "ignorant"), it is more probable that Vicky would remain silent.



      Is it valid to compare these two results to consider the relative probability of "know" and "silent"?



      I.e.:



      P(silent|know) < P(silent|ignorant)


      therefore it is more probable that Vicky's silence comes from "ignorant", and less probable that it comes from "know".



      I have seen people do this comparison, and conclude that Vicky more probably does not know of any affair. Is that a valid inference?



      Part 2: Probability that Vicky is ignorant



      Assuming that the inference of Part 1 is correct, we may run the same kind of problem again:



      Vicky is ignorant of any affair. If David had an affair, there is a 30% chance Vicky would remain ignorant. If David did not have an affair, there is (obviously) a 100% chance Vicky would remain ignorant.



      P(ignorant|affair) = 0.3
      P(ignorant|no-affair) = 1


      Is it valid to then compare these to determine the relative probability of "affair" and "no-affair"?



      P(ignorant|affair) < P(ignorant|no-affair) 


      Thus, it is more probable that Vicky remains ignorant on "no-affair" than on "affair" - so "no-affair" is more probable.



      Again - is this a valid inference?



      What's the point?



      Leaving aside all the other theoretical/methodological questions, someone could take this as a probabilistic argument to show David probably did not have an affair:



      "silent" is more probably derived from "ignorant" than "know".
      "ignorant" is more probably derived from "no-affair" than "affair"
      therefore, the fact of "silence" allows the conclusion "no affair" is more probable than "affair".


      In other words, the fact that Vicky hasn't said anything means that on the balance of probability (excluding all other factors), David has probably not had an affair.



      So - is all that valid, or is there some sort of illegitimate transfer or comparison happening here?







      probability logic conditional-probability






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 7 '18 at 3:10









      Alex Alex

      82




      82






















          1 Answer
          1






          active

          oldest

          votes


















          0












          $begingroup$

          In a word, yes this is a valid line of reasoning and is the essence of hypothesis testing and Bayesian updating. Everything else being equal, Vicky staying silent is a data point that increases (or in the extreme case at least doesn't decrease) the probability that David didn't have an affair. But the crucial aspect is the "everything else being equal." You are implicitly assuming that David having and not having an affair was roughly 50/50 before gathering evidence from Vicky. If you saw with your own eyes that David was kissing Mary, then you are already 100% sure he did have an affair. Whether Vicky stays silent or not will not affect your assessment of the situation in this case.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks. Can you (or someone else) clarify what you mean by "in the extreme case at least doesn't decrease the probability"?
            $endgroup$
            – Alex
            Dec 7 '18 at 8:01










          • $begingroup$
            I'm referring to the case when you know 100% David had an affair, or in other words the probability he didn't have an affair is 0. In such a case, Vicky's staying silent cannot increase the probability he didn't have an affair because you already know the situation with absolute certainty.
            $endgroup$
            – maridia
            Dec 7 '18 at 9:21











          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%2f3029410%2fis-it-valid-to-draw-event-probability-conclusions-by-comparison-of-calculations%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









          0












          $begingroup$

          In a word, yes this is a valid line of reasoning and is the essence of hypothesis testing and Bayesian updating. Everything else being equal, Vicky staying silent is a data point that increases (or in the extreme case at least doesn't decrease) the probability that David didn't have an affair. But the crucial aspect is the "everything else being equal." You are implicitly assuming that David having and not having an affair was roughly 50/50 before gathering evidence from Vicky. If you saw with your own eyes that David was kissing Mary, then you are already 100% sure he did have an affair. Whether Vicky stays silent or not will not affect your assessment of the situation in this case.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks. Can you (or someone else) clarify what you mean by "in the extreme case at least doesn't decrease the probability"?
            $endgroup$
            – Alex
            Dec 7 '18 at 8:01










          • $begingroup$
            I'm referring to the case when you know 100% David had an affair, or in other words the probability he didn't have an affair is 0. In such a case, Vicky's staying silent cannot increase the probability he didn't have an affair because you already know the situation with absolute certainty.
            $endgroup$
            – maridia
            Dec 7 '18 at 9:21
















          0












          $begingroup$

          In a word, yes this is a valid line of reasoning and is the essence of hypothesis testing and Bayesian updating. Everything else being equal, Vicky staying silent is a data point that increases (or in the extreme case at least doesn't decrease) the probability that David didn't have an affair. But the crucial aspect is the "everything else being equal." You are implicitly assuming that David having and not having an affair was roughly 50/50 before gathering evidence from Vicky. If you saw with your own eyes that David was kissing Mary, then you are already 100% sure he did have an affair. Whether Vicky stays silent or not will not affect your assessment of the situation in this case.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks. Can you (or someone else) clarify what you mean by "in the extreme case at least doesn't decrease the probability"?
            $endgroup$
            – Alex
            Dec 7 '18 at 8:01










          • $begingroup$
            I'm referring to the case when you know 100% David had an affair, or in other words the probability he didn't have an affair is 0. In such a case, Vicky's staying silent cannot increase the probability he didn't have an affair because you already know the situation with absolute certainty.
            $endgroup$
            – maridia
            Dec 7 '18 at 9:21














          0












          0








          0





          $begingroup$

          In a word, yes this is a valid line of reasoning and is the essence of hypothesis testing and Bayesian updating. Everything else being equal, Vicky staying silent is a data point that increases (or in the extreme case at least doesn't decrease) the probability that David didn't have an affair. But the crucial aspect is the "everything else being equal." You are implicitly assuming that David having and not having an affair was roughly 50/50 before gathering evidence from Vicky. If you saw with your own eyes that David was kissing Mary, then you are already 100% sure he did have an affair. Whether Vicky stays silent or not will not affect your assessment of the situation in this case.






          share|cite|improve this answer









          $endgroup$



          In a word, yes this is a valid line of reasoning and is the essence of hypothesis testing and Bayesian updating. Everything else being equal, Vicky staying silent is a data point that increases (or in the extreme case at least doesn't decrease) the probability that David didn't have an affair. But the crucial aspect is the "everything else being equal." You are implicitly assuming that David having and not having an affair was roughly 50/50 before gathering evidence from Vicky. If you saw with your own eyes that David was kissing Mary, then you are already 100% sure he did have an affair. Whether Vicky stays silent or not will not affect your assessment of the situation in this case.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 7 '18 at 3:40









          maridiamaridia

          1,065113




          1,065113












          • $begingroup$
            Thanks. Can you (or someone else) clarify what you mean by "in the extreme case at least doesn't decrease the probability"?
            $endgroup$
            – Alex
            Dec 7 '18 at 8:01










          • $begingroup$
            I'm referring to the case when you know 100% David had an affair, or in other words the probability he didn't have an affair is 0. In such a case, Vicky's staying silent cannot increase the probability he didn't have an affair because you already know the situation with absolute certainty.
            $endgroup$
            – maridia
            Dec 7 '18 at 9:21


















          • $begingroup$
            Thanks. Can you (or someone else) clarify what you mean by "in the extreme case at least doesn't decrease the probability"?
            $endgroup$
            – Alex
            Dec 7 '18 at 8:01










          • $begingroup$
            I'm referring to the case when you know 100% David had an affair, or in other words the probability he didn't have an affair is 0. In such a case, Vicky's staying silent cannot increase the probability he didn't have an affair because you already know the situation with absolute certainty.
            $endgroup$
            – maridia
            Dec 7 '18 at 9:21
















          $begingroup$
          Thanks. Can you (or someone else) clarify what you mean by "in the extreme case at least doesn't decrease the probability"?
          $endgroup$
          – Alex
          Dec 7 '18 at 8:01




          $begingroup$
          Thanks. Can you (or someone else) clarify what you mean by "in the extreme case at least doesn't decrease the probability"?
          $endgroup$
          – Alex
          Dec 7 '18 at 8:01












          $begingroup$
          I'm referring to the case when you know 100% David had an affair, or in other words the probability he didn't have an affair is 0. In such a case, Vicky's staying silent cannot increase the probability he didn't have an affair because you already know the situation with absolute certainty.
          $endgroup$
          – maridia
          Dec 7 '18 at 9:21




          $begingroup$
          I'm referring to the case when you know 100% David had an affair, or in other words the probability he didn't have an affair is 0. In such a case, Vicky's staying silent cannot increase the probability he didn't have an affair because you already know the situation with absolute certainty.
          $endgroup$
          – maridia
          Dec 7 '18 at 9:21


















          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.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3029410%2fis-it-valid-to-draw-event-probability-conclusions-by-comparison-of-calculations%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

          Marschland

          Dieringhausen