Is it possible for an event A to be independent from event B, but not the other way around?












1












$begingroup$


I was wondering, if event $A$ is independent from event $B$, would $B$ also be independent of event $A$? My original thought was that it should be independent, but then I realized if $A$ is independent from $B$ then we have: $$P(A|B)=P(A)label{1}tag{1}$$ and for $B$ to be independent from $A$ we need to have: $$P(B|A)=P(B)label{2}tag{2}$$ but in $ref{1}$ if $P(A)=0$ then $ref{2}$ doesn't make sense, so then $B$ wouldn't be independent from $A$?



Thank you










share|cite|improve this question







New contributor




Mashpa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$

















    1












    $begingroup$


    I was wondering, if event $A$ is independent from event $B$, would $B$ also be independent of event $A$? My original thought was that it should be independent, but then I realized if $A$ is independent from $B$ then we have: $$P(A|B)=P(A)label{1}tag{1}$$ and for $B$ to be independent from $A$ we need to have: $$P(B|A)=P(B)label{2}tag{2}$$ but in $ref{1}$ if $P(A)=0$ then $ref{2}$ doesn't make sense, so then $B$ wouldn't be independent from $A$?



    Thank you










    share|cite|improve this question







    New contributor




    Mashpa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      1












      1








      1





      $begingroup$


      I was wondering, if event $A$ is independent from event $B$, would $B$ also be independent of event $A$? My original thought was that it should be independent, but then I realized if $A$ is independent from $B$ then we have: $$P(A|B)=P(A)label{1}tag{1}$$ and for $B$ to be independent from $A$ we need to have: $$P(B|A)=P(B)label{2}tag{2}$$ but in $ref{1}$ if $P(A)=0$ then $ref{2}$ doesn't make sense, so then $B$ wouldn't be independent from $A$?



      Thank you










      share|cite|improve this question







      New contributor




      Mashpa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      I was wondering, if event $A$ is independent from event $B$, would $B$ also be independent of event $A$? My original thought was that it should be independent, but then I realized if $A$ is independent from $B$ then we have: $$P(A|B)=P(A)label{1}tag{1}$$ and for $B$ to be independent from $A$ we need to have: $$P(B|A)=P(B)label{2}tag{2}$$ but in $ref{1}$ if $P(A)=0$ then $ref{2}$ doesn't make sense, so then $B$ wouldn't be independent from $A$?



      Thank you







      probability-theory independence






      share|cite|improve this question







      New contributor




      Mashpa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question







      New contributor




      Mashpa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|cite|improve this question




      share|cite|improve this question






      New contributor




      Mashpa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      asked 2 hours ago









      MashpaMashpa

      273




      273




      New contributor




      Mashpa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.





      New contributor





      Mashpa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      Mashpa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






















          3 Answers
          3






          active

          oldest

          votes


















          3












          $begingroup$

          $P(A mid B) = P(A)$ should not be taken as the definition of independence, $P(A cap B) = P(A)P(B)$ should be taken as the definition of independence. From this we can prove $P(A mid B) = P(A)$ as a corollary, provided that $P(B) > 0$.






          share|cite|improve this answer









          $endgroup$





















            3












            $begingroup$

            $P(A|B)=P(A)$ is not the correct definition of independence. The correct definition is $P(Acap B)=P(A)P(B)$. These definitions are equivalent if $P(B)>0$. With the correct definition there is symmetry between $A$ and $B$ so $A$ independent of $B$ is same as $B$ independent of $A$






            share|cite|improve this answer









            $endgroup$





















              3












              $begingroup$

              $P(Bmid A)$ is undefined when $P(A)=0$, so you can’t draw any conclusions about independence of the two events from it. That one reason why (despite what the Wikipedia page on conditional probability might imply) the fundamental definition of independence of two events uses their joint probability: $A$ and $B$ are independent iff $P(Acap B)=P(A)P(B)$. This definition is symmetric.






              share|cite|improve this answer









              $endgroup$














                Your Answer








                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
                });


                }
                });






                Mashpa is a new contributor. Be nice, and check out our Code of Conduct.










                draft saved

                draft discarded


















                StackExchange.ready(
                function () {
                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3195328%2fis-it-possible-for-an-event-a-to-be-independent-from-event-b-but-not-the-other%23new-answer', 'question_page');
                }
                );

                Post as a guest















                Required, but never shown

























                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                3












                $begingroup$

                $P(A mid B) = P(A)$ should not be taken as the definition of independence, $P(A cap B) = P(A)P(B)$ should be taken as the definition of independence. From this we can prove $P(A mid B) = P(A)$ as a corollary, provided that $P(B) > 0$.






                share|cite|improve this answer









                $endgroup$


















                  3












                  $begingroup$

                  $P(A mid B) = P(A)$ should not be taken as the definition of independence, $P(A cap B) = P(A)P(B)$ should be taken as the definition of independence. From this we can prove $P(A mid B) = P(A)$ as a corollary, provided that $P(B) > 0$.






                  share|cite|improve this answer









                  $endgroup$
















                    3












                    3








                    3





                    $begingroup$

                    $P(A mid B) = P(A)$ should not be taken as the definition of independence, $P(A cap B) = P(A)P(B)$ should be taken as the definition of independence. From this we can prove $P(A mid B) = P(A)$ as a corollary, provided that $P(B) > 0$.






                    share|cite|improve this answer









                    $endgroup$



                    $P(A mid B) = P(A)$ should not be taken as the definition of independence, $P(A cap B) = P(A)P(B)$ should be taken as the definition of independence. From this we can prove $P(A mid B) = P(A)$ as a corollary, provided that $P(B) > 0$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 2 hours ago









                    bitesizebobitesizebo

                    1,78828




                    1,78828























                        3












                        $begingroup$

                        $P(A|B)=P(A)$ is not the correct definition of independence. The correct definition is $P(Acap B)=P(A)P(B)$. These definitions are equivalent if $P(B)>0$. With the correct definition there is symmetry between $A$ and $B$ so $A$ independent of $B$ is same as $B$ independent of $A$






                        share|cite|improve this answer









                        $endgroup$


















                          3












                          $begingroup$

                          $P(A|B)=P(A)$ is not the correct definition of independence. The correct definition is $P(Acap B)=P(A)P(B)$. These definitions are equivalent if $P(B)>0$. With the correct definition there is symmetry between $A$ and $B$ so $A$ independent of $B$ is same as $B$ independent of $A$






                          share|cite|improve this answer









                          $endgroup$
















                            3












                            3








                            3





                            $begingroup$

                            $P(A|B)=P(A)$ is not the correct definition of independence. The correct definition is $P(Acap B)=P(A)P(B)$. These definitions are equivalent if $P(B)>0$. With the correct definition there is symmetry between $A$ and $B$ so $A$ independent of $B$ is same as $B$ independent of $A$






                            share|cite|improve this answer









                            $endgroup$



                            $P(A|B)=P(A)$ is not the correct definition of independence. The correct definition is $P(Acap B)=P(A)P(B)$. These definitions are equivalent if $P(B)>0$. With the correct definition there is symmetry between $A$ and $B$ so $A$ independent of $B$ is same as $B$ independent of $A$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 2 hours ago









                            Kavi Rama MurthyKavi Rama Murthy

                            76.4k53370




                            76.4k53370























                                3












                                $begingroup$

                                $P(Bmid A)$ is undefined when $P(A)=0$, so you can’t draw any conclusions about independence of the two events from it. That one reason why (despite what the Wikipedia page on conditional probability might imply) the fundamental definition of independence of two events uses their joint probability: $A$ and $B$ are independent iff $P(Acap B)=P(A)P(B)$. This definition is symmetric.






                                share|cite|improve this answer









                                $endgroup$


















                                  3












                                  $begingroup$

                                  $P(Bmid A)$ is undefined when $P(A)=0$, so you can’t draw any conclusions about independence of the two events from it. That one reason why (despite what the Wikipedia page on conditional probability might imply) the fundamental definition of independence of two events uses their joint probability: $A$ and $B$ are independent iff $P(Acap B)=P(A)P(B)$. This definition is symmetric.






                                  share|cite|improve this answer









                                  $endgroup$
















                                    3












                                    3








                                    3





                                    $begingroup$

                                    $P(Bmid A)$ is undefined when $P(A)=0$, so you can’t draw any conclusions about independence of the two events from it. That one reason why (despite what the Wikipedia page on conditional probability might imply) the fundamental definition of independence of two events uses their joint probability: $A$ and $B$ are independent iff $P(Acap B)=P(A)P(B)$. This definition is symmetric.






                                    share|cite|improve this answer









                                    $endgroup$



                                    $P(Bmid A)$ is undefined when $P(A)=0$, so you can’t draw any conclusions about independence of the two events from it. That one reason why (despite what the Wikipedia page on conditional probability might imply) the fundamental definition of independence of two events uses their joint probability: $A$ and $B$ are independent iff $P(Acap B)=P(A)P(B)$. This definition is symmetric.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered 2 hours ago









                                    amdamd

                                    32k21053




                                    32k21053






















                                        Mashpa is a new contributor. Be nice, and check out our Code of Conduct.










                                        draft saved

                                        draft discarded


















                                        Mashpa is a new contributor. Be nice, and check out our Code of Conduct.













                                        Mashpa is a new contributor. Be nice, and check out our Code of Conduct.












                                        Mashpa is a new contributor. Be nice, and check out our Code of Conduct.
















                                        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%2f3195328%2fis-it-possible-for-an-event-a-to-be-independent-from-event-b-but-not-the-other%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

                                        Liste der Baudenkmale in Friedland (Mecklenburg)

                                        Single-Malt-Whisky

                                        Czorneboh