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






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.







$endgroup$

















    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






    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.







    $endgroup$















      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






      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.







      $endgroup$




      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




      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)

                                        Image
                                        In der Liste der Baudenkmale in Friedland sind alle denkmalgeschützten Bauten der Stadt Friedland (Mecklenburg-Vorpommern) und ihrer Ortsteile aufgelistet. Grundlage ist die Veröffentlichung der Denkmalliste des Landkreises Mecklenburgische Seenplatte (Auszug) vom 28. September 2016. Inhaltsverzeichnis 1 Baudenkmale nach Ortsteilen 1.1 Friedland 1.2 Bresewitz 1.3 Brohm 1.4 Cosa 1.5 Dishley 1.6 Dishley-Ausbau 1.7 Eichhorst 1.8 Glienke 1.9 Heinrichshöh 1.10 Heinrichswalde 1.11 Jatzke 1.12 Jatzke-Meierei 1.13 Liepen 1.14 Ramelow 1.15 Schwanbeck 2 Vorgesehener Denkmalbereich 3 Quelle 4 Weblinks Baudenkmale nach Ortsteilen | Friedland | ID Lage Bezeichnung Beschreibung Bild 281 An der Kleinbahn 2 Windmühle mit Umgang 282 Anklamer Straße alter Friedhof mit Portalwand 282 Anklamer Straße drei Erbbegräbnissen und Gräbern u. a. H.A. Riemann 284
                                        Read more

                                        Single-Malt-Whisky

                                        Image
                                        Fünf verschiedene schottische Single-Malt-Whiskys Als Single-Malt-Whisky oder kürzer Single Malt werden Whiskys bezeichnet, die zwei besondere Bedingungen erfüllen: Sie stammen zum ersten aus einer einzigen Brennerei, sind also kein Verschnitt aus mehreren Whiskysorten (daher Single , deutsch: einzeln). Zum zweiten wird als Getreide ausschließlich gemälzte Gerste verwendet (deswegen Malt , deutsch: Malz). Der Zusatz Single Malt wird als Prädikat verstanden, die so ausgezeichneten Whiskys sind in aller Regel besonders hochwertig. Hinzu kommt ihre besonders lange Lagerung, die Reifungszeit währt oft ein Jahrzehnt und länger. Inhaltsverzeichnis 1 Allgemeines 2 Herkunftsländer 2.1 Schottland 2.2 Irland 2.3 Japan 2.4 Kontinentaleuropa 3 Produktionsschritte 3.1 Mälzen 3.2 Brauen 3.3 Brennen 3.4 Reifung und Finish 3.5 Abfüllung 4 Altersangaben 5 Lagerung von Whiskyflaschen 6 Geschmacksklassifikation 7 Sonstig
                                        Read more

                                        Czorneboh

                                        Image
                                        Dieser Artikel erläutert den Berg Czorneboh, für die Gottheit siehe Czorneboh (Gottheit). Czorneboh Čornobóh Berggasthof mit Turm Höhe 555,7  m ü. NHN [1] Lage Freistaat Sachsen, Deutschland Gebirge Lausitzer Bergland Dominanz 15,4 km →  Kottmar Koordinaten 51° 7′ 11″  N , 14° 31′ 31″  O 51.119722222222 14.525194444444 555.7 Koordinaten: 51° 7′ 11″  N , 14° 31′ 31″  O Gestein Granodiorit Besonderheiten Aussichtsturm Czorneboh Der Czorneboh , obersorbisch Čornobóh , ist ein Berg zwischen Hochkirch und Cunewalde in der Oberlausitz. Er gehört zu der südöstlich von Bautzen gelegenen Czorneboh-Bergkette und ist mit einer Höhe von 555,7  m ü. NHN [1] der höchste Punkt dieses Ausläufers des Lausitzer Berglandes. Er leitet als Landschaftselement die wellige Granitlandschaft der Lausitzer Vorbergzone ein. Der Gipfel des Czorneboh liegt in der Gemarkung Meschwitz (Gemeinde Hochkirch), die Czornebohbaude jedoch
                                        Read more