Are pairwise mutually exclusive events the same as mutually exclusive events?












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Larson (1982) defining the probability axioms talks about "mutually exclusive" events, while Poirier (1995) about "$A_1, A_2, ldots$ as a sequence of pairwise mutually exclusive events events in the sigma-algebra $tilde A$."



I suppose that the two notions are equivalent (they both refer two disjoint sets), right? Does this make adding the word "pairwise" superfluous on behalf of Poirier?



Is there any other context out of probability that makes this distinction (using the word pair-wise) meaningful? According to wikipedia, in Logic, "pairwise mutually exclusive" means that both propositions cannot be true simultaneously, in contrast to just mutually exclusivity that means that if one is true, then the other cannot be true.






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    Larson (1982) defining the probability axioms talks about "mutually exclusive" events, while Poirier (1995) about "$A_1, A_2, ldots$ as a sequence of pairwise mutually exclusive events events in the sigma-algebra $tilde A$."



    I suppose that the two notions are equivalent (they both refer two disjoint sets), right? Does this make adding the word "pairwise" superfluous on behalf of Poirier?



    Is there any other context out of probability that makes this distinction (using the word pair-wise) meaningful? According to wikipedia, in Logic, "pairwise mutually exclusive" means that both propositions cannot be true simultaneously, in contrast to just mutually exclusivity that means that if one is true, then the other cannot be true.






    probability-theory terminology







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    bumped to the homepage by Community♦ 2 days ago


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      Larson (1982) defining the probability axioms talks about "mutually exclusive" events, while Poirier (1995) about "$A_1, A_2, ldots$ as a sequence of pairwise mutually exclusive events events in the sigma-algebra $tilde A$."



      I suppose that the two notions are equivalent (they both refer two disjoint sets), right? Does this make adding the word "pairwise" superfluous on behalf of Poirier?



      Is there any other context out of probability that makes this distinction (using the word pair-wise) meaningful? According to wikipedia, in Logic, "pairwise mutually exclusive" means that both propositions cannot be true simultaneously, in contrast to just mutually exclusivity that means that if one is true, then the other cannot be true.






      probability-theory terminology







      share|cite|improve this question















      Larson (1982) defining the probability axioms talks about "mutually exclusive" events, while Poirier (1995) about "$A_1, A_2, ldots$ as a sequence of pairwise mutually exclusive events events in the sigma-algebra $tilde A$."



      I suppose that the two notions are equivalent (they both refer two disjoint sets), right? Does this make adding the word "pairwise" superfluous on behalf of Poirier?



      Is there any other context out of probability that makes this distinction (using the word pair-wise) meaningful? According to wikipedia, in Logic, "pairwise mutually exclusive" means that both propositions cannot be true simultaneously, in contrast to just mutually exclusivity that means that if one is true, then the other cannot be true.






      probability-theory terminology




      probability-theory terminology






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      edited Apr 27 '16 at 17:15









      Carl Mummert

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      asked Jun 15 '15 at 20:18









      Konstantinos

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          They don't mean the same thing, but it is a short jump to show they are equivalent. A set of events, $A_1,...,A_n$ are mutually exclusive if the occurrence of one of them implies that the other $n-1$ events can't happen. It's immediate that the events are pairwise mutually exclusive. The other direction is immediate as well.



          They aren't a priori the same, but it's immediate that they're equivalent.






          share|cite|improve this answer





























            0














            Sets $A_1, A_2, ...$ are mutually disjoint if



            $$bigcap_{i=1}^{infty} A_i = emptyset $$



            Sets $A_1, A_2, ...$ are pairwise disjoint if



            $$A_i cap A_j = emptyset forall i ne j$$



            Apparently, most texts use 'disjoint' to refer to 'pairwise disjoint'. Whenever a text uses 'pairwise disjoint', we can assume 'disjoint' refers to 'mutually disjoint'




            In our case, 'mutually exclusive' by Larson means the same thing as 'pairwise mutually exclusive'. It depends on the text I guess.




            See my questions:



            Do Kolmogorov's axioms really need only disjointness rather than pairwise disjointness?



            http://meta.math.stackexchange.com/questions/21560/should-these-be-simply-disjoint-instead-of-pairwise-disjoint






            share|cite|improve this answer























            • Repeating a pure invention of your own does not make it less ludicrous.
              – Did
              Nov 20 '15 at 0:42



















            0














            What about vicious cycles? For example, suppose you have three binary random variables $A$, $B$, and $C$ that each take values $0$ or $1$. In this example, $A$ and $B$ are mutually exclusive if whenever $A = 0$, $B = 1$, and similarly for $C$.



            $A$, $B$, and $C$ cannot all be pairwise mutually exclusive.



            However, they can be "globally" mutually exclusive in the sense that, for example, if $A = 0$, then $B = 1$ and $C = 1$, and so on for the other variables.



            I think that this counterexample proves that pairwise mutual exclusivity is not equivalent to mutual exclusivity.






            share|cite|improve this answer





















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              0














              They don't mean the same thing, but it is a short jump to show they are equivalent. A set of events, $A_1,...,A_n$ are mutually exclusive if the occurrence of one of them implies that the other $n-1$ events can't happen. It's immediate that the events are pairwise mutually exclusive. The other direction is immediate as well.



              They aren't a priori the same, but it's immediate that they're equivalent.






              share|cite|improve this answer


























                0














                They don't mean the same thing, but it is a short jump to show they are equivalent. A set of events, $A_1,...,A_n$ are mutually exclusive if the occurrence of one of them implies that the other $n-1$ events can't happen. It's immediate that the events are pairwise mutually exclusive. The other direction is immediate as well.



                They aren't a priori the same, but it's immediate that they're equivalent.






                share|cite|improve this answer
























                  0












                  0








                  0






                  They don't mean the same thing, but it is a short jump to show they are equivalent. A set of events, $A_1,...,A_n$ are mutually exclusive if the occurrence of one of them implies that the other $n-1$ events can't happen. It's immediate that the events are pairwise mutually exclusive. The other direction is immediate as well.



                  They aren't a priori the same, but it's immediate that they're equivalent.






                  share|cite|improve this answer












                  They don't mean the same thing, but it is a short jump to show they are equivalent. A set of events, $A_1,...,A_n$ are mutually exclusive if the occurrence of one of them implies that the other $n-1$ events can't happen. It's immediate that the events are pairwise mutually exclusive. The other direction is immediate as well.



                  They aren't a priori the same, but it's immediate that they're equivalent.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jun 15 '15 at 20:29







                  user223391






























                      0














                      Sets $A_1, A_2, ...$ are mutually disjoint if



                      $$bigcap_{i=1}^{infty} A_i = emptyset $$



                      Sets $A_1, A_2, ...$ are pairwise disjoint if



                      $$A_i cap A_j = emptyset forall i ne j$$



                      Apparently, most texts use 'disjoint' to refer to 'pairwise disjoint'. Whenever a text uses 'pairwise disjoint', we can assume 'disjoint' refers to 'mutually disjoint'




                      In our case, 'mutually exclusive' by Larson means the same thing as 'pairwise mutually exclusive'. It depends on the text I guess.




                      See my questions:



                      Do Kolmogorov's axioms really need only disjointness rather than pairwise disjointness?



                      http://meta.math.stackexchange.com/questions/21560/should-these-be-simply-disjoint-instead-of-pairwise-disjoint






                      share|cite|improve this answer























                      • Repeating a pure invention of your own does not make it less ludicrous.
                        – Did
                        Nov 20 '15 at 0:42
















                      0














                      Sets $A_1, A_2, ...$ are mutually disjoint if



                      $$bigcap_{i=1}^{infty} A_i = emptyset $$



                      Sets $A_1, A_2, ...$ are pairwise disjoint if



                      $$A_i cap A_j = emptyset forall i ne j$$



                      Apparently, most texts use 'disjoint' to refer to 'pairwise disjoint'. Whenever a text uses 'pairwise disjoint', we can assume 'disjoint' refers to 'mutually disjoint'




                      In our case, 'mutually exclusive' by Larson means the same thing as 'pairwise mutually exclusive'. It depends on the text I guess.




                      See my questions:



                      Do Kolmogorov's axioms really need only disjointness rather than pairwise disjointness?



                      http://meta.math.stackexchange.com/questions/21560/should-these-be-simply-disjoint-instead-of-pairwise-disjoint






                      share|cite|improve this answer























                      • Repeating a pure invention of your own does not make it less ludicrous.
                        – Did
                        Nov 20 '15 at 0:42














                      0












                      0








                      0






                      Sets $A_1, A_2, ...$ are mutually disjoint if



                      $$bigcap_{i=1}^{infty} A_i = emptyset $$



                      Sets $A_1, A_2, ...$ are pairwise disjoint if



                      $$A_i cap A_j = emptyset forall i ne j$$



                      Apparently, most texts use 'disjoint' to refer to 'pairwise disjoint'. Whenever a text uses 'pairwise disjoint', we can assume 'disjoint' refers to 'mutually disjoint'




                      In our case, 'mutually exclusive' by Larson means the same thing as 'pairwise mutually exclusive'. It depends on the text I guess.




                      See my questions:



                      Do Kolmogorov's axioms really need only disjointness rather than pairwise disjointness?



                      http://meta.math.stackexchange.com/questions/21560/should-these-be-simply-disjoint-instead-of-pairwise-disjoint






                      share|cite|improve this answer














                      Sets $A_1, A_2, ...$ are mutually disjoint if



                      $$bigcap_{i=1}^{infty} A_i = emptyset $$



                      Sets $A_1, A_2, ...$ are pairwise disjoint if



                      $$A_i cap A_j = emptyset forall i ne j$$



                      Apparently, most texts use 'disjoint' to refer to 'pairwise disjoint'. Whenever a text uses 'pairwise disjoint', we can assume 'disjoint' refers to 'mutually disjoint'




                      In our case, 'mutually exclusive' by Larson means the same thing as 'pairwise mutually exclusive'. It depends on the text I guess.




                      See my questions:



                      Do Kolmogorov's axioms really need only disjointness rather than pairwise disjointness?



                      http://meta.math.stackexchange.com/questions/21560/should-these-be-simply-disjoint-instead-of-pairwise-disjoint







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Apr 13 '17 at 12:20









                      Community♦

                      1




                      1










                      answered Nov 20 '15 at 0:34









                      BCLC

                      1




                      1












                      • Repeating a pure invention of your own does not make it less ludicrous.
                        – Did
                        Nov 20 '15 at 0:42


















                      • Repeating a pure invention of your own does not make it less ludicrous.
                        – Did
                        Nov 20 '15 at 0:42
















                      Repeating a pure invention of your own does not make it less ludicrous.
                      – Did
                      Nov 20 '15 at 0:42




                      Repeating a pure invention of your own does not make it less ludicrous.
                      – Did
                      Nov 20 '15 at 0:42











                      0














                      What about vicious cycles? For example, suppose you have three binary random variables $A$, $B$, and $C$ that each take values $0$ or $1$. In this example, $A$ and $B$ are mutually exclusive if whenever $A = 0$, $B = 1$, and similarly for $C$.



                      $A$, $B$, and $C$ cannot all be pairwise mutually exclusive.



                      However, they can be "globally" mutually exclusive in the sense that, for example, if $A = 0$, then $B = 1$ and $C = 1$, and so on for the other variables.



                      I think that this counterexample proves that pairwise mutual exclusivity is not equivalent to mutual exclusivity.






                      share|cite|improve this answer


























                        0














                        What about vicious cycles? For example, suppose you have three binary random variables $A$, $B$, and $C$ that each take values $0$ or $1$. In this example, $A$ and $B$ are mutually exclusive if whenever $A = 0$, $B = 1$, and similarly for $C$.



                        $A$, $B$, and $C$ cannot all be pairwise mutually exclusive.



                        However, they can be "globally" mutually exclusive in the sense that, for example, if $A = 0$, then $B = 1$ and $C = 1$, and so on for the other variables.



                        I think that this counterexample proves that pairwise mutual exclusivity is not equivalent to mutual exclusivity.






                        share|cite|improve this answer
























                          0












                          0








                          0






                          What about vicious cycles? For example, suppose you have three binary random variables $A$, $B$, and $C$ that each take values $0$ or $1$. In this example, $A$ and $B$ are mutually exclusive if whenever $A = 0$, $B = 1$, and similarly for $C$.



                          $A$, $B$, and $C$ cannot all be pairwise mutually exclusive.



                          However, they can be "globally" mutually exclusive in the sense that, for example, if $A = 0$, then $B = 1$ and $C = 1$, and so on for the other variables.



                          I think that this counterexample proves that pairwise mutual exclusivity is not equivalent to mutual exclusivity.






                          share|cite|improve this answer












                          What about vicious cycles? For example, suppose you have three binary random variables $A$, $B$, and $C$ that each take values $0$ or $1$. In this example, $A$ and $B$ are mutually exclusive if whenever $A = 0$, $B = 1$, and similarly for $C$.



                          $A$, $B$, and $C$ cannot all be pairwise mutually exclusive.



                          However, they can be "globally" mutually exclusive in the sense that, for example, if $A = 0$, then $B = 1$ and $C = 1$, and so on for the other variables.



                          I think that this counterexample proves that pairwise mutual exclusivity is not equivalent to mutual exclusivity.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Aug 22 '17 at 9:14









                          gideonite

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