Take A1,A2,··· ,An a partition of Ω with∪n i=1Ai = Ω and Ai∩Aj =∅for i , j. Assume P(Ai) > 0...












-1














Please help me to solve this matter:
Take A1,A2,··· ,An a partition of Ω with∪n i=1Ai = Ω and Ai∩Aj =∅for i , j. Assume P(Ai) > 0 for all i. Prove the Bayes formula,
P(Ai|B) =
P(B|Ai)P(Ai)P n j=1 P(B|Aj)P(Aj)










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put on hold as off-topic by Did, Paul Frost, Cesareo, mrtaurho, amWhy Jan 4 at 14:11


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, Paul Frost, Cesareo, mrtaurho, amWhy

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    -1














    Please help me to solve this matter:
    Take A1,A2,··· ,An a partition of Ω with∪n i=1Ai = Ω and Ai∩Aj =∅for i , j. Assume P(Ai) > 0 for all i. Prove the Bayes formula,
    P(Ai|B) =
    P(B|Ai)P(Ai)P n j=1 P(B|Aj)P(Aj)










    share|cite|improve this question







    New contributor




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











    put on hold as off-topic by Did, Paul Frost, Cesareo, mrtaurho, amWhy Jan 4 at 14:11


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, Paul Frost, Cesareo, mrtaurho, amWhy

    If this question can be reworded to fit the rules in the help center, please edit the question.
















      -1












      -1








      -1







      Please help me to solve this matter:
      Take A1,A2,··· ,An a partition of Ω with∪n i=1Ai = Ω and Ai∩Aj =∅for i , j. Assume P(Ai) > 0 for all i. Prove the Bayes formula,
      P(Ai|B) =
      P(B|Ai)P(Ai)P n j=1 P(B|Aj)P(Aj)










      share|cite|improve this question







      New contributor




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











      Please help me to solve this matter:
      Take A1,A2,··· ,An a partition of Ω with∪n i=1Ai = Ω and Ai∩Aj =∅for i , j. Assume P(Ai) > 0 for all i. Prove the Bayes formula,
      P(Ai|B) =
      P(B|Ai)P(Ai)P n j=1 P(B|Aj)P(Aj)







      bayesian






      share|cite|improve this question







      New contributor




      Nicolas Cloet 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




      Nicolas Cloet 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




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









      asked Jan 4 at 1:07









      Nicolas Cloet

      11




      11




      New contributor




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





      New contributor





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






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




      put on hold as off-topic by Did, Paul Frost, Cesareo, mrtaurho, amWhy Jan 4 at 14:11


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, Paul Frost, Cesareo, mrtaurho, amWhy

      If this question can be reworded to fit the rules in the help center, please edit the question.




      put on hold as off-topic by Did, Paul Frost, Cesareo, mrtaurho, amWhy Jan 4 at 14:11


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, Paul Frost, Cesareo, mrtaurho, amWhy

      If this question can be reworded to fit the rules in the help center, please edit the question.






















          1 Answer
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          0














          $P(A_i|B)=frac{P(A_icap B)}{P(B)}=frac{P(A_icap B)}{sum_{j=1}^nP(A_jcap B)}=frac{P(B|A_i)P(A_i)}{sum_{j=1}^nP(B|A_j)P(A_j)}$






          share|cite|improve this answer





















          • Could anyone confirm?
            – Nicolas Cloet
            Jan 4 at 1:25










          • @NicolasCloet Cannot you confirm or infirm yourself?
            – Did
            Jan 4 at 8:22


















          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          0














          $P(A_i|B)=frac{P(A_icap B)}{P(B)}=frac{P(A_icap B)}{sum_{j=1}^nP(A_jcap B)}=frac{P(B|A_i)P(A_i)}{sum_{j=1}^nP(B|A_j)P(A_j)}$






          share|cite|improve this answer





















          • Could anyone confirm?
            – Nicolas Cloet
            Jan 4 at 1:25










          • @NicolasCloet Cannot you confirm or infirm yourself?
            – Did
            Jan 4 at 8:22
















          0














          $P(A_i|B)=frac{P(A_icap B)}{P(B)}=frac{P(A_icap B)}{sum_{j=1}^nP(A_jcap B)}=frac{P(B|A_i)P(A_i)}{sum_{j=1}^nP(B|A_j)P(A_j)}$






          share|cite|improve this answer





















          • Could anyone confirm?
            – Nicolas Cloet
            Jan 4 at 1:25










          • @NicolasCloet Cannot you confirm or infirm yourself?
            – Did
            Jan 4 at 8:22














          0












          0








          0






          $P(A_i|B)=frac{P(A_icap B)}{P(B)}=frac{P(A_icap B)}{sum_{j=1}^nP(A_jcap B)}=frac{P(B|A_i)P(A_i)}{sum_{j=1}^nP(B|A_j)P(A_j)}$






          share|cite|improve this answer












          $P(A_i|B)=frac{P(A_icap B)}{P(B)}=frac{P(A_icap B)}{sum_{j=1}^nP(A_jcap B)}=frac{P(B|A_i)P(A_i)}{sum_{j=1}^nP(B|A_j)P(A_j)}$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 4 at 1:11









          John_Wick

          1,436111




          1,436111












          • Could anyone confirm?
            – Nicolas Cloet
            Jan 4 at 1:25










          • @NicolasCloet Cannot you confirm or infirm yourself?
            – Did
            Jan 4 at 8:22


















          • Could anyone confirm?
            – Nicolas Cloet
            Jan 4 at 1:25










          • @NicolasCloet Cannot you confirm or infirm yourself?
            – Did
            Jan 4 at 8:22
















          Could anyone confirm?
          – Nicolas Cloet
          Jan 4 at 1:25




          Could anyone confirm?
          – Nicolas Cloet
          Jan 4 at 1:25












          @NicolasCloet Cannot you confirm or infirm yourself?
          – Did
          Jan 4 at 8:22




          @NicolasCloet Cannot you confirm or infirm yourself?
          – Did
          Jan 4 at 8:22



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