What is the intuition or proof behind the conditional Bayes' theorem?












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$begingroup$


In the book "Probability and statistics" by Morris H. DeGroot and Mark J. Schervish, on page 80, the conditional version of Bayes' theorem is given with no explanation:




$$Pr(B_imid A cap C) = dfrac{Pr(B_imid C) cdot Pr(Amid B_i cap C)}{sum (Pr(B_imid C)cdot Pr(Amid B_i cap C))}$$




What is the intuition or mathematical proof behind this?










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$endgroup$

















    3












    $begingroup$


    In the book "Probability and statistics" by Morris H. DeGroot and Mark J. Schervish, on page 80, the conditional version of Bayes' theorem is given with no explanation:




    $$Pr(B_imid A cap C) = dfrac{Pr(B_imid C) cdot Pr(Amid B_i cap C)}{sum (Pr(B_imid C)cdot Pr(Amid B_i cap C))}$$




    What is the intuition or mathematical proof behind this?










    share|cite|improve this question











    $endgroup$















      3












      3








      3


      3



      $begingroup$


      In the book "Probability and statistics" by Morris H. DeGroot and Mark J. Schervish, on page 80, the conditional version of Bayes' theorem is given with no explanation:




      $$Pr(B_imid A cap C) = dfrac{Pr(B_imid C) cdot Pr(Amid B_i cap C)}{sum (Pr(B_imid C)cdot Pr(Amid B_i cap C))}$$




      What is the intuition or mathematical proof behind this?










      share|cite|improve this question











      $endgroup$




      In the book "Probability and statistics" by Morris H. DeGroot and Mark J. Schervish, on page 80, the conditional version of Bayes' theorem is given with no explanation:




      $$Pr(B_imid A cap C) = dfrac{Pr(B_imid C) cdot Pr(Amid B_i cap C)}{sum (Pr(B_imid C)cdot Pr(Amid B_i cap C))}$$




      What is the intuition or mathematical proof behind this?







      statistics proof-writing bayes-theorem






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      share|cite|improve this question













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      edited Jan 19 at 1:25









      nbro

      2,48863475




      2,48863475










      asked Aug 11 '14 at 23:26









      ZhuluZhulu

      3711612




      3711612






















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          $begingroup$

          It's directly equivalent to:
          $$begin{align}
          Pr(B_imid A) & = frac{Pr(B_i cap A)}{Pr(A)}
          \[1ex] &= frac{Pr(B_i)cdot Pr(Amid B_i)}{sum_k Pr(B_k)cdotPr(Amid B_k)}
          end{align}$$
          But with conditioning when given $C$:
          $$begin{align}
          Pr(B_imid Acap C) & = frac{Pr(B_i cap Amid C)}{Pr(Amid C)}
          \[1ex] &= frac{Pr(B_i mid C)cdotPr(Amid B_icap C)}{sum_k Pr(B_kmid C)cdotPr(Amid B_kcap C)}
          end{align}$$




          How do you know equation 3 is true?




          $begin{align}
          Pr(B_imid Acap C) & = frac{Pr(B_icap A cap C)}{Pr(Acap C)} & text{by definition}
          \[1ex] & = frac{Pr(B_icap Amid C)cdot Pr(C)}{Pr(Amid C)cdotPr(C)} & text{by the same}
          \[1ex] & = frac{Pr(B_icap Amid C)}{Pr(Amid C)} & text{by cancelation}
          end{align}$




          Also, can you explain how you derived the denominator in equation 4?




          By the Law of Total Probability. If the set of ${B_k: kin {1..n}}$ partitions the sample space then the measure of event $A$ is the sum of the products of the measure of the partition and the measure of the event conditional on the partition.



          $$begin{align}Pr(A) quad & = Pr(B_1)cdotPr(Amid B_1) + cdots + Pr(B_k)cdotPr(Amid B_k) + cdots Pr(B_n)cdotPr(Amid B_n)
          \ & = sum_{k=1}^n Pr(B_k)cdotPr(Amid B_k)
          end{align}$$




          I understand everything you said except the actual steps transforming equation three to equation four. Are you treating Pr(A|C) as sum(Pr(A|C|Bk)Pr(Bk)).




          Yes. Thuswise:
          $$begin{align}
          Pr(Amid C) quad & = frac{Pr(Acap C)}{Pr(C)} & text{by conditional probability}
          \[1ex] & = frac{sum_k Pr(Acap Ccap B_k)}{Pr(C)} & text{by total probability}
          \[1ex] & = frac{sum_k Pr(B_kcap C)cdotPr(Amid Ccap B_k)}{Pr(C)} & text{by conditional probability}
          \[1ex] & = sum_k frac{Pr(B_kcap C)}{Pr(C)}cdotPr(Amid Ccap B_k) & text{by rearrangement}
          \[1ex] & = sum_k Pr(B_kmid C)cdotPr(Amid Ccap B_k) & text{by conditional probability}
          end{align}$$






          share|cite|improve this answer











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            3












            $begingroup$

            It's directly equivalent to:
            $$begin{align}
            Pr(B_imid A) & = frac{Pr(B_i cap A)}{Pr(A)}
            \[1ex] &= frac{Pr(B_i)cdot Pr(Amid B_i)}{sum_k Pr(B_k)cdotPr(Amid B_k)}
            end{align}$$
            But with conditioning when given $C$:
            $$begin{align}
            Pr(B_imid Acap C) & = frac{Pr(B_i cap Amid C)}{Pr(Amid C)}
            \[1ex] &= frac{Pr(B_i mid C)cdotPr(Amid B_icap C)}{sum_k Pr(B_kmid C)cdotPr(Amid B_kcap C)}
            end{align}$$




            How do you know equation 3 is true?




            $begin{align}
            Pr(B_imid Acap C) & = frac{Pr(B_icap A cap C)}{Pr(Acap C)} & text{by definition}
            \[1ex] & = frac{Pr(B_icap Amid C)cdot Pr(C)}{Pr(Amid C)cdotPr(C)} & text{by the same}
            \[1ex] & = frac{Pr(B_icap Amid C)}{Pr(Amid C)} & text{by cancelation}
            end{align}$




            Also, can you explain how you derived the denominator in equation 4?




            By the Law of Total Probability. If the set of ${B_k: kin {1..n}}$ partitions the sample space then the measure of event $A$ is the sum of the products of the measure of the partition and the measure of the event conditional on the partition.



            $$begin{align}Pr(A) quad & = Pr(B_1)cdotPr(Amid B_1) + cdots + Pr(B_k)cdotPr(Amid B_k) + cdots Pr(B_n)cdotPr(Amid B_n)
            \ & = sum_{k=1}^n Pr(B_k)cdotPr(Amid B_k)
            end{align}$$




            I understand everything you said except the actual steps transforming equation three to equation four. Are you treating Pr(A|C) as sum(Pr(A|C|Bk)Pr(Bk)).




            Yes. Thuswise:
            $$begin{align}
            Pr(Amid C) quad & = frac{Pr(Acap C)}{Pr(C)} & text{by conditional probability}
            \[1ex] & = frac{sum_k Pr(Acap Ccap B_k)}{Pr(C)} & text{by total probability}
            \[1ex] & = frac{sum_k Pr(B_kcap C)cdotPr(Amid Ccap B_k)}{Pr(C)} & text{by conditional probability}
            \[1ex] & = sum_k frac{Pr(B_kcap C)}{Pr(C)}cdotPr(Amid Ccap B_k) & text{by rearrangement}
            \[1ex] & = sum_k Pr(B_kmid C)cdotPr(Amid Ccap B_k) & text{by conditional probability}
            end{align}$$






            share|cite|improve this answer











            $endgroup$


















              3












              $begingroup$

              It's directly equivalent to:
              $$begin{align}
              Pr(B_imid A) & = frac{Pr(B_i cap A)}{Pr(A)}
              \[1ex] &= frac{Pr(B_i)cdot Pr(Amid B_i)}{sum_k Pr(B_k)cdotPr(Amid B_k)}
              end{align}$$
              But with conditioning when given $C$:
              $$begin{align}
              Pr(B_imid Acap C) & = frac{Pr(B_i cap Amid C)}{Pr(Amid C)}
              \[1ex] &= frac{Pr(B_i mid C)cdotPr(Amid B_icap C)}{sum_k Pr(B_kmid C)cdotPr(Amid B_kcap C)}
              end{align}$$




              How do you know equation 3 is true?




              $begin{align}
              Pr(B_imid Acap C) & = frac{Pr(B_icap A cap C)}{Pr(Acap C)} & text{by definition}
              \[1ex] & = frac{Pr(B_icap Amid C)cdot Pr(C)}{Pr(Amid C)cdotPr(C)} & text{by the same}
              \[1ex] & = frac{Pr(B_icap Amid C)}{Pr(Amid C)} & text{by cancelation}
              end{align}$




              Also, can you explain how you derived the denominator in equation 4?




              By the Law of Total Probability. If the set of ${B_k: kin {1..n}}$ partitions the sample space then the measure of event $A$ is the sum of the products of the measure of the partition and the measure of the event conditional on the partition.



              $$begin{align}Pr(A) quad & = Pr(B_1)cdotPr(Amid B_1) + cdots + Pr(B_k)cdotPr(Amid B_k) + cdots Pr(B_n)cdotPr(Amid B_n)
              \ & = sum_{k=1}^n Pr(B_k)cdotPr(Amid B_k)
              end{align}$$




              I understand everything you said except the actual steps transforming equation three to equation four. Are you treating Pr(A|C) as sum(Pr(A|C|Bk)Pr(Bk)).




              Yes. Thuswise:
              $$begin{align}
              Pr(Amid C) quad & = frac{Pr(Acap C)}{Pr(C)} & text{by conditional probability}
              \[1ex] & = frac{sum_k Pr(Acap Ccap B_k)}{Pr(C)} & text{by total probability}
              \[1ex] & = frac{sum_k Pr(B_kcap C)cdotPr(Amid Ccap B_k)}{Pr(C)} & text{by conditional probability}
              \[1ex] & = sum_k frac{Pr(B_kcap C)}{Pr(C)}cdotPr(Amid Ccap B_k) & text{by rearrangement}
              \[1ex] & = sum_k Pr(B_kmid C)cdotPr(Amid Ccap B_k) & text{by conditional probability}
              end{align}$$






              share|cite|improve this answer











              $endgroup$
















                3












                3








                3





                $begingroup$

                It's directly equivalent to:
                $$begin{align}
                Pr(B_imid A) & = frac{Pr(B_i cap A)}{Pr(A)}
                \[1ex] &= frac{Pr(B_i)cdot Pr(Amid B_i)}{sum_k Pr(B_k)cdotPr(Amid B_k)}
                end{align}$$
                But with conditioning when given $C$:
                $$begin{align}
                Pr(B_imid Acap C) & = frac{Pr(B_i cap Amid C)}{Pr(Amid C)}
                \[1ex] &= frac{Pr(B_i mid C)cdotPr(Amid B_icap C)}{sum_k Pr(B_kmid C)cdotPr(Amid B_kcap C)}
                end{align}$$




                How do you know equation 3 is true?




                $begin{align}
                Pr(B_imid Acap C) & = frac{Pr(B_icap A cap C)}{Pr(Acap C)} & text{by definition}
                \[1ex] & = frac{Pr(B_icap Amid C)cdot Pr(C)}{Pr(Amid C)cdotPr(C)} & text{by the same}
                \[1ex] & = frac{Pr(B_icap Amid C)}{Pr(Amid C)} & text{by cancelation}
                end{align}$




                Also, can you explain how you derived the denominator in equation 4?




                By the Law of Total Probability. If the set of ${B_k: kin {1..n}}$ partitions the sample space then the measure of event $A$ is the sum of the products of the measure of the partition and the measure of the event conditional on the partition.



                $$begin{align}Pr(A) quad & = Pr(B_1)cdotPr(Amid B_1) + cdots + Pr(B_k)cdotPr(Amid B_k) + cdots Pr(B_n)cdotPr(Amid B_n)
                \ & = sum_{k=1}^n Pr(B_k)cdotPr(Amid B_k)
                end{align}$$




                I understand everything you said except the actual steps transforming equation three to equation four. Are you treating Pr(A|C) as sum(Pr(A|C|Bk)Pr(Bk)).




                Yes. Thuswise:
                $$begin{align}
                Pr(Amid C) quad & = frac{Pr(Acap C)}{Pr(C)} & text{by conditional probability}
                \[1ex] & = frac{sum_k Pr(Acap Ccap B_k)}{Pr(C)} & text{by total probability}
                \[1ex] & = frac{sum_k Pr(B_kcap C)cdotPr(Amid Ccap B_k)}{Pr(C)} & text{by conditional probability}
                \[1ex] & = sum_k frac{Pr(B_kcap C)}{Pr(C)}cdotPr(Amid Ccap B_k) & text{by rearrangement}
                \[1ex] & = sum_k Pr(B_kmid C)cdotPr(Amid Ccap B_k) & text{by conditional probability}
                end{align}$$






                share|cite|improve this answer











                $endgroup$



                It's directly equivalent to:
                $$begin{align}
                Pr(B_imid A) & = frac{Pr(B_i cap A)}{Pr(A)}
                \[1ex] &= frac{Pr(B_i)cdot Pr(Amid B_i)}{sum_k Pr(B_k)cdotPr(Amid B_k)}
                end{align}$$
                But with conditioning when given $C$:
                $$begin{align}
                Pr(B_imid Acap C) & = frac{Pr(B_i cap Amid C)}{Pr(Amid C)}
                \[1ex] &= frac{Pr(B_i mid C)cdotPr(Amid B_icap C)}{sum_k Pr(B_kmid C)cdotPr(Amid B_kcap C)}
                end{align}$$




                How do you know equation 3 is true?




                $begin{align}
                Pr(B_imid Acap C) & = frac{Pr(B_icap A cap C)}{Pr(Acap C)} & text{by definition}
                \[1ex] & = frac{Pr(B_icap Amid C)cdot Pr(C)}{Pr(Amid C)cdotPr(C)} & text{by the same}
                \[1ex] & = frac{Pr(B_icap Amid C)}{Pr(Amid C)} & text{by cancelation}
                end{align}$




                Also, can you explain how you derived the denominator in equation 4?




                By the Law of Total Probability. If the set of ${B_k: kin {1..n}}$ partitions the sample space then the measure of event $A$ is the sum of the products of the measure of the partition and the measure of the event conditional on the partition.



                $$begin{align}Pr(A) quad & = Pr(B_1)cdotPr(Amid B_1) + cdots + Pr(B_k)cdotPr(Amid B_k) + cdots Pr(B_n)cdotPr(Amid B_n)
                \ & = sum_{k=1}^n Pr(B_k)cdotPr(Amid B_k)
                end{align}$$




                I understand everything you said except the actual steps transforming equation three to equation four. Are you treating Pr(A|C) as sum(Pr(A|C|Bk)Pr(Bk)).




                Yes. Thuswise:
                $$begin{align}
                Pr(Amid C) quad & = frac{Pr(Acap C)}{Pr(C)} & text{by conditional probability}
                \[1ex] & = frac{sum_k Pr(Acap Ccap B_k)}{Pr(C)} & text{by total probability}
                \[1ex] & = frac{sum_k Pr(B_kcap C)cdotPr(Amid Ccap B_k)}{Pr(C)} & text{by conditional probability}
                \[1ex] & = sum_k frac{Pr(B_kcap C)}{Pr(C)}cdotPr(Amid Ccap B_k) & text{by rearrangement}
                \[1ex] & = sum_k Pr(B_kmid C)cdotPr(Amid Ccap B_k) & text{by conditional probability}
                end{align}$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Aug 12 '14 at 2:43

























                answered Aug 11 '14 at 23:57









                Graham KempGraham Kemp

                88.1k43579




                88.1k43579






























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