conditional probability - conditioned twice












1












$begingroup$


Let $p(cdot)$ be a discreet probability function and $A,B,C$ be events. What does $p(A|B|C)$ mean? Is this the same as $p(A|B,C)$ or is it:



if we treat $D=A|B$ as another event, $p(D|C) = frac{p(Dcap C)}{p(C)}$?



How would $p(D cap C)$ be computed then with $D$ defined earlier?






probability







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Where did you see this? This is the first time I am seeing that double conditioning notation.
    $endgroup$
    – Ufuk Can Bicici
    Sep 11 '14 at 13:27






  • 1




    $begingroup$
    The notation p(A|B|C) does not exist, except in a couple of math.se questions asking about it.
    $endgroup$
    – Did
    Oct 6 '14 at 13:41
















1












$begingroup$


Let $p(cdot)$ be a discreet probability function and $A,B,C$ be events. What does $p(A|B|C)$ mean? Is this the same as $p(A|B,C)$ or is it:



if we treat $D=A|B$ as another event, $p(D|C) = frac{p(Dcap C)}{p(C)}$?



How would $p(D cap C)$ be computed then with $D$ defined earlier?






probability







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Where did you see this? This is the first time I am seeing that double conditioning notation.
    $endgroup$
    – Ufuk Can Bicici
    Sep 11 '14 at 13:27






  • 1




    $begingroup$
    The notation p(A|B|C) does not exist, except in a couple of math.se questions asking about it.
    $endgroup$
    – Did
    Oct 6 '14 at 13:41














1












1








1





$begingroup$


Let $p(cdot)$ be a discreet probability function and $A,B,C$ be events. What does $p(A|B|C)$ mean? Is this the same as $p(A|B,C)$ or is it:



if we treat $D=A|B$ as another event, $p(D|C) = frac{p(Dcap C)}{p(C)}$?



How would $p(D cap C)$ be computed then with $D$ defined earlier?






probability







share|cite|improve this question









$endgroup$




Let $p(cdot)$ be a discreet probability function and $A,B,C$ be events. What does $p(A|B|C)$ mean? Is this the same as $p(A|B,C)$ or is it:



if we treat $D=A|B$ as another event, $p(D|C) = frac{p(Dcap C)}{p(C)}$?



How would $p(D cap C)$ be computed then with $D$ defined earlier?






probability




probability






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Sep 11 '14 at 13:22









user157279user157279

547




547








  • 1




    $begingroup$
    Where did you see this? This is the first time I am seeing that double conditioning notation.
    $endgroup$
    – Ufuk Can Bicici
    Sep 11 '14 at 13:27






  • 1




    $begingroup$
    The notation p(A|B|C) does not exist, except in a couple of math.se questions asking about it.
    $endgroup$
    – Did
    Oct 6 '14 at 13:41














  • 1




    $begingroup$
    Where did you see this? This is the first time I am seeing that double conditioning notation.
    $endgroup$
    – Ufuk Can Bicici
    Sep 11 '14 at 13:27






  • 1




    $begingroup$
    The notation p(A|B|C) does not exist, except in a couple of math.se questions asking about it.
    $endgroup$
    – Did
    Oct 6 '14 at 13:41








1




1




$begingroup$
Where did you see this? This is the first time I am seeing that double conditioning notation.
$endgroup$
– Ufuk Can Bicici
Sep 11 '14 at 13:27




$begingroup$
Where did you see this? This is the first time I am seeing that double conditioning notation.
$endgroup$
– Ufuk Can Bicici
Sep 11 '14 at 13:27




1




1




$begingroup$
The notation p(A|B|C) does not exist, except in a couple of math.se questions asking about it.
$endgroup$
– Did
Oct 6 '14 at 13:41




$begingroup$
The notation p(A|B|C) does not exist, except in a couple of math.se questions asking about it.
$endgroup$
– Did
Oct 6 '14 at 13:41










3 Answers
3






active

oldest

votes


















0












$begingroup$

I think $$frac{Pleft(Acap Bmid Cright)}{Pleft(Bmid Cright)}=frac{Pleft(Acap Bcap Cright)}{Pleft(Cright)}/frac{Pleft(Bcap Cright)}{Pleft(Cright)}=frac{Pleft(Acap Bcap Cright)}{Pleft(Bcap Cright)}=Pleft(Amid Bcap Cright)$$
has the best 'chances' here :).






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Sum Rule:



    $$p(a) = sum_{b}p(a,b)$$



    As the sum rule holds for any probability distribution it trivially holds for the distribution a|c:



    $$p(a|c) = sum_{b}p(b, a|c)$$



    For the product rule:



    $$p(a,b)=p(a|b)p(b)$$ but does $$p(a,b|c)=p(a|b|c)p(b|c)$$?



    Well presumably, they are equal if $$p(a|b|c) = p(a|b,c)$$ - but does this equality hold?



    Yes: Aim to prove: $$p(a|b|c) = p(a|b,c)$$



    $$p(a,b|c) = p(a|b|c)p(b|c)$$



    $$p(a|b|c) = dfrac{p(a,b|c)}{p(b|c)} = dfrac{ dfrac{p(a,b,c)}{p(c)}} {dfrac{p(b,c)}{p(c)}} = dfrac{p(a,b,c)}{p(b,c)} = dfrac{p(a|b,c)p(b,c)}{p(b,c)} = p(a|b,c)$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      For others' benefit, I note that this answer (and the topic) is very similar to this other answer and topic.
      $endgroup$
      – davidlowryduda♦
      Jan 23 at 21:21



















    -1












    $begingroup$

    Yes $mathbb{P}(A|B|C) = mathbb{P}(A|B,C)$



    Since
    $$mathbb{P}(A,B)=mathbb{P}(A|B)cdotmathbb{P}(B)$$
    Remember that $mathbb{P}(cdot |C)$ also is a probability in the same space,
    $$mathbb{P}(A,B|C)=mathbb{P}(A|B|C)cdotmathbb{P}(B|C)$$
    Then
    $$mathbb{P}(A|B|C) = frac{mathbb{P}(A,B|C)}{mathbb{P}(B|C)}=frac{mathbb{P}(A,B,C)}{mathbb{P}(B,C)}=mathbb{P}(A|B,C)$$






    share|cite|improve this answer









    $endgroup$













      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      I think $$frac{Pleft(Acap Bmid Cright)}{Pleft(Bmid Cright)}=frac{Pleft(Acap Bcap Cright)}{Pleft(Cright)}/frac{Pleft(Bcap Cright)}{Pleft(Cright)}=frac{Pleft(Acap Bcap Cright)}{Pleft(Bcap Cright)}=Pleft(Amid Bcap Cright)$$
      has the best 'chances' here :).






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        I think $$frac{Pleft(Acap Bmid Cright)}{Pleft(Bmid Cright)}=frac{Pleft(Acap Bcap Cright)}{Pleft(Cright)}/frac{Pleft(Bcap Cright)}{Pleft(Cright)}=frac{Pleft(Acap Bcap Cright)}{Pleft(Bcap Cright)}=Pleft(Amid Bcap Cright)$$
        has the best 'chances' here :).






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          I think $$frac{Pleft(Acap Bmid Cright)}{Pleft(Bmid Cright)}=frac{Pleft(Acap Bcap Cright)}{Pleft(Cright)}/frac{Pleft(Bcap Cright)}{Pleft(Cright)}=frac{Pleft(Acap Bcap Cright)}{Pleft(Bcap Cright)}=Pleft(Amid Bcap Cright)$$
          has the best 'chances' here :).






          share|cite|improve this answer









          $endgroup$



          I think $$frac{Pleft(Acap Bmid Cright)}{Pleft(Bmid Cright)}=frac{Pleft(Acap Bcap Cright)}{Pleft(Cright)}/frac{Pleft(Bcap Cright)}{Pleft(Cright)}=frac{Pleft(Acap Bcap Cright)}{Pleft(Bcap Cright)}=Pleft(Amid Bcap Cright)$$
          has the best 'chances' here :).







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Sep 11 '14 at 13:59









          drhabdrhab

          103k545136




          103k545136























              0












              $begingroup$

              Sum Rule:



              $$p(a) = sum_{b}p(a,b)$$



              As the sum rule holds for any probability distribution it trivially holds for the distribution a|c:



              $$p(a|c) = sum_{b}p(b, a|c)$$



              For the product rule:



              $$p(a,b)=p(a|b)p(b)$$ but does $$p(a,b|c)=p(a|b|c)p(b|c)$$?



              Well presumably, they are equal if $$p(a|b|c) = p(a|b,c)$$ - but does this equality hold?



              Yes: Aim to prove: $$p(a|b|c) = p(a|b,c)$$



              $$p(a,b|c) = p(a|b|c)p(b|c)$$



              $$p(a|b|c) = dfrac{p(a,b|c)}{p(b|c)} = dfrac{ dfrac{p(a,b,c)}{p(c)}} {dfrac{p(b,c)}{p(c)}} = dfrac{p(a,b,c)}{p(b,c)} = dfrac{p(a|b,c)p(b,c)}{p(b,c)} = p(a|b,c)$$






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                For others' benefit, I note that this answer (and the topic) is very similar to this other answer and topic.
                $endgroup$
                – davidlowryduda♦
                Jan 23 at 21:21
















              0












              $begingroup$

              Sum Rule:



              $$p(a) = sum_{b}p(a,b)$$



              As the sum rule holds for any probability distribution it trivially holds for the distribution a|c:



              $$p(a|c) = sum_{b}p(b, a|c)$$



              For the product rule:



              $$p(a,b)=p(a|b)p(b)$$ but does $$p(a,b|c)=p(a|b|c)p(b|c)$$?



              Well presumably, they are equal if $$p(a|b|c) = p(a|b,c)$$ - but does this equality hold?



              Yes: Aim to prove: $$p(a|b|c) = p(a|b,c)$$



              $$p(a,b|c) = p(a|b|c)p(b|c)$$



              $$p(a|b|c) = dfrac{p(a,b|c)}{p(b|c)} = dfrac{ dfrac{p(a,b,c)}{p(c)}} {dfrac{p(b,c)}{p(c)}} = dfrac{p(a,b,c)}{p(b,c)} = dfrac{p(a|b,c)p(b,c)}{p(b,c)} = p(a|b,c)$$






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                For others' benefit, I note that this answer (and the topic) is very similar to this other answer and topic.
                $endgroup$
                – davidlowryduda♦
                Jan 23 at 21:21














              0












              0








              0





              $begingroup$

              Sum Rule:



              $$p(a) = sum_{b}p(a,b)$$



              As the sum rule holds for any probability distribution it trivially holds for the distribution a|c:



              $$p(a|c) = sum_{b}p(b, a|c)$$



              For the product rule:



              $$p(a,b)=p(a|b)p(b)$$ but does $$p(a,b|c)=p(a|b|c)p(b|c)$$?



              Well presumably, they are equal if $$p(a|b|c) = p(a|b,c)$$ - but does this equality hold?



              Yes: Aim to prove: $$p(a|b|c) = p(a|b,c)$$



              $$p(a,b|c) = p(a|b|c)p(b|c)$$



              $$p(a|b|c) = dfrac{p(a,b|c)}{p(b|c)} = dfrac{ dfrac{p(a,b,c)}{p(c)}} {dfrac{p(b,c)}{p(c)}} = dfrac{p(a,b,c)}{p(b,c)} = dfrac{p(a|b,c)p(b,c)}{p(b,c)} = p(a|b,c)$$






              share|cite|improve this answer









              $endgroup$



              Sum Rule:



              $$p(a) = sum_{b}p(a,b)$$



              As the sum rule holds for any probability distribution it trivially holds for the distribution a|c:



              $$p(a|c) = sum_{b}p(b, a|c)$$



              For the product rule:



              $$p(a,b)=p(a|b)p(b)$$ but does $$p(a,b|c)=p(a|b|c)p(b|c)$$?



              Well presumably, they are equal if $$p(a|b|c) = p(a|b,c)$$ - but does this equality hold?



              Yes: Aim to prove: $$p(a|b|c) = p(a|b,c)$$



              $$p(a,b|c) = p(a|b|c)p(b|c)$$



              $$p(a|b|c) = dfrac{p(a,b|c)}{p(b|c)} = dfrac{ dfrac{p(a,b,c)}{p(c)}} {dfrac{p(b,c)}{p(c)}} = dfrac{p(a,b,c)}{p(b,c)} = dfrac{p(a|b,c)p(b,c)}{p(b,c)} = p(a|b,c)$$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Jan 23 at 20:23









              Oliver GoldsteinOliver Goldstein

              111




              111












              • $begingroup$
                For others' benefit, I note that this answer (and the topic) is very similar to this other answer and topic.
                $endgroup$
                – davidlowryduda♦
                Jan 23 at 21:21


















              • $begingroup$
                For others' benefit, I note that this answer (and the topic) is very similar to this other answer and topic.
                $endgroup$
                – davidlowryduda♦
                Jan 23 at 21:21
















              $begingroup$
              For others' benefit, I note that this answer (and the topic) is very similar to this other answer and topic.
              $endgroup$
              – davidlowryduda♦
              Jan 23 at 21:21




              $begingroup$
              For others' benefit, I note that this answer (and the topic) is very similar to this other answer and topic.
              $endgroup$
              – davidlowryduda♦
              Jan 23 at 21:21











              -1












              $begingroup$

              Yes $mathbb{P}(A|B|C) = mathbb{P}(A|B,C)$



              Since
              $$mathbb{P}(A,B)=mathbb{P}(A|B)cdotmathbb{P}(B)$$
              Remember that $mathbb{P}(cdot |C)$ also is a probability in the same space,
              $$mathbb{P}(A,B|C)=mathbb{P}(A|B|C)cdotmathbb{P}(B|C)$$
              Then
              $$mathbb{P}(A|B|C) = frac{mathbb{P}(A,B|C)}{mathbb{P}(B|C)}=frac{mathbb{P}(A,B,C)}{mathbb{P}(B,C)}=mathbb{P}(A|B,C)$$






              share|cite|improve this answer









              $endgroup$


















                -1












                $begingroup$

                Yes $mathbb{P}(A|B|C) = mathbb{P}(A|B,C)$



                Since
                $$mathbb{P}(A,B)=mathbb{P}(A|B)cdotmathbb{P}(B)$$
                Remember that $mathbb{P}(cdot |C)$ also is a probability in the same space,
                $$mathbb{P}(A,B|C)=mathbb{P}(A|B|C)cdotmathbb{P}(B|C)$$
                Then
                $$mathbb{P}(A|B|C) = frac{mathbb{P}(A,B|C)}{mathbb{P}(B|C)}=frac{mathbb{P}(A,B,C)}{mathbb{P}(B,C)}=mathbb{P}(A|B,C)$$






                share|cite|improve this answer









                $endgroup$
















                  -1












                  -1








                  -1





                  $begingroup$

                  Yes $mathbb{P}(A|B|C) = mathbb{P}(A|B,C)$



                  Since
                  $$mathbb{P}(A,B)=mathbb{P}(A|B)cdotmathbb{P}(B)$$
                  Remember that $mathbb{P}(cdot |C)$ also is a probability in the same space,
                  $$mathbb{P}(A,B|C)=mathbb{P}(A|B|C)cdotmathbb{P}(B|C)$$
                  Then
                  $$mathbb{P}(A|B|C) = frac{mathbb{P}(A,B|C)}{mathbb{P}(B|C)}=frac{mathbb{P}(A,B,C)}{mathbb{P}(B,C)}=mathbb{P}(A|B,C)$$






                  share|cite|improve this answer









                  $endgroup$



                  Yes $mathbb{P}(A|B|C) = mathbb{P}(A|B,C)$



                  Since
                  $$mathbb{P}(A,B)=mathbb{P}(A|B)cdotmathbb{P}(B)$$
                  Remember that $mathbb{P}(cdot |C)$ also is a probability in the same space,
                  $$mathbb{P}(A,B|C)=mathbb{P}(A|B|C)cdotmathbb{P}(B|C)$$
                  Then
                  $$mathbb{P}(A|B|C) = frac{mathbb{P}(A,B|C)}{mathbb{P}(B|C)}=frac{mathbb{P}(A,B,C)}{mathbb{P}(B,C)}=mathbb{P}(A|B,C)$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered May 9 '18 at 6:39









                  Jhon Kevin Astoquillca AguilarJhon Kevin Astoquillca Aguilar

                  897




                  897






























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