Is $P(X|A,B)=frac{P(A|X)P(B|X)P(X)}{P(A)P(B)}$ when $A$ and $B$ are conditionally indepdent?












2














I am trying to evaluate $P(X|A,B)$ given that $A$ and $B$ are conditionally independent given $X$. I know that this can be expressed as,



$$P(X|A,B)=frac{P(A,B,X)}{P(A,B)}$$



Using the conditional independence of $A$ and $B$ given $X$ this is,
$$frac{P(A|B,X)P(B,X)}{P(A)P(B)}=frac{P(A|X)P(B,X)}{P(A)P(B)}$$
Using the chain rule this is
$$frac{P(A|X)P(B|X)P(X)}{P(A)P(B)}$$



Is this correct? Does this hold with conditional independence of $A$ and $B$ given $X$, or does it also require independence alone of $A$ and $B$?






probability probability-theory bayes-theorem







share|cite|improve this question




















  • 1




    You split $P(A, B) = P(A)P(B)$ -- that is independence of $A, B$, not conditional independence.
    – Mees de Vries
    Nov 20 '18 at 14:01
















2














I am trying to evaluate $P(X|A,B)$ given that $A$ and $B$ are conditionally independent given $X$. I know that this can be expressed as,



$$P(X|A,B)=frac{P(A,B,X)}{P(A,B)}$$



Using the conditional independence of $A$ and $B$ given $X$ this is,
$$frac{P(A|B,X)P(B,X)}{P(A)P(B)}=frac{P(A|X)P(B,X)}{P(A)P(B)}$$
Using the chain rule this is
$$frac{P(A|X)P(B|X)P(X)}{P(A)P(B)}$$



Is this correct? Does this hold with conditional independence of $A$ and $B$ given $X$, or does it also require independence alone of $A$ and $B$?






probability probability-theory bayes-theorem







share|cite|improve this question




















  • 1




    You split $P(A, B) = P(A)P(B)$ -- that is independence of $A, B$, not conditional independence.
    – Mees de Vries
    Nov 20 '18 at 14:01














2












2








2







I am trying to evaluate $P(X|A,B)$ given that $A$ and $B$ are conditionally independent given $X$. I know that this can be expressed as,



$$P(X|A,B)=frac{P(A,B,X)}{P(A,B)}$$



Using the conditional independence of $A$ and $B$ given $X$ this is,
$$frac{P(A|B,X)P(B,X)}{P(A)P(B)}=frac{P(A|X)P(B,X)}{P(A)P(B)}$$
Using the chain rule this is
$$frac{P(A|X)P(B|X)P(X)}{P(A)P(B)}$$



Is this correct? Does this hold with conditional independence of $A$ and $B$ given $X$, or does it also require independence alone of $A$ and $B$?






probability probability-theory bayes-theorem







share|cite|improve this question















I am trying to evaluate $P(X|A,B)$ given that $A$ and $B$ are conditionally independent given $X$. I know that this can be expressed as,



$$P(X|A,B)=frac{P(A,B,X)}{P(A,B)}$$



Using the conditional independence of $A$ and $B$ given $X$ this is,
$$frac{P(A|B,X)P(B,X)}{P(A)P(B)}=frac{P(A|X)P(B,X)}{P(A)P(B)}$$
Using the chain rule this is
$$frac{P(A|X)P(B|X)P(X)}{P(A)P(B)}$$



Is this correct? Does this hold with conditional independence of $A$ and $B$ given $X$, or does it also require independence alone of $A$ and $B$?






probability probability-theory bayes-theorem




probability probability-theory bayes-theorem






share|cite|improve this question















share|cite|improve this question













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edited Nov 20 '18 at 13:58

























asked Nov 19 '18 at 23:59









user617643

226




226








  • 1




    You split $P(A, B) = P(A)P(B)$ -- that is independence of $A, B$, not conditional independence.
    – Mees de Vries
    Nov 20 '18 at 14:01














  • 1




    You split $P(A, B) = P(A)P(B)$ -- that is independence of $A, B$, not conditional independence.
    – Mees de Vries
    Nov 20 '18 at 14:01








1




1




You split $P(A, B) = P(A)P(B)$ -- that is independence of $A, B$, not conditional independence.
– Mees de Vries
Nov 20 '18 at 14:01




You split $P(A, B) = P(A)P(B)$ -- that is independence of $A, B$, not conditional independence.
– Mees de Vries
Nov 20 '18 at 14:01










2 Answers
2






active

oldest

votes


















0














You could save some steps using Bayes' formula:
$$
P(X mid A, B) = frac{P(A, B mid X)P(X)}{P(A, B)}
$$
Using the conditional independence, this is
$$
frac{P(A mid X) P(B mid X) P(X)}{P(A, B)}
$$
But as has been pointed out, you cannot factor the denominator without assuming independence of $A$ and $B$






share|cite|improve this answer





















  • Thanks! I would upvote your answer but due to my low reputation the system does not permit me.
    – user617643
    Nov 20 '18 at 14:29



















0














Almost correct:
$$P(X|A,B) = frac{P(A|X)P(B|X)P(X)}{P(A,B)}$$



You know that A and B are conditionally independent given X but not that A and B are independent (that cannot be assumed).






share|cite|improve this answer





















  • Thanks! I would upvote your answer but due to my low reputation the system does not permit me.
    – user617643
    Nov 20 '18 at 14:29











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









0














You could save some steps using Bayes' formula:
$$
P(X mid A, B) = frac{P(A, B mid X)P(X)}{P(A, B)}
$$
Using the conditional independence, this is
$$
frac{P(A mid X) P(B mid X) P(X)}{P(A, B)}
$$
But as has been pointed out, you cannot factor the denominator without assuming independence of $A$ and $B$






share|cite|improve this answer





















  • Thanks! I would upvote your answer but due to my low reputation the system does not permit me.
    – user617643
    Nov 20 '18 at 14:29
















0














You could save some steps using Bayes' formula:
$$
P(X mid A, B) = frac{P(A, B mid X)P(X)}{P(A, B)}
$$
Using the conditional independence, this is
$$
frac{P(A mid X) P(B mid X) P(X)}{P(A, B)}
$$
But as has been pointed out, you cannot factor the denominator without assuming independence of $A$ and $B$






share|cite|improve this answer





















  • Thanks! I would upvote your answer but due to my low reputation the system does not permit me.
    – user617643
    Nov 20 '18 at 14:29














0












0








0






You could save some steps using Bayes' formula:
$$
P(X mid A, B) = frac{P(A, B mid X)P(X)}{P(A, B)}
$$
Using the conditional independence, this is
$$
frac{P(A mid X) P(B mid X) P(X)}{P(A, B)}
$$
But as has been pointed out, you cannot factor the denominator without assuming independence of $A$ and $B$






share|cite|improve this answer












You could save some steps using Bayes' formula:
$$
P(X mid A, B) = frac{P(A, B mid X)P(X)}{P(A, B)}
$$
Using the conditional independence, this is
$$
frac{P(A mid X) P(B mid X) P(X)}{P(A, B)}
$$
But as has been pointed out, you cannot factor the denominator without assuming independence of $A$ and $B$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 20 '18 at 14:26









Slug Pue

2,15111020




2,15111020












  • Thanks! I would upvote your answer but due to my low reputation the system does not permit me.
    – user617643
    Nov 20 '18 at 14:29


















  • Thanks! I would upvote your answer but due to my low reputation the system does not permit me.
    – user617643
    Nov 20 '18 at 14:29
















Thanks! I would upvote your answer but due to my low reputation the system does not permit me.
– user617643
Nov 20 '18 at 14:29




Thanks! I would upvote your answer but due to my low reputation the system does not permit me.
– user617643
Nov 20 '18 at 14:29











0














Almost correct:
$$P(X|A,B) = frac{P(A|X)P(B|X)P(X)}{P(A,B)}$$



You know that A and B are conditionally independent given X but not that A and B are independent (that cannot be assumed).






share|cite|improve this answer





















  • Thanks! I would upvote your answer but due to my low reputation the system does not permit me.
    – user617643
    Nov 20 '18 at 14:29
















0














Almost correct:
$$P(X|A,B) = frac{P(A|X)P(B|X)P(X)}{P(A,B)}$$



You know that A and B are conditionally independent given X but not that A and B are independent (that cannot be assumed).






share|cite|improve this answer





















  • Thanks! I would upvote your answer but due to my low reputation the system does not permit me.
    – user617643
    Nov 20 '18 at 14:29














0












0








0






Almost correct:
$$P(X|A,B) = frac{P(A|X)P(B|X)P(X)}{P(A,B)}$$



You know that A and B are conditionally independent given X but not that A and B are independent (that cannot be assumed).






share|cite|improve this answer












Almost correct:
$$P(X|A,B) = frac{P(A|X)P(B|X)P(X)}{P(A,B)}$$



You know that A and B are conditionally independent given X but not that A and B are independent (that cannot be assumed).







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 20 '18 at 14:15









DavidPM

16618




16618












  • Thanks! I would upvote your answer but due to my low reputation the system does not permit me.
    – user617643
    Nov 20 '18 at 14:29


















  • Thanks! I would upvote your answer but due to my low reputation the system does not permit me.
    – user617643
    Nov 20 '18 at 14:29
















Thanks! I would upvote your answer but due to my low reputation the system does not permit me.
– user617643
Nov 20 '18 at 14:29




Thanks! I would upvote your answer but due to my low reputation the system does not permit me.
– user617643
Nov 20 '18 at 14:29


















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