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?











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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







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    You split $P(A, B) = P(A)P(B)$ -- that is independence of $A, B$, not conditional independence.
    – Mees de Vries
    2 days ago















up vote
2
down vote

favorite












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









New contributor




user617643 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • 1




    You split $P(A, B) = P(A)P(B)$ -- that is independence of $A, B$, not conditional independence.
    – Mees de Vries
    2 days ago













up vote
2
down vote

favorite









up vote
2
down vote

favorite











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









New contributor




user617643 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











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









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edited 2 days ago





















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asked Nov 19 at 23:59









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  • 1




    You split $P(A, B) = P(A)P(B)$ -- that is independence of $A, B$, not conditional independence.
    – Mees de Vries
    2 days ago














  • 1




    You split $P(A, B) = P(A)P(B)$ -- that is independence of $A, B$, not conditional independence.
    – Mees de Vries
    2 days ago








1




1




You split $P(A, B) = P(A)P(B)$ -- that is independence of $A, B$, not conditional independence.
– Mees de Vries
2 days ago




You split $P(A, B) = P(A)P(B)$ -- that is independence of $A, B$, not conditional independence.
– Mees de Vries
2 days ago










2 Answers
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0
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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
    2 days ago


















up vote
0
down vote













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








New contributor




DavidPM is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • Thanks! I would upvote your answer but due to my low reputation the system does not permit me.
    – user617643
    2 days ago











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

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active

oldest

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active

oldest

votes








up vote
0
down vote



accepted










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
    2 days ago















up vote
0
down vote



accepted










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
    2 days ago













up vote
0
down vote



accepted







up vote
0
down vote



accepted






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 2 days ago









Slug Pue

2,14911020




2,14911020












  • Thanks! I would upvote your answer but due to my low reputation the system does not permit me.
    – user617643
    2 days ago


















  • Thanks! I would upvote your answer but due to my low reputation the system does not permit me.
    – user617643
    2 days ago
















Thanks! I would upvote your answer but due to my low reputation the system does not permit me.
– user617643
2 days ago




Thanks! I would upvote your answer but due to my low reputation the system does not permit me.
– user617643
2 days ago










up vote
0
down vote













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








New contributor




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


















  • Thanks! I would upvote your answer but due to my low reputation the system does not permit me.
    – user617643
    2 days ago















up vote
0
down vote













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








New contributor




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


















  • Thanks! I would upvote your answer but due to my low reputation the system does not permit me.
    – user617643
    2 days ago













up vote
0
down vote










up vote
0
down vote









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








New contributor




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









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








New contributor




DavidPM 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 answer



share|cite|improve this answer






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answered 2 days ago









DavidPM

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365




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  • Thanks! I would upvote your answer but due to my low reputation the system does not permit me.
    – user617643
    2 days ago


















  • Thanks! I would upvote your answer but due to my low reputation the system does not permit me.
    – user617643
    2 days ago
















Thanks! I would upvote your answer but due to my low reputation the system does not permit me.
– user617643
2 days ago




Thanks! I would upvote your answer but due to my low reputation the system does not permit me.
– user617643
2 days ago










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