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

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edited Nov 20 '18 at 13:58

asked Nov 19 '18 at 23:59

user617643

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
  

  
  
  
  

add a comment | 

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

edited Nov 20 '18 at 13:58

asked Nov 19 '18 at 23:59

user617643

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
  

  
  
  
  

add a comment | 

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

edited Nov 20 '18 at 13:58

asked Nov 19 '18 at 23:59

user617643

226

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

edited Nov 20 '18 at 13:58

asked Nov 19 '18 at 23:59

user617643

226

share|cite|improve this question

edited Nov 20 '18 at 13:58

asked Nov 19 '18 at 23:59

user617643

226

share|cite|improve this question

share|cite|improve this question

edited Nov 20 '18 at 13:58

edited Nov 20 '18 at 13:58

edited Nov 20 '18 at 13:58

asked Nov 19 '18 at 23:59

user617643

226

asked Nov 19 '18 at 23:59

user617643

226

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
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  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

add a comment | 

2 Answers
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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

answered Nov 20 '18 at 14:26

Slug Pue

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
  

  
  
  
  

add a comment | 

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

answered Nov 20 '18 at 14:15

DavidPM

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
  

  
  
  
  

add a comment | 

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2 Answers
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active

oldest

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2 Answers
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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

answered Nov 20 '18 at 14:26

Slug Pue

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
  

  
  
  
  

add a comment | 

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

answered Nov 20 '18 at 14:26

Slug Pue

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
  

  
  
  
  

add a comment | 

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

answered Nov 20 '18 at 14:26

Slug Pue

2,15111020

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

answered Nov 20 '18 at 14:26

Slug Pue

2,15111020

share|cite|improve this answer

share|cite|improve this answer

answered Nov 20 '18 at 14:26

Slug Pue

2,15111020

answered Nov 20 '18 at 14:26

Slug Pue

2,15111020

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
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  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

add a comment | 

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

answered Nov 20 '18 at 14:15

DavidPM

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
  

  
  
  
  

add a comment | 

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

answered Nov 20 '18 at 14:15

DavidPM

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
  

  
  
  
  

add a comment | 

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

answered Nov 20 '18 at 14:15

DavidPM

16618

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

answered Nov 20 '18 at 14:15

DavidPM

16618

share|cite|improve this answer

share|cite|improve this answer

answered Nov 20 '18 at 14:15

DavidPM

16618

answered Nov 20 '18 at 14:15

DavidPM

16618

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
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  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

add a comment | 

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