Conditional probability by joint distribution

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I have read the following equation:

$$P(E_1|H)*P(E_2|H)*P(H) = P(E_1,E_2,H) $$

I don't understand why this equation is true. I know that

$$P(E_1|H)*P(H) = P(E_1,H)$$

thus:
$$P(E_1|H)*P(E_2|H)*P(H)=P(E_1,H)*P(E_2|H)$$
but why:
$$P(E_1|H)*P(E_2|H)*P(H) = P(E_1,E_2,H) $$
As a extra information it is also given that:

$$P(E_1|H,E_2) = P(E_1|H)$$

probability stochastic-calculus

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edited Jul 25 at 17:36

asked Jul 25 at 15:11

Mo Prog

1307

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How/When did you use the "extra information"?
  – Did
  Jul 25 at 17:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  ((This is not a question of stochastic-calculus at all.))
  – Did
  Jul 25 at 17:58
  
  

  
  
  
  

add a comment | 

up vote
-1
down vote

favorite

I have read the following equation:

$$P(E_1|H)*P(E_2|H)*P(H) = P(E_1,E_2,H) $$

I don't understand why this equation is true. I know that

$$P(E_1|H)*P(H) = P(E_1,H)$$

thus:
$$P(E_1|H)*P(E_2|H)*P(H)=P(E_1,H)*P(E_2|H)$$
but why:
$$P(E_1|H)*P(E_2|H)*P(H) = P(E_1,E_2,H) $$
As a extra information it is also given that:

$$P(E_1|H,E_2) = P(E_1|H)$$

probability stochastic-calculus

share|cite|improve this question

edited Jul 25 at 17:36

asked Jul 25 at 15:11

Mo Prog

1307

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How/When did you use the "extra information"?
  – Did
  Jul 25 at 17:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  ((This is not a question of stochastic-calculus at all.))
  – Did
  Jul 25 at 17:58
  
  

  
  
  
  

add a comment | 

up vote
-1
down vote

favorite

up vote
-1
down vote

favorite

I have read the following equation:

$$P(E_1|H)*P(E_2|H)*P(H) = P(E_1,E_2,H) $$

I don't understand why this equation is true. I know that

$$P(E_1|H)*P(H) = P(E_1,H)$$

thus:
$$P(E_1|H)*P(E_2|H)*P(H)=P(E_1,H)*P(E_2|H)$$
but why:
$$P(E_1|H)*P(E_2|H)*P(H) = P(E_1,E_2,H) $$
As a extra information it is also given that:

$$P(E_1|H,E_2) = P(E_1|H)$$

probability stochastic-calculus

share|cite|improve this question

edited Jul 25 at 17:36

asked Jul 25 at 15:11

Mo Prog

1307

I have read the following equation:

$$P(E_1|H)*P(E_2|H)*P(H) = P(E_1,E_2,H) $$

I don't understand why this equation is true. I know that

$$P(E_1|H)*P(H) = P(E_1,H)$$

thus:
$$P(E_1|H)*P(E_2|H)*P(H)=P(E_1,H)*P(E_2|H)$$
but why:
$$P(E_1|H)*P(E_2|H)*P(H) = P(E_1,E_2,H) $$
As a extra information it is also given that:

$$P(E_1|H,E_2) = P(E_1|H)$$

probability stochastic-calculus

share|cite|improve this question

edited Jul 25 at 17:36

asked Jul 25 at 15:11

Mo Prog

1307

share|cite|improve this question

share|cite|improve this question

edited Jul 25 at 17:36

edited Jul 25 at 17:36

edited Jul 25 at 17:36

asked Jul 25 at 15:11

Mo Prog

1307

asked Jul 25 at 15:11

Mo Prog

1307

asked Jul 25 at 15:11

Mo Prog

1307

1307

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How/When did you use the "extra information"?
  – Did
  Jul 25 at 17:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  ((This is not a question of stochastic-calculus at all.))
  – Did
  Jul 25 at 17:58
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How/When did you use the "extra information"?
  – Did
  Jul 25 at 17:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  ((This is not a question of stochastic-calculus at all.))
  – Did
  Jul 25 at 17:58
  
  

  
  
  
  

How/When did you use the "extra information"?
– Did
Jul 25 at 17:57

How/When did you use the "extra information"?
– Did
Jul 25 at 17:57

((This is not a question of stochastic-calculus at all.))
– Did
Jul 25 at 17:58

((This is not a question of stochastic-calculus at all.))
– Did
Jul 25 at 17:58

add a comment | 

1 Answer
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2
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accepted

We are given that $$P(E_1|H,E_2)=P(E_1|H)$$

hence,

$$fracP(E_1,H,E_2)P(H,E_2)=P(E_1|H)$$

$$P(E_1,H,E_2)=P(E_1|H)P(H,E_2)$$

but we have $P(H,E_2)=P(H)P(E_2|H)$, hence the result.

share|cite|improve this answer

answered Jul 25 at 15:15

Siong Thye Goh

77.3k134794

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

We are given that $$P(E_1|H,E_2)=P(E_1|H)$$

hence,

$$fracP(E_1,H,E_2)P(H,E_2)=P(E_1|H)$$

$$P(E_1,H,E_2)=P(E_1|H)P(H,E_2)$$

but we have $P(H,E_2)=P(H)P(E_2|H)$, hence the result.

share|cite|improve this answer

answered Jul 25 at 15:15

Siong Thye Goh

77.3k134794

add a comment | 

up vote
2
down vote



accepted

We are given that $$P(E_1|H,E_2)=P(E_1|H)$$

hence,

$$fracP(E_1,H,E_2)P(H,E_2)=P(E_1|H)$$

$$P(E_1,H,E_2)=P(E_1|H)P(H,E_2)$$

but we have $P(H,E_2)=P(H)P(E_2|H)$, hence the result.

share|cite|improve this answer

answered Jul 25 at 15:15

Siong Thye Goh

77.3k134794

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

We are given that $$P(E_1|H,E_2)=P(E_1|H)$$

hence,

$$fracP(E_1,H,E_2)P(H,E_2)=P(E_1|H)$$

$$P(E_1,H,E_2)=P(E_1|H)P(H,E_2)$$

but we have $P(H,E_2)=P(H)P(E_2|H)$, hence the result.

share|cite|improve this answer

answered Jul 25 at 15:15

Siong Thye Goh

77.3k134794

We are given that $$P(E_1|H,E_2)=P(E_1|H)$$

hence,

$$fracP(E_1,H,E_2)P(H,E_2)=P(E_1|H)$$

$$P(E_1,H,E_2)=P(E_1|H)P(H,E_2)$$

but we have $P(H,E_2)=P(H)P(E_2|H)$, hence the result.

share|cite|improve this answer

answered Jul 25 at 15:15

Siong Thye Goh

77.3k134794

share|cite|improve this answer

share|cite|improve this answer

answered Jul 25 at 15:15

Siong Thye Goh

77.3k134794

answered Jul 25 at 15:15

Siong Thye Goh

77.3k134794

answered Jul 25 at 15:15

Siong Thye Goh

77.3k134794

77.3k134794

add a comment | 

add a comment | 

 

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