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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  • 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















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





















  • 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













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













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












share|cite|improve this question




share|cite|improve this question








edited Jul 25 at 17:36
























asked Jul 25 at 15:11









Mo Prog

1307




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  • 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










  • ((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











1 Answer
1






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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.






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    1 Answer
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    active

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    1 Answer
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    active

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    active

    oldest

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    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

























      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























        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













        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













        share|cite|improve this answer



        share|cite|improve this answer











        answered Jul 25 at 15:15









        Siong Thye Goh

        77.3k134794




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