0
$\begingroup$

I have a Markov chain where:

A->B->C->D

I am told that P(A,C,D|B) = P(A|B)P(C,D|B)

I am unable to prove why this is the case. Why is this as such?

$\endgroup$
0
$\begingroup$

P(A,C,D|B) = P(A,B,C,D)/P(B) = P(A)P(B|A)P(C|B)P(D|C)/P(B)

= [P(A)P(B|A)/P(B)] [P(C|B)P(D|C)]

= [P(A,B)/P(B)] [P(C,D|B)]

= P(A|B)P(C,D|B)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.