Prove if two variables are conditionally independent
$begingroup$
I have 4 random variables A, B, C, D.
I know that the joint is
$$p(A,B,C,D) = p(A)p(B|C,A)p(C)p(D|B,C) $$
And I want to prove (if true) that
$$ A perp D | B $$
I have tried this:
$$p(A,D,B) = sum_C p(A,B,C,D)
= sum_C p(A)p(B|C,A)p(C)p(D|B,C)
= sum_C p(A)p(B,C|A)p(D,C|B)
= p(A)p(B|A)p(D|B)$$
Then
$$p(A,D|B) = frac{p(A,D,B)}{p(B)} = frac{p(A)p(B|A)}{p(B)}p(D|B) = p(A|B) p(D|B)$$
My question is: is the marginalization correct? I know it works for $p(X)=sum_Y p(X,Y)$, but does it work in the same way in this case?
independence conditional-probability bayes-theorem
asked Dec 7 '18 at 12:56
XemnasXemnas
11
$endgroup$
add a comment |
$begingroup$
I have 4 random variables A, B, C, D.
I know that the joint is
$$p(A,B,C,D) = p(A)p(B|C,A)p(C)p(D|B,C) $$
And I want to prove (if true) that
$$ A perp D | B $$
I have tried this:
$$p(A,D,B) = sum_C p(A,B,C,D)
= sum_C p(A)p(B|C,A)p(C)p(D|B,C)
= sum_C p(A)p(B,C|A)p(D,C|B)
= p(A)p(B|A)p(D|B)$$
Then
$$p(A,D|B) = frac{p(A,D,B)}{p(B)} = frac{p(A)p(B|A)}{p(B)}p(D|B) = p(A|B) p(D|B)$$
My question is: is the marginalization correct? I know it works for $p(X)=sum_Y p(X,Y)$, but does it work in the same way in this case?
independence conditional-probability bayes-theorem
asked Dec 7 '18 at 12:56
XemnasXemnas
11
$endgroup$
$begingroup$
If you are asking about the step $$p(A,D,B) = sum_C p(A,B,C,D) $$ then yes, it is rock solid.
$endgroup$
– Did
Dec 7 '18 at 13:58
$begingroup$
The whole step? So $p(A,D,B) = p(A)p(B|A)p(D|B)$ ? I am not sure because maybe forobtaining those 2 joints $p(B,C|A)p(D,C|B)$ we would need two $p(C)$ and not only one?
$endgroup$
– Xemnas
Dec 7 '18 at 16:44
add a comment |
$begingroup$
I have 4 random variables A, B, C, D.
I know that the joint is
$$p(A,B,C,D) = p(A)p(B|C,A)p(C)p(D|B,C) $$
And I want to prove (if true) that
$$ A perp D | B $$
I have tried this:
$$p(A,D,B) = sum_C p(A,B,C,D)
= sum_C p(A)p(B|C,A)p(C)p(D|B,C)
= sum_C p(A)p(B,C|A)p(D,C|B)
= p(A)p(B|A)p(D|B)$$
Then
$$p(A,D|B) = frac{p(A,D,B)}{p(B)} = frac{p(A)p(B|A)}{p(B)}p(D|B) = p(A|B) p(D|B)$$
My question is: is the marginalization correct? I know it works for $p(X)=sum_Y p(X,Y)$, but does it work in the same way in this case?
independence conditional-probability bayes-theorem
asked Dec 7 '18 at 12:56
XemnasXemnas
11
$endgroup$
I have 4 random variables A, B, C, D.
I know that the joint is
$$p(A,B,C,D) = p(A)p(B|C,A)p(C)p(D|B,C) $$
And I want to prove (if true) that
$$ A perp D | B $$
I have tried this:
$$p(A,D,B) = sum_C p(A,B,C,D)
= sum_C p(A)p(B|C,A)p(C)p(D|B,C)
= sum_C p(A)p(B,C|A)p(D,C|B)
= p(A)p(B|A)p(D|B)$$
Then
$$p(A,D|B) = frac{p(A,D,B)}{p(B)} = frac{p(A)p(B|A)}{p(B)}p(D|B) = p(A|B) p(D|B)$$
My question is: is the marginalization correct? I know it works for $p(X)=sum_Y p(X,Y)$, but does it work in the same way in this case?
independence conditional-probability bayes-theorem
independence conditional-probability bayes-theorem
asked Dec 7 '18 at 12:56
XemnasXemnas
11
asked Dec 7 '18 at 12:56
XemnasXemnas
11
asked Dec 7 '18 at 12:56
XemnasXemnas
11
asked Dec 7 '18 at 12:56
XemnasXemnas
11
asked Dec 7 '18 at 12:56
XemnasXemnas
11
11
$begingroup$
If you are asking about the step $$p(A,D,B) = sum_C p(A,B,C,D) $$ then yes, it is rock solid.
$endgroup$
– Did
Dec 7 '18 at 13:58
$begingroup$
The whole step? So $p(A,D,B) = p(A)p(B|A)p(D|B)$ ? I am not sure because maybe forobtaining those 2 joints $p(B,C|A)p(D,C|B)$ we would need two $p(C)$ and not only one?
$endgroup$
– Xemnas
Dec 7 '18 at 16:44
add a comment |
$begingroup$
If you are asking about the step $$p(A,D,B) = sum_C p(A,B,C,D) $$ then yes, it is rock solid.
$endgroup$
– Did
Dec 7 '18 at 13:58
$begingroup$
The whole step? So $p(A,D,B) = p(A)p(B|A)p(D|B)$ ? I am not sure because maybe forobtaining those 2 joints $p(B,C|A)p(D,C|B)$ we would need two $p(C)$ and not only one?
$endgroup$
– Xemnas
Dec 7 '18 at 16:44
$begingroup$
If you are asking about the step $$p(A,D,B) = sum_C p(A,B,C,D) $$ then yes, it is rock solid.
$endgroup$
– Did
Dec 7 '18 at 13:58
$begingroup$
If you are asking about the step $$p(A,D,B) = sum_C p(A,B,C,D) $$ then yes, it is rock solid.
$endgroup$
– Did
Dec 7 '18 at 13:58
$begingroup$
The whole step? So $p(A,D,B) = p(A)p(B|A)p(D|B)$ ? I am not sure because maybe forobtaining those 2 joints $p(B,C|A)p(D,C|B)$ we would need two $p(C)$ and not only one?
$endgroup$
– Xemnas
Dec 7 '18 at 16:44
$begingroup$
The whole step? So $p(A,D,B) = p(A)p(B|A)p(D|B)$ ? I am not sure because maybe forobtaining those 2 joints $p(B,C|A)p(D,C|B)$ we would need two $p(C)$ and not only one?
$endgroup$
– Xemnas
Dec 7 '18 at 16:44
add a comment |
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$begingroup$
If you are asking about the step $$p(A,D,B) = sum_C p(A,B,C,D) $$ then yes, it is rock solid.
$endgroup$
– Did
Dec 7 '18 at 13:58
$begingroup$
The whole step? So $p(A,D,B) = p(A)p(B|A)p(D|B)$ ? I am not sure because maybe forobtaining those 2 joints $p(B,C|A)p(D,C|B)$ we would need two $p(C)$ and not only one?
$endgroup$
– Xemnas
Dec 7 '18 at 16:44