Does $a \perp\hspace{-1.3ex}\perp b \mid d$ imply $p(a,b,c\mid d) = p(a,c\mid d)\ p(b,c\mid d)$?
2026-04-12 16:56:17.1776012977
Conditional independence and factorization
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No. Assume for example that $c=a$, then for every $(x,y,z)$, $$ P[a=x,b=y,c=x\mid d=z]=P[a=x,b=y\mid d=z], $$ while $$ P[a=x,c=x\mid d=z]\cdot P[b=y,c=x\mid d=z]=P[a=x\mid d=z]\cdot P[a=x,b=y\mid d=z], $$ hence, in general, the LHS is greater than the RHS.