In the following slide:enter image description here it seems how it got A and B are independent is rather circular. The reason is it assumes $p(a,b,c) = p(a)p(b)p(c|a,b)$ and then marginalizes over $c$ to get $p(a,b) = p(a)p(b)$, while $p(a,b,c) = p(a,b)p(c|a,b) = p(a|b)p(b)p(c|a,b)$ instead. Am I correct? If yes then what is the correct way to reach that conclusion?

  • $\begingroup$ I think $A$ and $B$ being independent follows from the diagram itself. If you then marginalize $C$ out you still get them as independent. The interesting thing will happen if you condition on $C$. Then $A$ and $B$, whilst remaining independent, could appear as dependent to you, since you will select pairs of $a$ and $b$ based on specific $c$. $\endgroup$
    – Cryo
    Mar 9 at 9:14
  • $\begingroup$ @Cryo 'I think A and B being independent follows from the diagram itself' if that is the case then it shouldn't try to prove it (the middle part and the second equation) $\endgroup$
    – Sam
    Mar 12 at 9:22


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