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Consider the following potential on three nodes.

$$\psi(x_1,x_2,x_3) = f_a(x_1,x_2)f_b(x_2,x_3)f_c(x_1,x_3)$$

represented by the following factor graph:

A factor graph of X_1, X_2 and X_3

Now the notes claim that we can represent this factor graph as both a Bayesian network and a Markov networks as follows.

enter image description here

The representation is given here:

enter image description here

I'm failing to see the connection with the extra three nodes $Z_1,Z_2,Z_3$? What would the break down into clique potentials of the probability distribution looks like? Since there are only 2-cliques, the theory as I know it suggests that $$p(x_1,x_2,x_3,Z_1,Z_2,Z_3) = \frac 1 Z \psi_1(x_1,Z_1) \psi_2(Z_1,x_2) \psi_3(x_2,Z_2) \psi_4(Z_2,x_3) \psi_5(x_3,Z_3) \psi_6(Z_3,x_1)$$ where $Z$ is the normalising constant. However I'm confused as to what $\psi(Z_1)$ would represent and how it equals $f_a(x_1,x_2)$? Can anyone fill in the missing pieces of the construction?

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If you look at the Appendix of this article you might find a clearer explanation of the transformation steps in order to get the Markov Field representation equivalent to a factor graph. The appendix even generalizes to higher-order cliques using additional auxiliary variables. You can apply it to your example...

But maybe for pairwise cliques, the transformation is can be clear without auxiliary variables. The joint distribution you propose is wrong, you miss the "evidence" terms so that:

$$ p(x_1, x_2, x_3, z_1, z_2, z_3) = \frac{1}{Z} \psi(x_1,z_1)\psi(z_1,x_2)\psi(z_1) \psi(x_2,z_2)\psi(z_2,x_3)\psi(z_2)\\ \phantom{aaaa}\psi(x_1,z_3)\psi(z_3,x_3)\psi(z_3). $$

As explained in your post and in my linked article, to get the equivalence between this Markov field distribution and the factor graph we need to have that the pairwise cliques are indicator functions and the single variable cliques are equal to the function value (e.g. $\psi(z_a) = f_a(x_1,x_2)$).

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