When we stopped last time, we were in the middle of proving the Hammersley-Clifford Theorem. Today we will finish the proof, and talk more about undirected models.
A positive distribution $p(x) > 0$ satisfies the CIs (conditional independencies) of an undirected graph iff $p$ can be represented as
$$ p(x) = \frac{1}{Z} \prod_{c\in C} \phi_c(x_c). $$
where $C$ is the set of all (maximal) cliques in the graph and
$$ Z = \sum_{x} \prod_{c \in C} \phi_c(x_c). $$
In the last lecture we covered the first three steps of the proof of Hammersley-Clifford Theorem.
Define $x^* = (0,0,\dots,0)$ and $Q(x) = \ln(p(x) / p(x^*)).$
We can write $Q$ uniquely as
$$
Q(x) = \sum_i x_iG_i(x_i) + \sum_{i<j} x_ix_jG_{i,j}(x_i, x_j) + \sum_{i<j<k} x_ix_jx_k G_{i,j,k}(x_i, x_j, x_k) + \dots + x_1x_2\dots x_DG_{1,2,\dots,D}(x_1,x_2,\dots,x_D).
$$
Claim: We can choose $G$ uniquely to make the left hand side (LHS) and the right hand side (RHS) of the equation equal. Note that we are assuming the variable are discrete.
Let's define
$$ ⁍ $$
which means that only the $i^{th}$ component of $x$ is set to zero. Then we have
$$ \exp(Q(x) - Q(x^i)) = \frac{p(x)}{p(x^i)} = \frac{p(\mathsf{x}i = x_i|x{-i})}{p(\mathsf{x}i=0 | x{-i})}. $$