 Positive and negative phases of Boltzmann learning

Split the units into visible units $v$, whose patterns occur in the training data, and hidden units $z$. Let $q(v)$ be the empirical probability distribution and maximize the average log-likelihood $\ell ={\sum }_{v}q(v)\log p_{\beta }(v)$, where $p_{\beta }(v)={\sum }_z{p}_{\beta }(v,z)$. The positive phase clamps $v$ to data and averages over $q(v)p_{\beta }(z\mid v)$; the negative phase releases all units and averages over $p_{\beta }(v,z)$. The displayed difference uses these two expectations. When both units are visible, the positive pair product is supplied directly by the data.

For a proof, write $\log p_{\beta }(v)=\log {\sum }_z{e}^{-\beta E(v,z)}-\log Z$. Differentiating each finite sum gives, respectively, $-\beta ⟨\partial E{⟩}_{z\mid v}$ and $-\beta ⟨\partial E{⟩}_{v,z}$. Since $\partial E/\partial w_{ij}=-x_i{x}_j$, subtraction proves the rule. Similarly ${\partial }_{{\theta }_i}\ell =-\beta (⟨x_i{⟩}_{+}-⟨x_i{⟩}_{-})$. A small positive step along these gradients increases log-likelihood. The negative term suppresses associations overproduced by the model; strengthening observed pairs alone misses this correction. The identity requires equilibrium expectations; estimates from insufficiently mixed sampling can give a biased update.