# Energy function of RBM

The Hammersley-Clifford Theorem tells us that the distribution of a RBM must be Gibbs since it is Markov Random Field, but how to prove that its energy function must be of the form:

$$E = -\sum_{i,j} w_{ij} \, v_i \, h_j -\sum_i \alpha_i \, v_i - \sum_i \beta_i \, h_i$$

In this case the (maximal) cliques are rather trivial: just the pairs of nodes (again, take a look at the graph). Each pair has a corresponding term in the sum. When raising the exponential to $E$, the sum turns to a product of positive functions (the exponential is a positive function), where each function depends only on two variables corresponding to each of the cliques.
• It's convenient because it's linear. The linearity is particularly useful when you want to find gradients with respect to the coefficients ($w$, $\alpha$, and $\beta$ in your notation). The gradients are useful for optimizing the coefficients. It's meaningful because the it has a relationship to the inverse covariances among nodes. Commented Jun 11, 2013 at 20:05
• Actually, the relationship to the covariance may only apply if the nodes have Gaussian distributions--I'll have to think about it for a minute. But @juampa's point about factorizing is a very good one: finding a way to factorize the distribution makes just about everything in RBMs easier; if you know $h$, it's trivial write down the distribution of $v$ and vice versa. That wouldn't be so straightforward if the parameterization were different. Commented Jun 11, 2013 at 20:11