the probability distribution (softmax function):
\[p(x)=\frac{\exp(-E(x))}{\sum\limits_x{\exp(-E(x))}}\]
when there are hidden units,
\[P(x)=\sum\limits_h{P(x,h)}=\frac{1}{\sum_x\exp(-E(x))}\sum\limits_h{\exp(-E(x,h))}\]
now, we define the free energy function:
\[F(x)=-\log \sum\limits_h \exp(-E(x,h))\]
so that,
\[\sum\limits_h \exp(-E(x,h))=-\exp( F(x))\]
now, we rewrite the probability distribution for simpilification:
\[P(x)=\frac{\exp(-F(x))}{\sum_x{\exp(-F(x))}}\]
then, we define the overall cost function:
\[l(\theta,D)=-\frac{1}{N}\sum\limits_{x^{(i)} \in D}{\log p(x^{(i)})}\]
we firstly calculate the parcial gradient of $\log p(x)$ with respect to $\theta$:
\[-\log p(x)=F(x) + \log\left(\sum\limits_x{\exp(-F(x))}\right)\]
\[-\frac{\partial \log p(x)}{\partial \theta}=\frac{\partial F(x)}{\partial \theta}-\sum\limits_{\tilde x}{p(\tilde x)\frac{\partial F(\tilde x)}{\partial \theta}}\]
note that, the gradient contains two terms, which is called the positive phase and the negative phase. The first term increase the probability of training data, and the second term decrease the probability of samples generated by the model.
It‘s difficult to determine this gradient analytically, as we can‘t calculate $E_P[\frac{\partial F(x)}{\partial \theta}]$. So we might estimate the expectation using sample method.
we would like elements $\tilde x$ of $\mathcal{N}$ to be sampled according to $P(\tilde x)$, where $\mathcal{N}$ is called negative particles.
Given that, the gradient can then be written as:
\[ - \frac{\partial \log p(x)}{\partial \theta}\approx \frac{\partial F(x)}{\partial \theta} - \frac{1}{|\mathcal{N}|} \sum\limits_{\tilde x \in \mathcal{N}}\frac{\partial F(\tilde x)}{\partial \theta}\]
the energy function $E(v,h)$ of RBM is defined as :
\[E(v,h)=-b‘v-c‘h-h‘Wv\]
where
原文：http://www.cnblogs.com/ZJUT-jiangnan/p/5466814.html