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While going through the Wikipedia article about Boltzmann machines, I read the following on the probabilities of global states of Boltzmann machines in the "Equilibrium state" section (http://en.wikipedia.org/wiki/Boltzmann_machine#Equilibrium_state):

The network is run by repeatedly choosing a unit and setting its state according to the above formula. After running for long enough at a certain temperature, the probability of a global state of the network will depend only upon that global state's energy, according to a Boltzmann distribution. This means that log-probabilities of global states become linear in their energies.

The probability of a global state is usually defined as: $$ P(x) = \frac{e^{-Energy(x)}}{Z}, $$ where the partition function Z is: $$ Z = \sum_u{e^{-Energy(u)}}. $$ However, taking log of P(x): $$ \log{P(x)} = -Energy(x) - \log{(\sum_u{e^{-Energy(u)}})},$$ we arrive at an expression of the log probability of a global state which is not linear.

Could you tell me where I'm wrong?

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When I thought about it more, the expression for the log probability of a global state is linear in the energy of the state, because $- \log{(\sum_u{e^{-Energy(u)}})}$ is a constant term (some of $ e^{-Energy(u)}$ for all states) that doesn't change for different states.

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