Boltzmann machine probability of neuron flipping proof

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I'm trying to prove the following property of a Boltzmann machine. Let $x_i$ be $1$ or $-1,$ and $\Delta E_i$ be the energy change resulting from the flip:

$$P(x_i \to -x_i) = \frac{1}{1 + \exp\left(\frac{\Delta E_i} T \right)}$$

I've tried modifying the derivation in the "Probability of a unit's state" section of this Wikipedia page: https://en.wikipedia.org/wiki/Boltzmann_machine (the derivation is for something similar).

How does one represent $P(x_i \to -x_i)$ in terms of $P(x_i)$ and $P(-x_i)$? Is trying to find this a promising direction to take? If not, what different direction can be taken instead? Please don't answer this with a full proof, just guidance.

UPDATE: Below is the information I've been given:

  • For symmetric connections, $E(x) = -\frac{1}{2}\sum_i\sum_j w_{ij}x_ix_j$
  • When system is in thermal equilibrium, probability of state x is $P(x) = \frac{1}{Z}exp(-\frac{E(x)}{T})$
    • where $Z = \sum_i exp(-\frac{E_i}{T})$
  • I've also been given equations on probability of training samples, but it doesn't seem relevant

Also, I've corrected the original equation. It originally had a denominator of $1 + \exp\left(-\frac{\Delta E_i} T \right)$.