[SciPy-user] Maximum entropy distribution for Ising model - setup?

James Coughlan coughlan at ski.org
Fri Oct 27 15:11:39 CDT 2006


Hi,

I've done some max. ent. modeling, although I'm unfamiliar with the 
scipy maxentropy package. A few points:

1.) If N=10 spins suffices, then there are only 2^10=1024 possible 
configurations of your model, so you can calculate Z and thus the 
average spin <si> (and spin product <si*sj>) values exactly by hand, and 
completely bypass the maxentropy package. On the other hand, maybe scipy 
maxentropy can save you a lot of work!

2.) If you are doing max. ent. modeling, then you are given *empirical* 
(measured) values of <si>_{emp} and <si*sj>_{emp} and are trying to find 
model parameters hi and Jij to match these values, i.e. <si>=<si>_{emp} 
and <si*sj>=<si*sj>_{emp} where the LHS's are the averages w.r.t. the 
model. You can solve for the model parameters by iterating these 
gradient descent equations:

hi_{new}=hi_{old} + K * (<si> - <si>_{emp})
Jij_{new}=Jij_{old}+ K' * (<si*sj> - <si*sj>_{emp})

where K and K' are pos. "step size" constants. (On the RHS, <si> and 
<si*sj> are w.r.t. hi_{old} and Jij_{old}.)

But this process can be slow. It can be sped up by choosing K, K' 
adaptively; maybe the maxentropy package does this more intelligently.

3.) In a typical Ising model, interactions are nearest neighbor, which 
means that Jij is sparse (i.e. it only =1 for neighboring spins i and 
j). But in principal your model can certainly handle arbitrary Jij. 
However, it might be easier (and less prone to overfitting) if you keep 
Jij sparse.

Hope this helps.

Best,

James


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