I Restricted Boltzmann machine uniqueness

AI Thread Summary
The discussion centers on the uniqueness of distributions modeled by restricted Boltzmann machines (RBMs). The user inquires whether a specific configuration of biases and weights can yield the same probability distribution as another configuration. A participant suggests that by setting certain biases to zero, it may be possible to exchange weights and binary vectors to achieve the same distribution. However, there is uncertainty about whether this method is permissible within the framework of RBMs. The conversation highlights the complexities of parameter configurations in modeling distributions with RBMs.
Jufa
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I am dealing with restricted Boltzmann machines to model distributuins in my final degree project and some question has come to my mind.
A restricted Boltzmann machine with v visible binary neurons and h hidden neurons models a distribution in the following manner:

## f_i= e^{ \sum_k b[k] \sigma^i[k] + \sum_s \log(c[ s ] + e^{\sum_k w[ s ][ k ] b[ k ] })} ##
## Z = \sum_i f_i ##
## p_i = f_i/Z ##
Where b[ k ] and c[ s ] are, respectively, the k-th and the s-th bias of, again respectively, the visible and hidden layer.
w[ s ][k] is the component s, k of the weight matrix of the network.
"i" here refers to a certain binary vector with components ##\sigma^i[k]##.
My question is:
Given a certain restricted Boltzmann machine (i.e. a certain set of biases and weights) that models a certain distribution ##p_i##, is it possible to find another configuration (i.e. a different set of parameters and weights) such that it gives the same distribution?
Thanks in advance.
 
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If some moderator can explain why my text is strikethrough I would appreciate it.
 
We are attempting to fix the problem. I got rid of the strike through text but now the latex Mathjax isn’t rendering. We moved to a new version of forum software with a new post editor and perhaps it has injected or requires some special way to enter math expressions.
 
Testing mathjax ##e=mc^2## now And ##2+2=4## in base 10.
 
I think I have reset the corrected original version. Please refresh your browsers.

Hint: Do not use ,,,. If you do not "cheat" as I did here, the interpreter takes it for the command tag BEGIN bold / underline / strikeout / italic.
 
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Jufa said:
I am dealing with restricted Boltzmann machines to model distributuins in my final degree project and some question has come to my mind.
A restricted Boltzmann machine with v visible binary neurons and h hidden neurons models a distribution in the following manner:

## f_i= e^{ \sum_k b[k] \sigma^i[k] + \sum_s \log(c[ s ] + e^{\sum_k w[ s ][ k ] b[ k ] })} ##
## Z = \sum_i f_i ##
## p_i = f_i/Z ##
Where b[ k ] and c[ s ] are, respectively, the k-th and the s-th bias of, again respectively, the visible and hidden layer.
w[ s ][k] is the component s, k of the weight matrix of the network.
"i" here refers to a certain binary vector with components ##\sigma^i[k]##.
My question is:
Given a certain restricted Boltzmann machine (i.e. a certain set of biases and weights) that models a certain distribution ##p_i##, is it possible to find another configuration (i.e. a different set of parameters and weights) such that it gives the same distribution?
Thanks in advance.
I’m not very familiar with the topic, but trivially, if you set ##c[ s]=0##, then you get the same ##f_i## by exchanging the ##w[ s][k]## and the ##\sigma^i[k]##. So for instance,
$$f_{i,1}=\exp{\left(\sum_k{b[k]\sigma_1^i[k]}+\sum_s{ \log{e^{\sum_k{w_1[ s][k]b[k]}}}}\right)}$$
$$f_{i,2}=\exp{\left(\sum_k{b[k]\sigma_2^i[k]}+\sum_s{ \log{e^{\sum_k{w_2[ s][k]b[k]}}}}\right)}$$
Then let ##\sigma_1^i[k]=w_2[ s][k]## and ##\sigma_2^i[k]=w_1[ s][k]##.
That gives you the same probability distributions but I don’t know if that’s allowed with restricted Boltzmann machines.
 
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