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The 'conventional' configuration of RBMs are Binary-Binary and Gaussian-Binary (and sometimes Binary-Gaussian) units.

Although it is possible for both the visible and hidden units to be gaussian, wouldn't a Gaussian-Gaussian RBM just resemble a linear model, since there is no non-linearity in the networks units anymore? Thus, stacking them would not not have the same benefit as for, say, Binary-Binary RBMs. And when using them for dimensionality reduction, a simple PCA would achieve better results?

Am I missing any significant points in the training of RBMs or are Gaussian-Gaussian RBMs just that limited?

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2 Answers 2

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First, notice that you can fix the variance of the hidden units to 1, since the weight matrix will scale them arbitrarily.

Then:

  • If you learn the variance of each visible unit, you get factor analysis
  • If the variance of the visible units is tied, you get is PPCA (of the demeaned data)
  • If you fix the variance of the visible units to a small value, you get pure PCA.

In the last two cases, the weight matrix, $W$ will correspond to the leading eigenvectors of the data correlation matrix, up to a rotation.

Stacking several layers is not equivalent to having a larger layer. The distribution of all the units is still jointly Gaussian, but the connectivity restricts the covariance matrix to a certain structure.

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  • $\begingroup$ Doesn't the weight matrix between two Gaussian nodes $x, y$ only scale along the covariances between elements of $x$ and $y$? It can't affect the covariance between different elements within $x$. $\endgroup$
    – Neil G
    Commented Apr 20, 2015 at 20:58
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Yes, with just the means (without the covariances) being transmitted, it is a linear model. Gaussian nodes are useful when representing Gaussian inputs. I don't think they are useful as latent variables.

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    $\begingroup$ Perhaps the anonymous down-voter would mind sharing his/her reasons here, so that I, and others, can learn and improve? $\endgroup$
    – Neil G
    Commented May 14, 2015 at 1:02

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