I have a question about the last step of the proof of conditional probability of the restricted boltzman machine in the deeplearningbook.org

Below are the derivation from the book: enter image description here enter image description here enter image description here enter image description here Shouldn't equation 20.15 be

$P(h|v) = \displaystyle\prod_{j=1}^{nh}\sigma((2h_j-1)\times(c_i+\textbf{W}_{:,j}^\intercal \textbf{v}))$ (my own derived formula)
instead of
$P(h|v) = \displaystyle\prod_{j=1}^{nh}\sigma((2\textbf{h}-1)\odot(\textbf{c}+\textbf{W}^\intercal \textbf{v}))_j$ (in the deeplearningbook.org)

What would that j even refer to in the original question in the textbook anyway?

  • $\begingroup$ Can you include more information from the text so we can have more context? $\endgroup$ – Michael R. Chernick Jan 12 '17 at 6:16
  • $\begingroup$ @MichaelChernick I have updated the question. Is it more clear? Would you be able to help me answer the question? $\endgroup$ – MrDeep Jan 15 '17 at 4:03
  • $\begingroup$ Thank you for the additional information. Some of the notation is strange to me and I am not familiar with Boltzman machines and deep learning. But we do have experts here that could probably explain where you are going wrong. $\endgroup$ – Michael R. Chernick Jan 15 '17 at 7:35

$j$ indexes the hidden units in the RBM.

Assuming you meant $c_j$ instead of $c_i$ in your derived formula, it is equivalent to the provided formula.

$$ \begin{align} &\prod_j \sigma((2h-1) \odot (c + W^T v))_j \\ = &\prod_j \sigma((2h-1)_j \cdot (c + W^T v)_j) \\ = &\prod_j \sigma((2h_j-1) (c_j + (W^T)_j v)) \\ = &\prod_j \sigma((2h_j-1) (c_j + W_{:,j}^T v)) \\ \end{align} $$


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