Speech Synthesis: Gaussian DBMs

Feb 02, 2014

How to generate random speech sequences using gaussian DBMs

This semester I’m taking Yoshua Bengio’s representation learning class (IFT6266). In addition to formal evaluation, we’re also evaluated in the context of a big class project, in which we compete against each other to find the best solution to a machine learning problem. We’re to maintain a blog detailing our progress, and we can cite or be cited by other students, in analogy to what’s done in actual research.

Suppose you have a good, real-valued representation of audio frames and you wish to learn the distribution of the next audio frame conditioned on the previous one. The following DBM can achieve that:

Gaussian DBM

It is a three-layered DBM whose first and last layers are gaussian and whose hidden layer is binary.

Let’s start with the energy function:

\[ E(\mathbf{x}^t, \mathbf{h}, \mathbf{x}^{t+1}) = E_{bias}(\mathbf{x}^t, \mathbf{h}, \mathbf{x}^{t+1}) + E_{interact}(\mathbf{x}^t, \mathbf{h}, \mathbf{x}^{t+1}) \]


\[ E_{bias}(\mathbf{x}^t, \mathbf{h}, \mathbf{x}^{t+1}) = \frac{1}{2}(\mathbf{x}^t - \mathbf{b})^T(\mathbf{x}^t - \mathbf{b})

  • \mathbf{c}^T\mathbf{h}
  • \frac{1}{2}(\mathbf{x}^{t+1} - \mathbf{d})^T(\mathbf{x}^{t+1} - \mathbf{d}) \]


\[ E_{interact}(\mathbf{x}^t, \mathbf{h}, \mathbf{x}^{t+1}) =

  • \mathbf{h}^T\mathbf{W}\mathbf{x}^t
  • (\mathbf{x}^{t+1})^T\mathbf{U}\mathbf{h} \]

Sparing you the algebraic details, conditional probabilities for this model are given by

\[ \begin{split} p(\mathbf{x}^t \mid \mathbf{h}, \mathbf{x^{t+1}}) &= \mathcal{N}(\mathbf{x}^t \mid \mathbf{b} + \mathbf{W}^T\mathbf{h}, \mathbf{I}), \\\ p(\mathbf{h} \mid \mathbf{x}^t, \mathbf{x^{t+1}}) &= \text{sigmoid}(\mathbf{c} + \mathbf{W}\mathbf{x}^t + \mathbf{U}^T\mathbf{x}^{t+1}), \\\ p(\mathbf{x^{t+1}} \mid \mathbf{x}^t, \mathbf{h}) &= \mathcal{N}(\mathbf{x}^{t+1} \mid \mathbf{d} + \mathbf{U}\mathbf{h}, \mathbf{I}) \end{split} \]

and the gradient of the negative log-likelihood (NLL) of \(p(\mathbf{x}^{t+1} \mid \mathbf{x}^t)\) is given by

\[ \begin{split} \frac{\partial}{\partial \theta} -\log p(\mathbf{x}^t \mid \mathbf{x^{t+1}}) = &\mathbb{E}_{p(\mathbf{h} \mid \mathbf{x}^t, \mathbf{x}^{t+1})} \left[ \frac{\partial}{\partial \theta} E(\mathbf{x}^t, \mathbf{h}, \mathbf{x}^{t+1}) \right] \\\ - &\mathbb{E}_{p(\mathbf{h}, \mathbf{x}^{t+1} \mid \mathbf{x}^t)} \left[ \frac{\partial}{\partial \theta} E(\mathbf{x}^t, \mathbf{h}, \mathbf{x}^{t+1}) \right] \end{split} \]

For the positive phase, you sample from \(\mathbf{h}\) given \( \mathbf{x}_t \) and \( \mathbf{x}_{t+1} \) and for the negative phase, you sample from \( \mathbf{h} \) and \( \mathbf{x}_{t+1} \) given \( \mathbf{x}_t \).