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Monotonic RNN-T Loss

Theory

Loss function

The monotonic RNN-T loss can be written as

$$L = -\log p(a_1^S \mid x_1^T) = -\log \sum_{y_1^T : a_1^S} p(y_1^T \mid x_1^T) = -\log \sum_{y_1^T : a_1^S} \prod_{t=1}^T p_t(y_t \mid x_1^T, a(y_1^{t-1}))$$

where $S$ is the number of labels, $T$ is the number of time-steps, $a_1^S$ is the ground-truth label sequence, $x_1^T$ are the acoustic features, $y_1^T$ is the set of alignments of $a_1^S$ as a result of inserting blank symbols and $a( y_1^t)$ is a function that simplifies the history (usually removal of blanks and sometimes truncation to only the one or two most recent symbols). For simplicity, $x_1^T$ will be omitted from the dependencies from now on.

Forward-backward

The loss and gradients can be computed using the forward-backward-algorithm. For this, a forward variable

$$\alpha(t, s) = \sum_{y_1^t : a_1^s} \prod_{t'=1}^t p_{t'}(y_{t'} \mid a(y_1^{t'-1}))$$

and a backward variable

$$\beta(t, s) = \sum_{y_t^T : a_s^S} \prod_{t'=t}^T p_{t'}(y_{t'} \mid a(a_1^s || y_{t'}^T))$$

are introduced. These have the property

$$L = -\log \alpha(T, S) = -\log \beta(1, 0)$$

and adhere to the recursive equations

$$\alpha(t, s) = p_t(\epsilon \mid a(a_1^s)) \cdot \alpha(t-1, s) + p_t(a_s \mid a(a_1^{s-1})) \cdot \alpha(t-1, s-1)$$

and

$$\beta(t, s) = p_t(\epsilon \mid a(a_1^s)) \cdot \beta(t+1, s) + p_t(a_{s+1} \mid a(a_1^s)) \cdot \beta(t+1, s+1)$$

(excluding edge cases).

Gradients

For the gradients it is straightforward to prove that for any $t$

$$p(a_1^S) = \sum_{s=0}^S \alpha(t, s) \cdot \beta(t + 1, s)$$

And thus

$\frac{\partial p(a_1^S)}{\partial p_t(y \mid a(a_1^s))}$

$=\frac{\partial}{\partial p_t(y \mid a_1^s)} \left( \sum_{s'} \alpha(t, s') \cdot \beta(t+1, s') \right)$

$=\frac{\partial}{\partial p_t(y \mid a_1^s)} \left( \sum_{s'} \left( p_t(\epsilon \mid a(a_1^{s'})) \cdot \alpha(t-1, s') + p_t(a_{s'} \mid a(a_1^{s'-1})) \cdot \alpha(t-1, s'-1)\right) \cdot \beta(t+1, s') \right)$

$= \alpha(t-1, s) \cdot \beta(t+1, s)$ if $y = \epsilon$

$= \alpha(t-1, s) \cdot \beta(t+1, s+1)$ if $y = a_{s+1}$ and

$= 0$ otherwise.

which means for the overall gradient

$\frac{\partial L}{\partial p_t(y \mid a(a_1^s))}$

$= - \frac{\alpha(t-1, s) \cdot \beta(t+1, s)}{p(a_1^S)}$ if $y = \epsilon$

$= - \frac{\alpha(t-1, s) \cdot \beta(t+1, s+1)}{p(a_1^S)}$ if $y = a_{s+1}$

$= 0$ otherwise.

For expressing the derivative directly with respect to the logits $z_1^V$ where $p_t(y \mid a(a_1^s)) = \frac{e^{z_y}}{\sum_v e^{z_v}}$ we can derive with some calculation that

$\frac{\partial L}{\partial z_y}$

$= - \frac{\alpha(t-1, s) \cdot p(\epsilon \mid a(a_1^s)) \left(\beta(t+1, s) - \beta(t, s) \right)}{p(a_1^S)}$ if $y =\epsilon$

$= - \frac{\alpha(t-1, s) \cdot p(\epsilon \mid a(a_1^s)) \left(\beta(t+1, s+1) - \beta(t, s) \right)}{p(a_1^S)}$ if $y = a_{s+1}$

$= - \frac{\alpha(t-1, s) \cdot p(\epsilon \mid a(a_1^s)) \left(-\beta(t, s)\right)}{p(a_1^S)}$ otherwise

Example

Assume the following model posteriors $p_t(y \mid a(a_1^s))$ with $T = 4$, $S = 2$ and number of classes $V = 3$ with blank-index $0$.

// t = 1
0.6, 0.3, 0.1,  // s = 0
0.7, 0.1, 0.2,  // s = 1
0.5, 0.1, 0.4,  // s = 2

// t = 2
0.5, 0.4, 0.1,  // s = 0
0.5, 0.1, 0.4,  // s = 1
0.8, 0.1, 0.1,  // s = 2

// t = 3
0.4, 0.3, 0.3,  // s = 0
0.5, 0.1, 0.4,  // s = 1
0.7, 0.2, 0.1,  // s = 2

// t = 4
0.8, 0.1, 0.1,  // s = 0
0.3, 0.1, 0.6,  // s = 1
0.8, 0.1, 0.1   // s = 2

For $a_1^S = [1, 2]$ the valid alignments $y_1^T$ are as follows (with "." denoting blank):

  • . . 1 2
  • . 1 . 2
  • . 1 2 .
  • 1 . . 2
  • 1 . 2 .
  • 1 2 . .

The 6 paths have probabilities of

  • 0.6 * 0.5 * 0.3 * 0.6 = 0.0540
  • 0.6 * 0.4 * 0.5 * 0.6 = 0.0720
  • 0.6 * 0.4 * 0.4 * 0.8 = 0.0768
  • 0.3 * 0.5 * 0.5 * 0.6 = 0.0450
  • 0.3 * 0.5 * 0.4 * 0.8 = 0.0480
  • 0.3 * 0.4 * 0.7 * 0.8 = 0.0672

wich sum to a total of 0.363, i.e. -1.0134 in log space

The alphas then are as follows in probability and log space:

  • a(0, 0) = 1.0 -> 0.0
  • a(1, 0) = 0.6 -> -0.51
  • a(1, 1) = 0.3 -> -1.20
  • a(2, 0) = 0.5 * a(1, 0) = 0.3 -> -1.20
  • a(2, 1) = 0.5 * a(1, 1) + 0.4 * a(1, 0) = 0.39 -> -0.94
  • a(2, 2) = 0.4 * a(1, 1) = 0.12 -> -2.12
  • a(3, 1) = 0.5 * a(2, 1) + 0.3 * a(2, 0) = 0.285 -> -1.26
  • a(3, 2) = 0.7 * a(2, 2) + 0.4 * a(2, 1) = 0.24 -> -1.43
  • a(4, 2) = 0.8 * a(3, 2) + 0.6 * a(3, 1) = 0.363 -> -1.01

And the betas are as follows in probability and log space:

  • b(5, 2) = 1.0 -> 0.0
  • b(4, 2) = 0.8 -> -0.22
  • b(4, 1) = 0.6 -> -0.51
  • b(3, 2) = 0.7 * b(4, 2) = 0.56 -> -0.58
  • b(3, 1) = 0.5 * b(4, 1) + 0.4 * b(4, 2) = 0.62 -> -0.48
  • b(3, 0) = 0.3 * b(4, 1) = 0.18 -> -1.71
  • b(2, 1) = 0.5 * b(3, 1) + 0.4 * b(3, 2) = 0.534 -> -0.63
  • b(2, 0) = 0.5 * b(3, 0) + 0.4 * b(3, 1) = 0.338 -> -1.08
  • b(1, 0) = 0.6 * b(2, 0) + 0.3 * b(2, 1) = 0.363 -> -1.01

As we can see $\alpha(4, 2) = -1.01 = \beta(0, 1)$ is the overall log-likelihood.

Now, the gradients with respect to all the logits can be computed as

// t = 1
0.04, -0.14, 0.1,  // s = 0
0.0, 0.0, 0.0,  // s = 1
0.0, 0.0, 0.0,  // s = 2

// t = 2
0.13, -0.19, 0.06,  // s = 0
-0.04, 0.04, -0.01,  // s = 1
0.0, 0.0, 0.0,  // s = 2

// t = 3
0.06, -0.1, 0.04,  // s = 0
0.01, 0.07, -0.08,  // s = 1
-0.06, 0.04, 0.02,  // s = 2

// t = 4
0.0, 0.0, 0.0,  // s = 0
0.14, 0.05, -0.19,  // s = 1
-0.11, 0.05, 0.05   // s = 2