# Embedding derivative derivation

Sometimes you can't use automatic differentiation to train a model and then you have to do the derivation yourself. This derivation is for a simple two-layer model where the input layer is followed by an embedding layer which is followed by a fully connected softmax layer. It is based on some old and new matrix algebra (and calculus) useful for statistics.

## Softmax model

A single layer softmax model is defined like

$p(\mathbf{y}|\mathbf{x}, \Theta) = \frac{1}{Z} \prod_{c=1}^C \exp\{\mathbf{x}^T \Theta_c \}^{y_c} ,\\ Z=\sum_{c=1}^C \exp\{\mathbf{x}^T \Theta_c \} ,$

where

• $$C$$ is the number of classes,
• $$\mathbf{y}$$ is the vector of one-hot encoded labels,
• $$\mathbf{x}$$ is the vector of features,
• $$\Theta$$ is the matrix of parameters ($$\Theta_c$$ being the cth column), and
• $$Z$$ is the normalising constant.

## Softmax with embeddings

Now we want $$\mathbf{x}$$ to represent a sequence of words/characters. One way to do that is to concatenate a bunch of one-hot encoded vectors

$\mathbf{x}_{MV\times1} = [ \mathbf{x}_1^T\ \mathbf{x}_2^T \dots \mathbf{x}_M^M ]^T$

where each $$\mathbf{x}_i$$ is a $$V \times 1$$ vector that chooses an element out of the vocabulary.

We want each 1 in the input vector to select a row from the embedding matrix $$E_{V\times H}$$, where $$H$$ is the dimension of each embedding vector. So, $$\mathbf{x}^T R$$, where $$R$$ is our mystery matrix, should equal

$[\mathbf{x}_1^T E \ \ \mathbf{x}_2^T E \dots \mathbf{x}_M^T E ].$

This happens when $R = \begin{bmatrix} E & 0 & \ldots & 0 \\ 0 & E & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \ldots & E \end{bmatrix}_{MV\times MV} ,$

which is just the Kronecker product $$I \otimes E$$.

Now if we define

$\Theta_c = (I\otimes E)W_c,$

and substitute into the softmax definition above we get

$p(\mathbf{y}|\mathbf{x}, W, E) = \frac{1}{Z} \prod_{c=1}^C \exp\{\mathbf{x}^T (I\otimes E)W_c \}^{y_c} ,$

where

• $$V$$ is the number of embeddings (the vocabulary size),
• $$M$$ is the number of embeddings per data point,
• $$H$$ is the dimension of each embedding,
• $$\mathbf{y}_{C\times 1}$$ is the vector of one-hot encoded labels,
• $$\mathbf{x}_{VM\times 1}$$ is the vector of one-hot encoded features,
• $$I_{M\times M}$$ is an identity matrix,
• $$E_{V\times H}$$ is the latent embedding matrix, and
• $${W_c}_{MH\times 1}, \quad c \in 1 .. C \quad$$ are vectors of weights.

## Log likelihood

The log likelihood - our objective function - is

\begin{align} \mathcal{L} &= \sum_{n=1}^N \log p(\mathbf{y}_n|\mathbf{x}_n W E) \\ &= \sum_{n=1}^N \sum_{c=1}^C {y_{nc} \mathbf{x}_n^T (I\otimes E)W_c } - \log Z_n , \end{align}

## Derivative with respect to the weights

The differential in terms of $$W_c$$ is then

\begin{align} d\mathcal{L} &= \sum_{n=1}^N {y_{nc} \mathbf{x}_n^T (I\otimes E)dW_c } - \frac{1}{Z_n} d Z_n \\ &= \sum_{n=1}^N {y_{nc} \mathbf{x}_n^T (I\otimes E)dW_c } - \frac{1}{Z_n} d\exp\{\mathbf{x}_n^T (I\otimes E)W_c \} \\ &= \sum_{n=1}^N {y_{nc} \mathbf{x}_n^T (I\otimes E)dW_c } - \frac{1}{Z_n} \exp\{\mathbf{x}_n^T (I\otimes E)W_c \} \mathbf{x}_n^T (I\otimes E)dW_c \end{align}

which implies

$\frac{d\mathcal{L}}{dW_c} = \sum_{n=1}^N \left( y_{nc} - \frac{1}{Z_n} \exp\{\mathbf{x}_n^T (I\otimes E)W_c \} \right) \mathbf{x}_n^T (I\otimes E)$

## Derivative with respect to the embeddings

Here we use the trace operator to rearrange elements so that the differential is last. To rearrange the Kronecker product we also need the commutation matrix $$K_{p,q}$$ and the vector transposition operator (vec-transpose) $$(\cdot)^{(q)}$$.

\begin{align} d\mathcal{L} =& d\mathrm{tr} ( \sum_{n=1}^N \sum_{c=1}^C {y_{nc} \mathbf{x}_n^T (I\otimes E)W_c } ) - d\mathrm{tr} ( \frac{1}{Z_n} d Z_n ) \\ =& \sum_{n=1}^N \sum_{c=1}^C y_{nc} d \mathrm{tr} \left( W_c \mathbf{x}_n^T K_{M,V} (E\otimes I) K_{H, M} I \right) - \frac{1}{Z_n} \mathrm{tr} \left( d \sum_{c=1}^C \exp\{\mathbf{x}_n^T (I\otimes E) W_c \} \right) \\ =& \sum_{n=1}^N \sum_{c=1}^C y_{nc} d \mathrm{tr} \left( \left( W_c \mathbf{x}_n^T \right)^{(M)} \left( K_{M,V} (E\otimes I) K_{H, M} I \right)^{(M)} \right) \\ &- \frac{1}{Z_n} \exp\{\mathbf{x}_n^T d(I\otimes E) W_c \} d \mathrm{tr} \left( \mathbf{x}_n^T K_{M,V} (E\otimes I) K_{H, M} W_c \right) \\ =& \sum_{n=1}^N \sum_{c=1}^C d \mathrm{tr} \left( K_{H, M}^{(M)T} (I\otimes I) \left( W_c \mathbf{x}_n^T K_{M,V} \right)^{(M)T} E \right) \left(y_{nc} - \frac{1}{Z_n} \exp\{\mathbf{x}_n^T (I\otimes E) W_c \} \right) \\ =& \sum_{n=1}^N \sum_{c=1}^C \left(y_{nc} - \frac{1}{Z_n} \exp\{\mathbf{x}_n^T (I\otimes E) W_c \} \right) \mathrm{tr} \left( K_{H, M}^{(M)T} (I\otimes I) \left( W_c \mathbf{x}_n^T K_{M,V} \right)^{(M)T} dE \right) . \end{align}

The derivative is then

$\frac{d \mathcal{L}}{d E} = \sum_{n=1}^N \sum_{c=1}^C \left(y_{nc} - \frac{1}{Z_n} \exp\{\mathbf{x}_n^T (I\otimes E) W_c \} \right) K_{H, M}^{(M)T} \left( W_c \mathbf{x}_n^T K_{M,V} \right)^{(M)T} .$

Well, that seems to be the answer! The $$K_{H, M}^{(M)T} \left( W_c \mathbf{x}_n^T K_{M,V} \right)^{(M)T}$$ terms are basically just the outer product of the features and weights rearranged into the shape of $$E^T$$.

Next I'll show a naive implementation in numpy and some of the character embeddings it learned.

Written on December 30, 2015