Deriving the Learning Correction

For gradient descent, we need to derive the update to the matrix \({\bf A}\) based on training on a set of our data, \(({\bf x}^k, {\bf y}^k)\).

Important

The derivation we do here is specific to our choice of loss function, \(\mathcal{L}(A_{ij})\) and activation function, \(g(\xi)\).

Let’s start with our cost function:

\[\mathcal{L}(A_{ij}) = \sum_{i=1}^{N_\mathrm{out}} (z_i - y_i^k)^2 = \sum_{i=1}^{N_\mathrm{out}} \Biggl [ g\biggl (\underbrace{\sum_{j=1}^{N_\mathrm{in}} A_{ij} x^k_j}_{\equiv \alpha_i} \biggr ) - y^k_i \Biggr ]^2\]

where we’ll refer to the product \({\boldsymbol \alpha} \equiv {\bf Ax}\) to help simplify notation.

We can compute the derivative with respect to a single matrix element, \(A_{pq}\) by applying the chain rule:

\[\frac{\partial \mathcal{L}}{\partial A_{pq}} = 2 \sum_{i=1}^{N_\mathrm{out}} (z_i - y^k_i) \left . \frac{\partial g}{\partial \xi} \right |_{\xi=\alpha_i} \frac{\partial \alpha_i}{\partial A_{pq}}\]

with

\[\frac{\partial \alpha_i}{\partial A_{pq}} = \sum_{j=1}^{N_\mathrm{in}} \frac{\partial A_{ij}}{\partial A_{pq}} x^k_j = \sum_{j=1}^{N_\mathrm{in}} \delta_{ip} \delta_{jq} x^k_j = \delta_{ip} x^k_q\]

and for \(g(\xi)\), we will assume the sigmoid function,so

\[\frac{\partial g}{\partial \xi} = \frac{\partial}{\partial \xi} \frac{1}{1 + e^{-\xi}} =- (1 + e^{-\xi})^{-2} (- e^{-\xi}) = g(\xi) \frac{e^{-\xi}}{1+ e^{-\xi}} = g(\xi) (1 - g(\xi))\]

which gives us:

\[\begin{align*} \frac{\partial \mathcal{L}}{\partial A_{pq}} &= 2 \sum_{i=1}^{N_\mathrm{out}} (z_i - y^k_i) z_i (1 - z_i) \delta_{ip} x^k_q \\ &= 2 (z_p - y^k_p) z_p (1- z_p) x^k_q \end{align*}\]

where we used the fact that the \(\delta_{ip}\) means that only a single term contributes to the sum.

Note that:

• \(e_p^k \equiv (z_p - y_p^k)\) is the error on the output layer, and the correction is proportional to the error (as we would expect).

• The \(k\) superscripts here remind us that this is the result of only a single pair of data from the training set.

Now \({\bf z}\) and \({\bf y}^k\) are all vectors of size \(N_\mathrm{out} \times 1\) and \({\bf x}^k\) is a vector of size \(N_\mathrm{in} \times 1\), so we can write this expression for the matrix as a whole as:

\[\frac{\partial f}{\partial {\bf A}} = 2 ({\bf z} - {\bf y}^k) \circ {\bf z} \circ (1 - {\bf z}) \cdot ({\bf x}^k)^\intercal\]

where the operator \(\circ\) represents element-by-element multiplication (the Hadamard product).

Performing the update

We could do the update like we just saw with our gradient descent example: take a single data point, \(({\bf x}^k, {\bf y}^k)\) and do the full minimization, continually estimating the correction, \(\partial \mathcal{L}/\partial {\bf A}\) and updating \({\bf A}\) until we reach a minimum. The problem with this is that \(({\bf x}^k, {\bf y}^k)\) is only one point in our training data, and there is no guarantee that if we minimize completely with point \(k\) that we will also be a minimum with point \(k+1\).

Instead we take multiple passes through the training data (called epochs) and apply only a single push in the direction that gradient descent suggests, scaled by a learning rate \(\eta\).

The overall minimization appears as:

* Loop over epochs

• Loop over the training data, \(\{ ({\bf x}^0, {\bf y}^0), ({\bf x}^1, {\bf y}^1), \ldots \}\). We’ll refer to the current training pair as \(({\bf x}^k, {\bf y}^k)\)

  • Propagate \({\bf x}^k\) through the network, getting the output \({\bf z} = g({\bf A x}^k)\)

  • Compute the error on the output layer, \({\bf e}^k = {\bf z} - {\bf y}^k\)

  • Update the matrix \({\bf A}\) according to:

    \[{\bf A} \leftarrow {\bf A} - 2 \,\eta\, {\bf e}^k \circ {\bf z} \circ (1 - {\bf z}) \cdot ({\bf x}^k)^\intercal\]