SOFTMAX ACTIVATION:


This is a good resource.

In multiclass classification networks the softmax function:

The last hidden layer produces output values forming a vector \(\vec x = \mathbf x\). The output neuronal layer is meant to classify among \(K=1,\dots,k\) categories with a SoftMax activation function assigning conditional probabilities (given \(\mathbf x\)) to each one the \(K\) categories. In each node in the final (or ouput) layer the pre-activated values (logit values) will consist of the scalar products \(\mathbf{w}_j^\top\mathbf{x}\), where \(\mathbf w_j\in\{\mathbf{w}_1, \mathbf{w}_2,\dots,\mathbf{w}_k\}\). In other words, each category, \(k\) will have a different vector of weights pointing at it, determining the contribution of each element in the output of the previous layer (including a bias), encapsulated in \(\mathbf x\). However, the activation of this final layer will not take place element-wise (as for example with a sigmoid function in each neuron), but rather through the application of a SoftMax function, which will map a vector in \(\mathbb R^k\) to a vector of \(K\) elements in \([0,1]\). Here is a made-up NN to classify colors:

The softmax as

\[ \sigma(j)=\frac{\exp(\mathbf{w}_j^\top \mathbf x)}{\sum_{k=1}^K \exp(\mathbf{w}_k^\top\mathbf x)}=\frac{\exp(z_j)}{\sum_{k=1}^K \exp(z_k)}\]

This will result in a normalization of the output adding up to \(1\), interpretable as a probability mass functionn.

From Wikipedia:

"In probability theory, the output of the softmax function can be used to represent a categorical distribution – that is, a probability distribution over K different possible outcomes."

Notice that there is truly no strict need for an activation function. As in this post:

"At the end of a network, you can either use nothing (logits) and get a multi parameter regression." 

So why do it?

Advantages:

  1. softmax is optimal for maximum-likelihood estimation of the model parameters.

  2. The properties of softmax (all output values in the range (0, 1) and sum up to 1.0) make it suitable for a probabilistic interpretation that’s very useful in machine learning.

  3. Softmax normalization is a way of reducing the influence of extreme values or outliers in the data without removing data points from the set.


Although we can use mean squared error, cross-entropy is the preferred loss function for classification NN with softmax activation in the last layer. It is given by the function:

\[\begin{eqnarray} C = -\frac{1}{k} \sum_{k=1}^K \left[y_k \log (\sigma(z_k)) + (1-y_k ) \log (1-\sigma(z_k)) \right] \end{eqnarray}\]

As explained here, average cross-entropy (ACE) would be calculated as:

computed       | targets              | correct?
-----------------------------------------------
0.3  0.3  0.4  | 0  0  1 (democrat)   | yes
0.3  0.4  0.3  | 0  1  0 (republican) | yes
0.1  0.2  0.7  | 1  0  0 (other)      | no

\[-\left( (\log(0.3)\times 0) + (\log (0.3)\times 0) + (\log (0.4)\times1) \right) = -\log(0.4)\]

With one-hot encoding, the \(y\) vector will eliminate all elements, except for one (the correct value for the example). The expression \(y_k\log \hat y_k + (1-y_k)\log(1-\hat y_k)\) will become \(\log \hat y_k\) so that if the calculated probability for that category is close to \(1\), the loss function will approach zero, whereas, if the probabily (output of the softmax for that category) is close to zero, the loss will tend to infinity.


Derivative of the softmax function with respect to the logit \((z_j =\mathbf W_j^\top \cdot \mathbf x)\):

Computing the

\[\frac{\partial}{\partial z_i}\sigma(j)=\frac{\partial}{\partial z_i}\frac{\exp(z_j)}{\sum_{k=1}^K \exp(z_k)}\]

The derivative of \(\sum_{k=1}^K \exp(z_k)\) with respect to any \(z_i\) will be \(\exp(z_i)\). As for the numerator, \(\exp(z_j)\) the derivative will be \(\exp(z_i)\) if and only if \(z_i = z_j\); otherwise the derivative is \(0\).

If \(i = j\), and using the quotient rule,

\[\begin{align}\frac{\partial}{\partial z_i}\frac{\exp(z_j)}{\sum_{k=1}^K \exp(z_k)} &= \frac{\mathbf \exp(z_j)\sum_{k=1}^K \exp(z_k)\quad - \exp(z_i)\exp(z_j)}{\left[\sum_{k=1}^K \exp(z_k)\right]^2}\\[2ex] &= \frac{\exp(z_j)}{\sum_{k=1}^K \exp(z_k)}\frac{\sum_{k=1}^K \exp(z_k)-\exp(z_i)}{\sum_{k=1}^K \exp(z_k)}\\[2ex] &=\sigma(z_j)\,(1 - \sigma(z_i)) \end{align}\]

If on the other hand, \(i \neq j\):

\[\begin{align}\frac{\partial}{\partial z_i}\frac{\exp(z_j)}{\sum_{k=1}^K \exp(z_k)} &= \frac{0\quad - \exp(z_i)\exp( z_j)}{\left[\sum_{k=1}^K \exp(z_k)\right]^2}\\[2ex] &= - \frac{\exp(z_j)}{\sum_{k=1}^K \exp(z_k)}\frac{\exp(z_i)}{\sum_{k=1}^K \exp(z_k)}\\[2ex] &=-\sigma(z_j)\,\sigma(z_i) \end{align}\]

These two scenarios can be brought together as

\[\frac{\partial}{\partial z_i} \sigma(z_j)= \sigma(z_j)\left(\delta_{ij}-\sigma(z_i)\right)\]


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