# 3.4. Softmax Regression¶

Over the last two sections we worked through how to implement a linear regression model, both from scratch and using Gluon to automate most of the repetitive work like allocating and initializing parameters, defining loss functions, and implementing optimizers.

Regression is the hammer we reach for when we want to answer how much? or how many? questions. If you want to predict the number of dollars (the price) at which a house will be sold, or the number of wins a baseball team might have, or the number of days that a patient will remain hospitalized before being discharged, then you’re probably looking for a regression model.

In reality, we’re more often interested in making categorical assignments.

• Does this email belong in the spam folder or the inbox*?
• How likely is this customer to sign up for subscription service?*
• What is the object in the image (donkey, dog, cat, rooster, etc.)?
• Which object is a customer most likely to purchase?

When we’re interested in either assigning datapoints to categories or assessing the probability that a category applies, we call this task classification. The issue with the models that we studied so far is that they cannot be applied to problems of probability estimation.

## 3.4.1. Classification Problems¶

Let’s start with an admittedly somewhat contrived image problem where the input image has a height and width of 2 pixels and the color is grayscale. Thus, each pixel value can be represented by a scalar. We record the four pixels in the image as $$x_1, x_2, x_3, x_4$$. We assume that the actual labels of the images in the training data set are “cat”, “chicken” or “dog” (assuming that the three animals can be represented by 4 pixels).

To represent these labels we have two choices. Either we set $$y \in \{1, 2, 3\}$$, where the integers represent {dog, cat, chicken} respectively. This is a great way of storing such information on a computer. It would also lend itself rather neatly to regression, but the ordering of outcomes imposes some quite unnatural ordering. In our toy case, this would presume that cats are more similar to chickens than to dogs, at least mathematically. It doesn’t work so well in practice either, which is why statisticians invented an alternative approach: one hot encoding via

$y \in \{(1, 0, 0), (0, 1, 0), (0, 0, 1)\}$

That is, $$y$$ is viewed as a three-dimensional vector, where $$(1,0,0)$$ corresponds to “cat”, $$(0,1,0)$$ to “chicken” and $$(0,0,1)$$ to “dog”.

### 3.4.1.1. Network Architecture¶

If we want to estimate multiple classes, we need multiple outputs, matching the number of categories. This is one of the main differences to regression. Because there are 4 features and 3 output animal categories, the weight contains 12 scalars ($$w$$ with subscripts) and the bias contains 3 scalars ($$b$$ with subscripts). We compute these three outputs, $$o_1, o_2$$, and $$o_3$$, for each input:

\begin{split}\begin{aligned} o_1 &= x_1 w_{11} + x_2 w_{21} + x_3 w_{31} + x_4 w_{41} + b_1,\\ o_2 &= x_1 w_{12} + x_2 w_{22} + x_3 w_{32} + x_4 w_{42} + b_2,\\ o_3 &= x_1 w_{13} + x_2 w_{23} + x_3 w_{33} + x_4 w_{43} + b_3. \end{aligned}\end{split}

The neural network diagram below depicts the calculation above. Like linear regression, softmax regression is also a single-layer neural network. Since the calculation of each output, $$o_1, o_2$$, and $$o_3$$, depends on all inputs, $$x_1$$, $$x_2$$, $$x_3$$, and $$x_4$$, the output layer of the softmax regression is also a fully connected layer.

### 3.4.1.2. Softmax Operation¶

The chosen notation is somewhat verbose. In vector form we arrive at $$\mathbf{o} = \mathbf{W} \mathbf{x} + \mathbf{b}$$, which is much more compact to write and code. Since the classification problem requires discrete prediction output, we can use a simple approach to treat the output value $$o_i$$ as the confidence level of the prediction category $$i$$. We can choose the class with the largest output value as the predicted output, which is output $$\operatorname*{argmax}_i o_i$$. For example, if $$o_1$$, $$o_2$$, and $$o_3$$ are 0.1, 10, and 0.1, respectively, then the prediction category is 2, which represents “chicken”.

However, there are two problems with using the output from the output layer directly. On the one hand, because the range of output values from the output layer is uncertain, it is difficult for us to visually judge the meaning of these values. For instance, the output value 10 from the previous example indicates a level of “very confident” that the image category is “chicken”. That is because its output value is 100 times that of the other two categories. However, if $$o_1=o_3=10^3$$, then an output value of 10 means that the chance for the image category to be chicken is very low. On the other hand, since the actual label has discrete values, the error between these discrete values and the output values from an uncertain range is difficult to measure.

We could try forcing the outputs to correspond to probabilities, but there’s no guarantee that on new (unseen) data the probabilities would be nonnegative, let alone sum up to 1. For this kind of discrete value prediction problem, statisticians have invented classification models such as (softmax) logistic regression. Unlike linear regression, the output of softmax regression is subjected to a nonlinearity which ensures that the sum over all outcomes always adds up to 1 and that none of the terms is ever negative. The nonlinear transformation works as follows:

$\hat{\mathbf{y}} = \mathrm{softmax}(\mathbf{o}) \text{ where } \hat{y}_i = \frac{\exp(o_i)}{\sum_j \exp(o_j)}$

It is easy to see $$\hat{y}_1 + \hat{y}_2 + \hat{y}_3 = 1$$ with $$0 \leq \hat{y}_i \leq 1$$ for all $$i$$. Thus, $$\hat{y}$$ is a proper probability distribution and the values of $$o$$ now assume an easily quantifiable meaning. Note that we can still find the most likely class by

$\hat{\imath}(\mathbf{o}) = \operatorname*{argmax}_i o_i = \operatorname*{argmax}_i \hat y_i$

So, the softmax operation does not change the prediction category output but rather it gives the outputs $$\mathbf{o}$$ proper meaning. Summarizing it all in vector notation we get $${\mathbf{o}}^{(i)} = \mathbf{W} {\mathbf{x}}^{(i)} + {\mathbf{b}}$$ where $${\hat{\mathbf{y}}}^{(i)} = \mathrm{softmax}({\mathbf{o}}^{(i)})$$.

### 3.4.1.3. Vectorization for Minibatches¶

To improve computational efficiency further, we usually carry out vector calculations for mini-batches of data. Assume that we are given a mini-batch $$\mathbf{X}$$ of examples with dimensionality $$d$$ and batch size $$n$$. Moreover, assume that we have $$q$$ categories (outputs). Then the minibatch features $$\mathbf{X}$$ are in $$\mathbb{R}^{n \times d}$$, weights $$\mathbf{W} \in \mathbb{R}^{d \times q}$$ and the bias satisfies $$\mathbf{b} \in \mathbb{R}^q$$.

\begin{split}\begin{aligned} \mathbf{O} &= \mathbf{X} \mathbf{W} + \mathbf{b} \\ \hat{\mathbf{Y}} & = \mathrm{softmax}(\mathbf{O}) \end{aligned}\end{split}

This accelerates the dominant operation: $$\mathbf{W} \mathbf{X}$$ from a matrix-vector to a matrix-matrix product. The softmax itself can be computed by exponentiating all entries in $$\mathbf{O}$$ and then normalizing them by the sum appropriately.

## 3.4.2. Loss Function¶

Now that we have some mechanism for outputting probabilities, we need to transform this into a measure of how accurate things are, i.e. we need a loss function. For this we use the same concept that we already encountered in linear regression, namely likelihood maximization.

### 3.4.2.1. Log-Likelihood¶

The softmax function maps $$\mathbf{o}$$ into a vector of probabilities corresponding to various outcomes, such as $$p(y=\mathrm{cat}|\mathbf{x})$$. This allows us to compare the estimates with reality, simply by checking how well it predicted what we observe.

$p(Y|X) = \prod_{i=1}^n p(y^{(i)}|x^{(i)}) \text{ and thus } -\log p(Y|X) = \sum_{i=1}^n -\log p(y^{(i)}|x^{(i)})$

Minimizing $$-\log p(Y|X)$$ corresponds to predicting things well. This yields the loss function (we dropped the superscript $$(i)$$ to avoid notation clutter):

$l = -\log p(y|x) = - \sum_j y_j \log \hat{y}_j$

Here we used that by construction $$\hat{y} = \mathrm{softmax}(\mathbf{o})$$ and moreover, that the vector $$\mathbf{y}$$ consists of all zeroes but for the correct label, such as $$(1, 0, 0)$$. Hence the the sum over all coordinates $$j$$ vanishes for all but one term. Since all $$\hat{y}_j$$ are probabilities, their logarithm is never larger than $$0$$. Consequently, the loss function is minimized if we correctly predict $$y$$ with certainty, i.e. if $$p(y|x) = 1$$ for the correct label.

### 3.4.2.2. Softmax and Derivatives¶

Since the Softmax and the corresponding loss are so common, it is worth while understanding a bit better how it is computed. Plugging $$o$$ into the definition of the loss $$l$$ and using the definition of the softmax we obtain:

$l = -\sum_j y_j \log \hat{y}_j = \sum_j y_j \log \sum_k \exp(o_k) - \sum_j y_j o_j = \log \sum_k \exp(o_k) - \sum_j y_j o_j$

To understand a bit better what is going on, consider the derivative with respect to $$o$$. We get

$\partial_{o_j} l = \frac{\exp(o_j)}{\sum_k \exp(o_k)} - y_j = \mathrm{softmax}(\mathbf{o})_j - y_j = \Pr(y = j|x) - y_j$

In other words, the gradient is the difference between what the model thinks should happen, as expressed by the probability $$p(y|x)$$, and what actually happened, as expressed by $$y$$. In this sense, it is very similar to what we saw in regression, where the gradient was the difference between the observation $$y$$ and estimate $$\hat{y}$$. This seems too much of a coincidence, and indeed, it isn’t. In any exponential family model the gradients of the log-likelihood are given by precisely this term. This fact makes computing gradients a lot easier in practice.

### 3.4.2.3. Cross-Entropy Loss¶

Now consider the case where we don’t just observe a single outcome but maybe, an entire distribution over outcomes. We can use the same representation as before for $$y$$. The only difference is that rather than a vector containing only binary entries, say $$(0, 0, 1)$$, we now have a generic probability vector, say $$(0.1, 0.2, 0.7)$$. The math that we used previously to define the loss $$l$$ still works out fine, just that the interpretation is slightly more general. It is the expected value of the loss for a distribution over labels.

$l(\mathbf{y}, \hat{\mathbf{y}}) = - \sum_j y_j \log \hat{y}_j$

This loss is called the cross-entropy loss. It is one of the most commonly used ones for multiclass classification. To demystify its name we need some information theory. The following section can be skipped if needed.

## 3.4.3. Information Theory Basics¶

Information theory deals with the problem of encoding, decoding, transmitting and manipulating information (aka data), preferentially in as concise form as possible.

### 3.4.3.1. Entropy¶

A key concept is how many bits of information (or randomness) are contained in data. It can be measured as the entropy of a distribution $$p$$ via

$H[p] = \sum_j - p(j) \log p(j)$

One of the fundamental theorems of information theory states that in order to encode data drawn randomly from the distribution $$p$$ we need at least $$H[p]$$ ‘nats’ to encode it. If you wonder what a ‘nat’ is, it is the equivalent of bit but when using a code with base $$e$$ rather than one with base 2. One nat is $$\frac{1}{\log(2)} \approx 1.44$$ bit. $$H[p] / 2$$ is often also called the binary entropy.

To make this all a bit more theoretical consider the following: $$p(1) = \frac{1}{2}$$ whereas $$p(2) = p(3) = \frac{1}{4}$$. In this case we can easily design an optimal code for data drawn from this distribution, by using 0 to encode 1, 10 for 2 and 11 for 3. The expected number of bit is $$1.5 = 0.5 * 1 + 0.25 * 2 + 0.25 * 2$$. It is easy to check that this is the same as the binary entropy $$H[p] / \log 2$$.

### 3.4.3.2. Kullback Leibler Divergence¶

One way of measuring the difference between two distributions arises directly from the entropy. Since $$H[p]$$ is the minimum number of bits that we need to encode data drawn from $$p$$, we could ask how well it is encoded if we pick the ‘wrong’ distribution $$q$$. The amount of extra bits that we need to encode $$q$$ gives us some idea of how different these two distributions are. Let us compute this directly - recall that to encode $$j$$ using an optimal code for $$q$$ would cost $$-\log q(j)$$ nats, and we need to use this in $$p(j)$$ of all cases. Hence we have

$D(p\|q) = -\sum_j p(j) \log q(j) - H[p] = \sum_j p(j) \log \frac{p(j)}{q(j)}$

Note that minimizing $$D(p\|q)$$ with respect to $$q$$ is equivalent to minimizing the cross-entropy loss. This can be seen directly by dropping $$H[p]$$ which doesn’t depend on $$q$$. We thus showed that softmax regression tries the minimize the surprise (and thus the number of bits) we experience when seeing the true label $$y$$ rather than our prediction $$\hat{y}$$.

## 3.4.4. Model Prediction and Evaluation¶

After training the softmax regression model, given any example features, we can predict the probability of each output category. Normally, we use the category with the highest predicted probability as the output category. The prediction is correct if it is consistent with the actual category (label). In the next part of the experiment, we will use accuracy to evaluate the model’s performance. This is equal to the ratio between the number of correct predictions and the total number of predictions.

## 3.4.5. Summary¶

• We introduced the softmax operation which takes a vector maps it into probabilities.
• Softmax regression applies to classification problems. It uses the probability distribution of the output category in the softmax operation.
• Cross entropy is a good measure of the difference between two probability distributions. It measures the number of bits needed to encode the data given our model.

## 3.4.6. Exercises¶

1. Show that the Kullback-Leibler divergence $$D(p\|q)$$ is nonnegative for all distributions $$p$$ and $$q$$. Hint - use Jensen’s inequality, i.e. use the fact that $$-\log x$$ is a convex function.
2. Show that $$\log \sum_j \exp(o_j)$$ is a convex function in $$o$$.
3. We can explore the connection between exponential families and the softmax in some more depth
• Compute the second derivative of the cross entropy loss $$l(y,\hat{y})$$ for the softmax.
• Compute the variance of the distribution given by $$\mathrm{softmax}(o)$$ and show that it matches the second derivative computed above.
4. Assume that we three classes which occur with equal probability, i.e. the probability vector is $$(\frac{1}{3}, \frac{1}{3}, \frac{1}{3})$$.
• What is the problem if we try to design a binary code for it? Can we match the entropy lower bound on the number of bits?
• Can you design a better code. Hint - what happens if we try to encode two independent observations? What if we encode $$n$$ observations jointly?
5. Softmax is a misnomer for the mapping introduced above (but everyone in deep learning uses it). The real softmax is defined as $$\mathrm{RealSoftMax}(a,b) = \log (\exp(a) + \exp(b))$$.
• Prove that $$\mathrm{RealSoftMax}(a,b) > \mathrm{max}(a,b)$$.
• Prove that this holds for $$\lambda^{-1} \mathrm{RealSoftMax}(\lambda a, \lambda b)$$, provided that $$\lambda > 0$$.
• Show that for $$\lambda \to \infty$$ we have $$\lambda^{-1} \mathrm{RealSoftMax}(\lambda a, \lambda b) \to \mathrm{max}(a,b)$$.
• What does the soft-min look like?
• Extend this to more than two numbers.