1
$\begingroup$

In the last layer of  'CNNs'   it is common to use softmax activation functions for multi-class classification.I would like to know if is it necessary using a softmax activation function when creating a CNN for image classification task, and does it have nothing to do with the optimizer used to train the model ?

$\endgroup$
6
  • 1
    $\begingroup$ "Necessary" for what purpose? What problem are you trying to solve, and how does softmax fit into solving that problem? This post develops why softmax is used. Does this answer your question? Why or why not? stats.stackexchange.com/questions/419751/… $\endgroup$
    – Sycorax
    Aug 26 '20 at 20:27
  • $\begingroup$ To be more clear, I am using a CNN for image classification using the CIFAR10 dataset, My CNN contains 3 fully connected layers .I have applied Relu activation function on both 1st and 2nd one ,I was wondering if I have to use a softmax on the 3rd layer to have a proper model for classifying these images.(knowing that i used the same model without any activation function on the 3rd fully connected layer and i got a pretty good result) $\endgroup$
    – Ka_
    Aug 26 '20 at 22:03
  • $\begingroup$ So one way to think about this particular model is that you're assigning a probability to each of $k$ classes, and softmax is nice for the reasons outlined in the other post. If you're not using a probability model (perhaps using softmax or another activation), but just returning any real number, it doesn't seem like that would play nicely with common probability-based losses like binary cross-entropy because $\log(x) \forall x\le 0$ is not a real number. $\endgroup$
    – Sycorax
    Aug 26 '20 at 22:27
  • $\begingroup$ But working on a different scale can work if you adjust the loss function. See: stats.stackexchange.com/questions/475589/… $\endgroup$
    – Sycorax
    Aug 26 '20 at 22:27
  • 1
    $\begingroup$ $-\log(x)$ for positive $x$ decreases without bound. This means the model never stops training: it can always lower the loss by increasing the weights and biases to push $x$ to be larger. When using softmax, $x$ is a probability & the loss achieves the minimum of 0 at $x=1$. Also, softmax forces a tradeoff, but ReLUs don't, so ReLUs give a small loss by just returning a vector of very large constants, regardless of the input! Seems bad! $\endgroup$
    – Sycorax
    Aug 27 '20 at 14:35
1
$\begingroup$

Softmax outputs a probability vector. That means that

  • the elements are nonnegative and
  • the elements sum to 1.

To train a classification model with $m \ge 3$ classes, the standard approach is to use softmax as the final activation with multinomial cross-entropy loss. For a single instance, the loss is

$$ \begin{align} \mathcal{L} &= -\sum_{j=1}^m y_j \log(p_j) \end{align} $$

where $y$ is a vector with one value of 1 and the rest zero and $p_j$ are our predicted probabilities from the softmax. If the single value of 1 in $y$ is at index $k$, then the loss achieves a minimum value of 0 when $p_k = 1$. When $p_k=1$, this implies that the rest of the $p_{j\neq k}$ are all 0 (because $p$ is a vector of probabilities, so the total is 1).

In a comment, OP proposes using ReLU instead of softmax. However, there are some problems with this proposal.

  1. You can still encounter $\log(0)$, because ReLU can return zeros. (But this is not fatal, because we can "patch" it; a strictly positive ReLU activation like $\text{ReLU}(x)+\epsilon$ for some small $\epsilon>0$ avoids this.)

  2. For ReLUs, the sum of $p$ can be any nonnegative value. This is not a probability. Because $-\log(p_k)$ decreases without bound as $p_k$ increases, the model will never stop training. (But this isn't fatal; penalizing the weights and biases or or otherwise constraining them will prevent them from drifting away to $\pm\infty$.) On the other hand, for softmax, the largest $p_k$ can ever be is 1, so minimum loss is 0.

  3. ReLU does not force a tradeoff among the units, whereas softmax does. What this means is that if you use softmax want to increase the value of $p_k$, you have to decrease $\sum_{i\neq k} p_i$. The loss will be high whenever $p$ and $y$ are different. By contrast, the ReLU model can just return some vector of constants and have the same loss, no matter what the label is. Consider the three-class case where the correct prediction is the second class, we have $$\mathcal{L}=-0\times \log(c)-1\times\log(c)-0\times\log(c)=-\log(c).$$ Likewise, this same loss of is obtained for the same $p$ and any label vector $y$.

Clearly, (3) is fatal because the model has no useful information about which class is the most likely. A model that can always reduce the loss by ignoring the input entirely is a bogus model.

The key detail about softmax is that it forces a tradeoff among the values of $p$, because assigning any probability to the incorrect class is penalized. The only softmax model which has 0 multinomial cross-entropy loss is the model that assigns probability of 1 to the correct class for all instances.


Softmax isn't the only function you could use. A function like

$$ f(z)_i = \frac{\text{softplus}(z_i)}{\sum_i \text{softplus}(z_i)} $$ where the softplus function is

$$ \text{softplus}(x)=\log(1+\exp(x)) $$ could also work for a multi-class classification model because $f$ is

  • positive (avoids divide by zero),
  • non-negative and sums to 1 (is a probability), and
  • monotonic increasing.

We care about monotonicity because we want the property that large $z_i$ imply large probabilities. A non-monotonic function like squaring or absolute value means that we predict a certain class for very large or very small values. See: Why is softmax function used to calculate probabilities although we can divide each value by the sum of the vector?

$\endgroup$
1
  • 1
    $\begingroup$ Thank you so much for your explanation .It seems clear for me now . $\endgroup$
    – Ka_
    Aug 27 '20 at 15:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.