What's the relationship between an SVM and hinge loss?

Question

My colleague and I are trying to wrap our heads around the difference between logistic regression and an SVM. Clearly they are optimizing different objective functions. Is an SVM as simple as saying it's a discriminative classifier that simply optimizes the hinge loss? Or is it more complex than that? How do the support vectors come into play? What about the slack variables? Why can't you have deep SVM's the way you can't you have a deep neural network with sigmoid activation functions?

Answer 1

I will answer one thing at at time

Is an SVM as simple as saying it's a discriminative classifier that simply optimizes the hinge loss?

SVM is simply a linear classifier, optimizing hinge loss with L2 regularization.

Or is it more complex than that?

No, it is "just" that, however there are different ways of looking at this model leading to complex, interesting conclusions. In particular, this specific choice of loss function leads to extremely efficient kernelization, which is not true for log loss (logistic regression) nor mse (linear regression). Furthermore you can show very important theoretical properties, such as those related to Vapnik-Chervonenkis dimension reduction leading to smaller chance of overfitting.

Intuitively look at these three common losses:

• hinge: max(0, 1-py)
• log: y log p
• mse: (p-y)^2

Only the first one has the property that once something is classified correctly - it has 0 penalty. All the remaining ones still penalize your linear model even if it classifies samples correctly. Why? Because they are more related to regression than classification they want a perfect prediction, not just correct.

How do the support vectors come into play?

Support vectors are simply samples placed near the decision boundary (losely speaking). For linear case it does not change much, but as most of the power of SVM lies in its kernelization - there SVs are extremely important. Once you introduce kernel, due to hinge loss, SVM solution can be obtained efficiently, and support vectors are the only samples remembered from the training set, thus building a non-linear decision boundary with the subset of the training data.

What about the slack variables?

This is just another definition of the hinge loss, more usefull when you want to kernelize the solution and show the convexivity.

Why can't you have deep SVM's the way you can't you have a deep neural network with sigmoid activation functions?

You can, however as SVM is not a probabilistic model, its training might be a bit tricky. Furthermore whole strength of SVM comes from efficiency and global solution, both would be lost once you create a deep network. However there are such models, in particular SVM (with squared hinge loss) is nowadays often choice for the topmost layer of deep networks - thus the whole optimization is actually a deep SVM. Adding more layers in between has nothing to do with SVM or other cost - they are defined completely by their activations, and you can for example use RBF activation function, simply it has been shown numerous times that it leads to weak models (to local features are detected).

To sum up:

• there are deep SVMs, simply this is a typical deep neural network with SVM layer on top.
• there is no such thing as putting SVM layer "in the middle", as the training criterion is actually only applied to the output of the network.
• using of "typical" SVM kernels as activation functions is not popular in deep networks due to their locality (as opposed to very global relu or sigmoid)