I'm currently looking at the unconstrained primal form of the one-vs-all classifier

$$\sum\limits_{i=1}^{N_I} \sum\limits_{k=1,\atop k \neq y_i}^{N_K} L(1+ \mathbf{w_k}\cdot\mathbf{x_i}-\mathbf{w_{y_i}}\cdot\mathbf{x_i})$$


$N_I$ is the number of instances,
$N_K$ is the number of classes,
$N_F$ is the number of features,
$X$ is a $N_K \times N_F$ data matrix,
$y$ is a vector of class labels,
$W$ is an $N_K \times N_I$ matrix where each corresponds to the weights for the hyperplane splitting one class from the rest,
$L$ is some arbitrary loss function.

My understanding is that the functional above tries to find a hyperplane for each class that maximizes the distance between the samples within the associated class to all other samples. If the hyperplanes are correctly positioned then $\mathbf{w_k}\cdot\mathbf{x_i}$ should always be negative, $\mathbf{w_{y_i}}\cdot\mathbf{x_i}$ should always be positive and our loss function should come back fairly low.

I'm trying to implement this using the hinge loss which I believe in this case will end up being


However, in the above couldn't we end up with a situation where the hyperplanes classify all samples as belonging to every class. For example, if we are looking at the hyperplane seperating class 1 from all other classes, provided that $1+\mathbf{w_k}\cdot\mathbf{x_i}<\mathbf{w_{y_i}}\cdot\mathbf{x_i}$ then the incurred loss will be 0 despite $\mathbf{x_i}$ being classified as the wrong class.

Where have I gone wrong? Or does it not matter whether $\mathbf{w_k}\cdot\mathbf{x_i}$ is negative or positive provided that $\mathbf{w_{y_i}}\cdot\mathbf{x_i}$ ends up with a higher score? I have a feeling that my use of the hinge function as I've described here is incorrect but my use of Google today has only led to more confusion.

On a related note, why is there a 1 in the functional above? I would think that it would have little impact.


2 Answers 2


Your post seems to be mostly correct.

The way that multiclass linear classifiers are set up is that an example, $x$, is classified by the hyperplane that give the highest score: $\underset{k}{\mathrm{argmax}\,} w_k \cdot x$. It doesn't matter if these scores are positive or negative.

If the hinge loss for a particular example is zero, then this means that the example is correctly classified. To see this, the hinge loss will be zero when $1+w_{k}\cdot x_i<w_{y_i}\cdot x_i \;\forall k$. This is a stronger condition than $w_{k}\cdot x_i<w_{y_i}\cdot x_i \;\forall k$, which would indicate that example $i$ was correctly classified as $y_i$.

The 1 in the hinge loss is related to the "margin" of the classifier.

The hinge loss encourages scores from the correct class, $w_{y_i}\cdot x_i$ to not only be higher that scores from all the other classes, $w_k\cdot x_i$, but to be higher than these scores by an additive factor.

We can use the value 1 for the margin because the distance of a point from a hyperplane is scaled by the magnitude of the linear weights: $\frac{w}{|w|}\cdot x$ is the distance of $x$ from the hyperplane with normal vector $w$. Since the weights are the same for all points in the dataset, it only matters that the scaling factor—1—is the same for all data points.

Also, it may make things easier to understand if you parameterize the loss function as $L(x,y;w)$. You currently have the loss functions as a function of the linear margin, and this is not necessarily the case.

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    – cardinal
    Mar 4, 2012 at 19:05
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    – cardinal
    Mar 4, 2012 at 19:18

You are missing the binary outcome/label (that can take value of +1 and -1 for a given class) in the loss function: max(0, 1 - y*(w*x)) (see details below).

Overall, I think the specification above (both the notation and the loss function) over-complicates one-vs-all - instead one could just take a particular class, construct +1/-1 outcome y as well as the corresponding data matrix X (with Nf columns and Ni rows) and parameter vector w for that class, and write the corresponding hinge loss function for a classical binary classifier for that class: sum(max(0, 1 - y*(w*x))) where the sum is over all data instances, x is a row of X that corresponds to a particular instance. One does need "1" in the hinge loss function (since y*(w*x) >= 1 corresponds to the correct model prediction as far as the loss function is concerned).


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