CS229 Lecture Notes
Tengyu Ma and Andrew Ng October 7, 2020
Part V
Kernel Methods
1.1 Feature maps
Recall that in our discussion about linear regression, we considered the prob- lem of predicting the price of a house (denoted by y) from the living area of the house (denoted by x), and we fit a linear function of x to the training data. What if the price y can be more accurately represented as a non-linear function of x? In this case, we need a more expressive family of models than linear models.
We start by considering fitting cubic functions y = θ3×3 +θ2×2 +θ1x+θ0. It turns out that we can view the cubic function as a linear function over the a different set of feature variables (defined below). Concretely, let the function φ : R → R4 be defined as
1
φ(x)= x ∈R4. (1)
x2 x3
Let θ ∈ R4 be the vector containing θ0, θ1, θ2, θ3 as entries. Then we can rewrite the cubic function in x as:
θ3×3 +θ2×2 +θ1x+θ0 = θTφ(x)
Thus, a cubic function of the variable x can be viewed as a linear function
over the variables φ(x). To distinguish between these two sets of variables, 1
2
in the context of kernel methods, we will call the “original” input value the input attributes of a problem (in this case, x, the living area). When the original input is mapped to some new set of quantities φ(x), we will call those new quantities the features variables. (Unfortunately, different authors use different terms to describe these two things in different contexts.) We will call φ a feature map, which maps the attributes to the features.
1.2 LMS (least mean squares) with features
We will derive the gradient descent algorithm for fitting the model θT φ(x). First recall that for ordinary least square problem where we were to fit θT x, the batch gradient descent update is (see the first lecture note for its deriva- tion):
n ( i ) ( i ) ( i ) y −hθ(x ) x
θ:=θ+α
:=θ+α y−θxx. (2)
i=1
Let φ : Rd → Rp be a feature map that maps attribute x (in Rd) to the features φ(x) in Rp. (In the motivating example in the previous subsection, we have d = 1 and p = 4.) Now our goal is to fit the function θTφ(x), with θ being a vector in Rp instead of Rd. We can replace all the occurrences of x(i) in the algorithm above by φ(x(i)) to obtain the new update:
i=1
i=1
n ( i ) T ( i ) ( i )
n ( i ) T ( i ) ( i )
y −θ φ(x ) φ(x ) (3)
θ:=θ+α
Similarly, the corresponding stochastic gradient descent update rule is
θ := θ + α y(i) − θT φ(x(i)) φ(x(i)) (4) 1.3 LMS with the kernel trick
The gradient descent update, or stochastic gradient update above becomes computationally expensive when the features φ(x) is high-dimensional. For example, consider the direct extension of the feature map in equation (1) to
3
high-dimensional input x: suppose x ∈ Rd, and let φ(x) be the vector that contains all the monomials of x with degree ≤ 3
1
x1 x2
. x2
1
x1x2
φ(x)=x1x3 . .
x2x1 .
x31 x 21 x 2
.
The dimension of the features φ(x) is on the order of d3.1
hibitively long vector for computational purpose — when d = 1000, each update requires at least computing and storing a 10003 = 109 dimensional vector, which is 106 times slower than the update rule for for ordinary least squares updates (2).
It may appear at first that such d3 runtime per update and memory usage are inevitable, because the vector θ itself is of dimension p ≈ d3, and we may need to update every entry of θ and store it. However, we will introduce the kernel trick with which we will not need to store θ explicitly, and the runtime can be significantly improved.
For simplicity, we assume the initialize the value θ = 0, and we focus on the iterative update (3). The main observation is that at any time, θ can be represented as a linear combination of the vectors φ(x(1)), . . . , φ(x(n)). Indeed, we can show this inductively as follows. At initialization, θ = 0 = ni=1 0 · φ(x(i)). Assume at some point, θ can be represented as
n i=1
1Here, for simplicity, we include all the monomials with repetitions (so that, e.g., x1x2x3 and x2x3x1 both appear in φ(x)). Therefore, there are totally 1 + d + d2 + d3 entries in φ(x).
θ =
βiφ(x(i)) (6)
This is a pro-
(5)
4 for some β1,…,βn ∈ R. Then we claim that in the next round, θ is still a
linear combination of φ(x(1)), . . . , φ(x(n)) because
θ:=θ+α
n ( i ) T ( i ) ( i ) y −θ φ(x ) φ(x )
= =
βiφ(x(i)) + α y(i) − θT φ(x(i)) i=1
(i) T (i) (i) (βi+αy −θφ(x))φ(x)
φ(x(i))
i=1
n n
i=1 n
(7)
i=1 new βi
You may realize that our general strategy is to implicitly represent the p- dimensional vector θ by a set of coefficients β1, . . . , βn. Towards doing this, we derive the update rule of the coefficients β1, . . . , βn. Using the equation above, we see that the new βi depends on the old one via
βi := βi + α y(i) − θT φ(x(i)) (8) Here we still have the old θ on the RHS of the equation. Replacing θ by
θ = nj=1 βjφ(x(j)) gives
∀i∈{1,…,n},βi :=βi +α y(i) −
n βjφ(x(j))Tφ(x(i))
j=1
We often rewrite φ(x(j))T φ(x(i)) as ⟨φ(x(j)), φ(x(i))⟩ to emphasize that it’s the inner product of the two feature vectors. Viewing βi’s as the new representa- tion of θ, we have successfully translated the batch gradient descent algorithm into an algorithm that updates the value of β iteratively. It may appear that at every iteration, we still need to compute the values of ⟨φ(x(j)), φ(x(i))⟩ for all pairs of i,j, each of which may take roughly O(p) operation. However, two important properties come to rescue:
1. We can pre-compute the pairwise inner products ⟨φ(x(j)), φ(x(i))⟩ for all pairs of i, j before the loop starts.
2. For the feature map φ defined in (5) (or many other interesting fea- ture maps), computing ⟨φ(x(j)),φ(x(i))⟩ can be efficient and does not
5 necessarily require computing φ(x(i)) explicitly. This is because:
⟨φ(x),φ(z)⟩ = 1 + xizi + i=1
xixjzizj +
xixjxkzizjzk
(9)
d
i,j∈{1,…,d}
d d 2 d 3
= 1 + xizi + xizi + i=1 i=1
=1+⟨x,z⟩+⟨x,z⟩2 +⟨x,z⟩3
xizi i=1
Therefore, to compute ⟨φ(x),φ(z)⟩, we can first compute ⟨x,z⟩ with O(d) time and then take another constant number of operations to com- pute 1 + ⟨x,z⟩ + ⟨x,z⟩2 + ⟨x,z⟩3.
As you will see, the inner products between the features ⟨φ(x), φ(z)⟩ are essential here. We define the Kernel corresponding to the feature map φ as a function that maps X × X → R satisfying: 2
K(x, z) ≜ ⟨φ(x), φ(z)⟩ (10) To wrap up the discussion, we write the down the final algorithm as
follows:
1. Compute all the values K(x(i), x(j)) ≜ ⟨φ(x(i)), φ(x(j))⟩ using equa-
tion (9) for all i,j ∈ {1,…,n}. Set β := 0. 2. Loop:
n
∀i∈{1,…,n},βi :=βi +α y(i) −
Or in vector notation, letting K be the n × n matrix with Kij =
K(x(i), x(j)), we have
β := β + α(⃗y − Kβ)
With the algorithm above, we can update the representation β of the vector θ efficiently with O(n2) time per update. Finally, we need to show
2Recall that X is the space of the input x. In our running example, X = Rd
i,j,k∈{1,…,d}
j=1
βjK(x(i),x(j)) (11)
6 that the knowledge of the representation β suffices to compute the prediction
θT φ(x). Indeed, we have θT φ(x) =
n i=1
βiφ(x(i))T φ(x) =
n i=1
βiK(x(i), x) (12)
You may realize that fundamentally all we need to know about the feature map φ(·) is encapsulated in the corresponding kernel function K(·,·). We will expand on this in the next section.
1.4 Properties of kernels
In the last subsection, we started with an explicitly defined feature map φ, which induces the kernel function K(x, z) ≜ ⟨φ(x), φ(z)⟩. Then we saw that the kernel function is so intrinsic so that as long as the kernel function is defined, the whole training algorithm can be written entirely in the language of the kernel without referring to the feature map φ, so can the prediction of a test example x (equation (12).)
Therefore, it would be tempted to define other kernel function K(·,·) and run the algorithm (11). Note that the algorithm (11) does not need to explicitly access the feature map φ, and therefore we only need to ensure the existence of the feature map φ, but do not necessarily need to be able to explicitly write φ down.
What kinds of functions K(·, ·) can correspond to some feature map φ? In other words, can we tell if there is some feature mapping φ so that K(x, z) = φ(x)T φ(z) for all x, z?
If we can answer this question by giving a precise characterization of valid kernel functions, then we can completely change the interface of selecting feature maps φ to the interface of selecting kernel function K. Concretely, we can pick a function K, verify that it satisfies the characterization (so that there exists a feature map φ that K corresponds to), and then we can run update rule (11). The benefit here is that we don’t have to be able to compute φ or write it down analytically, and we only need to know its existence. We will answer this question at the end of this subsection after we go through several concrete examples of kernels.
Suppose x, z ∈ Rd, and let’s first consider the function K(·, ·) defined as: K(x, z) = (xT z)2.
We can also write this as
K(x,z) =
=
=
d i=1
d j=1
xizi d d
xjzj
i=1 j=1 d
i,j =1
xixjzizj (xixj )(zizj )
Thus, we see that K(x,z) = ⟨φ(x),φ(z)⟩ is the kernel function that corre- sponds to the the feature mapping φ given (shown here for the case of d = 3)
by
xx 11
x1x2 x1x3 x2x1
φ(x)=x2x2 . x2x3 x3x1 x3x2
x3 x3
Revisiting the computational efficiency perspective of kernel, note that whereas calculating the high-dimensional φ(x) requires O(d2) time, finding K(x,z) takes only O(d) time—linear in the dimension of the input attributes.
For another related example, also consider K(·,·) defined by K(x,z) = (xTz+c)2
d d√√2 = (xixj)(zizj) + ( 2cxi)( 2czi) + c .
i,j =1 i=1
(Check this yourself.) This function K is a kernel function that corresponds
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to the feature mapping (again shown for d = 3) xx
11
x1x2
x1x3 x2x1 x2x2 x2x3
φ(x)= x3x1 , x3x2
xx √3 3
√2cx1 √2cx2
and the parameter c controls the relative weighting between the xi (first order) and the xixj (second order) terms.
More broadly, the kernel K(x, z) = (xT z + c)k corresponds to a feature mapping to an d+k feature space, corresponding of all monomials of the
k
form xi1xi2 …xik that are up to order k. However, despite working in this
O(dk)-dimensional space, computing K(x,z) still takes only O(d) time, and hence we never need to explicitly represent feature vectors in this very high dimensional feature space.
Kernels as similarity metrics. Now, let’s talk about a slightly different view of kernels. Intuitively, (and there are things wrong with this intuition, but nevermind), if φ(x) and φ(z) are close together, then we might expect K(x, z) = φ(x)T φ(z) to be large. Conversely, if φ(x) and φ(z) are far apart— say nearly orthogonal to each other—then K(x, z) = φ(x)T φ(z) will be small. So, we can think of K(x,z) as some measurement of how similar are φ(x) and φ(z), or of how similar are x and z.
Given this intuition, suppose that for some learning problem that you’re working on, you’ve come up with some function K(x, z) that you think might be a reasonable measure of how similar x and z are. For instance, perhaps you chose
||x−z||2 K(x,z)=exp − 2σ2 .
2cx3 c
8
This is a reasonable measure of x and z’s similarity, and is close to 1 when x and z are close, and near 0 when x and z are far apart. Does there exist
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a feature map φ such that the kernel K defined above satisfies K(x,z) = φ(x)T φ(z)? In this particular example, the answer is yes. This kernel is called the Gaussian kernel, and corresponds to an infinite dimensional feature mapping φ. We will give a precise characterization about what properties a function K needs to satisfy so that it can be a valid kernel function that corresponds to some feature map φ.
Necessary conditions for valid kernels. Suppose for now that K is indeed a valid kernel corresponding to some feature mapping φ, and we will first see what properties it satisfies. Now, consider some finite set of n points (not necessarily the training set) {x(1), . . . , x(n)}, and let a square, n-by-n matrix K be defined so that its (i,j)-entry is given by Kij = K(x(i),x(j)). This matrix is called the kernel matrix. Note that we’ve overloaded the notation and used K to denote both the kernel function K(x,z) and the kernel matrix K, due to their obvious close relationship.
Now, if K is a valid kernel, then Kij = K(x(i), x(j)) = φ(x(i))T φ(x(j)) = φ(x(j))T φ(x(i)) = K(x(j), x(i)) = Kji, and hence K must be symmetric. More- over, letting φk(x) denote the k-th coordinate of the vector φ(x), we find that for any vector z, we have
T z Kz =
ziKijzj
(i) T (j)
= = =
=
≥
ziφ(x ) φ(x )zj (i) (j)
ij
ij
zi φk(x )φk(x )zj ijk
kij
2 zi φk (x(i) )
ki
(i) (j)
zi φk (x )φk (x )zj
0.
The second-to-last step uses the fact that aiaj = ( ai)2 for ai =
i,j i
ziφk(x(i)). Since z was arbitrary, this shows that K is positive semi-definite
(K ≥ 0).
Hence, we’ve shown that if K is a valid kernel (i.e., if it corresponds to
some feature mapping φ), then the corresponding kernel matrix K ∈ Rn×n is symmetric positive semidefinite.
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Sufficient conditions for valid kernels. More generally, the condition above turns out to be not only a necessary, but also a sufficient, condition for K to be a valid kernel (also called a Mercer kernel). The following result is due to Mercer.3
Theorem (Mercer). Let K : Rd × Rd → R be given. Then for K to be a valid (Mercer) kernel, it is necessary and sufficient that for any {x(1), . . . , x(n)}, (n < ∞), the corresponding kernel matrix is symmetric pos- itive semi-definite.
Given a function K, apart from trying to find a feature mapping φ that corresponds to it, this theorem therefore gives another way of testing if it is a valid kernel. You’ll also have a chance to play with these ideas more in problem set 2.
In class, we also briefly talked about a couple of other examples of ker- nels. For instance, consider the digit recognition problem, in which given an image (16x16 pixels) of a handwritten digit (0-9), we have to figure out which digit it was. Using either a simple polynomial kernel K(x, z) = (xT z)k or the Gaussian kernel, SVMs were able to obtain extremely good perfor- mance on this problem. This was particularly surprising since the input attributes x were just 256-dimensional vectors of the image pixel intensity values, and the system had no prior knowledge about vision, or even about which pixels are adjacent to which other ones. Another example that we briefly talked about in lecture was that if the objects x that we are trying to classify are strings (say, x is a list of amino acids, which strung together form a protein), then it seems hard to construct a reasonable, “small” set of features for most learning algorithms, especially if different strings have dif- ferent lengths. However, consider letting φ(x) be a feature vector that counts the number of occurrences of each length-k substring in x. If we’re consid- ering strings of English letters, then there are 26k such strings. Hence, φ(x) is a 26k dimensional vector; even for moderate values of k, this is probably too big for us to efficiently work with. (e.g., 264 ≈ 460000.) However, using (dynamic programming-ish) string matching algorithms, it is possible to ef- ficiently compute K(x, z) = φ(x)T φ(z), so that we can now implicitly work in this 26k-dimensional feature space, but without ever explicitly computing feature vectors in this space.
3Many texts present Mercer’s theorem in a slightly more complicated form involving L2 functions, but when the input attributes take values in Rd, the version given here is equivalent.
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Application of kernel methods: We’ve seen the application of kernels to linear regression. In the next part, we will introduce the support vector machines to which kernels can be directly applied. dwell too much longer on it here. In fact, the idea of kernels has significantly broader applicability than linear regression and SVMs. Specifically, if you have any learning algorithm that you can write in terms of only inner products ⟨x,z⟩ between input attribute vectors, then by replacing this with K(x,z) where K is a kernel, you can “magically” allow your algorithm to work efficiently in the high dimensional feature space corresponding to K. For instance, this kernel trick can be applied with the perceptron to derive a kernel perceptron algorithm. Many of the algorithms that we’ll see later in this class will also be amenable to this method, which has come to be known as the “kernel trick.”
Part VI
Support Vector Machines
This set of notes presents the Support Vector Machine (SVM) learning al- gorithm. SVMs are among the best (and many believe are indeed the best) “off-the-shelf” supervised learning algorithms. To tell the SVM story, we’ll need to first talk about margins and the idea of separating data with a large “gap.” Next, we’ll talk about the optimal margin classifier, which will lead us into a digression on Lagrange duality. We’ll also see kernels, which give a way to apply SVMs efficiently in very high dimensional (such as infinite- dimensional) feature spaces, and finally, we’ll close off the story with the SMO algorithm, which gives an efficient implementation of SVMs.
2 Margins: Intuition
We’ll start our story on SVMs by talking about margins. This section will give the intuitions about margins and about the “confidence” of our predic- tions; these ideas will be made formal in Section 4.
Consider logistic regression, where the probability p(y = 1|x; θ) is mod- eled by hθ(x) = g(θTx). We then predict “1” on an input x if and only if hθ(x) ≥ 0.5, or equivalently, if and only if θTx ≥ 0. Consider a positive training example (y = 1). The larger θT x is, the larger also is hθ(x) = p(y = 1|x; θ), and thus also the higher our degree of “confidence” that the label is 1. Thus, informally we can think of our prediction as being very confident that
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y = 1 if θT x ≫ 0. Similarly, we think of logistic regression as confidently predicting y = 0, if θT x ≪ 0. Given a training set, again informally it seems that we’d have found a good fit to the training data if we can find θ so that θT x(i) ≫ 0 whenever y(i) = 1, and θT x(i) ≪ 0 whenever y(i) = 0, since this would reflect a very confident (and correct) set of classifications for all the training examples. This seems to be a nice goal to aim for, and we’ll soon formalize this idea using the notion of functional margins.
For a different type of intuition, consider the following figure, in which x’s represent positive training examples, o’s denote negative training examples, a decision boundary (this is the line given by the equation θT x = 0, and is also called the separating hyperplane) is also shown, and three points have also been labeled A, B and C.
A
B C
Notice that the point A is very far from the decision boundary. If we are asked to make a prediction for the value of y at A, it seems we should be quite confident that y = 1 there. Conversely, the point C is very close to the decision boundary, and while it’s on the side of the decision boundary on which we would predict y = 1, it seems likely that just a small change to the decision boundary could easily have caused out prediction to be y = 0. Hence, we’re much more confident about our prediction at A than at C. The point B lies in-between these two cases, and more broadly, we see that if a point is far from the separating hyperplane, then we may be significantly more confident in our predictions. Again, informally we think it would be nice if, given a training set, we manage to find a decision boundary that allows us to make all correct and confident (meaning far from the decision
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boundary) predictions on the training examples. We’ll formalize this later using the notion of geometric margins.
3 Notation
To make our discussion of SVMs easier, we’ll first need to introduce a new notation for talking about classification. We will be considering a linear classifier for a binary classification problem with labels y and features x. From now, we’ll use y ∈ {−1, 1} (instead of {0, 1}) to denote the class labels. Also, rather than parameterizing our linear classifier with the vector θ, we will use parameters w, b, and write our classifier as
hw,b(x) = g(wT x + b).
Here, g(z) = 1 if z ≥ 0, and g(z) = −1 otherwise. This “w,b” notation allows us to explicitly treat the intercept term b separately from the other parameters. (We also drop the convention we had previously of letting x0 = 1 be an extra coordinate in the input feature vector.) Thus, b takes the role of what was previously θ0, and w takes the role of [θ1 . . . θd]T .
Note also that, from our definition of g above, our classifier will directly predict either 1 or −1 (cf. the perceptron algorithm), without first going through the intermediate step of estimating p(y = 1) (which is what logistic regression does).
4 Functional and geometric margins
Let’s formalize the notions of the functional and geometric margins. Given a training example (x(i), y(i)), we define the functional margin of (w, b) with respect to the training example as
γˆ(i) = y(i)(wT x(i) + b).
Note that if y(i) = 1, then for the functional margin to be large (i.e., for our prediction to be confident and correct), we need wT x(i) + b to be a large positive number. Conversely, if y(i) = −1, then for the functional margin to be large, we need wT x(i) + b to be a large negative number. Moreover, if y(i)(wT x(i) + b) > 0, then our prediction on this example is correct. (Check this yourself.) Hence, a large functional margin represents a confident and a correct prediction.
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For a linear classifier with the choice of g given above (taking values in {−1, 1}), there’s one property of the functional margin that makes it not a very good measure of confidence, however. Given our choice of g, we note that if we replace w with 2w and b with 2b, then since g(wT x+b) = g(2wT x+2b), this would not change hw,b(x) at all. I.e., g, and hence also hw,b(x), depends only on the sign, but not on the magnitude, of wT x + b. However, replacing (w,b) with (2w,2b) also results in multiplying our functional margin by a factor of 2. Thus, it seems that by exploiting our freedom to scale w and b, we can make the functional margin arbitrarily large without really changing anything meaningful. Intuitively, it might therefore make sense to impose some sort of normalization condition such as that ||w||2 = 1; i.e., we might replace (w,b) with (w/||w||2,b/||w||2), and instead consider the functional margin of (w/||w||2,b/||w||2). We’ll come back to this later.
Given a training set S = {(x(i),y(i));i = 1,…,n}, we also define the function margin of (w, b) with respect to S as the smallest of the functional margins of the individual training examples. Denoted by γˆ, this can therefore be written:
γˆ = min γˆ(i). i=1,…,n
Next, let’s talk about geometric margins. Consider the picture below:
Aw
γ (i) B
The decision boundary corresponding to (w, b) is shown, along with the vector w. Note that w is orthogonal (at 90◦) to the separating hyperplane. (You should convince yourself that this must be the case.) Consider the point at A, which represents the input x(i) of some training example with
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label y(i) = 1. Its distance to the decision boundary, γ(i), is given by the line segment AB.
How can we find the value of γ(i)? Well, w/||w|| is a unit-length vector pointing in the same direction as w. Since A represents x(i), we therefore find that the point B is given by x(i) − γ(i) · w/||w||. But this point lies on the decision boundary, and all points x on the decision boundary satisfy the equation wT x + b = 0. Hence,
Solving for γ(i) yields γ(i) =
w
wT x(i) −γ(i)
wTx(i)+b w T b ||w|| = ||w|| x(i) + ||w||.
||w||
+b=0.
This was worked out for the case of a positive training example at A in the figure, where being on the “positive” side of the decision boundary is good. More generally, we define the geometric margin of (w,b) with respect to a training example (x(i), y(i)) to be
T γ(i) = y(i) w x(i) + b .
||w|| ||w||
Note that if ||w|| = 1, then the functional margin equals the geometric margin—this thus gives us a way of relating these two different notions of margin. Also, the geometric margin is invariant to rescaling of the parame- ters; i.e., if we replace w with 2w and b with 2b, then the geometric margin does not change. This will in fact come in handy later. Specifically, because of this invariance to the scaling of the parameters, when trying to fit w and b to training data, we can impose an arbitrary scaling constraint on w without changing anything important; for instance, we can demand that ||w|| = 1, or |w1| = 5, or |w1 +b|+|w2| = 2, and any of these can be satisfied simply by rescaling w and b.
Finally, given a training set S = {(x(i), y(i)); i = 1, . . . , n}, we also define the geometric margin of (w,b) with respect to S to be the smallest of the geometric margins on the individual training examples:
γ = min γ(i). i=1,…,n
5 The optimal margin classifier
Given a training set, it seems from our previous discussion that a natural desideratum is to try to find a decision boundary that maximizes the (ge- ometric) margin, since this would reflect a very confident set of predictions on the training set and a good “fit” to the training data. Specifically, this will result in a classifier that separates the positive and the negative training examples with a “gap” (geometric margin).
For now, we will assume that we are given a training set that is linearly separable; i.e., that it is possible to separate the positive and negative ex- amples using some separating hyperplane. How will we find the one that achieves the maximum geometric margin? We can pose the following opti- mization problem:
maxγ,w,b γ
s.t. y(i)(wTx(i) +b) ≥ γ, i = 1,…,n
||w|| = 1.
I.e., we want to maximize γ, subject to each training example having func- tional margin at least γ. The ||w|| = 1 constraint moreover ensures that the functional margin equals to the geometric margin, so we are also guaranteed that all the geometric margins are at least γ. Thus, solving this problem will result in (w, b) with the largest possible geometric margin with respect to the training set.
If we could solve the optimization problem above, we’d be done. But the “||w|| = 1” constraint is a nasty (non-convex) one, and this problem certainly isn’t in any format that we can plug into standard optimization software to solve. So, let’s try transforming the problem into a nicer one. Consider:
Here, we’re going to maximize γˆ/||w||, subject to the functional margins all
being at least γˆ. Since the geometric and functional margins are related by
γ = γˆ/||w|, this will give us the answer we want. Moreover, we’ve gotten rid
of the constraint ||w|| = 1 that we didn’t like. The downside is that we now
have a nasty (again, non-convex) objective γˆ function; and, we still don’t ||w||
have any off-the-shelf software that can solve this form of an optimization problem.
γˆ ||w||
maxγˆ,w,b
s.t. y(i)(wTx(i) +b) ≥ γˆ, i = 1,…,n
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17
Let’s keep going. Recall our earlier discussion that we can add an arbi- trary scaling constraint on w and b without changing anything. This is the key idea we’ll use now. We will introduce the scaling constraint that the functional margin of w, b with respect to the training set must be 1:
γˆ = 1 .
Since multiplying w and b by some constant results in the functional margin being multiplied by that same constant, this is indeed a scaling constraint, and can be satisfied by rescaling w, b. Plugging this into our problem above, and noting that maximizing γˆ/||w|| = 1/||w|| is the same thing as minimizing ||w||2, we now have the following optimization problem:
min s.t.
1 ||w||2 2
y(i)(wTx(i) +b) ≥ 1, i = 1,…,n
w,b
We’ve now transformed the problem into a form that can be efficiently solved. The above is an optimization problem with a convex quadratic ob- jective and only linear constraints. Its solution gives us the optimal mar- gin classifier. This optimization problem can be solved using commercial quadratic programming (QP) code.4
While we could call the problem solved here, what we will instead do is make a digression to talk about Lagrange duality. This will lead us to our optimization problem’s dual form, which will play a key role in allowing us to use kernels to get optimal margin classifiers to work efficiently in very high dimensional spaces. The dual form will also allow us to derive an efficient algorithm for solving the above optimization problem that will typically do much better than generic QP software.
6 Lagrange duality (optional reading)
Let’s temporarily put aside SVMs and maximum margin classifiers, and talk about solving constrained optimization problems.
Consider a problem of the following form: minw f (w)
s.t. hi(w) = 0, i = 1,…,l.
4You may be familiar with linear programming, which solves optimization problems that have linear objectives and linear constraints. QP software is also widely available, which allows convex quadratic objectives and linear constraints.
18
Some of you may recall how the method of Lagrange multipliers can be used to solve it. (Don’t worry if you haven’t seen it before.) In this method, we define the Lagrangian to be
and set L’s partial derivatives to zero:
∂L=0; ∂L=0,
∂wi ∂βi
and solve for w and β.
In this section, we will generalize this to constrained optimization prob-
lems in which we may have inequality as well as equality constraints. Due to time constraints, we won’t really be able to do the theory of Lagrange duality justice in this class,5 but we will give the main ideas and results, which we will then apply to our optimal margin classifier’s optimization problem.
Consider the following, which we’ll call the primal optimization problem:
minw f (w)
s.t. gi(w) ≤ 0, i = 1,…,k
hi(w) = 0, i = 1,…,l.
To solve it, we start by defining the generalized Lagrangian
L(w, β) = f(w) +
Here, the βi’s are called the Lagrange multipliers. We would then find
k L(w, α, β) = f(w) +
l i=1
αigi(w) +
Here, the αi’s and βi’s are the Lagrange multipliers. Consider the quantity
θP (w) = max L(w, α, β). α,β : αi≥0
Here, the “P” subscript stands for “primal.” Let some w be given. If w violates any of the primal constraints (i.e., if either gi(w) > 0 or hi(w) ̸= 0 for some i), then you should be able to verify that
θP(w) = max f(w) + α,β : αi≥0
= ∞.
k i=1
αigi(w) +
l i=1
βihi(w) (13) (14)
i=1
l i=1
βihi(w)
βihi(w).
5Readers interested in learning more about this topic are encouraged to read, e.g., R. T. Rockarfeller (1970), Convex Analysis, Princeton University Press.
19 Conversely, if the constraints are indeed satisfied for a particular value of w,
then θP (w) = f (w). Hence,
f(w) if w satisfies primal constraints
Thus, θP takes the same value as the objective in our problem for all val- ues of w that satisfies the primal constraints, and is positive infinity if the constraints are violated. Hence, if we consider the minimization problem
θP (w) = ∞ otherwise.
min θP (w) = min max w w α,β:αi≥0
L(w, α, β),
we see that it is the same problem (i.e., and has the same solutions as) our original, primal problem. For later use, we also define the optimal value of the objective to be p∗ = minw θP(w); we call this the value of the primal problem.
Now, let’s look at a slightly different problem. We define
θD(α, β) = min L(w, α, β). w
Here, the “D” subscript stands for “dual.” Note also that whereas in the definition of θP we were optimizing (maximizing) with respect to α,β, here we are minimizing with respect to w.
We can now pose the dual optimization problem:
max θD(α, β) = max min L(w, α, β).
α,β : αi≥0 α,β : αi≥0 w
This is exactly the same as our primal problem shown above, except that the order of the “max” and the “min” are now exchanged. We also define the optimal value of the dual problem’s objective to be d∗ = maxα,β : αi≥0 θD(w).
How are the primal and the dual problems related? It can easily be shown that
d∗ = max minL(w,α,β) ≤ min max α,β:αi≥0 w w α,β:αi≥0
L(w,α,β) = p∗.
(You should convince yourself of this; this follows from the “max min” of a function always being less than or equal to the “min max.”) However, under certain conditions, we will have
d∗ = p∗,
so that we can solve the dual problem in lieu of the primal problem. Let’s see what these conditions are.
20
Suppose f and the gi’s are convex,6 and the hi’s are affine.7 Suppose further that the constraints gi are (strictly) feasible; this means that there exists some w so that gi(w) < 0 for all i.
Under our above assumptions, there must exist w∗, α∗, β∗ so that w∗ is the solution to the primal problem, α∗,β∗ are the solution to the dual problem, and moreover p∗ = d∗ = L(w∗, α∗, β∗). Moreover, w∗, α∗ and β∗ satisfy the Karush-Kuhn-Tucker (KKT) conditions, which are as follows:
∂ L(w∗, α∗, β∗) ∂wi
∂ L(w∗, α∗, β∗) ∂βi
α i∗ g i ( w ∗ ) gi(w∗)
α∗
= 0, = 0,
= 0, ≤ 0, ≥ 0,
i = 1,...,d (15) i = 1,...,l (16)
i = 1,...,k (17) i = 1,...,k (18) i = 1,...,k (19)
Moreover, if some w∗,α∗,β∗ satisfy the KKT conditions, then it is also a solution to the primal and dual problems.
We draw attention to Equation (17), which is called the KKT dual complementarity condition. Specifically, it implies that if αi∗ > 0, then gi(w∗) = 0. (I.e., the “gi(w) ≤ 0” constraint is active, meaning it holds with equality rather than with inequality.) Later on, this will be key for showing that the SVM has only a small number of “support vectors”; the KKT dual complementarity condition will also give us our convergence test when we talk about the SMO algorithm.
7 Optimal margin classifiers
Note: The equivalence of optimization problem (20) and the optimization problem (24), and the relationship between the primary and dual variables in equation (22) are the most important take home messages of this section.
6When f has a Hessian, then it is convex if and only if the Hessian is positive semi- definite. For instance, f(w) = wTw is convex; similarly, all linear (and affine) functions are also convex. (A function f can also be convex without being differentiable, but we won’t need those more general definitions of convexity here.)
7I.e., there exists ai, bi, so that hi(w) = aTi w + bi. “Affine” means the same thing as linear, except that we also allow the extra intercept term bi.
21
Previously, we posed the following (primal) optimization problem for find- ing the optimal margin classifier:
1 ||w||2 (20) 2
y(i)(wTx(i) +b) ≥ 1, i = 1,…,n We can write the constraints as
gi(w) = −y(i)(wTx(i) +b)+1 ≤ 0.
We have one such constraint for each training example. Note that from the KKT dual complementarity condition, we will have αi > 0 only for the train- ing examples that have functional margin exactly equal to one (i.e., the ones corresponding to constraints that hold with equality, gi(w) = 0). Consider the figure below, in which a maximum margin separating hyperplane is shown by the solid line.
min s.t.
w,b
The points with the smallest margins are exactly the ones closest to the decision boundary; here, these are the three points (one negative and two pos- itive examples) that lie on the dashed lines parallel to the decision boundary. Thus, only three of the αi’s—namely, the ones corresponding to these three training examples—will be non-zero at the optimal solution to our optimiza- tion problem. These three points are called the support vectors in this problem. The fact that the number of support vectors can be much smaller than the size the training set will be useful later.
Let’s move on. Looking ahead, as we develop the dual form of the prob- lem, one key idea to watch out for is that we’ll try to write our algorithm
22
in terms of only the inner product ⟨x(i),x(j)⟩ (think of this as (x(i))Tx(j)) between points in the input feature space. The fact that we can express our algorithm in terms of these inner products will be key when we apply the kernel trick.
When we construct the Lagrangian for our optimization problem we have:
i=1
Note that there’re only “αi” but no “βi” Lagrange multipliers, since the problem has only inequality constraints.
Let’s find the dual form of the problem. To do so, we need to first minimize L(w,b,α) with respect to w and b (for fixed α), to get θD, which we’ll do by setting the derivatives of L with respect to w and b to zero. We have:
This implies that
L(w,b,α)=2||w|| −
αi y (w x +b)−1 . (21)
1 2 n ( i ) T ( i )
∇wL(w, b, α) = w − n
w =
As for the derivative with respect to b, we obtain
∂ n
1 n
i=1
αiy(i) = 0. (23) the Lagrangian (Equation 21), and simplify, we get
∂bL(w, b, α) =
If we take the definition of w in Equation (22) and plug that back into
n L(w,b,α)=
n i=1
αiy(i).
Recall that we got to the equation above by minimizing L with respect to w and b. Putting this together with the constraints αi ≥ 0 (that we always had)
y(i)y(j)αiαj(x(i))Tx(j) −b
But from Equation (23), the last term must be zero, so we obtain
αi −2
n 1n
L(w, b, α) = αi − 2 y(i)y(j)αiαj (x(i))T x(j). i=1 i,j =1
i=1
i,j=1
n i=1
αiy(i)x(i) = 0
αiy(i)x(i). (22)
i=1
23 and the constraint (23), we obtain the following dual optimization problem:
n 1n
maxα W (α) = αi − 2 y(i)y(j)αiαj ⟨x(i), x(j)⟩. (24)
i=1 i,j =1 s.t. αi ≥0, i=1,…,n
n
αiy(i) = 0,
i=1
You should also be able to verify that the conditions required for p∗ = d∗ and the KKT conditions (Equations 15–19) to hold are indeed satisfied in our optimization problem. Hence, we can solve the dual in lieu of solving the primal problem. Specifically, in the dual problem above, we have a maximization problem in which the parameters are the αi’s. We’ll talk later about the specific algorithm that we’re going to use to solve the dual problem, but if we are indeed able to solve it (i.e., find the α’s that maximize W(α) subject to the constraints), then we can use Equation (22) to go back and find the optimal w’s as a function of the α’s. Having found w∗, by considering the primal problem, it is also straightforward to find the optimal value for the intercept term b as
∗ maxi:y(i)=−1 w∗T x(i) + mini:y(i)=1 w∗T x(i)
b=− 2 . (25)
(Check for yourself that this is correct.)
Before moving on, let’s also take a more careful look at Equation (22),
which gives the optimal value of w in terms of (the optimal value of) α. Suppose we’ve fit our model’s parameters to a training set, and now wish to make a prediction at a new point input x. We would then calculate wT x + b, and predict y = 1 if and only if this quantity is bigger than zero. But using (22), this quantity can also be written:
wTx+b = =
Hence, if we’ve found the αi’s, in order to make a prediction, we have to calculate a quantity that depends only on the inner product between x and the points in the training set. Moreover, we saw earlier that the αi’s will all
n T αiy(i)x(i)
i=1 n
i=1
x+b (26) αiy(i)⟨x(i), x⟩ + b. (27)
24
be zero except for the support vectors. Thus, many of the terms in the sum above will be zero, and we really need to find only the inner products between x and the support vectors (of which there is often only a small number) in order calculate (27) and make our prediction.
By examining the dual form of the optimization problem, we gained sig- nificant insight into the structure of the problem, and were also able to write the entire algorithm in terms of only inner products between input feature vectors. In the next section, we will exploit this property to apply the ker- nels to our classification problem. The resulting algorithm, support vector machines, will be able to efficiently learn in very high dimensional spaces.
8 Regularization and the non-separable case (optional reading)
The derivation of the SVM as presented so far assumed that the data is linearly separable. While mapping data to a high dimensional feature space via φ does generally increase the likelihood that the data is separable, we can’t guarantee that it always will be so. Also, in some cases it is not clear that finding a separating hyperplane is exactly what we’d want to do, since that might be susceptible to outliers. For instance, the left figure below shows an optimal margin classifier, and when a single outlier is added in the upper-left region (right figure), it causes the decision boundary to make a dramatic swing, and the resulting classifier has a much smaller margin.
To make the algorithm work for non-linearly separable datasets as well as be less sensitive to outliers, we reformulate our optimization (using l1
regularization) as follows:
1 2 n
minγ,w,b 2||w|| +C
s.t. y(i)(wTx(i) +b) ≥ 1−ξi,
ξi
ξi ≥0, i=1,…,n.
i=1
Thus, examples are now permitted to have (functional) margin less than 1, and if an example has functional margin 1 − ξi (with ξ > 0), we would pay a cost of the objective function being increased by Cξi. The parameter C controls the relative weighting between the twin goals of making the ||w||2 small (which we saw earlier makes the margin large) and of ensuring that most examples have functional margin at least 1.
As before, we can form the Lagrangian:
1T n n (i)T n
L(w,b,ξ,α,r)=2w w+C ξi− i=1
i=1
αi y (x w+b)−1+ξi − riξi. i=1
Here, the αi’s and ri’s are our Lagrange multipliers (constrained to be ≥ 0). We won’t go through the derivation of the dual again in detail, but after setting the derivatives with respect to w and b to zero as before, substituting them back in, and simplifying, we obtain the following dual form of the problem:
n 1n
maxα W (α) = αi − 2 y(i)y(j)αiαj ⟨x(i), x(j)⟩
i=1 i,j =1
s.t. 0≤αi ≤C, i=1,…,n
n
αiy(i) = 0,
i=1
As before, we also have that w can be expressed in terms of the αi’s as given in Equation (22), so that after solving the dual problem, we can continue to use Equation (27) to make our predictions. Note that, somewhat surprisingly, in adding l1 regularization, the only change to the dual problem is that what was originally a constraint that 0 ≤ αi has now become 0 ≤ αi ≤ C. The calculation for b∗ also has to be modified (Equation 25 is no longer valid); see the comments in the next section/Platt’s paper.
Also, the KKT dual-complementarity conditions (which in the next sec- tion will be useful for testing for the convergence of the SMO algorithm)
i = 1,…,n
25
are:
26
αi =0 ⇒ y(i)(wTx(i) +b)≥1 (28) αi=C ⇒ y(i)(wTx(i)+b)≤1 (29) 0<αi
αnew = αnew,unclipped if L ≤ αnew,unclipped ≤ H 222
L if αnew,unclipped < L 2
Finally, having found the αnew, we can use Equation (34) to go back and find 2
the optimal value of αnew. 1
There’re a couple more details that are quite easy but that we’ll leave you to read about yourself in Platt’s paper: One is the choice of the heuristics used to select the next αi, αj to update; the other is how to update b as the SMO algorithm is run.