mhjensen
659 lines
22 KiB
HTML
659 lines
22 KiB
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{'highest level': 2,
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'sections': [('Support Vector Machines, overarching aims', 2, None, '___sec0'),
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('Hyperplanes and all that', 2, None, '___sec1'),
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('What is a hyperplane?', 2, None, '___sec2'),
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('A $p$-dimensional space of features', 2, None, '___sec3'),
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('The two-dimensional case', 2, None, '___sec4'),
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('Getting into the details', 2, None, '___sec5'),
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('First attempt at a minimization approach', 2, None, '___sec6'),
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('Solving the equations', 2, None, '___sec7'),
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('A better approach', 2, None, '___sec8'),
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('A quick reminder on Lagrangian multipliers',
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2,
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None,
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'___sec9'),
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('Adding the muliplier', 2, None, '___sec10'),
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('Setting up the problem', 2, None, '___sec11'),
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('The problem to solve', 2, None, '___sec12'),
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('The last steps', 2, None, '___sec13'),
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('A soft classifier', 2, None, '___sec14'),
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('Soft optmization problem', 2, None, '___sec15'),
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('Kernels and non-linearity', 2, None, '___sec16')]}
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<!-- ------------------- main content ---------------------- -->
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<center><h1>Data Analysis and Machine Learning: Support Vector Machines</h1></center> <!-- document title -->
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<p>
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<!-- author(s): Morten Hjorth-Jensen -->
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<center>
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<b>Morten Hjorth-Jensen</b> [1, 2]
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</center>
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<p>
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<!-- institution(s) -->
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<center>[1] <b>Department of Physics, University of Oslo</b></center>
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<center>[2] <b>Department of Physics and Astronomy and National Superconducting Cyclotron Laboratory, Michigan State University</b></center>
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<br>
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<p>
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<center><h4>Nov 5, 2018</h4></center> <!-- date -->
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<br>
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<p>
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<!-- !split --><br><br><br><br><br><br><br><br><br><br>
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<h2 id="___sec0">Support Vector Machines, overarching aims </h2>
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<p>
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A Support Vector Machine (SVM) is a very powerful and versatile
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Machine Learning model, capable of performing linear or nonlinear
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classification, regression, and even outlier detection. It is one of
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the most popular models in Machine Learning, and anyone interested in
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Machine Learning should have it in their toolbox. SVMs are
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particularly well suited for classification of complex but small-sized or
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medium-sized datasets.
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<p>
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The case with two well-separated classes only can be understood in an intuitive way in terms of lines in a two-dimensional space separating the two classes (see figure below).
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<p>
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The basic mathematics behind the SVM is however less familiar to most of us.
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It relies on the definition of hyperplanes and the
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definition of a <b>margin</b> which separates classes (in case of
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classification problems) of variables. It is also used for regression
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problems.
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<p>
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With SVMs we distinguish between hard margin and soft margins. The latter introduces a so-called softening parameter to be discussed below.
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We distinguish also between linear and non-linear approaches. The latter are the most frequent ones since it is rather unlikely that we can separate classes easily by say straight lines.
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<p>
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<!-- !split --><br><br><br><br><br><br><br><br><br><br>
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<h2 id="___sec1">Hyperplanes and all that </h2>
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<p>
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The theory behind support vector machines (SVM hereafter) is based on
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the mathematical description of so-called hyperplanes. Let us start
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with a two-dimensional case. This will also allow us to introduce our
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first SVM examples. These will be tailored to the case of two specific
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classes, as displayed in the figure here.
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<p>
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We assume here that our data set can be well separated into two
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domains, where a straight line does the job in the separating the two
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classes. Here the two classes are represented by either crosses or
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circles.
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<p>
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<!-- !split --><br><br><br><br><br><br><br><br><br><br>
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<h2 id="___sec2">What is a hyperplane? </h2>
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<p>
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The aim of the SVM algorithm is to find a hyperplane in an \( p \)-dimensional space, where \( p \) is the number of features that distinctly classifies the data points.
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<p>
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In a \( p \)-dimensional space, a hyperplane is what we call an affine subspace of dimension of \( p-1 \).
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As an example, in two dimension, a hyperplane is simply as straight line while in three dimensions it is
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a two-dimensional subspace, or stated simply, a plane.
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<p>
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In two dimensions, with the variables \( x_1 \) and \( x_2 \), the hyperplane is defined as
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$$
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b+w_1x_1+w_2x_2=0,
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$$
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where \( b \) is the intercept and \( w_1 \) and \( w_2 \) define the elements of a vector orthogonal to the line
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\( b+w_1x_1+w_2x_2=0 \).
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In two dimensions we define the vectors \( \boldsymbol{x} =[x1,x2] \) and \( \boldsymbol{w}=[w1,w2] \).
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We can then rewrite the above equation as
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$$
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\boldsymbol{w}^T\boldsymbol{x}+b=0.
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$$
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<p>
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<!-- !split --><br><br><br><br><br><br><br><br><br><br>
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<h2 id="___sec3">A \( p \)-dimensional space of features </h2>
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<p>
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We limit ourselves to two classes of outputs \( y_i \) and assign these classes the values \( y_i = \pm 1 \).
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In a \( p \)-dimensional space of say \( p \) features we have a hyperplane defines as
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$$
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b+wx_1+w_2x_2+\dots +w_px_p=0.
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$$
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If we define a
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matrix \( \boldsymbol{X}=\left[\boldsymbol{x}_1,\boldsymbol{x}_2,\dots, \boldsymbol{x}_p\right] \)
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of dimension \( n\times p \), where \( n \) represents the observations for each feature and each vector \( x_i \) is a column vector of the matrix \( \boldsymbol{X} \),
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$$
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\boldsymbol{x}_i = \begin{bmatrix} x_{i1} \\ x_{i2} \\ \dots \\ \dots \\ x_{ip} \end{bmatrix}.
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$$
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If the above condition is not met for a given vector \( \boldsymbol{x}_i \) we have
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$$
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b+w_1x_{i1}+w_2x_{i}2+\dots +w_px_{ip} >0,
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$$
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if our output \( y_i=1 \).
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In this case we say that \( \boldsymbol{x}_i \) lies on one of the sides of the hyperplane and if
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$$
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b+w_1x_{i1}+w_2x_{i}2+\dots +w_px_{ip} < 0,
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$$
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for the class of observations \( y_i=-1 \),
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then \( \boldsymbol{x}_i \) lies on the other side.
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<p>
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Equivalently, for the two classes of observations we have
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$$
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y_i\left(b+w_1x_{i1}+w_2x_{i}2+\dots +w_px_{ip}\right) > 0.
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$$
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<p>
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When we try to separate hyperplanes, if it exists, we can use it to construct a natural classifier: a test observation is assigned a given class depending on which side of the hyperplane it is located.
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<p>
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<!-- !split -->
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<h2 id="___sec4">The two-dimensional case </h2>
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<p>
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Let us try to develop our intuition about SVMs by limiting ourselves to a two-dimensional
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plane. To separate the two classes of data points, there are many
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possible lines (hyperplanes if you prefer a more strict naming)
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that could be chosen. Our objective is to find a
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plane that has the maximum margin, i.e the maximum distance between
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data points of both classes. Maximizing the margin distance provides
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some reinforcement so that future data points can be classified with
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more confidence.
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<p>
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What a linear classifier attempts to accomplish is to split the
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feature space into two half spaces by placing a hyperplane between the
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data points. This hyperplane will be our decision boundary. All
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points on one side of the plane will belong to class one and all points
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on the other side of the plane will belong to the second class two.
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<p>
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Unfortunately there are many ways in which we can place a hyperplane
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to divide the data. Below is an example of two candidate hyperplanes
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for our data sample.
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<p>
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<!-- !split --><br><br><br><br><br><br><br><br><br><br>
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<h2 id="___sec5">Getting into the details </h2>
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<p>
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Let us define the function
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$$
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f(x) = \boldsymbol{w}^T\boldsymbol{x}+b = 0,
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$$
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as the function that determines the line \( L \) that separates two classes (our two features), see the figure here.
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<p>
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Any point defined by \( \boldsymbol{x}_i \) and \( \boldsymbol{x}_2 \) on the line \( L \) will satisfy \( \boldsymbol{w}^T(\boldsymbol{x}_1-\boldsymbol{x}_2)=0 \).
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<p>
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The signed distance \( \delta \) from any point defined by a vector \( \boldsymbol{x} \) and a point \( \boldsymbol{x}_0 \) on the line \( L \) is then
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$$
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\delta = \frac{1}{\vert\vert \boldsymbol{w}\vert\vert}(\boldsymbol{w}^T\boldsymbol{x}+b).
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$$
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<p>
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<!-- !split --><br><br><br><br><br><br><br><br><br><br>
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<h2 id="___sec6">First attempt at a minimization approach </h2>
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<p>
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How do we find the parameter \( b \) and the vector \( \boldsymbol{w} \)? What we could
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do is to define a cost function which now contains the set of all
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misclassified points \( M \) and attempt to minimize this function
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$$
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C(\boldsymbol{w},b) = -\sum_{i\in M} y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b).
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$$
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<p>
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We could now for example define all values \( y_i =1 \) as misclassified in case we have \( \boldsymbol{w}^T\boldsymbol{x}_i+b < 0 \) and the opposite if we have \( y_i=-1 \). Taking the derivatives gives us
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$$
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\frac{\partial C}{\partial b} = -\sum_{i\in M} y_i,
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$$
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and
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$$
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\frac{\partial C}{\partial \boldsymbol{w}} = -\sum_{i\in M} y_ix_i.
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$$
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<p>
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<!-- !split --><br><br><br><br><br><br><br><br><br><br>
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<h2 id="___sec7">Solving the equations </h2>
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<p>
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We can now use the Newton-Raphson method or gradient descent to solve the equations
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$$
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b \leftarrow b +\eta \frac{\partial C}{\partial b},
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$$
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and
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$$
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\boldsymbol{w} \leftarrow \boldsymbol{w} +\eta \frac{\partial C}{\partial \boldsymbol{w}},
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$$
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where \( \eta \) is our by now well-known learning rate.
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<p>
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There are however problems with this approach, although it looks
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pretty straightforward to implement. In case we separate our data into
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two distinct classes, we may up with many possible lines, as indicated
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in the figure and shown by running the following program. For small
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gaps between the entries, we may also end up needing many iterations
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before the solutions converge and if the data cannot be separated
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properly into two distinct classes, we may not experience a converge
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at all.
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<p>
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<!-- !split --><br><br><br><br><br><br><br><br><br><br>
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<h2 id="___sec8">A better approach </h2>
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<p>
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A better approach is rather to try to define a large margin between
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the two classes (if they are well separated from the beginning).
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<p>
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Thus, we wish to find a margin \( M \) with \( \boldsymbol{w} \) normalized to
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\( \vert\vert \boldsymbol{w}\vert\vert =1 \) subject to the condition
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$$
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y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b) \geq M \hspace{0.1cm}\forall i=1,2,\dots, p.
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$$
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All points are thus at a signed distance from the decision boundary defined by the line \( L \). The parameters \( b \) and \( w_1 \) and \( w_2 \) define this line.
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<p>
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We seek thus the largest value \( M \) defined by
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$$
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\frac{1}{\vert \vert \boldsymbol{w}\vert\vert}y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b) \geq M \hspace{0.1cm}\forall i=1,2,\dots, n,
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$$
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or just
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$$
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y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b) \geq M\vert \vert \boldsymbol{w}\vert\vert \hspace{0.1cm}\forall i.
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$$
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If we scale the equation so that \( \vert \vert \boldsymbol{w}\vert\vert = 1/M \), we have to find the minimum of
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\( \boldsymbol{w}^T\boldsymbol{w}=\vert \vert \boldsymbol{w}\vert\vert \) (the norm) subject to the condition
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$$
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y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b) \geq 1 \hspace{0.1cm}\forall i.
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$$
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<p>
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We have thus defined our margin as the invers of the norm of \( \boldsymbol{w} \). We want to minimize the norm in order to have a as large as possible margin \( M \). Before we proceed, we need to remind ourselves about Lagrangian multipliers.
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<p>
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<!-- !split --><br><br><br><br><br><br><br><br><br><br>
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<h2 id="___sec9">A quick reminder on Lagrangian multipliers </h2>
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<p>
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Consider a function of three independent variables \( f(x,y,z) \) . For the function \( f \) to be an
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extreme we have
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$$
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df=0.
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$$
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A necessary and sufficient condition is
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$$
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\frac{\partial f}{\partial x} =\frac{\partial f}{\partial y}=\frac{\partial f}{\partial z}=0,
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$$
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due to
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$$
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df = \frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy+\frac{\partial f}{\partial z}dz.
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$$
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In many problems the variables \( x,y,z \) are often subject to constraints (such as those above for the margin)
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so that they are no longer all independent. It is possible at least in principle to use each
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constraint to eliminate one variable
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and to proceed with a new and smaller set of independent varables.
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<p>
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The use of so-called Lagrangian multipliers is an alternative technique when the elimination
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of variables is incovenient or undesirable. Assume that we have an equation of constraint on
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the variables \( x,y,z \)
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$$
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\phi(x,y,z) = 0,
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$$
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resulting in
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$$
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d\phi = \frac{\partial \phi}{\partial x}dx+\frac{\partial \phi}{\partial y}dy+\frac{\partial \phi}{\partial z}dz =0.
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$$
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Now we cannot set anymore
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$$
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\frac{\partial f}{\partial x} =\frac{\partial f}{\partial y}=\frac{\partial f}{\partial z}=0,
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$$
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if \( df=0 \) is wanted
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because there are now only two independent variables! Assume \( x \) and \( y \) are the independent
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variables.
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Then \( dz \) is no longer arbitrary.
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<p>
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<!-- !split --><br><br><br><br><br><br><br><br><br><br>
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<h2 id="___sec10">Adding the muliplier </h2>
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<p>
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However, we can add to
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$$
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df = \frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy+\frac{\partial f}{\partial z}dz,
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$$
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a multiplum of \( d\phi \), viz. \( \lambda d\phi \), resulting in
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$$
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df+\lambda d\phi = (\frac{\partial f}{\partial z}+\lambda
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\frac{\partial \phi}{\partial x})dx+(\frac{\partial f}{\partial y}+\lambda\frac{\partial \phi}{\partial y})dy+
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(\frac{\partial f}{\partial z}+\lambda\frac{\partial \phi}{\partial z})dz =0.
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$$
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Our multiplier is chosen so that
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$$
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\frac{\partial f}{\partial z}+\lambda\frac{\partial \phi}{\partial z} =0.
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$$
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<p>
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We need to remember that we took \( dx \) and \( dy \) to be arbitrary and thus we must have
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$$
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\frac{\partial f}{\partial x}+\lambda\frac{\partial \phi}{\partial x} =0,
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$$
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and
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$$
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\frac{\partial f}{\partial y}+\lambda\frac{\partial \phi}{\partial y} =0.
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$$
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When all these equations are satisfied, \( df=0 \). We have four unknowns, \( x,y,z \) and
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\( \lambda \). Actually we want only \( x,y,z \), \( \lambda \) needs not to be determined,
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it is therefore often called
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Lagrange's undetermined multiplier.
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If we have a set of constraints \( \phi_k \) we have the equations
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$$
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\frac{\partial f}{\partial x_i}+\sum_k\lambda_k\frac{\partial \phi_k}{\partial x_i} =0.
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$$
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<p>
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<!-- !split --><br><br><br><br><br><br><br><br><br><br>
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<h2 id="___sec11">Setting up the problem </h2>
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In order to solve the above problem, we define the following Lagrangian function to be minimized
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$$
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{\cal L}(\lambda,b,\boldsymbol{w})=\frac{1}{2}\boldsymbol{w}^T\boldsymbol{w}-\sum_{i=1}^n\lambda_i\left[y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b)-1\right],
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$$
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where \( \lambda_i \) is a so-called Lagrange multiplier subject to the condition \( \lambda_i \geq 0 \).
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<p>
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Taking the derivatives with respect to \( b \) and \( \boldsymbol{w} \) we obtain
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$$
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\frac{\partial {\cal L}}{\partial b} = -\sum_{i} \lambda_iy_i=0,
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$$
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and
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$$
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\frac{\partial {\cal L}}{\partial \boldsymbol{w}} = 0 = \boldsymbol{w}-\sum_{i} \lambda_iy_i\boldsymbol{x}_i.
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$$
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Inserting these constraints into the equation for \( {\cal L} \) we obtain
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$$
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{\cal L}=\sum_i\lambda_i-\frac{1}{2}\sum_{ij}^n\lambda_i\lambda_jy_iy_j\boldsymbol{x}_i^T\boldsymbol{x}_j,
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$$
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subject to the constraints \( \lambda_i\geq 0 \) and \( \sum_i\lambda_iy_i=0 \).
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We must in addition satisfy the <a href="https://en.wikipedia.org/wiki/Karush%E2%80%93Kuhn%E2%80%93Tucker_conditions" target="_blank">Karush-Kuhn-Tucker</a> (KKT) condition
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$$
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\lambda_i\left[y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b) -1\right] \hspace{0.1cm}\forall i.
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$$
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<ol>
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<li> If \( \lambda_i > 0 \), then \( y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b)=1 \) and we say that \( x_i \) is on the boundary.</li>
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<li> If \( y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b)> 1 \), we say \( x_i \) is not on the boundary and we set \( \lambda_i=0 \).</li>
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</ol>
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When \( \lambda_i > 0 \), the vectors \( \boldsymbol{x}_i \) are called support vectors. They are the vectors closest to the line (or hyperplane) and define the margin \( M \).
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<p>
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<!-- !split --><br><br><br><br><br><br><br><br><br><br>
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<h2 id="___sec12">The problem to solve </h2>
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<p>
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We can rewrite
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$$
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{\cal L}=\sum_i\lambda_i-\frac{1}{2}\sum_{ij}^n\lambda_i\lambda_jy_iy_j\boldsymbol{x}_i^T\boldsymbol{x}_j,
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$$
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and its constraints in terms of a matrix-vector problem where we minimize w.r.t. \( \lambda \) the following problem
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$$
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\frac{1}{2} \boldsymbol{\lambda}^T\begin{bmatrix} y_1y_1\boldsymbol{x}_1^T\boldsymbol{x}_1 & y_1y_2\boldsymbol{x}_1^T\boldsymbol{x}_2 & \dots & \dots & y_1y_n\boldsymbol{x}_1^T\boldsymbol{x}_n \\
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y_2y_1\boldsymbol{x}_2^T\boldsymbol{x}_1 & y_2y_2\boldsymbol{x}_2^T\boldsymbol{x}_2 & \dots & \dots & y_1y_n\boldsymbol{x}_2^T\boldsymbol{x}_n \\
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\dots & \dots & \dots & \dots & \dots \\
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\dots & \dots & \dots & \dots & \dots \\
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|
y_ny_1\boldsymbol{x}_n^T\boldsymbol{x}_1 & y_ny_2\boldsymbol{x}_n^T\boldsymbol{x}_2 & \dots & \dots & y_ny_n\boldsymbol{x}_n^T\boldsymbol{x}_n \\
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\end{bmatrix}\boldsymbol{\lambda}-\mathbb{1}\boldsymbol{\lambda},
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|
$$
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|
subject to \( \boldsymbol{y}^T\boldsymbol{\lambda}=0 \). Here we defined the vectors \( \boldsymbol{\lambda} =[\lambda_1,\lambda_2,\dots,\lambda_n] \) and
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\( \boldsymbol{y}=[y_1,y_2,\dots,y_n] \).
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<p>
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<!-- !split --><br><br><br><br><br><br><br><br><br><br>
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<h2 id="___sec13">The last steps </h2>
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<p>
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|
Solving the above problem, yields the values of \( \lambda_i \).
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To find the coefficients of your hyperplane we need simply to compute
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|
$$
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|
\boldsymbol{w}=\sum_{i} \lambda_iy_i\boldsymbol{x}_i.
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$$
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|
With our vector \( \boldsymbol{w} \) we can in turn find the value of the intercept \( b \) (here in two dimensions) via
|
|
$$
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y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b)=1,
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$$
|
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|
resulting in
|
|
$$
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|
b = \frac{1}{y_i}-\boldsymbol{w}^T\boldsymbol{x}_i,
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|
$$
|
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|
or if we write it out in terms of the support vectors only, with \( N_s \) being their number, we have
|
|
$$
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|
b = \frac{1}{N_s}\sum_{j\in N_s}\left(y_j-\sum_{i=1}^n\lambda_iy_i\boldsymbol{x}_i^T\boldsymbol{x}_j\right).
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|
$$
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|
With our hyperplane coefficients we can use our classifier to assign any observation by simply using
|
|
$$
|
|
y_i = \mathrm{sign}(\boldsymbol{w}^T\boldsymbol{x}_i+b).
|
|
$$
|
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|
Below we discuss how to find the optimal values of \( \lambda_i \). Before we proceed however, we discuss now the so-called soft classifier.
|
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|
<p>
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|
<!-- !split --><br><br><br><br><br><br><br><br><br><br>
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<h2 id="___sec14">A soft classifier </h2>
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|
<p>
|
|
Till now, the margin is strictly defined by the support vectors. This defines what is called a hard classifier, that is the margins are well defined.
|
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|
<p>
|
|
Suppose now that classes overlap in feature space, as shown in the
|
|
figure here. One way to deal with this problem before we define the
|
|
so-called <b>kernel approach</b>, is to allow a kind of slack in the sense
|
|
that we allow some points to be on the wrong side of the margin.
|
|
|
|
<p>
|
|
We introduce thus the so-called <b>slack</b> variables \( \boldsymbol{\xi} =[\xi_1,x_2,\dots,x_n] \) and
|
|
modify our previous equation
|
|
$$
|
|
y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b)=1,
|
|
$$
|
|
|
|
to
|
|
$$
|
|
y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b)=1-\xi_i,
|
|
$$
|
|
|
|
with the requirement \( \xi_i\geq 0 \). The total violation is now \( \sum_i\xi \).
|
|
The value \( \xi_i \) in the constraint the last constraint corresponds to the amount by which the prediction
|
|
\( y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b)=1 \) is on the wrong side of its margin. Hence by bounding the sum \( \sum_i \xi_i \),
|
|
we bound the total amount by which predictions fall on the wrong side of their margins.
|
|
|
|
<p>
|
|
Misclassifications occur when \( \xi_i > 1 \). Thus bounding the total sum by some value \( C \) bounds in turn the total number of
|
|
misclassifications.
|
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|
<p>
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|
<!-- !split --><br><br><br><br><br><br><br><br><br><br>
|
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|
<h2 id="___sec15">Soft optmization problem </h2>
|
|
|
|
<p>
|
|
This has in turn the consequences that we change our optmization problem to finding the minimum of
|
|
$$
|
|
{\cal L}=\frac{1}{2}\boldsymbol{w}^T\boldsymbol{w}-\sum_{i=1}^n\lambda_i\left[y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b)-(1-\xi_)\right]+C\sum_{i=1}^n\xi_i-\sum_{i=1}^n\gamma_i\xi_i,
|
|
$$
|
|
|
|
subject to
|
|
$$
|
|
y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b)=1-\xi_i \hspace{0.1cm}\forall i,
|
|
$$
|
|
|
|
with the requirement \( \xi_i\geq 0 \).
|
|
|
|
<p>
|
|
Taking the derivatives with respect to \( b \) and \( \boldsymbol{w} \) we obtain
|
|
$$
|
|
\frac{\partial {\cal L}}{\partial b} = -\sum_{i} \lambda_iy_i=0,
|
|
$$
|
|
|
|
and
|
|
$$
|
|
\frac{\partial {\cal L}}{\partial \boldsymbol{w}} = 0 = \boldsymbol{w}-\sum_{i} \lambda_iy_i\boldsymbol{x}_i,
|
|
$$
|
|
|
|
and
|
|
$$
|
|
\lambda_i = C-\gamma_i \hspace{0.1cm}\forall i.
|
|
$$
|
|
|
|
Inserting these constraints into the equation for \( {\cal L} \) we obtain the same equation as before
|
|
$$
|
|
{\cal L}=\sum_i\lambda_i-\frac{1}{2}\sum_{ij}^n\lambda_i\lambda_jy_iy_j\boldsymbol{x}_i^T\boldsymbol{x}_j,
|
|
$$
|
|
|
|
but now subject to the constraints \( \lambda_i\geq 0 \), \( \sum_i\lambda_iy_i=0 \) and \( 0\leq\lambda_i \leq C \).
|
|
We must in addition satisfy the Karush-Kuhn-Tucker condition which now reads
|
|
$$
|
|
\lambda_i\left[y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b) -(1-\xi_)\right]=0 \hspace{0.1cm}\forall i,
|
|
$$
|
|
|
|
$$
|
|
\gamma_i\xi_i = 0,
|
|
$$
|
|
|
|
and
|
|
$$
|
|
y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b) -(1-\xi_) \geq 0 \hspace{0.1cm}\forall i.
|
|
$$
|
|
|
|
<p>
|
|
<!-- !split --><br><br><br><br><br><br><br><br><br><br>
|
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|
<h2 id="___sec16">Kernels and non-linearity </h2>
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<!-- ------------------- end of main content --------------- -->
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<center style="font-size:80%">
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<!-- copyright --> © 1999-2018, Morten Hjorth-Jensen. Released under CC Attribution-NonCommercial 4.0 license
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</center>
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