annotate org/svm.org @ 158:043412f7a429

s.
author Robert McIntyre <rlm@mit.edu>
date Thu, 28 Jan 2016 09:36:35 -0800
parents dc521fea9219
children
rev   line source
rlm@71 1 #+TITLE: Notes on SVMs
rlm@71 2 #+AUTHOR:Dylan Holmes
rlm@71 3 #+SETUPFILE: ../../aurellem/org/setup.org
rlm@71 4 #+INCLUDE: ../../aurellem/org/level-0.org
rlm@71 5 #+MATHJAX: align:"left" mathml:t path:"http://www.aurellem.org/MathJax/MathJax.js"
rlm@71 6
rlm@71 7 * Question: How do you find $\vec{w}$, $b$, and the \alpha{}'s?
rlm@71 8
rlm@71 9 You should take a look at the class notes, for starters:
rlm@71 10 http://ai6034.mit.edu/fall12/images/SVM_and_Boosting.pdf
rlm@71 11
rlm@71 12 You can also take a look at @388 to see an example of (part of) a
rlm@71 13 worked problem.
rlm@71 14
rlm@71 15
rlm@71 16 When solving SVM problems, there are some useful equations to keep in
rlm@71 17 mind:
rlm@71 18
rlm@71 19 1. $\vec{w}\cdot \vec{x} + b = 0$ defines the boundary, and in
rlm@71 20 particular $\vec{w}\cdot \vec{x} + b \geq 0$ defines the positive
rlm@71 21 side of the boundary.
rlm@71 22 2. $\vec{w}\cdot \vec{x} + b = \pm 1$ defines the positive and
rlm@71 23 negative gutters.
rlm@71 24 3. The distance between the gutters (the width of the margin) is
rlm@71 25 $\text{margin-width} = \frac{2}{||\vec{w}||} =
rlm@71 26 \frac{2}{\sqrt{\vec{w}\cdot \vec{w}}}$
rlm@71 27 4. $\sum_{\text{training data}} y_i \alpha_i =
rlm@71 28 0$
rlm@71 29 5. $\sum_{\text{training data}} y_i \alpha_i \vec{x}_i = \vec{w}$
rlm@71 30
rlm@71 31 The last two equations are the ones that can help you find the
rlm@71 32 $\alpha_i$. You may also find it useful to think of the $\alpha_i$
rlm@71 33 as measuring \ldquo{}supportiveness\rdquo{}. This means, for
rlm@71 34 example, that: - $\alpha_i$ is zero for non-support vectors,
rlm@71 35 i.e. training points that do not determine the boundary, and which
rlm@71 36 would not affect the placement of the boundary if deleted - When you
rlm@71 37 compare two separate SVM problems, where the first has support
rlm@71 38 vectors that are far from the boundary, and the second has support
rlm@71 39 vectors very close to the boundary, the latter support vectors have
rlm@71 40 comparatively higher $\alpha_i$ values.
rlm@71 41
rlm@71 42 ** Can \alpha{} be zero for support vectors?
rlm@71 43 No, \alpha{} is zero for a point if and only if it's not a support
rlm@71 44 vector. You can see this because in equation (5), just the points
rlm@71 45 with nonzero alphas contribute to $\vec{w}$.
rlm@71 46
rlm@71 47 ** Are all points on the gutter support vectors? How do you tell whether a point is a support vector?
rlm@71 48
rlm@71 49 Not all points on the gutter are support vectors. To tell if a
rlm@71 50 training point is a support vector, imagine what would happen if you
rlm@71 51 deleted it --- would the decision boundary be different? If it would,
rlm@71 52 the point is a support vector. If it wouldn't, the point isn't a
rlm@71 53 support vector.
rlm@71 54
rlm@71 55 * 2011 Q2 Part A
rlm@71 56
rlm@71 57 The equation for the line you drew in part A is $y = 0$, and points
rlm@71 58 will be classified as positive if $y\geq 0$.
rlm@71 59
rlm@71 60 # shouldn't points be classified as positive only if leq 1 + b dot w x?
rlm@71 61 Now in general, the equation for the decision boundary is
rlm@71 62 $\vec{w}\cdot \vec{x} + b = 0$, and points $\vec{x}$ are
rlm@71 63 classified as positive if $\vec{w}\cdot \vec{x} + b \geq 0$. To
rlm@71 64 solve for $\vec{w}$ and $b$, you just manipulate your inequation
rlm@71 65 for the boundary into this form:
rlm@71 66
rlm@71 67 $$y\geq 0$$
rlm@71 68 $$0x + 1y + 0 \geq 0$$
rlm@71 69 $$\begin{bmatrix}0&amp;1\end{bmatrix}\cdot \begin{bmatrix}x\\y\end{bmatrix} + 0 \geq 0$$
rlm@71 70
rlm@71 71 And so we have:
rlm@71 72
rlm@71 73 $$\underbrace{\begin{bmatrix}0&amp;1\end{bmatrix}}_{\vec{w}}\cdot \underbrace{\begin{bmatrix}x\\y\end{bmatrix}}_{\vec{x}} + \underbrace{0}_{b} \geq 0$$
rlm@71 74
rlm@71 75 ...which is /almost/ right. The trouble is that we can multiply this
rlm@71 76 inequation by any positive constant $c>0$ and it will still be true
rlm@71 77 --- so, the right answer might not be the $\vec{w}$ and $b$ we found
rlm@71 78 above, but instead a /multiple/ of them. To fix this, we multiply the
rlm@71 79 inequation by $c>0$, and solve for $c$:
rlm@71 80
rlm@71 81 $$c\cdot (0x + 1y + 0) \geq c\cdot0$$
rlm@71 82 $$\begin{bmatrix}0&amp;c\end{bmatrix}\cdot \begin{bmatrix}x\\y\end{bmatrix} +
rlm@71 83 0 \geq 0$$
rlm@71 84
rlm@71 85 One formula that you can use to solve for $c$ is the margin-width
rlm@71 86 formula; by examining the lines you drew, you see that the margin
rlm@71 87 width is *4*. Now, in general, the equation for margin width is
rlm@71 88
rlm@71 89 $$\text{margin-width} = \frac{2}{||\vec{w}||} = \frac{2}{\sqrt{\vec{w}\cdot \vec{w}}}$$
rlm@71 90
rlm@71 91 So, for the second time, we use the technique of setting the quantity
rlm@71 92 you can see (margin width = 4) equal to the general formula (margin
rlm@71 93 width = 2/||w||) and solving for the parameters in the general
rlm@71 94 formula:
rlm@71 95
rlm@71 96 $$\frac{2}{||w||} = 4$$
rlm@71 97
rlm@71 98 $$\frac{2}{\sqrt{\vec{w}\cdot \vec{w}}} = 4$$
rlm@71 99
rlm@71 100 $$\frac{4}{\vec{w}\cdot \vec{w}} = 16$$ (squaring both sides)
rlm@71 101
rlm@71 102 $$\vec{w}\cdot \vec{w} = \frac{1}{4}$$
rlm@71 103
rlm@71 104 $$\begin{bmatrix}0&amp;c\end{bmatrix}\cdot \begin{bmatrix}0\\c\end{bmatrix}
rlm@71 105 = \frac{1}{4}$$
rlm@71 106
rlm@71 107 (This uses the expression for $\vec{w}$ we got after
rlm@71 108 we multiplied by $c$ above.)
rlm@71 109
rlm@71 110 $$\begin{eqnarray*}
rlm@71 111 c^2 &=& \frac{1}{4}\\
rlm@71 112 c &=& \pm \frac{1}{2}\\
rlm@71 113 c &=& \frac{1}{2}\end{eqnarray*}$$...because we require $c>0$ so that it doesn't flip the inequality.
rlm@71 114
rlm@71 115
rlm@71 116 Returning to our inequality, we have
rlm@71 117
rlm@71 118 $$\begin{eqnarray*}\begin{bmatrix}0&amp;c\end{bmatrix}\cdot \begin{bmatrix}x\\y\end{bmatrix} + 0 &\geq& 0\\
rlm@71 119 \begin{bmatrix}0&amp;\frac{1}{2}\end{bmatrix}\cdot \begin{bmatrix}x\\y\end{bmatrix} + 0 &\geq& 0\end{eqnarray*}$$
rlm@71 120
rlm@71 121 $$\underbrace{\begin{bmatrix}0&amp;\frac{1}{2}\end{bmatrix}}_{\vec{w}}\cdot \underbrace{\begin{bmatrix}x\\y\end{bmatrix}}_{\vec{x}} + \underbrace{0}_{b} \geq 0$$
rlm@71 122
rlm@71 123 And this is truly the right answer, now that we have solved for $c$.
rlm@71 124
rlm@71 125
rlm@71 126 * What's the use of gutters?
rlm@71 127 The gutters are the regions where you have no hard data supporting
rlm@71 128 your classification, but the decision boundary is, as its name
rlm@71 129 implies, the cutoff for making a decision between the two
rlm@71 130 classifications. As such, you would likely have lower confidence in
rlm@71 131 any decision within the gutters, but a classifier needs to make a
rlm@71 132 decision one way or the other for all possible test data (there's no
rlm@71 133 reason not to do so, as making no decision means you're automatically
rlm@71 134 wrong in all cases, whereas even a lousy classifier will get some of
rlm@71 135 them right).
rlm@71 136
rlm@71 137 ----
rlm@71 138
rlm@71 139 Here are two more ways to look at it. First of all, and simplest, SVMs
rlm@71 140 are supposed to find the boundary with the widest margin. The width of
rlm@71 141 the margin is just the perpendicular distance between the gutters, so
rlm@71 142 the gutters are important for that reason.
rlm@71 143
rlm@71 144 Second of all, if you attempt to define the SVM problem
rlm@71 145 geometrically, i.e. put it into a form that can be solved by a
rlm@71 146 computer, it looks something like the following.
rlm@71 147
rlm@71 148 #+BEGIN_QUOTE
rlm@71 149 Find a boundary --- which is defined by a normal vector
rlm@71 150 $\vec{w}$ and an offset $b$ --- such that:
rlm@71 151 1. The boundary \ldquo{} $\vec{w}\cdot \vec{x} + b = 0$ \rdquo{}
rlm@71 152 separates the positive and negative points
rlm@71 153 2. The width of the margin is as large as possible. Remember, the
rlm@71 154 margin width is twice the distance between the boundary and the
rlm@71 155 training point that is closest to the boundary; in other words,
rlm@71 156 $$\text{margin-width} = \min_{\text{training data}} 2 y_i
rlm@71 157 (\vec{w}\cdot \vec{x}_i + b)$$
rlm@71 158 #+END_QUOTE
rlm@71 159
rlm@71 160 Unfortunately, the problem as stated has no unique solution, since if
rlm@71 161 you find a satisfactory pair $\langle \vec{w}, b \rangle$ , then the
rlm@71 162 pair $\langle 3\vec{w}, 3b\rangle$ (for example) defines the same
rlm@71 163 boundary but has a larger margin. The problem is that any multiple of
rlm@71 164 a satisfactory pair yields another satisfactory pair. In essence,
rlm@71 165 we're just changing the units of measurement without materially
rlm@71 166 affecting the boundary.
rlm@71 167
rlm@71 168 We can /remove this additional degree of freedom by adding another
rlm@71 169 constraint/ to the problem which establishes a sense of scale. For
rlm@71 170 example, we could require $\vec{w}$ to be a /unit/ normal vector,
rlm@71 171 i.e. we could require that $||\vec{w}|| = 1$. This fixes the problem
rlm@71 172 and gives SVMs a unique solution.
rlm@71 173
rlm@71 174 In 6.034, and elsewhere, we use an alternate constraint; we impose the /gutter conditions/:
rlm@71 175
rlm@71 176 $\vec{w}\cdot \vec{x_+} + b = + 1$ for positive training points
rlm@71 177 $\vec{x_+}$.
rlm@71 178
rlm@71 179 $\vec{w}\cdot \vec{x_-} + b = -1$ for negative
rlm@71 180 training points $\vec{x_-}$.
rlm@71 181
rlm@71 182 which we often combine into a single inequality: $y_i (\vec{w}\cdot
rlm@71 183 \vec{x}_i + b) \geqslant 1$ for all training points $\langle
rlm@71 184 \vec{x}_i, y_i\rangle$.
rlm@71 185
rlm@71 186
rlm@71 187 This constraint is enough to eliminate the extra degree of freedom and give SVMs a unique solution. Moreover, you get a nice formula for the margin width by requiring that it holds:
rlm@71 188
rlm@71 189 $\text{margin-width} = \frac{2}{||\vec{w}||}$ because the gutter
rlm@71 190 conditions are enforced.
rlm@71 191
rlm@71 192
rlm@71 193 That's why gutters are important.