 # Support Vector Machines (and Kernel Methods in general)

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Support Vector Machines (and Kernel Methods in general)
Machine Learning March 23, 2010

Last Time Multilayer Perceptron/Logistic Regression Networks
Neural Networks Error Backpropagation

Today Support Vector Machines
Note: we’ll rely on some math from Optimality Theory that we won’t derive.

Maximum Margin Perceptron (and other linear classifiers) can lead to many equally valid choices for the decision boundary Are these really “equally valid”?

Max Margin How can we pick which is best?
Maximize the size of the margin. Small Margin Large Margin Are these really “equally valid”?

Support Vectors Support Vectors are those input points (vectors) closest to the decision boundary 1. They are vectors 2. They “support” the decision hyperplane

Support Vectors Define this as a decision problem
The decision hyperplane: No fancy math, just the equation of a hyperplane.

Support Vectors Aside: Why do some cassifiers use or
Simplicity of the math and interpretation. For probability density function estimation 0,1 has a clear correlate. For classification, a decision boundary of 0 is more easily interpretable than .5.

Support Vectors Define this as a decision problem
The decision hyperplane: Decision Function:

Support Vectors Define this as a decision problem
The decision hyperplane: Margin hyperplanes:

Support Vectors The decision hyperplane: Scale invariance

Support Vectors The decision hyperplane: Scale invariance

Support Vectors The decision hyperplane: Scale invariance
This scaling does not change the decision hyperplane, or the support vector hyperplanes. But we will eliminate a variable from the optimization The decision hyperplane: Scale invariance

What are we optimizing? We will represent the size of the margin in terms of w. This will allow us to simultaneously Identify a decision boundary Maximize the margin

How do we represent the size of the margin in terms of w?
There must at least one point that lies on each support hyperplanes Proof outline: If not, we could define a larger margin support hyperplane that does touch the nearest point(s).

How do we represent the size of the margin in terms of w?
There must at least one point that lies on each support hyperplanes Proof outline: If not, we could define a larger margin support hyperplane that does touch the nearest point(s).

How do we represent the size of the margin in terms of w?
There must at least one point that lies on each support hyperplanes Thus: And:

How do we represent the size of the margin in terms of w?
There must at least one point that lies on each support hyperplanes Thus: And:

How do we represent the size of the margin in terms of w?
The vector w is perpendicular to the decision hyperplane If the dot product of two vectors equals zero, the two vectors are perpendicular.

How do we represent the size of the margin in terms of w?
The margin is the projection of x1 – x2 onto w, the normal of the hyperplane.

Aside: Vector Projection

How do we represent the size of the margin in terms of w?
The margin is the projection of x1 – x2 onto w, the normal of the hyperplane. Projection: Size of the Margin:

Maximizing the margin Goal: maximize the margin
Linear Separability of the data by the decision boundary

Max Margin Loss Function
If constraint optimization then Lagrange Multipliers Optimize the “Primal”

Max Margin Loss Function
Optimize the “Primal” Partial wrt b

Max Margin Loss Function
Optimize the “Primal” Partial wrt w

Max Margin Loss Function
Optimize the “Primal” Partial wrt w \frac{\partial L(\vec{w},b)}{\partial \vec{w}}&=&0\\\vec{w} - \sum_{i=0}^{N-1}\alpha_it_i\vec{x_i}&=&0\\\vec{w}&=&\sum_{i=0}^{N-1}\alpha_it_i\vec{x_i}\\ Now have to find αi. Substitute back to the Loss function

Max Margin Loss Function
Construct the “dual” W(\alpha) = \sum_{i=0}^{N-1}\alpha_i - \frac{1}{2}\sum_{i,j=0}^{N-1}\alpha_i\alpha_j t_it_j(\vec{x_i}\cdot\vec{x_j})

Dual formulation of the error
Optimize this quadratic program to identify the lagrange multipliers and thus the weights There exist (extremely) fast approaches to quadratic optimization in both C, C++, Python, Java and R

If Q is positive semi definite, then f(x) is convex. If f(x) is convex, then there is a single maximum.

Support Vector Expansion
New decision Function Independent of the Dimension of x! When αi is non-zero then xi is a support vector When αi is zero xi is not a support vector

Kuhn-Tucker Conditions
In constraint optimization: At the optimal solution Constraint * Lagrange Multiplier = 0 \alpha_i(1-t_i(\vec{w}^T\vec{x_i} + b))=0 Only points on the decision boundary contribute to the solution!

Visualization of Support Vectors

Interpretability of SVM parameters
What else can we tell from alphas? If alpha is large, then the associated data point is quite important. It’s either an outlier, or incredibly important. But this only gives us the best solution for linearly separable data sets…

Basis of Kernel Methods
The decision process doesn’t depend on the dimensionality of the data. We can map to a higher dimensionality of the data space. Note: data points only appear within a dot product. The error is based on the dot product of data points – not the data points themselves.

Basis of Kernel Methods
Since data points only appear within a dot product. Thus we can map to another space through a replacement The error is based on the dot product of data points – not the data points themselves.

Learning Theory bases of SVMs
Theoretical bounds on testing error. The upper bound doesn’t depend on the dimensionality of the space The lower bound is maximized by maximizing the margin, γ, associated with the decision boundary.

Why we like SVMs They work Easily interpreted.
Good generalization Easily interpreted. Decision boundary is based on the data in the form of the support vectors. Not so in multilayer perceptron networks Principled bounds on testing error from Learning Theory (VC dimension)

SVM vs. MLP SVMs have many fewer parameters
SVM: Maybe just a kernel parameter MLP: Number and arrangement of nodes and eta learning rate SVM: Convex optimization task MLP: likelihood is non-convex -- local minima R(\theta)=\frac{1}{N}\sum_{n=0}^N\frac{1}{2}\left(y_n-g\left(\sum_k w_{kl}g\left(\sum_jw_{jk}g\left(\sum_iw_{ij}x_{n,i}\right) \right)\right)\right)^2

Soft margin classification
There can be outliers on the other side of the decision boundary, or leading to a small margin. Solution: Introduce a penalty term to the constraint function

Soft Max Dual Still Quadratic Programming!
W(\alpha) = \sum_{i=0}^{N-1}\alpha_i - \frac{1}{2}\sum_{i,j=0}^{N-1}t_it_j\alpha_i\alpha_j(x_i\cdot x_j)

Soft margin example Points are allowed within the margin, but cost is introduced. Hinge Loss

Probabilities from SVMs
Support Vector Machines are discriminant functions Discriminant functions: f(x)=c Discriminative models: f(x) = argmaxc p(c|x) Generative Models: f(x) = argmaxc p(x|c)p(c)/p(x) No (principled) probabilities from SVMs SVMs are not based on probability distribution functions of class instances.

Efficiency of SVMs Not especially fast. Training – n^3 Evaluation – n
Quadratic Programming efficiency Evaluation – n Need to evaluate against each support vector (potentially n)

Good Bye Next time: The Kernel “Trick” -> Kernel Methods or
How can we use SVMs that are not linearly separable?

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