Neural Networks - A biologically inspired model. The efficient backpropagation learning algorithm. Hidden layers. Overfitting - Fitting the data too well; fitting the noise. Deterministic noise versus stochastic noise. Regularization - Putting the brakes on fitting the noise. Hard and soft constraints. Augmented error and weight decay. Validation - Taking a peek out of sample. Model selection and data contamination.
Cross validation. Support Vector Machines - One of the most successful learning algorithms; getting a complex model at the price of a simple one.
Kernel Methods - Extending SVM to infinite-dimensional spaces using the kernel trick, and to non-separable data using soft margins. Failed to load latest commit information. Added problems 6. May 31, Added problem 7. Jun 16, Added problem 9. Jul 14, Initial commit. Jun 6, Appendix B Linear Algebra. Added exercises B. Jul 16, Added appendix C. Jul 18, Chapter 2 Training vs. Moved out of sub directories. Mar 7, Jul 19, Solutions to Chapter 1 The Learning Problem. Jul 9, Solutions to Chapter 2 Training versus Testing.
Solutions to Chapter 3 The Linear Model. Solutions to Chapter 4 Overfitting. Solutions to Chapter 5 Three Learning Principles. Solutions to Chapter 6 Similarity-Based Methods. Added problems 9. Jul 10, Solutions to Chapter 7 Neural Networks. Solutions to Chapter 8 Support Vector Machine.
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Consider two potential clients of this fingerprint system. D The moral of this example is that the choice of the error measure depends on how the system is going to be used. The right values depend on the application. An unauthorized person will gain access to a highly sensitive facility. All future revenue from this annoyed customer is lost. We need to specify the error values for a false accept and for a false reject.
In the supermarket and CIA scenarios. If the right person is accepted or an intruder is rejected. On the other hand. This should be reflected in a much higher cost for the false accept. False rejects. The other is that the weighted cost may be a difficult objective function for optimizers to work with.
One is that the user may not provide an error specification. We have already seen an example of this with the simple binary error used in this chapter. Assume we randomly picked all the y's according to the distribution P y I x over the entire input space X. This realization of P y I x i s effectively a target function. Remember the two questions of learning? With the same learning model.
If y is real-valued for example. Eout may be as close to Ein in the noisy case as it is in the If we use the same h to a pproximate a noisy version of f given by y f x. One can think of a noisy target as a deterministic target plus added noise. This situation can be readily modeled within the same framework that we have. This does not mean that learning a noisy target is as easy as learning a deterministic one. This view suggests that a deterministic target function can be considered a special case of a noisy target.
A data point x. Our entire analysis of the feasibility of learning applies to noisy target functions as well. While both distributions model probabilistic aspects of x and y. In Chapter 2. When you look at the ba l l it is black. You pick a bag at ra ndom a nd then pick one of the ba lls in that bag at random. I n more tha n two d i mensions. You now pick the second ba l l from that same bag.
For simplicity. Technical ly. One bag has 2 black ba l ls and the other has a black and a white ba l l. Problem 1. The fol lowing steps wil l guide you through the proof. What is the pro bability that this ba l l is also black? Be sure to mark the exa m ples from different classes d ifferently. Com pare you r resu lts with b. This problem leads you to explore the algorith m fu rther with data sets of d ifferent sizes a n d dimensions. I n t h e iterations of each experiment.
Compare you r resu lts with b. Plot a histogra m for the n u m ber of u pdates that the a lgorith m takes to converge. In practice. How many u pdates does the a lgorithm ta ke to converge? PLA converges more q uickly tha n the bound p suggests. Report the n u m ber of u pdates that the a lgorith m ta kes before converging. Compare you r results with b. Com ment on whether f is close to g. Plot the training data set. To get g.
Generate a test data set of size That is. Report the error on the test set. T h e algorithm a bove i s a variant of the so ca l led Adaline Adaptive Linear Neuron a lgorithm for perceptron learn ing. In this problem. For a given coin. Assume we have a n u mber of coins that generate different sa m ples independently.
One of the sim plest forms of that law is the Chebyshev Inequality. In P roblem 1. UN are iid random varia bles. On the same plot show the bound that wou ld be obtained usi ng the Hoeffding I neq u a lity. Remember that for a single coin. Eval u ate U s as a fun ction of s. For a fixed V of size N. We focus on the simple case of flipping a fair coin. For the two risk matrices in Exa mple 1.
You have now proved that i n a noiseless setting. This in-sa mple error should weight the different types of errors based on the risk matrix. What happens to you r two estimators hmean and hmed? Argue that for a ny two deterministic a lgorithms Ai a nd A2. Similar results can be proved for more genera l settings. Although these problems are not the exact ones that will appear on the exam. They are the 'training set' in your learning.
If the professor's goal is to help you do better in the exam. Chapter 2 Training versus Testing Before the final exam. We began the analysis of in-sample error in Chapter 1. Doing well in the exam is not the goal in and of itself. It expressly measures training performance. The same distinction between training and testing happens in learning from data. Eout is based on the performance over the entire input space X. The in sample error Ein. Such performance has the benefit of looking at the solutions and adjusting accordingly.
The goal is for you to learn the course material. The exam is merely a way to gauge how well you have learned the material.
We will also discuss the conceptual and practical implications of the contrast between training and testing. If the exam problems are known ahead of time. This is important for learning. Eout The mathematical results provide fundamental insights into learning from data. The Eout h 2: Ein h. To see that the Hoeffding Inequality implies 1. E also holds. Eout 2: Ein. We would like to replace with M as 1 Sometimes 'generalization error' is used another name for Eout. Not only do we want to know that the hypothesis g that we choose say the one with the best training error will continue to do well out of sample i.
E for all h E 1-l. To make it easier on the not-so-mathematically inclined. If 1-l is an infinite set. Pick a tolerance level 8. We will also make the contrast between a training set and a test set more precise. Generalization error. E direction of the bound assures us that we couldn't do much better because every hypothesis with a higher Ein than the g we have chosen will have a comparably higher Eout. Generalization is a key issue in learning. This can be rephrased as follows.
We may now 2Me 2NE2. Notice that the other side of IEout Ein l The error bound ln in 2. We have already discussed how the value of Ein does not always generalize to a similar value of Eout. A word of warning: this chapter is the heaviest in this book in terms of mathematical abstraction. Once we properly account for the overlaps of the different hypotheses.
In a typical learning model. To do this. The union bound says that the total area covered by If h1 is very similar to h2 for instance. We then over-estimated the probability using the union bound. The mathematical theory of generalization hinges on this observation. If the events B1.
If you take the perceptron model for instance. Such an N-tuple is called a dichotomy since it splits x1. The definition o f the growth function i s based on the number o f different hypotheses that 1-l can implement. The dichotomies generated by 1-l on these points are defined by 1-l x1. If h E 1-l is applied to a finite sample x1. The growth function is defined for a hypothesis set 1-l by where I I denotes the cardinality number of elements of a set.
For any 1-l. Definition 2. A larger 1-l x1. This signifies that 1-l is as diverse as can be on this particular sample. Each h E 1-l generates a dichotomy on x1. We will focus on binary target functions for the purpose of this analysis.
If 1-l is capable of generating all possible dichotomies on x1. Let x1. These three steps will yield the generalization bound that we need. To compute mH N. Let us find a formula for mH N in each of the following cases. In the case of 4 points. Figure 2. Example 2.
D Let us now illustrate how to compute mH N for some simple hypothesis sets. These examples will confirm the intuition that m1-l N grows faster when the hypothesis set 1-l becomes more complex. The dichotomy of red versus blue on the 3 colinear points in part a cannot be generated by a perceptron. At most 14 out of the possible 16 dichotomies on any 4 points can be generated. The most a perceptron can do on any 4 points is 14 dichotomies out of the possible One can verify that there are no 4 points that the perceptron can shatter.
Since this is the most we can get for any points. N which is allowed. Per the next :figure. If both end values fall in the same region. Notice that m1-l N grows as the square of of the 'simpler' positive ray case. As we vary a. To compute m1-l N in this case. If you connect the 1 points with a polygon.
For the dichotomies that have less than three 1 points. Adding up these possibilities. This does since it is defined based on the maximum 2. Each hypothesis is specified by the two end values of that interval. The dichotomy we get is decided Nil by which two regions contain the end values of the interval. To compute m1-l N. The dichotomy we get on the points is decided by which region contains the value a.
Getting a good bound on mH N will prove much easier than computing m1l N itself. If no data set of size k can be shattered by 1-l. D It is not practical to try to compute m11 N for every hypothesis set we use. Verify that m If k is a break point. In general. We now use the break point k to derive a bound on the growth function m11 N for all values of N. Exercise 2. We will exploit this idea to get a significant bound on m1-l N in general.
If m1-l N replaced M in Equa- tion 2. The fact that the bound is polynomial is crucial. A similar green box will tell you when rejoin. Since B N. To prove the polynomial bound. The notation B comes from ' Binomial' and the reason will become clear shortly. This bound will therefore apply to any 1-l. To evaluate B N.
The definition of B N. This means that we will generalize well given a sufficient number of examples. Absent a break point as is the case in the convex hypothesis example. We now assume N 2: 2 and k 2: 2 and try t o develop a recursion. A second different dichotomy must differ on at least one point and then that subset of size 1 would be shattered.
We list these dichotomies in the following table. XN in the table are labels for the N points of the dichotomy. Consider the dichotomies on xi. We collect these dichotomies in the set S1. The remaining dichotomies on the first N 1 points appear twice. Since no subset of k of these first N 1 points can - Let S1 have a rows.
St and S-. We collect these dichotomies in the set S2 which can be divided into two equal parts. We have chosen a convenient order in which to list the dichotomies. Consider the B N. Since the total number of rows in the table is B N. N0 and all k. If there existed such a subset. Assume the statement is true for all N Lemma 2. We can also use the recursion to bound B N. The proof is by induction on N.
II It turns out that B N. We have thus proved the induction step. The implication of Theorem 2. Theorem 2. For a given break point k. The RHS is polynomial in N of degree k. The form of the polynomial bound can be further simplified to make the dependency on dvc more salient. It is easy to see that no smaller break point exists since ti can shatter dvc points. It is also the best we can do using this line of reasoning.
The smaller the break point. We state a useful form here. If dvc i s the VC dimension o f ti. The Vapnik. Chervonenkis dimension of a hypothesis set ti. Note: you can use the break points you found in Exercise 2. Because of its significant role.
For any finite value of dvc. One implication of this discussion is that there is a division of models into two classes. Ein will be close to Eout. This is done in two steps. There is a logical difference in arguing that dvc is at least a certain value. The smaller dvc is. One way to gain insight about dvc is to try to compute it for learning models that we are familiar with. The 'bad models' have infinite dvc. The 'good models' have finite dvc. With a bad model. Perceptrons are one case where we can compute dvc exactly.
If we manage to do that. This is because dvc 2. If we were to directly replace M by mH N in 2. The perceptron case provides a nice intuition about the VC dimension. No set of N points can be shattered by 1-l. Diversity is not necessarily a good thing in the context of generalization. One can view the VC dimension as measuring the 'effective' number of parameters. The VC dimension measures these effective parameters or 'degrees of freedom' that enable the model to express a diverse set of hypotheses.
Wd In other models. This is consistent with Figure 2. In the case of perceptrons. Based only on this information. Conclude that there is some dichotomy that cannot be implemented. Any set of N points can be shattered by 1-l. There is a set of N points that cannot be shattered by 1-l. This means that some vector is a linear combination of all the other vectors. The more parameters a model has.
Show that the dimension of the perceptron with d 1 para m eters. Eout g :s. Each V is a point on that canvas. Let 's think of probabilities of different events as areas on that canvas. Let us think of this space as a 'canvas' Figure 2.
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