# Difference between revisions of "stat841f10"

## Proposal Fall 2010

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## Reference Textbook

The Elements of Statistical Learning: Data Mining, Inference, and Prediction, Second Edition, February 2009 Trevor Hastie, Robert Tibshirani, Jerome Friedman (3rd Edition is available)

## Classification - September 21, 2010

### Classification

Statistical classification, or simply known as classification, is an area of supervised learning that addresses the problem of how to systematically assign unlabeled (classes unknown) novel data to their labels (classes or groups or types) by using knowledge of their features (characteristics or attributes) that are obtained from observation and/or measurement. A classifier is a specific technique or method for performing classification. To classify new data, a classifier first uses labeled (classes are known) training data to train a model, and then it uses a function known as its classification rule to assign a label to each new data input after feeding the input's known feature values into the model to determine how much the input belongs to each class.

Classification has been an important task for people and society since the beginnings of history. According to this link, the earliest application of classification in human society was probably done by prehistory peoples for recognizing which wild animals were beneficial to people and which ones were harmful, and the earliest systematic use of classification was done by the famous Greek philosopher Aristotle (384 BC - 322 BC) when he, for example, grouped all living things into the two groups of plants and animals. Classification is generally regarded as one of four major areas of statistics, with the other three major areas being regression, clustering, and dimensionality reduction (feature extraction or manifold learning). Please be noted that some people consider classification to be a broad area that consists of both supervised and unsupervised methods of classifying data. In this view, as can be seen in this link, clustering is simply a special case of classification and it may be called unsupervised classification.

In classical statistics, classification techniques were developed to learn useful information using small data sets where there is usually not enough of data. When machine learning was developed after the application of computers to statistics, classification techniques were developed to work with very large data sets where there is usually too many data. A major challenge facing data mining using machine learning is how to efficiently find useful patterns in very large amounts of data. An interesting quote that describes this problem quite well is the following one made by the retired Yale University Librarian Rutherford D. Rogers, a link to a source of which can be found here.

       "We are drowning in information and starving for knowledge."
- Rutherford D. Rogers


In the Information Age, machine learning when it is combined with efficient classification techniques can be very useful for data mining using very large data sets. This is most useful when the structure of the data is not well understood but the data nevertheless exhibit strong statistical regularity. Areas in which machine learning and classification have been successfully used together include search and recommendation (e.g. Google, Amazon), automatic speech recognition and speaker verification, medical diagnosis, analysis of gene expression, drug discovery etc.

The formal mathematical definition of classification is as follows:

Definition: Classification is the prediction of a discrete random variable $\mathcal{Y}$ from another random variable $\mathcal{X}$, where $\mathcal{Y}$ represents the label assigned to a new data input and $\mathcal{X}$ represents the known feature values of the input.

A set of training data used by a classifier to train its model consists of $\,n$ independently and identically distributed (i.i.d) ordered pairs $\,\{(X_{1},Y_{1}), (X_{2},Y_{2}), \dots , (X_{n},Y_{n})\}$, where the values of the $\,ith$ training input's feature values $\,X_{i} = (\,X_{i1}, \dots , X_{id}) \in \mathcal{X} \subset \mathbb{R}^{d}$ is a d-dimensional vector and the label of the $\, ith$ training input is $\,Y_{i} \in \mathcal{Y}$ that can take a finite number of values. The classification rule used by a classifier has the form $\,h: \mathcal{X} \mapsto \mathcal{Y}$. After the model is trained, each new data input whose feature values is $\,x$ is given the label $\,\hat{Y}=h(x)$.

As an example, if we would like to classify some vegetables and fruits, then our training data might look something like the one shown in the following picture from Professor Ali Ghodsi's Fall 2010 STAT 841 slides.

After we have selected a classifier and then built our model using our training data, we could use the classifier's classification rule $\ h$ to classify any newly-given vegetable or fruit such as the one shown in the following picture from Professor Ali Ghodsi's Fall 2010 STAT 841 slides after first obtaining its feature values.

As another example, suppose we wish to classify newly-given fruits into apples and oranges by considering three features of a fruit that comprise its color, its diameter, and its weight. After selecting a classifier and constructing a model using training data $\,\{(X_{color, 1}, X_{diameter, 1}, X_{weight, 1}, Y_{1}), \dots , (X_{color, n}, X_{diameter, n}, X_{weight, n}, Y_{n})\}$, we could then use the classifier's classification rule $\,h$ to assign any newly-given fruit having known feature values $\,x = (\,x_{color}, x_{diameter} , x_{weight})$ the label $\, \hat{Y}=h(x) \in \mathcal{Y}= \{apple,orange\}$.

### Error rate

The empirical error rate (or training error rate) of a classifier having classification rule $\,h$ is defined as the frequency at which $\,h$ does not correctly classify the data inputs in the training set, i.e., it is defined as $\,\hat{L}_{n} = \frac{1}{n} \sum_{i=1}^{n} I(h(X_{i}) \neq Y_{i})$, where $\,I$ is an indicator variable and $\,I = \left\{\begin{matrix} 1 &\text{if } h(X_i) \neq Y_i \\ 0 &\text{if } h(X_i) = Y_i \end{matrix}\right.$. Here, $\,X_{i} \in \mathcal{X}$ and $\,Y_{i} \in \mathcal{Y}$ are the known feature values and the true class of the $\,ith$ training input, respectively.

The true error rate $\,L(h)$ of a classifier having classification rule $\,h$ is defined as the probability that $\,h$ does not correctly classify any new data input, i.e., it is defined as $\,L(h)=P(h(X) \neq Y)$. Here, $\,X \in \mathcal{X}$ and $\,Y \in \mathcal{Y}$ are the known feature values and the true class of that input, respectively.

In practice, the empirical error rate is obtained to estimate the true error rate, whose value is impossible to be known because the parameter values of the underlying process cannot be known but can only be estimated using available data. The empirical error rate, in practice, estimates the true error rate quite well in that, as mentioned here, it is an unbiased estimator of the true error rate.

### Bayes Classifier

A Bayes classifier is a simple probabilistic classifier based on applying Bayes' Theorem (from Bayesian statistics) with strong (naive) independence assumptions. A more descriptive term for the underlying probability model would be "independent feature model".

In simple terms, a Bayes classifier assumes that the presence (or absence) of a particular feature of a class is unrelated to the presence (or absence) of any other feature. For example, a fruit may be considered to be an apple if it is red, round, and about 4" in diameter. Even if these features depend on each other or upon the existence of the other features, a Bayes classifier considers all of these properties to independently contribute to the probability that this fruit is an apple.

Depending on the precise nature of the probability model, naive Bayes classifiers can be trained very efficiently in a supervised learning setting. In many practical applications, parameter estimation for Bayes models uses the method of maximum likelihood; in other words, one can work with the naive Bayes model without believing in Bayesian probability or using any Bayesian methods.

In spite of their design and apparently over-simplified assumptions, naive Bayes classifiers have worked quite well in many complex real-world situations. In 2004, analysis of the Bayesian classification problem has shown that there are some theoretical reasons for the apparently unreasonable efficacy of Bayes classifiers [1]. Still, a comprehensive comparison with other classification methods in 2006 showed that Bayes classification is outperformed by more current approaches, such as boosted trees or random forests[2].

An advantage of the naive Bayes classifier is that it requires a small amount of training data to estimate the parameters (means and variances of the variables) necessary for classification. Because independent variables are assumed, only the variances of the variables for each class need to be determined and not the entire covariance matrix.

After training its model using training data, the Bayes classifier classifies any new data input in two steps. First, it uses the input's known feature values and the Bayes formula to calculate the input's posterior probability of belonging to each class. Then, it uses its classification rule to place the input into the most-probable class, which is the one associated with the input's largest posterior probability.

In mathematical terms, for a new data input having feature values $\,(X = x)\in \mathcal{X}$, the Bayes classifier labels the input as $(Y = y) \in \mathcal{Y}$, such that the input's posterior probability $\,P(Y = y|X = x)$ is maximum over all of the members of $\mathcal{Y}$.

Suppose there are $\,k$ classes and we are given a new data input having feature values $\,x$. The following derivation shows how the Bayes classifier finds the input's posterior probability $\,P(Y = y|X = x)$ of belonging to each class $y \in \mathcal{Y}$.

\begin{align} P(Y=y|X=x) &= \frac{P(X=x|Y=y)P(Y=y)}{P(X=x)} \\ &=\frac{P(X=x|Y=y)P(Y=y)}{\Sigma_{\forall i \in \mathcal{Y}}P(X=x|Y=i)P(Y=i)} \end{align}

Here, $\,P(Y=y|X=x)$ is known as the posterior probability as mentioned above, $\,P(Y=y)$ is known as the prior probability, $\,P(X=x|Y=y)$ is known as the likelihood, and $\,P(X=x)$ is known as the evidence.

In the special case where there are two classes, i.e., $\, \mathcal{Y}=\{0, 1\}$, the Bayes classifier makes use of the function $\,r(x)=P\{Y=1|X=x\}$ which is the posterior probability of a new data input having feature values $\,x$ belonging to the class $\,Y = 1$. Following the above derivation for the posterior probabilities of a new data input, the Bayes classifier calculates $\,r(x)$ as follows:

\begin{align} r(x)&=P(Y=1|X=x) \\ &=\frac{P(X=x|Y=1)P(Y=1)}{P(X=x)}\\ &=\frac{P(X=x|Y=1)P(Y=1)}{P(X=x|Y=1)P(Y=1)+P(X=x|Y=0)P(Y=0)} \end{align}

The Bayes classifier's classification rule $\,h^*: \mathcal{X} \mapsto \mathcal{Y}$, then, is

$\, h^*(x)= \left\{\begin{matrix} 1 &\text{if } \hat r(x)\gt \frac{1}{2} \\ 0 &\mathrm{otherwise} \end{matrix}\right.$.

Here, $\,x$ is the feature values of a new data input and $\hat r(x)$ is the estimated value of the function $\,r(x)$ given by the Bayes classifier's model after feeding $\,x$ into the model. Still in this special case of two classes, the Bayes classifier's decision boundary is defined as the set $\,D(h)=\{x: P(Y=1|X=x)=P(Y=0|X=x)\}$. The decision boundary $\,D(h)$ essentially combines together the trained model and the decision function $\,h^*$, and it is used by the Bayes classifier to assign any new data input to a label of either $\,Y = 0$ or $\,Y = 1$ depending on which side of the decision boundary the input lies in. From this decision boundary, it is easy to see that, in the case where there are two classes, the Bayes classifier's classification rule can be re-expressed as

$\, h^*(x)= \left\{\begin{matrix} 1 &\text{if } P(Y=1|X=x)\gt P(Y=0|X=x) \\ 0 &\mathrm{otherwise} \end{matrix}\right.$.

Bayes Classification Rule Optimality Theorem The Bayes classifier is the optimal classifier in that it results in the least possible true probability of misclassification for any given new data input, i.e., for any generic classifier having classification rule $\,h$, it is always true that $\,L(h^*(x)) \le L(h(x))$. Here, $\,L$ represents the true error rate, $\,h^*$ is the Bayes classifier's classification rule, and $\,x$ is any given data input's feature values.

Although the Bayes classifier is optimal in the theoretical sense, other classifiers may nevertheless outperform it in practice. The reason for this is that various components which make up the Bayes classifier's model, such as the likelihood and prior probabilities, must either be estimated using training data or be guessed with a certain degree of belief. As a result, the estimated values of the components in the trained model may deviate quite a bit from their true population values, and this can ultimately cause the calculated posterior probabilities of inputs to deviate quite a bit from their true values. Estimation of all these probability functions, as likelihood, prior probability, and evidence function is a very expensive task, computationally, which also makes some other classifiers more favorable than Bayes classifier.

A detailed proof of this theorem is available here.

Defining the classification rule:

In the special case of two classes, the Bayes classifier can use three main approaches to define its classification rule $\,h^*$:

1) Empirical Risk Minimization: Choose a set of classifiers $\mathcal{H}$ and find $\,h^*\in \mathcal{H}$ that minimizes some estimate of the true error rate $\,L(h^*)$.
2) Regression: Find an estimate $\hat r$ of the function $x$ and define
$\, h^*(x)= \left\{\begin{matrix} 1 &\text{if } \hat r(x)\gt \frac{1}{2} \\ 0 &\mathrm{otherwise} \end{matrix}\right.$.
3) Density Estimation: Estimate $\,P(X=x|Y=0)$ from the $\,X_{i}$'s for which $\,Y_{i} = 0$, estimate $\,P(X=x|Y=1)$ from the $\,X_{i}$'s for which $\,Y_{i} = 1$, and estimate $\,P(Y = 1)$ as $\,\frac{1}{n} \sum_{i=1}^{n} Y_{i}$. Then, calculate $\,\hat r(x) = \hat P(Y=1|X=x)$ and define
$\, h^*(x)= \left\{\begin{matrix} 1 &\text{if } \hat r(x)\gt \frac{1}{2} \\ 0 &\mathrm{otherwise} \end{matrix}\right.$.

Typically, the Bayes classifier uses approach 3 to define its classification rule. These three approaches can easily be generalized to the case where the number of classes exceeds two.

Multi-class classification:

Suppose there are $\,k$ classes, where $\,k \ge 2$.

In the above discussion, we introduced the Bayes formula for this general case:

\begin{align} P(Y=y|X=x) &=\frac{P(X=x|Y=y)P(Y=y)}{\Sigma_{\forall i \in \mathcal{Y}}P(X=x|Y=i)P(Y=i)} \end{align}

which can re-worded as:

\begin{align} P(Y=y|X=x) &=\frac{f_y(x)\pi_y}{\Sigma_{\forall i \in \mathcal{Y}} f_i(x)\pi_i} \end{align}

Here, $\,f_y(x) = P(X=x|Y=y)$ is known as the likelihood function and $\,\pi_y = P(Y=y)$ is known as the prior probability.

In the general case where there are at least two classes, the Bayes classifier uses the following theorem to assign any new data input having feature values $\,x$ into one of the $\,k$ classes.

Theorem

Suppose that $\mathcal{Y}= \{1, \dots, k\}$, where $\,k \ge 2$. Then, the optimal classification rule is $\,h^*(x) = arg max_{i} P(Y=i|X=x)$, where $\,i \in \{1, \dots, k\}$.

Example: We are going to predict if a particular student will pass STAT 441/841. There are two classes represented by $\, \mathcal{Y}\in \{ 0,1 \}$, where 1 refers to pass and 0 refers to fail. Suppose that the prior probabilities are estimated or guessed to be $\,\hat P(Y = 1) = \hat P(Y = 0) = 0.5$. We have data on past student performances, which we shall use to train the model. For each student, we know the following:

Whether or not the student’s GPA was greater than 3.0 (G).
Whether or not the student had a strong math background (M).
Whether or not the student was a hard worker (H).
Whether or not the student passed or failed the course. Note: these are the known y values in the training data.

These known data are summarized in the following tables:

For each student, his/her feature values is $\, x = \{G, M, H\}$ and his or her class is $\, y \in \{0, 1\}$.

Suppose there is a new student having feature values $\, x = \{0, 1, 0\}$, and we would like to predict whether he/she would pass the course. $\,\hat r(x)$ is found as follows:

$\, \hat r(x) = P(Y=1|X =(0,1,0))=\frac{P(X=(0,1,0)|Y=1)P(Y=1)}{P(X=(0,1,0)|Y=0)P(Y=0)+P(X=(0,1,0)|Y=1)P(Y=1)}=\frac{0.05*0.5}{0.05*0.5+0.2*0.5}=\frac{0.025}{0.125}=\frac{1}{5}\lt \frac{1}{2}.$

The Bayes classifier assigns the new student into the class $\, h^*(x)=0$. Therefore, we predict that the new student would fail the course.

Naive Bayes Classifier:

The naive Bayes classifier is a special (simpler) case of the Bayes classifier. It uses an extra assumption: that the presence (or absence) of a particular feature of a class is unrelated to the presence (or absence) of any other feature. This assumption allows for an easier likelihood function $\,f_y(x)$ in the equation:

\begin{align} P(Y=y|X=x) &=\frac{f_y(x)\pi_y}{\Sigma_{\forall i \in \mathcal{Y}} f_i(x)\pi_i} \end{align}

The simper form of the likelihood function seen in the naive Bayes is:

\begin{align} f_y(x) = P(X=x|Y=y) = {\prod_{i=1}^{n} P(X_{i}=x_{i}|Y=y)} \end{align}

The Bayes classifier taught in class was not the naive Bayes classifier.

### Bayesian vs. Frequentist

The Bayesian view of probability and the frequentist view of probability are the two major schools of thought in the field of statistics regarding how to interpret the probability of an event.

The Bayesian view of probability states that, for any event E, event E has a prior probability that represents how believable event E would occur prior to knowing anything about any other event whose occurrence could have an impact on event E's occurrence. Theoretically, this prior probability is a belief that represents the baseline probability for event E's occurrence. In practice, however, event E's prior probability is unknown, and therefore it must either be guessed at or be estimated using a sample of available data. After obtaining a guessed or estimated value of event E's prior probability, the Bayesian view holds that the probability, that is, the believability of event E's occurrence, can always be made more accurate should any new information regarding events that are relevant to event E become available. The Bayesian view also holds that the accuracy for the estimate of the probability of event E's occurrence is higher as long as there are more useful information available regarding events that are relevant to event E. The Bayesian view therefore holds that there is no intrinsic probability of occurrence associated with any event. If one adherers to the Bayesian view, one can then, for instance, predict tomorrow's weather as having a probability of, say, $\,50%$ for rain. The Bayes classifier as described above is a good example of a classifier developed from the Bayesian view of probability. The earliest works that lay the framework for the Bayesian view of probability is accredited to Thomas Bayes (1702–1761).

In contrast to the Bayesian view of probability, the frequentist view of probability holds that there is an intrinsic probability of occurrence associated with every event to which one can carry out many, if not an infinite number, of well-defined independent random trials. In each trial for an event, the event either occurs or it does not occur. Suppose $n_x$ denotes the number of times that an event occurs during its trials and $n_t$ denotes the total number of trials carried out for the event. The frequentist view of probability holds that, in the long run, where the number of trials for an event approaches infinity, one could theoretically approach the intrinsic value of the event's probability of occurrence to any arbitrary degree of accuracy, i.e., :$P(x) = \lim_{n_t\rightarrow \infty}\frac{n_x}{n_t}$. In practice, however, one can only carry out a finite number of trials for an event and, as a result, the probability of the event's occurrence can only be approximated as $P(x) \approx \frac{n_x}{n_t}$. If one adherers to the frequentist view, one cannot, for instance, predict the probability that there would be rain tomorrow. This is because one cannot possibly carry out trials for any event that is set in the future. The founder of the frequentist school of thought is arguably the famous Greek philosopher Aristotle. In his work Rhetoric, Aristotle gave the famous line "the probable is that which for the most part happens".

More information regarding the Bayesian and the frequentist schools of thought are available here. Furthermore, an interesting and informative youtube video that explains the Bayesian and frequentist views of probability is available here.

## Linear and Quadratic Discriminant Analysis

First, we shall limit ourselves to the case where there are two classes, i.e. $\, \mathcal{Y}=\{0, 1\}$. In the above discussion, we introduced the Bayes classifier's decision boundary $\,D(h^*)=\{x: P(Y=1|X=x)=P(Y=0|X=x)\}$, which represents a hyperplane that determines the class of any new data input depending on which side of the hyperplane the input lies in. Now, we shall look at how to derive the Bayes classifier's decision boundary under certain assumptions of the data. Linear discriminant analysis (LDA) and quadratic discriminant analysis (QDA) are two of the most well-known ways for deriving the Bayes classifier's decision boundary, and we shall look at each of them in turn.

Let us denote the likelihood $\ P(X=x|Y=y)$ as $\ f_y(x)$ and the prior probability $\ P(Y=y)$ as $\ \pi_y$.

First, we shall examine LDA. As explained above, the Bayes classifier is optimal. However, in practice, the prior and conditional densities are not known. Under LDA, one gets around this problem by making the assumptions that both of the two classes have multivariate normal (Gaussian) distributions and the two classes have the same covariance matrix $\, \Sigma$. Under the assumptions of LDA, we have: $\ P(X=x|Y=y) = f_y(x) = \frac{1}{ (2\pi)^{d/2}|\Sigma|^{1/2} }\exp\left( -\frac{1}{2} (x - \mu_k)^\top \Sigma^{-1} (x - \mu_k) \right)$. Now, to derive the Bayes classifier's decision boundary using LDA, we equate $\, P(Y=1|X=x)$ to $\, P(Y=0|X=x)$ and proceed from there. The derivation of $\,D(h^*)$ is as follows:

$\,Pr(Y=1|X=x)=Pr(Y=0|X=x)$
$\,\Rightarrow \frac{Pr(X=x|Y=1)Pr(Y=1)}{Pr(X=x)}=\frac{Pr(X=x|Y=0)Pr(Y=0)}{Pr(X=x)}$ (using Bayes' Theorem)
$\,\Rightarrow Pr(X=x|Y=1)Pr(Y=1)=Pr(X=x|Y=0)Pr(Y=0)$ (canceling the denominators)
$\,\Rightarrow f_1(x)\pi_1=f_0(x)\pi_0$
$\,\Rightarrow \frac{1}{ (2\pi)^{d/2}|\Sigma|^{1/2} }\exp\left( -\frac{1}{2} (x - \mu_1)^\top \Sigma^{-1} (x - \mu_1) \right)\pi_1=\frac{1}{ (2\pi)^{d/2}|\Sigma|^{1/2} }\exp\left( -\frac{1}{2} (x - \mu_0)^\top \Sigma^{-1} (x - \mu_0) \right)\pi_0$
$\,\Rightarrow \exp\left( -\frac{1}{2} (x - \mu_1)^\top \Sigma^{-1} (x - \mu_1) \right)\pi_1=\exp\left( -\frac{1}{2} (x - \mu_0)^\top \Sigma^{-1} (x - \mu_0) \right)\pi_0$
$\,\Rightarrow -\frac{1}{2} (x - \mu_1)^\top \Sigma^{-1} (x - \mu_1) + \log(\pi_1)=-\frac{1}{2} (x - \mu_0)^\top \Sigma^{-1} (x - \mu_0) +\log(\pi_0)$ (taking the log of both sides).
$\,\Rightarrow \log(\frac{\pi_1}{\pi_0})-\frac{1}{2}\left( x^\top\Sigma^{-1}x + \mu_1^\top\Sigma^{-1}\mu_1 - 2x^\top\Sigma^{-1}\mu_1 - x^\top\Sigma^{-1}x - \mu_0^\top\Sigma^{-1}\mu_0 + 2x^\top\Sigma^{-1}\mu_0 \right)=0$ (expanding out)
$\,\Rightarrow \log(\frac{\pi_1}{\pi_0})-\frac{1}{2}\left( \mu_1^\top\Sigma^{-1} \mu_1-\mu_0^\top\Sigma^{-1}\mu_0 - 2x^\top\Sigma^{-1}(\mu_1-\mu_0) \right)=0$ (canceling out alike terms and factoring).

It is easy to see that, under LDA, the Bayes's classifier's decision boundary $\,D(h^*)$ has the form $\,ax+b=0$ and it is linear in $\,x$. This is where the word linear in linear discriminant analysis comes from.

LDA under the two-classes case can easily be generalized to the general case where there are $\,k \ge 2$ classes. In the general case, suppose we wish to find the Bayes classifier's decision boundary between the two classes $\,m$ and $\,n$, then all we need to do is follow a derivation very similar to the one shown above, except with the classes $\,1$ and $\,0$ being replaced by the classes $\,m$ and $\,n$. Following through with a similar derivation as the one shown above, one obtains the Bayes classifier's decision boundary $\,D(h^*)$ between classes $\,m$ and $\,n$ to be $\,\log(\frac{\pi_m}{\pi_n})-\frac{1}{2}\left( \mu_m^\top\Sigma^{-1} \mu_m-\mu_n^\top\Sigma^{-1}\mu_n - 2x^\top\Sigma^{-1}(\mu_m-\mu_n) \right)=0$ . In addition, for any two classes $\,m$ and $\,n$ for whom we would like to find the Bayes classifier's decision boundary using LDA, if $\,m$ and $\,n$ both have the same number of data, then, in this special case, the resulting decision boundary would lie exactly halfway between the centers (means) of $\,m$ and $\,n$.

The Bayes classifier's decision boundary for any two classes as derived using LDA looks something like the one that can be found in this link:

Although the assumption under LDA may not hold true for most real-world data, it nevertheless usually performs quite well in practice, where it often provides near-optimal classifications. For instance, the Z-Score credit risk model that was designed by Edward Altman in 1968 and revisited in 2000, is essentially a weighted LDA. This model has demonstrated a 85-90% success rate in predicting bankruptcy, and for this reason it is still in use today.

According to this link, some of the limitations of LDA include:

• LDA implicitly assumes that the data in each class has a Gaussian distribution.
• LDA implicitly assumes that the mean rather than the variance is the discriminating factor.
• LDA may over-fit the training data.

## Linear and Quadratic Discriminant Analysis cont'd - September 23, 2010

### LDA x QDA

Linear discriminant analysis[1] is a statistical method used to find the linear combination of features which best separate two or more classes of objects or events. It is widely applied in classifying diseases, positioning, product management, and marketing research. LDA assumes that the different classes have the same covariance matrix $\, \Sigma$.

Quadratic Discriminant Analysis[2], on the other hand, aims to find the quadratic combination of features. It is more general than linear discriminant analysis. Unlike LDA, QDA does not make the assumption that the different classes have the same covariance matrix $\, \Sigma$. Instead, QDA makes the assumption that each class $\, k$ has its own covariance matrix $\, \Sigma_k$.

The derivation of the Bayes classifier's decision boundary $\,D(h^*)$ under QDA is similar to that under LDA. Again, let us first consider the two-classes case where $\, \mathcal{Y}=\{0, 1\}$. This derivation is given as follows:

$\,Pr(Y=1|X=x)=Pr(Y=0|X=x)$
$\,\Rightarrow \frac{Pr(X=x|Y=1)Pr(Y=1)}{Pr(X=x)}=\frac{Pr(X=x|Y=0)Pr(Y=0)}{Pr(X=x)}$ (using Bayes' Theorem)
$\,\Rightarrow Pr(X=x|Y=1)Pr(Y=1)=Pr(X=x|Y=0)Pr(Y=0)$ (canceling the denominators)
$\,\Rightarrow f_1(x)\pi_1=f_0(x)\pi_0$
$\,\Rightarrow \frac{1}{ (2\pi)^{d/2}|\Sigma_1|^{1/2} }\exp\left( -\frac{1}{2} (x - \mu_1)^\top \Sigma_1^{-1} (x - \mu_1) \right)\pi_1=\frac{1}{ (2\pi)^{d/2}|\Sigma_0|^{1/2} }\exp\left( -\frac{1}{2} (x - \mu_0)^\top \Sigma_0^{-1} (x - \mu_0) \right)\pi_0$
$\,\Rightarrow \frac{1}{|\Sigma_1|^{1/2} }\exp\left( -\frac{1}{2} (x - \mu_1)^\top \Sigma_1^{-1} (x - \mu_1) \right)\pi_1=\frac{1}{|\Sigma_0|^{1/2} }\exp\left( -\frac{1}{2} (x - \mu_0)^\top \Sigma_0^{-1} (x - \mu_0) \right)\pi_0$ (by cancellation)
$\,\Rightarrow -\frac{1}{2}\log(|\Sigma_1|)-\frac{1}{2} (x - \mu_1)^\top \Sigma_1^{-1} (x - \mu_1)+\log(\pi_1)=-\frac{1}{2}\log(|\Sigma_0|)-\frac{1}{2} (x - \mu_0)^\top \Sigma_0^{-1} (x - \mu_0)+\log(\pi_0)$ (by taking the log of both sides)
$\,\Rightarrow \log(\frac{\pi_1}{\pi_0})-\frac{1}{2}\log(\frac{|\Sigma_1|}{|\Sigma_0|})-\frac{1}{2}\left( x^\top\Sigma_1^{-1}x + \mu_1^\top\Sigma_1^{-1}\mu_1 - 2x^\top\Sigma_1^{-1}\mu_1 - x^\top\Sigma_0^{-1}x - \mu_0^\top\Sigma_0^{-1}\mu_0 + 2x^\top\Sigma_0^{-1}\mu_0 \right)=0$ (by expanding out)
$\,\Rightarrow \log(\frac{\pi_1}{\pi_0})-\frac{1}{2}\log(\frac{|\Sigma_1|}{|\Sigma_0|})-\frac{1}{2}\left( x^\top(\Sigma_1^{-1}-\Sigma_0^{-1})x + \mu_1^\top\Sigma_1^{-1}\mu_1 - \mu_0^\top\Sigma_0^{-1}\mu_0 - 2x^\top(\Sigma_1^{-1}\mu_1-\Sigma_0^{-1}\mu_0) \right)=0$

It is easy to see that, under QDA, the decision boundary $\,D(h^*)$ has the form $\,ax^2+bx+c=0$ and it is quadratic in $\,x$. This is where the word quadratic in quadratic discriminant analysis comes from.

As is the case with LDA, QDA under the two-classes case can easily be generalized to the general case where there are $\,k \ge 2$ classes. In the general case, suppose we wish to find the Bayes classifier's decision boundary between the two classes $\,m$ and $\,n$, then all we need to do is follow a derivation very similar to the one shown above, except with the classes $\,1$ and $\,0$ being replaced by the classes $\,m$ and $\,n$. Following through with a similar derivation as the one shown above, one obtains the Bayes classifier's decision boundary $\,D(h^*)$ between classes $\,m$ and $\,n$ to be $\,\log(\frac{\pi_m}{\pi_n})-\frac{1}{2}\log(\frac{|\Sigma_m|}{|\Sigma_n|})-\frac{1}{2}\left( x^\top(\Sigma_m^{-1}-\Sigma_n^{-1})x + \mu_m^\top\Sigma_m^{-1}\mu_m - \mu_n^\top\Sigma_n^{-1}\mu_n - 2x^\top(\Sigma_m^{-1}\mu_m-\Sigma_n^{-1}\mu_n) \right)=0$.

### Summarizing LDA and QDA

We can summarize what we have learned so far into the following theorem.

Theorem:

Suppose that $\,Y \in \{1,\dots,K\}$, if $\,f_k(x) = Pr(X=x|Y=k)$ is Gaussian, the Bayes Classifier rule is

$\,h^*(x) = \arg\max_{k} \delta_k(x)$

where,

• In the case of LDA, which assumes that a common covariance matrix is shared by all classes, $\,\delta_k(x) = x^\top\Sigma^{-1}\mu_k - \frac{1}{2}\mu_k^\top\Sigma^{-1}\mu_k + log (\pi_k)$, and the Bayes classifier's decision boundary $\,D(h^*)$ is linear in $\,x$.
• In the case of QDA, which assumes that each class has its own covariance matrix, $\,\delta_k(x) = - \frac{1}{2}log(|\Sigma_k|) - \frac{1}{2}(x-\mu_k)^\top\Sigma_k^{-1}(x-\mu_k) + log (\pi_k)$, and the Bayes classifier's decision boundary $\,D(h^*)$ is quadratic in $\,x$.

Note $\,\arg\max_{k} \delta_k(x)$returns the set of k for which $\,\delta_k(x)$ attains its largest value.

### In practice

We need to estimate the prior, so in order to do this, we use the Maximum Likelihood estimates from the sample for $\,\pi,\mu_k,\Sigma_k$ in place of their true values, i.e.

File:estimation.png
Estimation of the probability of belonging to either class k or l

$\,\hat{\pi_k} = \hat{Pr}(y=k) = \frac{n_k}{n}$

$\,\hat{\mu_k} = \frac{1}{n_k}\sum_{i:y_i=k}x_i$

$\,\hat{\Sigma_k} = \frac{1}{n_k}\sum_{i:y_i=k}(x_i-\hat{\mu_k})(x_i-\hat{\mu_k})^\top$

Common covariance, denoted $\Sigma$, is defined as the weighted average of the covariance for each class.

In the case where we need a common covariance matrix, we get the estimate using the following equation:

$\,\Sigma=\frac{\sum_{r=1}^{k}(n_r\Sigma_r)}{\sum_{l=1}^{k}(n_l)}$

Where: $\,n_r$ is the number of data points in class r, $\,\Sigma_r$ is the covariance of class r and $\,n$ is the total number of data points, $\,k$ is the number of classes.

See the details about the estimation of covarience matrices.

### Computation For QDA And LDA

First, let us consider QDA, and examine each of the following two cases.

Case 1: (Example) $\, \Sigma_k = I$

$\, \Sigma_k = I$ for every class $\,k$ implies that our data is spherical. This means that the data of each class $\,k$ is distributed symmetrically around the center $\,\mu_k$, i.e. the isocontours are all circles.

We have:

$\,\delta_k = - \frac{1}{2}log(|I|) - \frac{1}{2}(x-\mu_k)^\top I(x-\mu_k) + log (\pi_k)$

We see that the first term in the above equation, $\,\frac{-1}{2}log(|I|)$, is zero since $\ |I|=1$. The second term contains $\, (x-\mu_k)^\top I(x-\mu_k) = (x-\mu_k)^\top(x-\mu_k)$, which is the squared Euclidean distance between $\,x$ and $\,\mu_k$. Therefore we can find the distance between a point and each center and adjust it with the log of the prior, $\,log(\pi_k)$. The class that has the minimum distance will maximize $\,\delta_k$. According to the theorem, we can then classify the point to a specific class $\,k$.

Case 2: (General Case) $\, \Sigma_k \ne I$

We can decompose this as:

$\, \Sigma_k = U_kS_kV_k^\top = U_kS_kU_k^\top$ (In general when $\,X=U_kS_kV_k^\top$, $\,U_k$ is the eigenvectors of $\,X_kX_k^T$ and $\,V_k$ is the eigenvectors of $\,X_k^\top X_k$. So if $\, X_k$ is symmetric, we will have $\, U_k=V_k$. Here $\, \Sigma_k$ is symmetric, because it is the covariance matrix of $X_k$) and the inverse of $\,\Sigma_k$ is

$\, \Sigma_k^{-1} = (U_kS_kU_k^\top)^{-1} = (U_k^\top)^{-1}S_k^{-1}U_k^{-1} = U_kS_k^{-1}U_k^\top$ (since $\,U_k$ is orthonormal)

So from the formula for $\,\delta_k$, the second term is

\begin{align} (x-\mu_k)^\top\Sigma_k^{-1}(x-\mu_k)&= (x-\mu_k)^\top U_kS_k^{-1}U_k^T(x-\mu_k)\\ & = (U_k^\top x-U_k^\top\mu_k)^\top S_k^{-1}(U_k^\top x-U_k^\top \mu_k)\\ & = (U_k^\top x-U_k^\top\mu_k)^\top S_k^{-\frac{1}{2}}S_k^{-\frac{1}{2}}(U_k^\top x-U_k^\top\mu_k) \\ & = (S_k^{-\frac{1}{2}}U_k^\top x-S_k^{-\frac{1}{2}}U_k^\top\mu_k)^\top I(S_k^{-\frac{1}{2}}U_k^\top x-S_k^{-\frac{1}{2}}U_k^\top \mu_k) \\ & = (S_k^{-\frac{1}{2}}U_k^\top x-S_k^{-\frac{1}{2}}U_k^\top\mu_k)^\top(S_k^{-\frac{1}{2}}U_k^\top x-S_k^{-\frac{1}{2}}U_k^\top \mu_k) \\ \end{align}

where we have the squared Euclidean distance between $\, S_k^{-\frac{1}{2}}U_k^\top x$ and $\, S_k^{-\frac{1}{2}}U_k^\top\mu_k$.

A transformation of all the data points can be done from $\,x$ to $\,x^*$ where $\, x^* \leftarrow S_k^{-\frac{1}{2}}U_k^\top x$.

A similar transformation of all the centers can be done from $\,\mu_k$ to $\,\mu_k^*$ where $\, \mu_k^* \leftarrow S_k^{-\frac{1}{2}}U_k^\top \mu_k$.

It is now possible to do classification with $\,x^*$ and $\,\mu_k^*$, treating them as in Case 1 above.

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| type = style | image = | imageright = | style = | textstyle = | text = This article may require cleanup to meet Wikicoursenote's quality standards. The specific problem is: The sentence above is right since by transforming$\,x$ to $\,x^*$ where $\, x^* \leftarrow S_k^{-\frac{1}{2}}U_k^\top x$, the new variable variance is $I$. Please improve this article if you can. (October 18 2010) | small = | smallimage = | smallimageright = | smalltext = }}

Note that when we have multiple classes, we also need to compute $\, log{|\Sigma_k|}$ respectively. Then we compute $\,\delta_k$ for QDA .

Note that when we have multiple classes, they must all have the same transformation, in another word, have same covariance $\,\Sigma_k$,else, ahead of time we would have to assume a data point belongs to one class or the other. All classes therefore need to have the same shape for classification to be applicable using this method. So this method works for LDA.

If the classes have different shapes, in another word, have different covariance $\,\Sigma_k$, can we use the same method to transform all data points $\,x$ to $\,x^*$?

The answer is NO. Consider that you have two classes with different shapes, then consider transforming them to the same shape. Given a data point, justify which class this point belongs to. The question is, which transformation can you use? For example, if you use the transformation of class A, then you have assumed that this data point belongs to class A.

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The answer is Yes. Consider that you have two classes with different shapes. Given a data point, justify which class this point belongs to. You just do the transformations corresponding to the 2 classes respectively, then you get $\,\delta_1 ,\delta_2$ ,then you determine which class the data point belongs to by comparing $\,\delta_1$ and $\,\delta_2$ .

In summary, to apply QDA on a data set $\,X$, in the general case where $\, \Sigma_k \ne I$ for each class $\,k$, one can proceed as follows:

Step 1: For each class $\,k$, apply singular value decomposition on $\,X_k$ to obtain $\,S_k$ and $\,U_k$.
Step 2: For each class $\,k$, transform each $\,x$ belonging to that class to $\,x^* = S_k^{-\frac{1}{2}}U_k^\top x$, and transform its center $\,\mu_k$ to $\,\mu_k^* = S_k^{-\frac{1}{2}}U_k^\top \mu_k$.
Step 3: For each data point $\,x \in X$, find the squared Euclidean distance between the transformed data point $\,x^*$ and the transformed center $\,\mu^*$ of each class, and assign $\,x$ to the class such that the squared Euclidean distance between $\,x^*$ and $\,\mu^*$ is the least over all of the classes.

Now, let us consider LDA. Here, one can derive a classification scheme that is quite similar to that shown above. The main difference is the assumption of a common variance across the classes, so we perform the Singular Value Decomposition once, as opposed to k times.

To apply LDA on a data set $\,X$, one can proceed as follows:

Step 1: Apply singular value decomposition on $\,X$ to obtain $\,S$ and $\,U$.
Step 2: For each $\,x \in X$, transform $\,x$ to $\,x^* = S^{-\frac{1}{2}}U^\top x$, and transform each center $\,\mu$ to $\,\mu^* = S^{-\frac{1}{2}}U^\top \mu$.
Step 3: For each data point $\,x \in X$, find the squared Euclidean distance between the transformed data point $\,x^*$ and the transformed center $\,\mu^*$ of each class, and assign $\,x$ to the class such that the squared Euclidean distance between $\,x^*$ and $\,\mu^*$ is the least over all of the classes.

Kernel QDA In actual data scenarios, it is generally true that QDA will provide a better classifier for the data then LDA because QDA does not assume that the covariance matrix for each class is identical, as LDA assumes. However, QDA still assumes that the class conditional distribution is Gaussian, which is not always the case in real-life scenarios. The link provided at the beginning of this paragraph describes a kernel-based QDA method which does not have the Gaussian distribution assumption.

### The Number of Parameters in LDA and QDA

Both LDA and QDA require us to estimate parameters. The more estimation we have to do, the less robust our classification algorithm will be.

LDA: Since we just need to compare the differences between one given class and remaining $\,K-1$ classes, totally, there are $\,K-1$ differences. For each of them, $\,a^{T}x+b$ requires $\,d+1$ parameters. Therefore, there are $\,(K-1)\times(d+1)$ parameters.

QDA: For each of the differences, $\,x^{T}ax + b^{T}x + c$ requires $\frac{1}{2}(d+1)\times d + d + 1 = \frac{d(d+3)}{2}+1$ parameters. Therefore, there are $(K-1)(\frac{d(d+3)}{2}+1)$ parameters.

A plot of the number of parameters that must be estimated, in terms of (K-1). The x-axis represents the number of dimensions in the data. As is easy to see, QDA is far less robust than LDA for high-dimensional data sets.

## Trick: Using LDA to do QDA - September 28, 2010

There is a trick that allows us to use the linear discriminant analysis (LDA) algorithm to generate as its output a quadratic function that can be used to classify data. This trick is similar to, but more primitive than, the Kernel trick that will be discussed later in the course.

Essentially, the trick involves adding one or more new features (i.e. new dimensions) that just contain our original data projected to that dimension. We then do LDA on our new higher-dimensional data. The answer provided by LDA can then be collapsed onto a lower dimension, giving us a quadratic answer.

### Motivation

Why would we want to use LDA over QDA? In situations where we have fewer data points, LDA turns out to be more robust.

If we look back at the equations for LDA and QDA, we see that in LDA we must estimate $\,\mu_1$, $\,\mu_2$ and $\,\Sigma$. In QDA we must estimate all of those, plus another $\,\Sigma$; the extra $\,\frac{d(d-1)}{2}$ estimations make QDA less robust with fewer data points.

### Theoretically

Suppose we can estimate some vector $\underline{w}^T$ such that

$y = \underline{w}^Tx$

where $\underline{w}$ is a d-dimensional column vector, and $x\ \epsilon\ \mathbb{R}^d$ (vector in d dimensions).

We also have a non-linear function $g(x) = y = x^Tvx + \underline{w}^Tx$ that we cannot estimate.

Using our trick, we create two new vectors, $\,\underline{w}^*$ and $\,x^*$ such that:

$\underline{w}^{*T} = [w_1,w_2,...,w_d,v_1,v_2,...,v_d]$

and

$x^{*T} = [x_1,x_2,...,x_d,{x_1}^2,{x_2}^2,...,{x_d}^2]$

We can then estimate a new function, $g^*(x,x^2) = y^* = \underline{w}^{*T}x^*$.

Note that we can do this for any $x$ and in any dimension; we could extend a $D \times n$ matrix to a quadratic dimension by appending another $D \times n$ matrix with the original matrix squared, to a cubic dimension with the original matrix cubed, or even with a different function altogether, such as a $\,sin(x)$ dimension. Pay attention, We don't do QDA with LDA. If we try QDA directly on this problem the resulting decision boundary will be different. Here we try to find a nonlinear boundary for a better possible boundary but it is different with general QDA method. We can call it nonlinear LDA.

### By Example

Let's use our trick to do a quadratic analysis of the 2_3 data using LDA.

>> load 2_3;
>> [U, sample] = princomp(X');
>> sample = sample(:,1:2);

We start off the same way, by using PCA to reduce the dimensionality of our data to 2.
>> X_star = zeros(400,4);
>> X_star(:,1:2) = sample(:,:);
>> for i=1:400
for j=1:2
X_star(i,j+2) = X_star(i,j)^2;
end
end

This projects our sample into two more dimensions by squaring our initial two dimensional data set.
>> group = ones(400,1);
>> group(201:400) = 2;
>> [class, error, POSTERIOR, logp, coeff] = classify(X_star, X_star, group, 'linear');
>> sum (class==group)
ans =
375

We can now display our results.
>> k = coeff(1,2).const;
>> l = coeff(1,2).linear;
>> f = sprintf('0 = %g+%g*x+%g*y+%g*(x)^2+%g*(y)^2', k, l(1), l(2),l(3),l(4));
>> ezplot(f,[min(sample(:,1)), max(sample(:,1)), min(sample(:,2)), max(sample(:,2))]);

The plot shows the quadratic decision boundary obtained using LDA in the four-dimensional space on the 2_3.mat data. Counting the blue and red points that are on the wrong side of the decision boundary, we can confirm that we have correctly classified 375 data points.
Not only does LDA give us a better result than it did previously, it actually beats QDA, which only correctly classified 371 data points for this data set. Continuing this procedure by adding another two dimensions with $x^4$ (i.e. we set X_star(i,j+2) = X_star(i,j)^4) we can correctly classify 376 points.

### LDA and QDA in Matlab

We have examined the theory behind Linear Discriminant Analysis (LDA) and Quadratic Discriminant Analysis (QDA) above; how do we use these algorithms in practice? Matlab offers us a function called classify that allows us to perform LDA and QDA quickly and easily.

In class, we were shown an example of using LDA and QDA on the 2_3 data that is used in the first assignment. The code below applies LDA to the same data set and reproduces that example, slightly modified, and explains each step.

>> load 2_3;
>> [U, sample] = princomp(X');
>> sample = sample(:,1:2);

First, we do principal component analysis (PCA) on the 2_3 data to reduce the dimensionality of the original data from 64 dimensions to 2. Doing this makes it much easier to visualize the results of the LDA and QDA algorithms.

>> plot (sample(1:200,1), sample(1:200,2), '.');
>> hold on;
>> plot (sample(201:400,1), sample(201:400,2), 'r.');

Recall that in the 2_3 data, the first 200 elements are images of the number two handwritten and the last 200 elements are images of the number three handwritten. This code sets up a plot of the data such that the points that represent a 2 are blue, while the points that represent a 3 are red.
See title and legend for information on adding the title and legend.
Before using classify we can set up a vector that contains the actual labels for our data, to train the classification algorithm. If we don't know the labels for the data, then the element in the group vector should be an empty string or NaN. (See grouping data for more information.)
>> group = ones(400,1);
>> group(201:400) = 2;

We can now classify our data.
>> [class, error, POSTERIOR, logp, coeff] = classify(sample, sample, group, 'linear');

The full details of this line can be examined in the Matlab help file linked above. What we care about are class, which contains the labels that the algorithm thinks that each data point belongs to, and coeff, which contains information about the line that the algorithm created to separate the data into the two classes.
We can see the efficacy of the algorithm by comparing class to group.
>> sum (class==group)
ans =
369

This compares the value in class to the value in group. The answer of 369 tells us that the algorithm correctly determined the classes of the points 369 times, out of a possible 400 data points. This gives us an empirical error rate of 0.0775.
We can see the line produced by LDA using coeff.
>> k = coeff(1,2).const;
>> l = coeff(1,2).linear;
>> f = sprintf('0 = %g+%g*x+%g*y', k, l(1), l(2));
>> ezplot(f, [min(sample(:,1)), max(sample(:,1)), min(sample(:,2)), max(sample(:,2))]);

Those familiar with the programming language C will find the sprintf line refreshingly familiar; those with no exposure to C are directed to Matlab's sprintf page. Essentially, this code sets up the equation of the line in the form 0 = a + bx + cy. We then use the ezplot function to plot the line.
The 2-3 data after LDA is performed. The line shows where the two classes are split.
Let's perform the same steps, except this time using QDA. The main difference with QDA is a slightly different call to classify, and a more complicated procedure to plot the line.
>> [class, error, POSTERIOR, logp, coeff] = classify(sample, sample, group, 'quadratic');
>> sum (class==group)
ans =
371
>> k = coeff(1,2).const;
>> l = coeff(1,2).linear;
>> f = sprintf('0 = %g+%g*x+%g*y+%g*x^2+%g*x*y+%g*y^2', k, l(1), l(2), q(1,1), q(1,2)+q(2,1), q(2,2));
>> ezplot(f, [min(sample(:,1)), max(sample(:,1)), min(sample(:,2)), max(sample(:,2))]);

The 2-3 data after QDA is performed. The curved line shows where QDA splits the two classes. Note that QDA is only correct in 2 more data points compared to LDA; we can see a blue point and a red point that lie on the correct side of the curve produced by QDA that do not lie on the correct side of the line produced by LDA.

classify can also be used with other discriminant analysis algorithms. The steps laid out above would only need to be modified slightly for those algorithms.

Recall: An analysis of the function of princomp in matlab.
In our assignment 1, we have learnt that how to perform Principal Component Analysis using SVD method. In fact, the matlab offers us a function called princomp which can perform PCA conveniently. From the matlab help file on princomp, you can find the details about this function. But here we will analyze the code of the function of princomp() in matlab to find something different when comparing with SVD method. The following is the code of princomp and explanations to some emphasized steps.

   function [pc, score, latent, tsquare] = princomp(x);
%   PRINCOMP Principal Component Analysis (centered and scaled data).
%   [PC, SCORE, LATENT, TSQUARE] = PRINCOMP(X) takes a data matrix X and
%   returns the principal components in PC, the so-called Z-scores in SC
%   ORES, the eigenvalues of the covariance matrix of X in LATENT,
%   and Hotelling's T-squared statistic for each data point in TSQUARE.
%   Reference: J. Edward Jackson, A User's Guide to Principal Components
%   John Wiley & Sons, Inc. 1991 pp. 1-25.
%   B. Jones 3-17-94
%   Copyright 1993-2002 The MathWorks, Inc.
%   $Revision: 2.9$  $Date: 2002/01/17 21:31:45$
[m,n] = size(x);       %  get the lengh of the rows and columns of matrix x.
r = min(m-1,n);        %  max possible rank of X
avg = mean(x);         %  the mean of every column of X
centerx = (x - avg(ones(m,1),:));
%  centers X by subtracting off column means
[U,latent,pc] = svd(centerx./sqrt(m-1),0);
%  "economy size" decomposition
score = centerx*pc;
%  the representation of X in the principal component space
if nargout < 3
return;
end
latent = diag(latent).^2;
if (r   latent = [latent(1:r); zeros(n-r,1)];
score(:,r+1:end) = 0;
end
if nargout < 4
return;
end
tmp = sqrt(diag(1./latent(1:r)))*score(:,1:r)';
tsquare = sum(tmp.*tmp)';


From the above code, we should pay attention to the following aspects when comparing with SVD method:

First, Rows of $\,X$ correspond to observations, columns to variables. When using princomp on 2_3 data in assignment 1, note that we take the transpose of $\,X$.

 >> load 2_3;
>> [U, score] = princomp(X');


Second, princomp centers X by subtracting off column means.

The third, when $\,X=UdV'$, princomp uses $\,V$ as coefficients for principal components, rather than $\,U$.

The following is an example to perform PCA using princomp and SVD respectively to get the same results.

SVD method
 >> load 2_3
>> mn=mean(X,2);
>> X1=X-repmat(mn,1,400);
>> [s d v]=svd(X1');
>> y=X1'*v;

princomp
 >>[U score]=princomp(X');


Then we can see that y=score, v=U.

useful resouces: LDA and QDA in Matlab[3],[4],[5]

## Reference

1. Harry Zhang. The optimality of naive bayes. FLAIRS Conference. AAAI Press, 2004

2. Rich Caruana and Alexandru N. Mizil. An empirical comparison of supervised learning algorithms. In ICML ’06: Proceedings of the 23rd international conference on Machine learning, pages 161–168, New York, NY, USA, 2006, ACM.

LDA:[6]

QDA:[8]

## Principal Component Analysis - September 30, 2010

### Rough definition

Keepings two important aspects of data analysis in mind:

• Reducing covariance in data
• Preserving information stored in data(Variance is a source of information)

Principal component analysis (PCA) is a dimensionality-reduction method invented by Karl Pearson in 1901 [9]. Depending on where this methodology is applied, other common names of PCA include the Karhunen–Loève transform (KLT) , the Hotelling transform, and the proper orthogonal decomposition (POD). PCA is the simplist eigenvector-based multivariate analysis. It reduces the dimensionality of the data by revealing the internal structure of the data in a way that best explains the variance in the data. To this end, PCA works by using a user-defined number of the most important directions of variation (dimensions or principal components) of the data to project the data onto these directions so as to produce a lower-dimensional representation of the original data. The resulting lower-dimensional representation of our data is usually much easier to visualize and it also exhibits the most informative aspects (dimensions) of our data whilst capturing as much of the variation exhibited by our data as it possibly could.

Furthermore, if one considers the lower dimensional representation produced by PCA as a least squares fit of our original data, then it can also be easily shown that this representation is the one that minimizes the reconstruction error of our data. It should be noted however, that one usually does not have control over which dimensions PCA deems to be the most informative for a given set of data, and thus one usually does not know which dimensions PCA selects to be the most informative dimensions in order to create the lower-dimensional representation.

Suppose $\,X$ is our data matrix containing $\,d$-dimensional data. The idea behind PCA is to apply singular value decomposition to $\,X$ to replace the rows of $\,X$ by a subset of it that captures as much of the variance in $\,X$ as possible. First, through the application of singular value decomposition to $\,X$, PCA obtains all of our data's directions of variation. These directions would also be ordered from left to right, with the leftmost directions capturing the most amount of variation in our data and the rightmost directions capturing the least amount. Then, PCA uses a subset of these directions to map our data from its original space to a lower-dimensional space.

By applying singular value decomposition to $\,X$, $\,X$ is decomposed as $\,X = U\Sigma V^T \,$. The $\,d$ columns of $\,U$ are the eigenvectors of $\,XX^T \,$. The $\,d$ columns of $\,V$ are the eigenvectors of $\,X^TX \,$. The $\,d$ diagonal values of $\,\Sigma$ are the square roots of the eigenvalues of $\,XX^T \,$ (also of $\,X^TX \,$), and they correspond to the columns of $\,U$ (also of $\,V$).

We are interested in $\,U$, whose $\,d$ columns are the $\,d$ directions of variation of our data. Ordered from left to right, the $\,ith$ column of $\,U$ is the $\,ith$ most informative direction of variation of our data. That is, the $\,ith$ column of $\,U$ is the $\,ith$ most effective column in terms of capturing the total variance exhibited by our data. A subset of the columns of $\,U$ is used by PCA to reduce the dimensionality of $\,X$ by projecting $\,X$ onto the columns of this subset. In practice, when we apply PCA to $\,X$ to reduce the dimensionality of $\,X$ from $\,d$ to $\,k$, where $k \lt d\,$, we would proceed as follows:

Step 1: Center $\,X$ so that it would have zero mean.
Step 2: Apply singular value decomposition to $\,X$ to obtain $\,U$.
Step 3: Suppose we denote the resulting $\,k$-dimensional representation of $\,X$ by $\,Y$. Then, $\,Y$ is obtained as $\,Y = U_k^TX$. Here, $\,U_k$ consists of the first (leftmost) $\,k$ columns of $\,U$ that correspond to the $\,k$ largest diagonal elements of $\,\Sigma$.

PCA takes a sample of d - dimensional vectors and produces an orthogonal(zero covariance) set of d 'Principal Components'. The first Principal Component is the direction of greatest variance in the sample. The second principal component is the direction of second greatest variance (orthogonal to the first component), etc.

Then we can preserve most of the variance in the sample in a lower dimension by choosing the first k Principle Components and approximating the data in k - dimensional space, which is easier to analyze and plot.

### Principal Components of handwritten digits

Suppose that we have a set of 130 images (28 by 23 pixels) of handwritten threes. {{

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| type = style | image = | imageright = | style = | textstyle = | text = This article may require cleanup to meet Wikicoursenote's quality standards. The specific problem is: If anyone can tell me where I can find the 2-3 data set, I would create the new image. In the mean time, I found a non-copyrighted image of different looking 3s online, but as you can see, it is not as nice as one we could make.. Please improve this article if you can. (September 6 2010) | small = | smallimage = | smallimageright = | smalltext = }} {{

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We can represent each image as a vector of length 644 ($644 = 23 \times 28$). Then we can represent the entire data set as a 644 by 130 matrix, shown below. Each column represents one image (644 rows = 644 pixels).

Using PCA, we can approximate the data as the product of two smaller matrices, which I will call $V \in M_{644,2}$ and $W \in M_{2,103}$. If we expand the matrix product then each image is approximated by a linear combination of the columns of V: $\hat{f}(\lambda) = \bar{x} + \lambda_1 v_1 + \lambda_2 v_2$, where $\lambda = [\lambda_1, \lambda_2]^T$ is a column of W.

To demonstrate this process, we can compare the images of 2s and 3s. We will apply PCA to the data, and compare the images of the labeled data. This is an example in classifying.

Don't worry about the constant term for now. The point is that we can represent an image using just 2 coefficients instead of 644. Also notice that the coefficients correspond to features of the handwritten digits. The picture below shows the first two principal components for the set of handwritten threes.

The first coefficient represents the width of the entire digit, and the second coefficient represents the slant of each handwritten digit.

### Derivation of the first Principle Component

For finding the direction of maximum variation, Let \begin{align}\textbf{w}\end{align} be an arbitrary direction, \begin{align}\textbf{x}\end{align} a data point, and \begin{align}\displaystyle u\end{align} be the length of the projection of \begin{align}\textbf{x}\end{align} in the direction \begin{align}\textbf{w}\end{align}.

\begin{align} \textbf{w} &= [w_1, \ldots, w_D]^T \\ \textbf{x} &= [x_1, \ldots, x_D]^T \\ u &= \frac{\textbf{w}^T \textbf{x}}{\sqrt{\textbf{w}^T\textbf{w}}} \end{align}

The direction \begin{align}\textbf{w}\end{align} is the same as \begin{align}c\textbf{w}\end{align}, for any scalar $c$, so without loss of generality we assume that:

\begin{align} |\textbf{w}| &= \sqrt{\textbf{w}^T\textbf{w}} = 1 \\ u &= \textbf{w}^T \textbf{x}. \end{align}

Let $x_1, \ldots, x_D$ be random variables, then we set our goal as to maximize the variance of $u$,

$\textrm{var}(u) = \textrm{var}(\textbf{w}^T \textbf{x}) = \textbf{w}^T \Sigma \textbf{w}.$

For a finite data set we replace the covariance matrix $\Sigma$ by $s$. The sample covariance matrix

$\textrm{var}(u) = \textbf{w}^T s\textbf{w} .$

The above mentioned variable is the variance of \begin{align}\displaystyle u \end{align} formed by the weight vector \begin{align}\textbf{w} \end{align}. The first principal component is the vector \begin{align}\textbf{w} \end{align} that maximizes the variance,

$\textrm{PC} = \underset{\textbf{w}}{\operatorname{arg\,max}} \, \left( \operatorname{var}(u) \right) = \underset{\textbf{w}}{\operatorname{arg\,max}} \, \left( \textbf{w}^T s \textbf{w} \right)$

where arg max denotes the value of \begin{align}\textbf{w} \end{align} that maximizes the function. Our goal is to find the weight \begin{align}\textbf{w} \end{align} that maximizes this variability, subject to a constraint. Since our function is convex, it has no maximum value. Therefore we need to add a constraint that restricts the length of \begin{align}\textbf{w} \end{align}. However, we are only interested in the direction of the variability, so the problem becomes

$\underset{\textbf{w}}{\operatorname{max}} \, \left( \textbf{w}^T s \textbf{w} \right)$

s.t. $\textbf{w}^T \textbf{w} = 1.$

Notice,

$\textbf{w}^T s \textbf{w} \leq \| \textbf{w}^T s \textbf{w} \| \leq \| s \| \| \textbf{w} \| = \| s \|.$

Therefore the variance is bounded, so the maximum exists. We find the this maximum using the method of Lagrange multipliers.

#### Lagrange Multiplier

Before we can proceed, we must review Lagrange Multipliers.

"The red line shows the constraint g(x,y) = c. The blue lines are contours of f(x,y). The point where the red line tangentially touches a blue contour is our solution." [Lagrange Multipliers, Wikipedia]

To find the maximum (or minimum) of a function $\displaystyle f(x,y)$ subject to constraints $\displaystyle g(x,y) = 0$, we define a new variable $\displaystyle \lambda$ called a Lagrange Multiplier and we form the Lagrangian,

$\displaystyle L(x,y,\lambda) = f(x,y) - \lambda g(x,y)$

If $\displaystyle (x^*,y^*)$ is the max of $\displaystyle f(x,y)$, there exists $\displaystyle \lambda^*$ such that $\displaystyle (x^*,y^*,\lambda^*)$ is a stationary point of $\displaystyle L$ (partial derivatives are 0).
In addition $\displaystyle (x^*,y^*)$ is a point in which functions $\displaystyle f$ and $\displaystyle g$ touch but do not cross. At this point, the tangents of $\displaystyle f$ and $\displaystyle g$ are parallel or gradients of $\displaystyle f$ and $\displaystyle g$ are parallel, such that:

$\displaystyle \nabla_{x,y } f = \lambda \nabla_{x,y } g$

where,
$\displaystyle \nabla_{x,y} f = (\frac{\partial f}{\partial x},\frac{\partial f}{\partial{y}}) \leftarrow$ the gradient of $\, f$
$\displaystyle \nabla_{x,y} g = (\frac{\partial g}{\partial{x}},\frac{\partial{g}}{\partial{y}}) \leftarrow$ the gradient of $\, g$

#### Example

Suppose we wish to maximise the function $\displaystyle f(x,y)=x-y$ subject to the constraint $\displaystyle x^{2}+y^{2}=1$. We can apply the Lagrange multiplier method on this example; the lagrangian is:

$\displaystyle L(x,y,\lambda) = x-y - \lambda (x^{2}+y^{2}-1)$

We want the partial derivatives equal to zero:

$\displaystyle \frac{\partial L}{\partial x}=1+2 \lambda x=0$

$\displaystyle \frac{\partial L}{\partial y}=-1+2\lambda y=0$

$\displaystyle \frac{\partial L}{\partial \lambda}=x^2+y^2-1$

Solving the system we obtain 2 stationary points: $\displaystyle (\sqrt{2}/2,-\sqrt{2}/2)$ and $\displaystyle (-\sqrt{2}/2,\sqrt{2}/2)$. In order to understand which one is the maximum, we just need to substitute it in $\displaystyle f(x,y)$ and see which one as the biggest value. In this case the maximum is $\displaystyle (\sqrt{2}/2,-\sqrt{2}/2)$.

#### Determining W

Back to the original problem, from the Lagrangian we obtain,

$\displaystyle L(\textbf{w},\lambda) = \textbf{w}^T S \textbf{w} - \lambda (\textbf{w}^T \textbf{w} - 1)$

If $\textbf{w}^T \textbf{w}$ is a unit vector then the second part of the equation is 0.

If $\textbf{w}^T \textbf{w}$ is not a unit vector then the second part of the equation increases. Thus decreasing overall $\displaystyle L(\textbf{w},\lambda)$. Maximization happens when $\textbf{w}^T \textbf{w} =1$

(Note that to take the derivative with respect to w below, $\textbf{w}^T S \textbf{w}$ can be thought of as a quadratic function of w, hence the 2sw below. For more matrix derivatives, see section 2 of the Matrix Cookbook)

Taking the derivative with respect to w, we get:

$\displaystyle \frac{\partial L}{\partial \textbf{w}} = 2S\textbf{w} - 2\lambda\textbf{w}$

Set $\displaystyle \frac{\partial L}{\partial \textbf{w}} = 0$, we get

$\displaystyle S\textbf{w}^* = \lambda^*\textbf{w}^*$

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| type = style | image = | imageright = | style = | textstyle = | text = This article may require cleanup to meet Wikicoursenote's quality standards. The specific problem is: It is good discussion, what will happen if we don't have distinct eigenvalues and eigenvectors? What does this situation mean?. Please improve this article if you can. (October 2010) | small = | smallimage = | smallimageright = | smalltext = }} {{

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| type = style | image = | imageright = | style = | textstyle = | text = This article may require cleanup to meet Wikicoursenote's quality standards. The specific problem is: If the eigenvalues are not distinct, I suppose we could still take the leftmost eigenvector by default. Not sure if this is the correct approach, so can anyone please explain further? Thanks. Please improve this article if you can. (October 2010) | small = | smallimage = | smallimageright = | smalltext = }} {{

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| type = style | image = | imageright = | style = | textstyle = | text = This article may require cleanup to meet Wikicoursenote's quality standards. The specific problem is: As U is the eigenvector of a symetric matrix, is it possible that we have 2 similar eigen vector?. Please improve this article if you can. (October 2010) | small = | smallimage = | smallimageright = | smalltext = }} {{

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| type = style | image = | imageright = | style = | textstyle = | text = This article may require cleanup to meet Wikicoursenote's quality standards. The specific problem is: $\lambda_1 \geq \lambda_2 \geq ... \geq \lambda_D$, I think it has no difference when eigenvalues are equal. Please improve this article if you can. (Nov 6 2010) | small = | smallimage = | smallimageright = | smalltext = }} {{

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| type = style | image = | imageright = | style = | textstyle = | text = This article may require cleanup to meet Wikicoursenote's quality standards. The specific problem is: Since we're only interested in the largest (left-most) $n$ eigenvalues, if other smaller eigenvalues are equal it does not really matter in PCA. If two eigenvalues are equal, this means that the data has the same amount of variation in each of the two directions that they correspond to. Thus, either one of the directions could be chosen without loss of generality to reduce the variation, unless there is some other consideration that would give an advantage to choosing one direction over another. Note also, that if ALL of the eigenvalues are the same then this means that the data is on the surface of a d-dimensional sphere (all directions have the same amount of variation). Is this correct?. Please improve this article if you can. (Nov 13 2010) | small = | smallimage = | smallimageright = | smalltext = }}

From the eigenvalue equation $\, \textbf{w}^*$ is an eigenvector of S and $\, \lambda^*$ is the corresponding eigenvalue of S. If we substitute $\displaystyle\textbf{w}^*$ in $\displaystyle \textbf{w}^T S\textbf{w}$ we obtain,

$\displaystyle\textbf{w}^{*T} S\textbf{w}^* = \textbf{w}^{*T} \lambda^* \textbf{w}^* = \lambda^* \textbf{w}^{*T} \textbf{w}^* = \lambda^*$

In order to maximize the objective function we choose the eigenvector corresponding to the largest eigenvalue. We choose the first PC, u1 to have the maximum variance
(i.e. capturing as much variability in $\displaystyle x_1, x_2,...,x_D$ as possible.) Subsequent principal components will take up successively smaller parts of the total variability.

D dimensional data will have D eigenvectors

$\lambda_1 \geq \lambda_2 \geq ... \geq \lambda_D$ where each $\, \lambda_i$ represents the amount of variation in direction $\, i$

so that

$Var(u_1) \geq Var(u_2) \geq ... \geq Var(u_D)$

Note that the Principal Components decompose the total variance in the data:

$\displaystyle \sum_{i = 1}^D Var(u_i) = \sum_{i = 1}^D \lambda_i = Tr(S) = Var(\sum_{i = 1}^n x_i)$

i.e. the sum of variations in all directions is the variation in the whole data

Example from class

We apply PCA to the noise data, making the assumption that the intrinsic dimensionality of the data is 10. We now try to compute the reconstructed images using the top 10 eigenvectors and plot the original and reconstructed images

The Matlab code is as follows:

 load noisy
who
size(X)
imagesc(reshape(X(:,1),20,28)')
colormap gray
imagesc(reshape(X(:,1),20,28)')
m_X=mean(X,2);
mm=repmat(m_X,1,300);
XX=X-mm;
[u s v] = svd(XX);
xHat = u(:,1:10)*s(1:10,1:10)*v(:,1:10)'; % use ten principal components
xHat=xHat+mm;
figure
imagesc(reshape(xHat(:,1000),20,28)') % here '1000' can be changed to different values, e.g. 105, 500, etc.
colormap gray


Running the above code gives us 2 images - the first one represents the noisy data - we can barely make out the face

The second one is the denoised image

As you can clearly see, more features can be distinguished from the picture of the de-noised face compared to the picture of the noisy face. This is because almost all of the noise in the noisy image is captured by the principal components (directions of variation) that capture the least amount of variation in the image, and these principal components were discarded when we used the few principal components that capture most of the image's variation to generate the image's lower-dimensional representation. If we took more principal components, at first the image would improve since the intrinsic dimensionality is probably more than 10. But if you include all the components you get the noisy image, so not all of the principal components improve the image. In general, it is difficult to choose the optimal number of components.

#### Application of PCA - Feature Extraction

One of the applications of PCA is to group similar data (e.g. images). There are generally two methods to do this. We can classify the data (i.e. give each data a label and compare different types of data) or cluster (i.e. do not label the data and compare output for classes).

Generally speaking, we can do this with the entire data set (if we have an 8X8 picture, we can use all 64 pixels). However, this is hard, and it is easier to use the reduced data and features of the data.

#### General PCA Algorithm

The PCA Algorithm is summarized as follows (taken from the Lecture Slides).

#### Algorithm

Recover basis: Calculate $XX^T =\Sigma_{i=1}^{n} x_i x_{i}^{T}$ and let $U=$ eigenvectors of $X X^T$ corresponding to the top $d$ eigenvalues.

Encoding training data: Let $Y=U^TX$ where $Y$ is a $d \times n$ matrix of encoding of the original data.

Reconstructing training data: $\hat{X}= UY=UU^TX$.

Encode set example: $y=U^T x$ where $y$ is a $d-$dimentional encoding of $x$.

Reconstruct test example: $\hat{x}= Uy=UU^Tx$.

Other Notes:

1. The mean of the data(X) must be 0. This means we may have to preprocess the data by subtracting off the mean(see detailsPCA in Wikipedia.)
2. Encoding the data means that we are projecting the data onto a lower dimensional subspace by taking the inner product. Encoding: $X_{D\times n} \longrightarrow Y_{d\times n}$ using mapping $\, U^T X_{D \times n}$.
3. When we reconstruct the training set, we are only using the top d dimensions.This will eliminate the dimensions that have lower variance (e.g. noise). Reconstructing: $\hat{X}_{D\times n}\longleftarrow Y_{d \times n}$ using mapping $\, U_dY_{d \times n}$, where $\,U_d$ contains the first (leftmost) $\,d$ columns of $\,U$.
4. We can compare the reconstructed test sample to the reconstructed training sample to classify the new data.

## Fisher's (Linear) Discriminant Analysis (FDA) - Two Class Problem - October 5, 2010

### Sir Ronald A. Fisher

Fisher's Discriminant Analysis (FDA), also known as Fisher's Linear Discriminant Analysis (LDA) in some sources, is a classical feature extraction technique. It was originally described in 1936 by Sir Ronald Aylmer Fisher, an English statistician and eugenicist who has been described as one of the founders of modern statistical science. His original paper describing FDA can be found here; a Wikipedia article summarizing the algorithm can be found here. In this paper Fisher used for the first time the term DISCRIMINANT FUNCTION. The term DISCRIMINANT ANALYSIS was introduced later by Fisher himself in a subsequent paper which can be found here.

### Contrasting FDA with PCA

As in PCA, the goal of FDA is to project the data in a lower dimension. You might ask, why was FDA invented when PCA already existed? There is a simple explanation for this that can be found here. PCA is an unsupervised method for classification, so it does not take into account the labels in the data. Suppose we have two clusters that have very different or even opposite labels from each other but are nevertheless positioned in a way such that they are very much parallel to each other and also very near to each other. In this case, most of the total variation of the data is in the direction of these two clusters. If we use PCA in cases like this, then both clusters would be projected onto the direction of greatest variation of the data to become sort of like a single cluster after projection. PCA would therefore mix up these two clusters that, in fact, have very different labels. What we need to do instead, in this cases like this, is to project the data onto a direction that is orthogonal to the direction of greatest variation of the data. This direction is in the least variation of the data. On the 1-dimensional space resulting from such a projection, we would then be able to effectively classify the data, because these two clusters would be perfectly or nearly perfectly separated from each other taking into account of their labels. This is exactly the idea behind FDA.

The main difference between FDA and PCA is that, in FDA, in contrast to PCA, we are not interested in retaining as much of the variance of our original data as possible. Rather, in FDA, our goal is to find a direction that is useful for classifying the data (i.e. in this case, we are looking for a direction that is most representative of a particular characteristic e.g. glasses vs. no-glasses). Suppose we have 2-dimensional data, then FDA would attempt to project the data of each class onto a point in such a way that the resulting two points would be as far apart from each other as possible. Intuitively, this basic idea behind FDA is the optimal way for separating each pair of classes along a certain direction.

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| type = style | image = | imageright = | style = | textstyle = | text = This article may require cleanup to meet Wikicoursenote's quality standards. The specific problem is: Just a thought: how relevant is "Dimensionality reduction techniques" to the concept of "subspace clustering"? As in subspace clustering, the goal is to find a set of features (relevant features, the concept is referred to as local feature relevance in the literature) in the high dimensional space, where potential subspaces accommodating different classes of data points can be defined. This means; the data points are dense when they are considered in a subset of dimensions (features).. Please improve this article if you can. (October 2010) | small = | smallimage = | smallimageright = | smalltext = }} {{

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| type = style | image = | imageright = | style = | textstyle = | text = This article may require cleanup to meet Wikicoursenote's quality standards. The specific problem is: If I'm not mistaken, classification techniques like FDA use labeled training data whereas clustering techniques use unlabeled training data instead. Any other input regarding this would be much appreciated. Thanks. Please improve this article if you can. (October 2010) | small = | smallimage = | smallimageright = | smalltext = }} {{

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| type = style | image = | imageright = | style = | textstyle = | text = This article may require cleanup to meet Wikicoursenote's quality standards. The specific problem is: An extension of clustering is subspace clustering in which different subspace are searched through to find the relavant and appropriate dimentions. High dimentional data sets are roughly equiedistant from each other, so feature selection methods are used to remove the irrelavant dimentions. These techniques do not keep the relative distance so PCA is not useful for these applications. It should be noted that subspace clustering localize their search unlike feature selection algorithms.for more information click here[10]. Please improve this article if you can. (October 2010) | small = | smallimage = | smallimageright = | smalltext = }}

The number of dimensions that we want to reduce the data to depends on the number of classes:
For a 2-classes problem, we want to reduce the data to one dimension (a line), $\displaystyle Z \in \mathbb{R}^{1}$
Generally, for a k-classes problem, we want to reduce the data to k-1 dimensions, $\displaystyle Z \in \mathbb{R}^{k-1}$

As we will see from our objective function, we want to maximize the separation of the classes and to minimize the within-variance of each class. That is, our ideal situation is that the individual classes are as far away from each other as possible, and at the same time the data within each class are as close to each other as possible (collapsed to a single point in the most extreme case).

The following diagram summarizes this goal.

In fact, the two examples above may represent the same data projected on two different lines.

### Distance Metric Learning VS FDA

In many fundamental machine learning problems, the Euclidean distances between data points do not represent the desired topology that we are trying to capture. Kernel methods address this problem by mapping the points into new spaces where Euclidean distances may be more useful. An alternative approach is to construct a Mahalanobis distance (quadratic Gaussian metric) over the input space and use it in place of Euclidean distances. This approach can be equivalently interpreted as a linear transformation of the original inputs, followed by Euclidean distance in the projected space. This approach has attracted a lot of recent interest.

Some of the proposed algorithms are iterative and computationally expensive. In the paper,"Distance Metric Learning VS FDA " written by our instructor, they propose a closed-form solution to one algorithm that previously required expensive semidefinite optimization. They provide a new problem setup in which the algorithm performs better or as well as some standard methods, but without the computational complexity. Furthermore, they show a strong relationship between these methods and the Fisher Discriminant Analysis (FDA). They also extend the approach by kernelizing it, allowing for non-linear transformations of the metric.

Example

In the paper "Distance Metric Learning VS FDA ", classification error rate for three of the six UCI datasets, each learned metric is projected onto a lowdimensional subspace, shown along the x axis are shown as below.

,

### FDA Goals

An intuitive description of FDA can be given by visualizing two clouds of data, as shown above. Ideally, we would like to collapse all of the data points in each cloud onto one point on some projected line, then make those two points as far apart as possible. In doing so, we make it very easy to tell which class a data point belongs to. In practice, it is not possible to collapse all of the points in a cloud to one point, but we attempt to make all of the points in a cloud close to each other while simultaneously far from the points in the other cloud.

#### Example in R

PCA and FDA primary dimension for normal multivariate data, using R.
>> X = matrix(nrow=400,ncol=2)
>> X[1:200,] = mvrnorm(n=200,mu=c(1,1),Sigma=matrix(c(1,1.5,1.5,3),2))
>> X[201:400,] = mvrnorm(n=200,mu=c(5,3),Sigma=matrix(c(1,1.5,1.5,3),2))
>> Y = c(rep("red",200),rep("blue",200))

Create 2 multivariate normal random variables with $\, \mu_1 = \left( \begin{array}{c}1 \\ 1 \end{array} \right), \mu_2 = \left( \begin{array}{c}5 \\ 3 \end{array} \right). ~\textrm{Cov} = \left( \begin{array}{cc} 1 & 1.5 \\ 1.5 & 3 \end{array} \right)$. Create Y, an index indicating which class they belong to.
>> s <- svd(X,nu=1,nv=1)

Calculate the singular value decomposition of X. The most significant direction is in s$v[,1], and is displayed as a black line. >> s2 <- lda(X,grouping=Y)  The lda function, given the group for each item, uses Fischer's Linear Discriminant Analysis (FLDA) to find the most discriminant direction. This can be found in s2$scaling.

Now that we've calculated the PCA and FLDA decompositions, we create a plot to demonstrate the differences between the two algorithms. FLDA is clearly better suited to discriminating between two classes whereas PCA is primarily good for reducing the number of dimensions when data is high-dimensional.

>> plot(X,col=Y,main="PCA vs. FDA example")

Plot the set of points, according to colours given in Y.
>> slope = s$v[2]/s$v[1]
>> intercept = mean(X[,2])-slope*mean(X[,1])
>> abline(a=intercept,b=slope)

Plot the main PCA direction, drawn through the mean of the dataset. Only the direction is significant.
>> slope2 = s2$scaling[2]/s2$scaling[1]
>> intercept2 = mean(X[,2])-slope2*mean(X[,1])
>> abline(a=intercept2,b=slope2,col="red")

Plot the FLDA direction, again through the mean.
>> legend(-2,7,legend=c("PCA","FDA"),col=c("black","red"),lty=1)

Labeling the lines directly on the graph makes it easier to interpret.

FDA projects the data into lower dimensional space, where the distances between the projected means are maximum and the within-class variances are minimum. There are two categories of classification problems:

1. Two-class problem

2. Multi-class problem (addressed next lecture)

### Two-class problem

In the two-class problem, we have the pre-knowledge that data points belong to two classes. Intuitively speaking points of each class form a cloud around the mean of the class, with each class having possibly different size. To be able to separate the two classes we must determine the class whose mean is closest to a given point while also accounting for the different size of each class, which is represented by the covariance of each class.

Assume $\underline{\mu_{1}}=\frac{1}{n_{1}}\displaystyle\sum_{i:y_{i}=1}\underline{x_{i}}$ and $\displaystyle\Sigma_{1}$, represent the mean and covariance of the 1st class, and $\underline{\mu_{2}}=\frac{1}{n_{2}}\displaystyle\sum_{i:y_{i}=2}\underline{x_{i}}$ and $\displaystyle\Sigma_{2}$ represent the mean and covariance of the 2nd class. We have to find a transformation which satisfies the following goals:

1.To make the means of these two classes as far apart as possible

In other words, the goal is to maximize the distance after projection between class 1 and class 2. This can be done by maximizing the distance between the means of the classes after projection. When projecting the data points to a one-dimensional space, all points will be projected to a single line; the line we seek is the one with the direction that achieves maximum separation of classes upon projection. If the original points are $\underline{x_{i}} \in \mathbb{R}^{d}$and the projected points are $\underline{w}^T \underline{x_{i}}$ then the mean of the projected points will be $\underline{w}^T \underline{\mu_{1}}$ and $\underline{w}^T \underline{\mu_{2}}$ for class 1 and class 2 respectively. The goal now becomes to maximize the Euclidean distance between projected means, $(\underline{w}^T\underline{\mu_{1}}-\underline{w}^T\underline{\mu_{2}})^T (\underline{w}^T\underline{\mu_{1}}-\underline{w}^T\underline{\mu_{2}})$. The steps of this maximization are given below.

2.We want to collapse all data points of each class to a single point, i.e., minimize the covariance within each class

Notice that the variance of the projected classes 1 and 2 are given by $\underline{w}^T\Sigma_{1}\underline{w}$ and $\underline{w}^T\Sigma_{2}\underline{w}$. The second goal is to minimize the sum of these two covariances (the summation of the two covariances is a valid covariance, satisfying the symmetry and positive semi-definite criteria).

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As is demonstrated below, both of these goals can be accomplished simultaneously.

Original points are $\underline{x_{i}} \in \mathbb{R}^{d}$
$\ \{ \underline x_1 \underline x_2 \cdot \cdot \cdot \underline x_n \}$

Projected points are $\underline{z_{i}} \in \mathbb{R}^{1}$ with $\underline{z_{i}} = \underline{w}^T \cdot\underline{x_{i}}$ where $\ z_i$ is a scalar

#### 1. Minimizing within-class variance

$\displaystyle \min_w (\underline{w}^T\sum_1\underline{w})$

$\displaystyle \min_w (\underline{w}^T\sum_2\underline{w})$

and this problem reduces to $\displaystyle \min_w (\underline{w}^T(\sum_1 + \sum_2)\underline{w})$
(where $\,\sum_1$ and $\,\sum_2$ are the covariance matrices of the 1st and 2nd classes of data, respectively)

Let $\displaystyle \ s_w=\sum_1 + \sum_2$ be the within-classes covariance. Then, this problem can be rewritten as $\displaystyle \min_w (\underline{w}^Ts_w\underline{w})$.

#### 2. Maximize the distance between the means of the projected data

$\displaystyle \max_w ||\underline{w}^T \mu_1 - \underline{w}^T \mu_2||^2,$

\begin{align} ||\underline{w}^T \mu_1 - \underline{w}^T \mu_2||^2 &= (\underline{w}^T \mu_1 - \underline{w}^T \mu_2)^T(\underline{w}^T \mu_1 - \underline{w}^T \mu_2)\\ &= (\mu_1^T\underline{w} - \mu_2^T\underline{w})(\underline{w}^T \mu_1 - \underline{w}^T \mu_2)\\ &= (\mu_1 - \mu_2)^T \underline{w} \underline{w}^T (\mu_1 - \mu_2) \\ &= ((\mu_1 - \mu_2)^T \underline{w})^{T} (\underline{w}^T (\mu_1 - \mu_2))^{T} \\ &= \underline{w}^T(\mu_1 - \mu_2)(\mu_1 - \mu_2)^T \underline{w} \end{align}

Note that in the last line above the order is rearranged clockwise because the answer is a scalar.

Let $\displaystyle s_B = (\mu_1 - \mu_2)(\mu_1 - \mu_2)^T$, the between-class covariance, then the goal is to $\displaystyle \max_w (\underline{w}^T(\mu_1 - \mu_2)(\mu_1 - \mu_2)^T\underline{w})$ or $\displaystyle \max_w (\underline{w}^Ts_B\underline{w})$.

### The Objective Function for FDA

We want an objective function which satisfies both of the goals outlined above (at the same time).

1. $\displaystyle \min_w (\underline{w}^T(\sum_1 + \sum_2)\underline{w})$ or $\displaystyle \min_w (\underline{w}^Ts_w\underline{w})$
2. $\displaystyle \max_w (\underline{w}^T(\mu_1 - \mu_2)(\mu_1 - \mu_2)^T\underline{w})$ or $\displaystyle \max_w (\underline{w}^Ts_B\underline{w})$

So, we construct our objective function as maximizing the ratio of the two goals brought above:

$\displaystyle \max_w \frac{(\underline{w}^T(\mu_1 - \mu_2)(\mu_1 - \mu_2)^T\underline{w})} {(\underline{w}^T(\sum_1 + \sum_2)\underline{w})}$

or equivalently,

$\displaystyle \max_w \frac{(\underline{w}^Ts_B\underline{w})}{(\underline{w}^Ts_w\underline{w})}$
One may argue that we can use subtraction for this purpose, while this approach is true but it can be shown it will need another scaling factor. Thus using this ratio is more efficient.

As the objective function is convex, and so it does not have a maximum. To get around this problem, we have to add the constraint that w must have unit length, and then solvethis optimization problem we form the lagrangian:

$\displaystyle L(\underline{w},\lambda) = \underline{w}^Ts_B\underline{w} - \lambda (\underline{w}^Ts_w\underline{w} -1)$

Then, we equate the partial derivative of L with respect to $\underline{w}$: $\displaystyle \frac{\partial L}{\partial \underline{w}}=2s_B \underline{w} - 2\lambda s_w \underline{w} = 0$

$s_B \underline{w} = \lambda s_w \underline{w}$
$s_w^{-1}s_B \underline{w}= \lambda\underline{w}$

This is in the form of generalized eigenvalue problem. Therefore, $\underline{w}$ is the largest eigenvector of $s_w^{-1}s_B$

This solution can be further simplified as follow:

$s_w^{-1}(\mu_1 - \mu_2)(\mu_1 - \mu_2)^T\underline{w} = \lambda\underline{w}$

Since $(\mu_1 - \mu_2)^T\underline{w}$ is a scalar then $s_w^{-1}(\mu_1 - \mu_2)$$\underline{w}$

This gives the direction of $\underline{w}$ without doing eigenvalue decomposition in the case of 2-class problem.

Note: In order for ${s_w}$ to have an inverse, it must have full rank. This can be achieved by ensuring that the number of data points $\,\ge$ the dimensionality of $\underline{x_{i}}$.

### FDA Using Matlab

Note: The following example was not actually mentioned in this lecture

We see now an application of the theory that we just introduced. Using Matlab, we find the principal components and the projection by Fisher Discriminant Analysis of two Bivariate normal distributions to show the difference between the two methods.

      %First of all, we generate the two data set:
% First data set X1
X1 = mvnrnd([1,1],[1 1.5; 1.5 3], 300);
%In this case:
mu_1=[1;1];
Sigma_1=[1 1.5; 1.5 3];
%where mu and sigma are the mean and covariance matrix.
% Second data set X2
X2 = mvnrnd([5,3],[1 1.5; 1.5 3], 300);
%Here mu_2=[5;3] and Sigma_2=[1 1.5; 1.5 3]
%The plot of the two distributions is:
plot(X1(:,1),X1(:,2),'.b'); hold on;
plot(X2(:,1),X2(:,2),'ob')



      %We compute the principal components:
% Combine data sets to map both into the same subspace
X=[X1;X2];
X=X';
% We used built-in PCA function in Matlab
[coefs, scores]=princomp(X);

plot([0 coefs(1,1)], [0 coefs(2,1)],'b')
plot([0 coefs(1,1)]*10, [0 coefs(2,1)]*10,'r')
sw=2*[1 1.5;1.5 3]   % sw=Sigma1+Sigma2=2*Sigma1
w=sw\[4; 2]       % calculate s_w^{-1}(mu2 - mu1)
plot ([0 w(1)], [0 w(2)],'g')


      %We now make the projection:
Xf=w'*X
figure
plot(Xf(1:300),1,'ob') %In this case, since it's a one dimension data, the plot is "Data Vs Indexes"
hold on
plot(Xf(301:600),1,'or')



      %We see that in the above picture that there is very little overlapping
Xp=coefs(:,1)'*X
figure
plot(Xp(1:300),1,'b')
hold on
plot(Xp(301:600),2,'or')



      %In this case there is an overlapping since we project the first principal component on [Xp=coefs(:,1)'*X]


### Some of FDA applications

There are many applications for FDA in many domains; a few examples are stated below:

• Speech/Music/Noise Classification in Hearing Aids

FDA can be used to enhance listening comprehension when the user goes from one sound environment to another different one. In practice, many people who require hearing aids do not wear them due in part to the nusiance of having to adjust the settings each time a user changes noise environments (for example, from a quiet walk in the to park to a crowded cafe). If the hearing aid itself could distinguish between the type of sound environment and automatically adjust its settings itself, many more people may be willing to wear and use the hearing aids. The paper referenced below examines the difference in using a classifier based on one level and three classes ("speech", "noisy" or "music" environments) and a classifier based on two levels with two classes each ("speech" versus "non-speech" and then for the "non-speech" group, between "noisy" and "music") and also includes a discussion about the feasibility of implementing these classifiers in the hearing aids. For more information review this paper by Alexandre et al. here.

• Application to Face Recognition

FDA can be used in face recognition for different situations. Instead of using the one-dimensional LDA where the data is transformed into long column vectors with less-than-full-rank covariance matrices for the within-class and between-class covariance matrices, several other approaches of using FDA are suggested here including a two-dimensional approach where the data is stored as a matrix rather than a column vector. In this case, the covariance matrices are full-rank. Details can be found in the paper by Kong et al. here.

• Palmprint Recognition

FDA is used in biometrics to implement an automated palmprint recognition system. See An Automated Palmprint Recognition System by Tee et al. here.

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Some other examples of applications of FDA can be found in the references 4, 5, 6, 7 and 8 as listed below.

### References

1. Kong, H.; Wang, L.; Teoh, E.K.; Wang, J.-G.; Venkateswarlu, R.; , "A framework of 2D Fisher discriminant analysis: application to face recognition with small number of training samples," Computer Vision and Pattern Recognition, 2005. CVPR 2005. IEEE Computer Society Conference on , vol.2, no., pp. 1083- 1088 vol. 2, 20-25 June 2005 doi: 10.1109/CVPR.2005.30 1

2. Enrique Alexandre, Roberto Gil-Pita, Lucas Cuadra, Lorena A´lvarez, Manuel Rosa-Zurera, "SPEECH/MUSIC/NOISE CLASSIFICATION IN HEARING AIDS USING A TWO-LAYER CLASSIFICATION SYSTEM WITH MSE LINEAR DISCRIMINANTS", 16th European Signal Processing Conference (EUSIPCO 2008), Lausanne, Switzerland, August 25-29, 2008, copyright by EURASIP, 2

3. Connie, Tee; Jin, Andrew Teoh Beng; Ong, Michael Goh Kah; Ling, David Ngo Chek; "An automated palmprint recognition system", Journal of Image and Vision Computing, 2005. 3

4. met, Francesca; Boqué, Ricard; Ferré, Joan; "Application of non-negative matrix factorization combined with Fisher's linear discriminant analysis for classification of olive oil excitation-emission fluorescence spectra", Journal of Chemometrics and Intelligent Laboratory Systems, 2006. 4

5. Chiang, Leo H.;Kotanchek, Mark E.;Kordon, Arthur K.; "Fault diagnosis based on Fisher discriminant analysis and support vector machines" Journal of Computers & Chemical Engineering, 2004 5

6. Yang, Jian ;Frangi, Alejandro F.; Yang, Jing-yu; "A new kernel Fisher discriminant algorithm with application to face recognition", 2004 6

7. Cawley, Gavin C.; Talbot, Nicola L. C.; "Efficient leave-one-out cross-validation of kernel fisher discriminant classifiers", Journal of Pattern Recognition , 2003 7

8. Kodipaka, S.; Vemuri, B.C.; Rangarajan, A.; Leonard, C.M.; Schmallfuss, I.; Eisenschenk, S.; "Kernel Fisher discriminant for shape-based classification in epilepsy" Hournal Medical Image Analysis, 2007. 8

## Fisher's (Linear) Discriminant Analysis (FDA) - Multi-Class Problem - October 7, 2010

#### Obtaining Covariance Matrices

The within-class covariance matrix $\mathbf{S}_{W}$ is easy to obtain:

\begin{align} \mathbf{S}_{W} = \sum_{i=1}^{k} \mathbf{S}_{i} \end{align}

where $\mathbf{S}_{i} = \frac{1}{n_{i}}\sum_{j: y_{j}=i}(\mathbf{x}_{j} - \mathbf{\mu}_{i})(\mathbf{x}_{j} - \mathbf{\mu}_{i})^{T}$ and $\mathbf{\mu}_{i} = \frac{\sum_{j: y_{j}=i}\mathbf{x}_{j}}{n_{i}}$.

However, the between-class covariance matrix $\mathbf{S}_{B}$ is not easy to compute directly. To bypass this problem we use the following method. We know that the total covariance $\,\mathbf{S}_{T}$ of a given set of data is constant and known, and we can also decompose this variance into two parts: the within-class variance $\mathbf{S}_{W}$ and the between-class variance $\mathbf{S}_{B}$ in a way that is similar to ANOVA. We thus have:

\begin{align} \mathbf{S}_{T} = \mathbf{S}_{W} + \mathbf{S}_{B} \end{align}

where the total variance is given by

\begin{align} \mathbf{S}_{T} = \frac{1}{n} \sum_{i}(\mathbf{x_{i}-\mu})(\mathbf{x_{i}-\mu})^{T} \end{align}

We can now get $\mathbf{S}_{B}$ from the relationship:

\begin{align} \mathbf{S}_{B} = \mathbf{S}_{T} - \mathbf{S}_{W} \end{align}

Actually, there is another way to obtain $\mathbf{S}_{B}$. Suppose the data contains $\, k$ classes, and each class $\, j$ contains $\, n_{j}$ data points. We denote the overall mean vector by

\begin{align} \mathbf{\mu} = \frac{1}{n}\sum_{i}\mathbf{x_{i}} = \frac{1}{n}\sum_{j=1}^{k}n_{j}\mathbf{\mu}_{j} \end{align}

Thus the total covariance matrix $\mathbf{S}_{T}$ is

\begin{align} \mathbf{S}_{T} = \frac{1}{n} \sum_{i}(\mathbf{x_{i}-\mu})(\mathbf{x_{i}-\mu})^{T} \end{align}

Thus we obtain

\begin{align} & \mathbf{S}_{T} = \sum_{i=1}^{k}\sum_{j: y_{j}=i}(\mathbf{x}_{j} - \mathbf{\mu}_{i} + \mathbf{\mu}_{i} - \mathbf{\mu})(\mathbf{x}_{j} - \mathbf{\mu}_{i} + \mathbf{\mu}_{i} - \mathbf{\mu})^{T} \\& = \sum_{i=1}^{k}\sum_{j: y_{j}=i}(\mathbf{x}_{j}-\mathbf{\mu}_{i})(\mathbf{x}_{j}-\mathbf{\mu}_{i})^{T}+ \sum_{i=1}^{k}\sum_{j: y_{j}=i}(\mathbf{\mu}_{i}-\mathbf{\mu})(\mathbf{\mu}_{i}-\mathbf{\mu})^{T} \\& = \mathbf{S}_{W} + \sum_{i=1}^{k} n_{i}(\mathbf{\mu}_{i}-\mathbf{\mu})(\mathbf{\mu}_{i}-\mathbf{\mu})^{T} \end{align}

Since the total covariance $\mathbf{S}_{T}$ is the sum of the within-class covariance $\mathbf{S}_{W}$ and the between-class covariance $\mathbf{S}_{B}$, we can denote the second term in the final line of the derivation above as the between-class covariance matrix $\mathbf{S}_{B}$, thus we obtain

\begin{align} \mathbf{S}_{B} = \sum_{i=1}^{k} n_{i}(\mathbf{\mu}_{i}-\mathbf{\mu})(\mathbf{\mu}_{i}-\mathbf{\mu})^{T} \end{align}

Recall that in the two class case problem, we have

\begin{align} & \mathbf{S}_{B}^* = (\mathbf{\mu}_{1}-\mathbf{\mu}_{2})(\mathbf{\mu}_{1}-\mathbf{\mu}_{2})^{T} \\ & = (\mathbf{\mu}_{1}-\mathbf{\mu}+\mathbf{\mu}-\mathbf{\mu}_{2})(\mathbf{\mu}_{1}-\mathbf{\mu}+\mathbf{\mu}-\mathbf{\mu}_{2})^{T} \\ & = ((\mathbf{\mu}_{1}-\mathbf{\mu})-(\mathbf{\mu}_{2}-\mathbf{\mu}))((\mathbf{\mu}_{1}-\mathbf{\mu})-(\mathbf{\mu}_{2}-\mathbf{\mu}))^{T} \\ & = ((\mathbf{\mu}_{1}-\mathbf{\mu})-(\mathbf{\mu}_{2}-\mathbf{\mu}))((\mathbf{\mu}_{1}-\mathbf{\mu})^{T}-(\mathbf{\mu}_{2}-\mathbf{\mu})^{T}) \\ & = (\mathbf{\mu}_{1}-\mathbf{\mu})(\mathbf{\mu}_{1}-\mathbf{\mu})^{T}-(\mathbf{\mu}_{1}-\mathbf{\mu})(\mathbf{\mu}_{2}-\mathbf{\mu})^{T}-(\mathbf{\mu}_{2}-\mathbf{\mu})(\mathbf{\mu}_{1}-\mathbf{\mu})^{T}+(\mathbf{\mu}_{2}-\mathbf{\mu})(\mathbf{\mu}_{2}-\mathbf{\mu})^{T} \end{align}

\begin{align} & \mathbf{S}_{B} = n_{1}(\mathbf{\mu}_{1}-\mathbf{\mu})(\mathbf{\mu}_{1}-\mathbf{\mu})^{T} + n_{2}(\mathbf{\mu}_{2}-\mathbf{\mu})(\mathbf{\mu}_{2}-\mathbf{\mu})^{T} \end{align}

Apparently, they are very similar.

Now, we are trying to find the optimal transformation. Basically, we have

\begin{align} \mathbf{z}_{i} = \mathbf{W}^{T}\mathbf{x}_{i}, i=1,2,...,k-1 \end{align}

where $\mathbf{z}_{i}$ is a $(k-1)\times 1$ vector, $\mathbf{W}$ is a $d\times (k-1)$ transformation matrix, i.e. $\mathbf{W} = [\mathbf{w}_{1}, \mathbf{w}_{2},..., \mathbf{w}_{k-1}]$, and $\mathbf{x}_{i}$ is a $d\times 1$ column vector.

Thus we obtain

\begin{align} & \mathbf{S}_{W}^{\ast} = \sum_{i=1}^{k}\sum_{j: y_{j}=i}(\mathbf{W}^{T}\mathbf{x}_{j}-\mathbf{W}^{T}\mathbf{\mu}_{i})(\mathbf{W}^{T}\mathbf{x}_{j}-\mathbf{W}^{T}\mathbf{\mu}_{i})^{T} \\ & = \sum_{i=1}^{k}\sum_{j: y_{j}=i}(\mathbf{W}^{T}(\mathbf{x}_{j}-\mathbf{\mu}_{i}))(\mathbf{W}^{T}(\mathbf{x}_{j}-\mathbf{\mu}_{i}))^{T} \\ & = \sum_{i=1}^{k}\sum_{j: y_{j}=i}(\mathbf{W}^{T}(\mathbf{x}_{j}-\mathbf{\mu}_{i}))((\mathbf{x}_{j}-\mathbf{\mu}_{i})^{T}\mathbf{W}) \\ & = \sum_{i=1}^{k}\sum_{j: y_{j}=i}\mathbf{W}^{T}(\mathbf{x}_{j}-\mathbf{\mu}_{i})(\mathbf{x}_{j}-\mathbf{\mu}_{i})^{T}\mathbf{W} \\ & = \mathbf{W}^{T}\left[\sum_{i=1}^{k}\sum_{j: y_{j}=i}(\mathbf{x}_{j}-\mathbf{\mu}_{i})(\mathbf{x}_{j}-\mathbf{\mu}_{i})^{T}\right]\mathbf{W} \\ & = \mathbf{W}^{T}\mathbf{S}_{W}\mathbf{W} \end{align}

Similarly, we obtain

\begin{align} & \mathbf{S}_{B}^{\ast} = \sum_{i=1}^{k}n_{i}(\mathbf{W}^{T}\mathbf{\mu}_{i}-\mathbf{W}^{T}\mathbf{\mu})(\mathbf{W}^{T}\mathbf{\mu}_{i}-\mathbf{W}^{T}\mathbf{\mu})^{T} \\ & = \sum_{i=1}^{k}n_{i}\mathbf{W}^{T}(\mathbf{\mu}_{i}-\mathbf{\mu})(\mathbf{\mu}_{i}-\mathbf{\mu})^{T}\mathbf{W} \\ & = \mathbf{W}^{T}\left[ \sum_{i=1}^{k}n_{i}(\mathbf{\mu}_{i}-\mathbf{\mu})(\mathbf{\mu}_{i}-\mathbf{\mu})^{T}\right]\mathbf{W} \\ & = \mathbf{W}^{T}\mathbf{S}_{B}\mathbf{W} \end{align}

Now, we use the following as our measure:

\begin{align} \sum_{i=1}^{k}n_{i}\|(\mathbf{W}^{T}\mathbf{\mu}_{i}-\mathbf{W}^{T}\mathbf{\mu})^{T}\|^{2} \end{align}

The solution for this question is that the columns of the transformation matrix $\mathbf{W}$ are exactly the eigenvectors that correspond to the largest $k-1$ eigenvalues with respect to

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\begin{align} \mathbf{S}_{W}^{-1}\mathbf{S}_{B}\mathbf{w}_{i} = \lambda_{i}\mathbf{w}_{i} \end{align}

Recall that the Frobenius norm of $X$ is

\begin{align} \|\mathbf{X}\|^2_{2} = Tr(\mathbf{X}^{T}\mathbf{X}) \end{align}

\begin{align} & \sum_{i=1}^{k}n_{i}\|(\mathbf{W}^{T}\mathbf{\mu}_{i}-\mathbf{W}^{T}\mathbf{\mu})^{T}\|^{2} \\ & = \sum_{i=1}^{k}n_{i}Tr[(\mathbf{W}^{T}\mathbf{\mu}_{i}-\mathbf{W}^{T}\mathbf{\mu})(\mathbf{W}^{T}\mathbf{\mu}_{i}-\mathbf{W}^{T}\mathbf{\mu})^{T}] \\ & = Tr[\sum_{i=1}^{k}n_{i}(\mathbf{W}^{T}\mathbf{\mu}_{i}-\mathbf{W}^{T}\mathbf{\mu})(\mathbf{W}^{T}\mathbf{\mu}_{i}-\mathbf{W}^{T}\mathbf{\mu})^{T}] \\ & = Tr[\sum_{i=1}^{k}n_{i}\mathbf{W}^{T}(\mathbf{\mu}_{i}-\mathbf{\mu})(\mathbf{\mu}_{i}-\mathbf{\mu})^{T}\mathbf{W}] \\ & = Tr[\mathbf{W}^{T}\sum_{i=1}^{k}n_{i}(\mathbf{\mu}_{i}-\mathbf{\mu})(\mathbf{\mu}_{i}-\mathbf{\mu})^{T}\mathbf{W}] \\ & = Tr[\mathbf{W}^{T}\mathbf{S}_{B}\mathbf{W}] \end{align}

Similarly, we can get $Tr[\mathbf{W}^{T}\mathbf{S}_{W}\mathbf{W}]$. Thus we have the following classic criterion function that Fisher used

\begin{align} \phi(\mathbf{W}) = \frac{Tr[\mathbf{W}^{T}\mathbf{S}_{B}\mathbf{W}]}{Tr[\mathbf{W}^{T}\mathbf{S}_{W}\mathbf{W}]} \end{align}

Similar to the two class case problem, we have:

max $Tr[\mathbf{W}^{T}\mathbf{S}_{B}\mathbf{W}]$ subject to $Tr[\mathbf{W}^{T}\mathbf{S}_{W}\mathbf{W}]=1$

To solve this optimization problem a Lagrange multiplier $\Lambda$, which actually is a $d \times d$ diagonal matrix, is introduced:

\begin{align} L(\mathbf{W},\Lambda) = Tr[\mathbf{W}^{T}\mathbf{S}_{B}\mathbf{W}] - \Lambda\left\{ Tr[\mathbf{W}^{T}\mathbf{S}_{W}\mathbf{W}] - 1 \right\} \end{align}

Differentiating with respect to $\mathbf{W}$ we obtain:

\begin{align} \frac{\partial L}{\partial \mathbf{W}} = (\mathbf{S}_{B} + \mathbf{S}_{B}^{T})\mathbf{W} - \Lambda (\mathbf{S}_{W} + \mathbf{S}_{W}^{T})\mathbf{W} \end{align}

Note that the $\mathbf{S}_{B}$ and $\mathbf{S}_{W}$ are both symmetric matrices, thus set the first derivative to zero, we obtain:

\begin{align} \mathbf{S}_{B}\mathbf{W} - \Lambda\mathbf{S}_{W}\mathbf{W}=0 \end{align}

Thus,

\begin{align} \mathbf{S}_{B}\mathbf{W} = \Lambda\mathbf{S}_{W}\mathbf{W} \end{align}

where

$\mathbf{\Lambda} = \begin{pmatrix} \lambda_{1} & & 0\\ &\ddots&\\ 0 & &\lambda_{d} \end{pmatrix}$

and $\mathbf{W} = [\mathbf{w}_{1}, \mathbf{w}_{2},..., \mathbf{w}_{k-1}]$.

As a matter of fact, $\mathbf{\Lambda}$ must have $\mathbf{k-1}$ nonzero eigenvalues, because $rank({S}_{W}^{-1}\mathbf{S}_{B})=k-1$.

Therefore, the solution for this question is as same as the previous case. The columns of the transformation matrix $\mathbf{W}$ are exactly the eigenvectors that correspond to the largest $k-1$ eigenvalues with respect to

\begin{align} \mathbf{S}_{W}^{-1}\mathbf{S}_{B}\mathbf{w}_{i} = \lambda_{i}\mathbf{w}_{i} \end{align}

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### Generalization of Fisher's Linear Discriminant Analysis

Fisher's Linear Discriminant Analysis (Fisher, 1936) is very popular among users of discriminant analysis. Some of the reasons for this are its simplicity and lack of necessity for strict assumptions. However, it has optimality properties only if the underlying distributions of the groups are multivariate normal. It is also easy to verify that the discriminant rule obtained can be very harmed by only a small number of outlying observations. Outliers are very hard to detect in multivariate data sets and even when they are detected simply discarding them is not the most efficient way of handling the situation. Therefore, there is a need for robust procedures that can accommodate the outliers and are not strongly affected by them. Then, a generalization of Fisher's linear discriminant algorithm [[11]] is developed to lead easily to a very robust procedure.

Also notice that LDA can be seen as a dimensionality reduction technique. In general k-class problems, we have k means which lie on a linear subspace with dimension k-1. Given a data point, we are looking for the closest class mean to this point. In LDA, we project the data point to the linear subspace and calculate distances within that subspace. If the dimensionality of the data, d, is much larger than the number of classes, k, then we have a considerable drop in dimensionality from d dimensions to k - 1 dimensions.

## Linear and Logistic Regression - October 12, 2010

### Linear Regression

Linear regression is an approach for modeling the response variable $\, y$ under the assumption that $\, y$ is a linear function of a set of explanatory variables $\,X$. Any observed deviation from this assumed linear relationship between $\, y$ and $\,X$ is attributed to an unobserved random variable $\, \epsilon$ that adds random noise.

In linear regression, the goal is use a set of training data $\{y_i,\, x_{i1}, \ldots, x_{id}\}, i=1, \ldots, n$ to find a linear combination $\,\beta^T = \begin{pmatrix}\beta_1 & \cdots & \beta_d & \beta_0\end{pmatrix}$ that best explains the variation in $\, y$. In $\,\beta$, $\,\beta_0$ is the intercept of the fitted line that approximates the assumed linear relationship between $\, y$ and $\,X$. $\,\beta_0$ enables this fitted line to be situated away from the origin. In classification, the goal is to classify data into groups so that group members are more similar within groups than between groups.

If the data is 2-dimensional, a model of $\, y$ as a function of $\,X$ constructed using training data under the assumption of linear regression typically looks like the one in the following figure:

The linear regression model is a very simple regression model. According to Bayes Classification we estimate the posterior probability as
$P( Y=k | X=x )= \frac{f_{k}(x)\pi_{k}}{\Sigma_{l}f_{l}(x)\pi_{l}}$

For the purpose of classification, the linear regression model assumes that the regression function $\,E(Y|X)$ is a linear combination of the inputs $\,X$.

That is, the full model under linear regression has the general form

\begin{align} y_i = \beta_1 x_{i1} + \cdots + \beta_d x_{id} + \beta_0 + \varepsilon_i = \beta^T x_i + \varepsilon_i, \qquad i = 1, \ldots, n, \end{align}

and the fitted model that can be used to estimate the response $\, y$ of any new data point has the form

\begin{align} \hat y_i = \beta_1 x_{i1} + \cdots + \beta_d x_{id} + \beta_0 = \beta^T x_i, \qquad i = 1, \ldots, n. \end{align}.

In matrix form, the full model can be expressed as

\begin{align} \mathbf{y} = \mathbf{X}^T \beta + \varepsilon \end{align}

and the fitted model can be expressed as

\begin{align} \hat \mathbf{y} = \mathbf{X}^T \beta \end{align}

Here, $\,\beta^T = \begin{pmatrix}\beta_1 & \cdots & \beta_d & \beta_0\end{pmatrix}$ is a $1 \times (d+1)$ vector and $\mathbf{X}= \begin{pmatrix} \mathbf{x}_{1} \cdots \mathbf{x}_{n}\\ 1 \cdots 1 \end{pmatrix}$ is a $(d+1) \times n$ matrix. Here, $\mathbf{x}_{i}$ is a $d \times 1$ vector.

Given the input data $\,\mathbf{x}_{1}, ..., \mathbf{x}_{n}$ and $\,y_{1}, ..., y_{n}$, our goal is to find $\,\beta^{T}$ such that the linear model fits the data while minimizing sum of squared errors using the Least Squares method. Note that vectors $\mathbf{x}_{i}$ could be numerical inputs, transformations of the original data, i.e. $\log \mathbf{x}_{i}$ or $\sin \mathbf{x}_{i}$, or basis expansions, i.e. $\mathbf{x}_{i}^{2}$ or $\mathbf{x}_{i}\times \mathbf{x}_{j}$.

To determine the values for $\,\beta^{T}$, we minimize the residual sum-of-squares

\begin{align} \mathrm{RSS}(\beta)=(\mathbf{y}-\mathbf{X}^T \beta)(\mathbf{y}-\mathbf{X}^T \beta)^{T} \end{align}

This is a quadratic function in $\,d+1$ parameters. The parameters that minimize the RSS can be determined by differentiating with respect to $\,\beta$. We then obtain

\begin{align} \frac{\partial \mathrm{RSS}}{\partial \beta} = -2\mathbf{X}(\mathbf{y}^{T}-\mathbf{X}^T \beta)^{T} \end{align}
\begin{align} \frac{\partial^{2}\mathrm{RSS}}{\partial \beta \partial \beta^{T}}=2\mathbf{X}\mathbf{X}^{T} \end{align}

Setting the first derivative to zero,

\begin{align} \mathbf{X}(\mathbf{y}-\mathbf{X}^{T}\hat{\beta})=0 \end{align}

we obtain the solution

\begin{align} \hat \beta = (\mathbf{X}\mathbf{X}^{T})^{-1}\mathbf{X}\mathbf{y} \end{align}

Thus the fitted values at the inputs are

\begin{align} \mathbf{\hat y} = \mathbf{X}^{T}\hat{\beta} = \mathbf{X}^{T}(\mathbf{X}\mathbf{X}^{T})^{-1}\mathbf{X}\mathbf{y} \end{align}

where $\mathbf{H} = \mathbf{X}^{T}(\mathbf{X}\mathbf{X}^{T})^{-1}\mathbf{X}$ is called the hat matrix.

A more efficient way to do this is by QR Factorization

$X^T = QR$ where Q is an orthonormal matrix and R is an upper triangular matrix

\begin{align} \hat{\beta} &=& ((QR){^T}(QR))^{-1}(QR)^{T}y \\ &=& ((R^{T}Q^{T}QR))^{-1}(QR)^{T}y \\ &=& (R^{T}R)^{-1}R^{T}Qy \\ &=& R^{-1}(R^{-T}R^{T})Qy \\ &=& R^{-1}Qy \end{align}

Therefore $\hat{\beta}$ can be solved for by solving $R\hat{\beta} = Qy$

• Note For classification purposes, this is not a correct model. Recall the following application of Bayes classifier:

$r(x)= P( Y=k | X=x )= \frac{f_{k}(x)\pi_{k}}{\Sigma_{l}f_{l}(x)\pi_{l}}$
It is clear that to make sense mathematically, $\displaystyle r(x)$ must be a value between 0 and 1 and must also sum up to 1. If this is estimated with the regression function $\displaystyle r(x)=E(Y|X=x)$ and $\mathbf{\hat\beta}$ is learned as above, then there is nothing that would restrict $\displaystyle r(x)$ to meet these two criteria. This is more direct approach to classification since it do not need to estimate $\ f_k(x)$ and $\ \pi_k$. $\ 1 \times P(Y=1|X=x)+0 \times P(Y=0|X=x)=E(Y|X)$. This model does not classify Y between 0 and 1, so it is not good but at times it can lead to a decent classifier. $\ y_i=\frac{1}{n_1}$ $\ \frac{-1}{n_2}$

### Logistic Regression

The logistic regression model arises from the desire to model the posterior probabilities of the $\displaystyle K$ classes via linear functions in $\displaystyle x$, while at the same time ensuring that they sum to one and remain in [0,1]. Logistic regression models are usually fit by maximum likelihood, using the conditional probabilities $\displaystyle Pr(Y|X)$. Since $\displaystyle Pr(Y|X)$ completely specifies the conditional distribution, the multinomial distribution is appropriate. This model is widely used in biostatistical applications for two classes. For instance: people survive or die, have a disease or not, have a risk factor or not.

#### logistic function

A logistic function or logistic curve is the most common of the sigmoid functions. Given below are five examples of sigmoid functions, with the first being the logistic function.

1. $y = \frac{1}{1+e^{-x}}$

2. $\frac{dy}{dx} = y(1-y)=\frac{e^{x}}{(1+e^{x})^{2}}$

3. $y(0) = \frac{1}{2}$

4. $\int y dx = ln(1 + e^{x})$

5. $y(x) = \frac{1}{2} + \frac{1}{4}x - \frac{1}{48}x^{3} + \frac{1}{48}x^{5} \cdots$

The logistic curve shows early exponential growth for negative t, which slows to linear growth of slope 1/4 near t = 0, then approaches y = 1 with an exponentially decaying gap.

An early application of the logistic function was due to Pierre-François Verhulst who, in 1838, used the logistic function to derive a logistic equation now known as the Verhulst equation to model population growth. Verhulst was inspired by Thomas Malthus's work An Essay on the Principle of Population, and his own work was published after reading Malthus' work. Independently of Verhulst, in 1925, Alfred J. Lotka again used the logistic function to derive a logistic equation to model population growth, and he referred to his equation as the law of population growth.

#### Intuition behind Logistic Regression

Recall that, for classification purposes, the linear regression model presented in the above section is not correct because it does not force $\,r(x)$ to be between 0 and 1 and also sum to 1. Consider the following log odds model (for two classes):

$\log\left(\frac{P(Y=1|X=x)}{P(Y=0|X=x)}\right)=\beta^Tx$

Calculating $\,P(Y=1|X=x)$ leads us to the logistic regression model, which as opposed to the linear regression model, allows the modelling of the posterior probabilities of the classes through linear methods and at the same time ensures that they sum to one and are between 0 and 1. It is a type of Generalized Linear Model (GLM).

#### The Logistic Regression Model

The logistic regression model for the two class case is defined as

Class 1

We have that

$P(Y=1 | X=x) =\frac{\exp(\underline{\beta}^T \underline{x})}{1+\exp(\underline{\beta}^T \underline{x})}=P(x;\underline{\beta})$

This is shown as the top figure on the right.

Class 0

We have that

$P(Y=0 | X=x) = 1-P(Y=1 | X=x)=1-\frac{\exp(\underline{\beta}^T \underline{x})}{1+\exp(\underline{\beta}^T \underline{x})}=\frac{1}{1+\exp(\underline{\beta}^T \underline{x})}$

This is shown as the bottom figure on the right.

#### Fitting a Logistic Regression

Logistic regression tries to fit a distribution. The common practice in statistics is to fit density function, posterior density of each class (Pr(Y|X), to data using maximum likelihood. The maximum likelihood estimate of $\underline\beta$, denoted $\hat \beta_{ML}$, maximizes the probability of observing the training data $\{y_i,\, x_{i1}, \ldots, x_{id}\}, i=1, \ldots, n$ from the known distribution. Combining $\displaystyle P(Y=1 | X=x)$ and $\displaystyle P(Y=0 | X=x)$ as follows, we can consider the two classes at the same time:

$p(\underline{x_{i}};\underline{\beta}) = \left(\frac{\exp(\underline{\beta}^T \underline{x_i})}{1+\exp(\underline{\beta}^T \underline{x_i})}\right)^{y_i} \left(\frac{1}{1+\exp(\underline{\beta}^T \underline{x_i})}\right)^{1-y_i}$

Assuming the data $\displaystyle {x_{1},...,x_{n}}$ is drawn independently, the likelihood function is

\begin{align} \mathcal{L}(\theta)&=p({x_{1},...,x_{n}};\theta)\\ &=\displaystyle p(x_{1};\theta) p(x_{2};\theta)... p(x_{n};\theta) \quad \mbox{(by independence and identical distribution)}\\ &= \prod_{i=1}^n p(x_{i};\theta) \end{align}

Since it is more convenient to work with the log-likelihood function, take the log of both sides, we get

$\displaystyle l(\theta)=\displaystyle \sum_{i=1}^n \log p(x_{i};\theta)$

So,

\begin{align} l(\underline\beta)&=\displaystyle\sum_{i=1}^n y_{i}\log\left(\frac{\exp(\underline{\beta}^T \underline{x_i})}{1+\exp(\underline{\beta}^T \underline{x_i})}\right)+(1-y_{i})\log\left(\frac{1}{1+\exp(\underline{\beta}^T\underline{x_i})}\right)\\ &= \displaystyle\sum_{i=1}^n y_{i}(\underline{\beta}^T\underline{x_i}-\log(1+\exp(\underline{\beta}^T\underline{x_i}))+(1-y_{i})(-\log(1+\exp(\underline{\beta}^T\underline{x_i}))\\ &= \displaystyle\sum_{i=1}^n y_{i}\underline{\beta}^T\underline{x_i}-y_{i} \log(1+\exp(\underline{\beta}^T\underline{x_i}))- \log(1+\exp(\underline{\beta}^T\underline{x_i}))+y_{i} \log(1+\exp(\underline{\beta}^T\underline{x_i}))\\ &=\displaystyle\sum_{i=1}^n y_{i}\underline{\beta}^T\underline{x_i}- \log(1+\exp(\underline{\beta}^T\underline{x_i}))\\ \end{align}

Note: The reader may find it useful to review vector derivatives before continuing.

To maximize the log-likelihood, set its derivative to 0.

\begin{align} \frac{\partial l}{\partial \underline{\beta}} &= \sum_{i=1}^n \left[{y_i} \underline{x}_i- \frac{\exp(\underline{\beta}^T \underline{x_i})}{1+\exp(\underline{\beta}^T \underline{x}_i)}\underline{x}_i\right]\\ &=\sum_{i=1}^n \left[{y_i} \underline{x}_i - p(\underline{x}_i;\underline{\beta})\underline{x}_i\right] \end{align}

There are n+1 nonlinear equations in $\beta$. The first column is a vector of 1's, and $\ \sum_{i=1}^n {y_i} =\sum_{i=1}^n p(\underline{x}_i;\underline{\beta})$ i.e. the expected number of class ones matches the observed number.

To solve this equation, the Newton-Raphson algorithm is used which requires the second derivative of the log-likelihood $\,l(\beta)$ with respect to $\,\beta$ in addition to the first derivative of $\,l(\beta)$ with respect to $\,\beta$. This is demonstrated in the next section.

#### Extension

• When we are dealing with a problem with more than two classes, we need to generalize our logistic regression to a Multinomial Logit model.
• Limitations of Logistic Regression:
1. We know that there is no assumptions made about the distributions of the features of the data (i.e. the explanatory variables). However, the features should not be highly correlated with one another because this could cause problems with estimation.
2. Large number of data points (i.e.the sample sizes) are required for logistic regression to provide sufficient estimates of the paramters in both classes. The more number of features/dimensions of the data, the larger the sample size required.
3. According to this source however, the only real limitation of logistic regression as compared to other types of regression such as linear regression is that the response variable $\,y$ can only take discrete values.

## Lecture summary

This lecture introduced logistic regression as a classification technique by using linear regression as a stepping-stone. Classification using models found by linear regression is discouraged, but linear regression provides insight into other forms of regression. However, one important difference between linear and logistic regression is that the former uses the Least-Squares technique to estimate parameters while the latter uses Maximum Likelihood Estimation for this task. Maximum Likelihood Estimation works by fitting a density function (in this case, a logistic function) that maximizes the probability of observing the training data. The lecture finishes by noting some caveats of using logistic regression.

## Logistic Regression Cont. - October 14, 2010

### Logistic Regression Model

Recall that in the last lecture, we learned the logistic regression model.

• $P(Y=1 | X=x)=P(\underline{x};\underline{\beta})=\frac{exp(\underline{\beta}^T \underline{x})}{1+exp(\underline{\beta}^T \underline{x})}$
• $P(Y=0 | X=x)=1-P(\underline{x};\underline{\beta})=\frac{1}{1+exp(\underline{\beta}^T \underline{x})}$

### Estimating Parameters $\underline{\beta}$

Criteria: find a $\underline{\beta}$ that maximizes the conditional likelihood of Y given X using the training data.

From above, we have the first derivative of the log-likelihood:

$\frac{\partial l}{\partial \underline{\beta}} = \sum_{i=1}^n \left[{y_i} \underline{x}_i- \frac{exp(\underline{\beta}^T \underline{x}_i)}{1+exp(\underline{\beta}^T \underline{x}_i)}\underline{x}_i\right]$ $=\sum_{i=1}^n \left[{y_i} \underline{x}_i - P(\underline{x}_i;\underline{\beta})\underline{x}_i\right]$

Newton-Raphson Algorithm:

If we want to find $\ x^*$ such that $\ f(x^*)=0$, we proceed by first arbitrarily picking a starting point $\,x^* = x^{old}$ and we iterate the following two steps until convergence, i.e. when $\, x^{new}$ is sufficiently close to $\, x^{old}$ using an arbitrary criterion of closeness:
Step 1: $\, x^{new} \leftarrow x^{old}-\frac {f(x^{old})}{f'(x^{old})}$

Step 2: $\, x^{old} \leftarrow x^{new}$

If $\ f'(x)=0$ , then we can replace the two steps above by the following two steps:
Step 1: $\ x^{new} \leftarrow x^{old}-\frac {f'(x^{old})}{f''(x^{old})}$

Step 2: $\ x^{old} \leftarrow x^{new}$

If we want to maximize or minimize $\ f(x)$, then we solve for the value of $\,x$ at which $\ f'(x)=0$ using the following iterative updating rule that generates $\ x^{new}$ from $\ x^{old}$:
$\ x^{new} \leftarrow x^{old}-\frac {f'(x^{old})}{f''(x^{old})}$

Using vector notation, the above rule can be written as

$X^{new} \leftarrow X^{old} - H^{-1}\nabla$
H is the Hessian matrix or second derivative matrix and $\,\nabla$ is the gradient or first derivative vector, and both H and $\,\nabla$ are evaluated at $\,X^{old}$

note: If the Hessian is not invertible the generalized inverse or pseudo inverse can be used

As shown above ,the Newton-Raphson algorithm requires the second-derivative or Hessian.

$\frac{\partial^{2} l}{\partial \underline{\beta} \partial \underline{\beta}^T }= \sum_{i=1}^n - \underline{x_i} \frac{(exp(\underline{\beta}^T\underline{x}_i) \underline{x}_i^T)(1+exp(\underline{\beta}^T \underline{x}_i))- exp(\underline{\beta}^T\underline{x}_i)\underline{x}_i^Texp(\underline{\beta}^T\underline{x}_i)}{(1+exp(\underline{\beta}^T \underline{x}_i))^2}$

(note: $\frac{\partial\underline{\beta}^T\underline{x}_i}{\partial \underline{\beta}^T}=\underline{x}_i^T$ you can check it here, it's a very useful website including a Matrix Reference Manual that you can find information about linear algebra and the properties of real and complex matrices.)

$=\sum_{i=1}^n \frac{(-\underline{x}_i exp(\underline{\beta}^T\underline{x}_i) \underline{x}_i^T)}{(1+exp(\underline{\beta}^T \underline{x}))(1+exp(\underline{\beta}^T \underline{x}))}$ (by cancellation)
$=\sum_{i=1}^n - \underline{x}_i \underline{x}_i^T P(\underline{x}_i;\underline{\beta}）)[1-P(\underline{x}_i;\underline{\beta})]）$(since $P(\underline{x}_i;\underline{\beta})=\frac{exp(\underline{\beta}^T \underline{x}_i)}{1+exp(\underline{\beta}^T \underline{x}_i)}$ and $1-P(\underline{x}_i;\underline{\beta})=\frac{1}{1+exp(\underline{\beta}^T \underline{x}_i)}$)

The same second derivative can be achieved if we reduce the occurrences of beta to 1 by the identity$\frac{a}{1+a}=1-\frac{1}{1+a}$

and then solving $\frac{\partial}{\partial \underline{\beta}^T}\sum_{i=1}^n \left[{y_i} \underline{x}_i-\left[1-\frac{1}{1+exp(\underline{\beta}^T \underline{x}_i)}\right]\underline{x}_i\right]$

In each of the iterative steps, starting with the existing $\,\underline{\beta}^{old}$ which is initialized with an arbitrarily chosen value, the Newton-Raphson updating rule for obtaining $\,\underline{\beta}^{new}$ is

$\,\underline{\beta}^{new}\leftarrow \,\underline{\beta}^{old}- (\frac{\partial ^2 l}{\partial \underline{\beta}\partial \underline{\beta}^T})^{-1}(\frac{\partial l}{\partial \underline{\beta}})$ where the derivatives are evaluated at $\,\underline{\beta}^{old}$

The iterations terminate when $\underline{\beta}^{new}$ is very close to $\underline{\beta}^{old}$ according to an arbitrarily defined criterion.

Each iteration can be described in matrix form.

• Let $\,\underline{Y}$ be the column vector of $\,y_i$. ($n\times1$)
• Let $\,X$ be the ${(d+1)}\times{n}$ input matrix.
• Let $\,\underline{P}$ be the ${n}\times{1}$ vector with $\,i$th element $P(\underline{x}_i;\underline{\beta}^{old})$.
• Let $\,W$ be an ${n}\times{n}$ diagonal matrix with $\,i,i$th element $P(\underline{x}_i;\underline{\beta}^{old})[1-P(\underline{x}_i;\underline{\beta}^{old})]$

then

$\frac{\partial l}{\partial \underline{\beta}} = X(\underline{Y}-\underline{P})$

$\frac{\partial ^2 l}{\partial \underline{\beta}\partial \underline{\beta}^T} = -XWX^T$

The Newton-Raphson step is

$\underline{\beta}^{new} \leftarrow \underline{\beta}^{old}+(XWX^T)^{-1}X(\underline{Y}-\underline{P})$

This equation is sufficient for computation of the logistic regression model. However, we can simplify further to uncover an interesting feature of this equation.

\begin{align} \underline{\beta}^{new} &= \,\underline{\beta}^{old}- (\frac{\partial ^2 l}{\partial \underline{\beta}\partial \underline{\beta}^T})^{-1}(\frac{\partial l}{\partial \underline{\beta}})\\ &= \,\underline{\beta}^{old}- (-XWX^T)^{-1}X(\underline{Y}-\underline{P})\\ &= \,(XWX^T)^{-1}(XWX^T)\underline{\beta}^{old}- (XWX^T)^{-1}(XWX^T)(-XWX^T)^{-1}X(\underline{Y}-\underline{P})\\ &= (XWX^T)^{-1}(XWX^T)\underline{\beta}^{old}+(XWX^T)^{-1}XWW^{-1}(\underline{Y}-\underline{P})\\ &=(XWX^T)^{-1}XW[X^T\underline{\beta}^{old}+W^{-1}(\underline{Y}-\underline{P})]\\ &=(XWX^T)^{-1}XWZ \end{align}

where $Z=X^{T}\underline{\beta}^{old}+W^{-1}(\underline{Y}-\underline{P})$

This is an adjusted response and it is solved repeatedly as $\, P$, $\, W$, and $\, Z$ are iteratively updated during the steps until convergence is achieved. This algorithm is called iteratively reweighted least squares because it solves the weighted least squares problem iteratively.

Recall that linear regression by least squares finds the following minimum: $\min_{\underline{\beta}}(\underline{y}-X^T \underline{\beta})^T(\underline{y}-X^T \underline{\beta})$

we have $\underline\hat{\beta}=(XX^T)^{-1}X\underline{y}$

Similarly, we can say that $\underline{\beta}^{new}$ is the solution of a weighted least square problem:

$\underline{\beta}^{new} \leftarrow arg \min_{\underline{\beta}}(Z-X^T \underline{\beta})W(Z-X^T \underline{\beta})$

#### Pseudo Code

First, initialize $\,\underline{\beta}^{old} \leftarrow 0$ and set $\,\underline{Y}$, the labels associated with the observations $\,i=1...n$. Then, in each iterative step, perform the following:

1. Compute $\,\underline{P}$ according to the equation $P(\underline{x}_i;\underline{\beta})=\frac{exp(\underline{\beta}^T \underline{x}_i)}{1+exp(\underline{\beta}^T \underline{x}_i)}$ for all $\,i=1...n$.
2. Compute the diagonal matrix $\,W$ by setting $\,W_{i,i}$ to $P(\underline{x}_i;\underline{\beta}）)[1-P(\underline{x}_i;\underline{\beta})]$ for all $\,i=1...n$.
3. Compute $Z \leftarrow X^T\underline{\beta}+W^{-1}(\underline{Y}-\underline{P})$.
4. $\underline{\beta}^{new} \leftarrow (XWX^T)^{-1}XWZ$.
5. If $\underline{\beta}^{new}$ is sufficiently close to $\underline{\beta}^{old}$ according to an arbitrarily defined criterion, then stop; otherwise, set $\,\underline{\beta}^{old} \leftarrow \underline{\beta}^{new}$ and another iterative step is made towards convergence between $\underline{\beta}^{new}$ and $\underline{\beta}^{old}$.

#### Classification

To implement classification, we should compute $\underline{\beta}^{T} x$. If $\underline{\beta}^{T} x \lt 0$, then $\, x$ belongs to class 0 , otherwise it belongs to class 1 .

### Comparison with Linear Regression

• Similarities
1. They both attempt to estimate $\,P(Y=k|X=x)$ (For logistic regression, we just mentioned about the case that $\,k=0$ or $\,k=1$ now).
2. They both have linear boundaries.
note:For linear regression, we assume the model is linear. The boundary is $P(Y=k|X=x)=\underline{\beta}^T\underline{x}_i+\underline{\beta}_0=0.5$ (linear)
For logistic regression, the boundary is $P(Y=k|X=x)=\frac{exp(\underline{\beta}^T \underline{x})}{1+exp(\underline{\beta}^T \underline{x})}=0.5 \Rightarrow exp(\underline{\beta}^T \underline{x})=1\Rightarrow \underline{\beta}^T \underline{x}=0$ (nonlinear)
• Differences

1. Linear regression: $\,P(Y=k|X=x)$ is linear function of $\,x$, $\,P(Y=k|X=x)$ is not guaranteed to fall between 0 and 1 and to sum up to 1. There exists a closed form solution for least squares.
2. Logistic regression: $\,P(Y=k|X=x)$ is a nonlinear function of $\,x$, and it is guaranteed to range from 0 to 1 and to sum up to 1. No closed form solution exists, so the Newton-Raphson algorithm is typically used to arrive at an estimate for the parameters.

### Comparison with LDA

1. The linear logistic model only consider the conditional distribution $\,P(Y=k|X=x)$. No assumption is made about $\,P(X=x)$.
2. The LDA model specifies the joint distribution of $\,X$ and $\,Y$.
3. Logistic regression maximizes the conditional likelihood of $\,Y$ given $\,X$: $\,P(Y=k|X=x)$
4. LDA maximizes the joint likelihood of $\,Y$ and $\,X$: $\,P(Y=k,X=x)$.
5. If $\,\underline{x}$ is d-dimensional,the number of adjustable parameter in logistic regression is $\,d$. The number of parameters grows linearly w.r.t dimension.
6. If $\,\underline{x}$ is d-dimensional,the number of adjustable parameter in LDA is $\,(2d)+d(d+1)/2+2=(d^2+5d+4)/2$. The number of parameters grows quadratically w.r.t dimension.
7. LDA estimate parameters more efficiently by using more information about data and samples without class labels can be also used in LDA.

Robustness:

1. Logistic regression relies on fewer assumptions, so it is generally felt to be more robust (Hastie, T., et al., 2009, p. 128). For high-dimensionality data, logistic regression is more accommodating.
2. Logistic regression is also more robust because it down-weights outliers, unlike LDA (Hastie, T., et al., 2009, p. 128).
3. In practice, Logistic regression and LDA often give similar results (Hastie, T., et al., 2009, p. 128).

Also in order to compare the results obtained by LDA, QDA and Logistic regression methods, following link can be used: http://www.cs.uwaterloo.ca/~a2curtis/courses/2005/ML-classification.pdf.

Many other advantages of logistic regression are explained here.

#### By example

Now we compare LDA and Logistic regression by an example. Again, we use them on the 2_3 data.

 >>load 2_3;
>>[U, sample] = princomp(X');
>>sample = sample(:,1:2);
>>plot (sample(1:200,1), sample(1:200,2), '.');
>>hold on;
>>plot (sample(201:400,1), sample(201:400,2), 'r.');

First, we do PCA on the data and plot the data points that represent 2 or 3 in different colors. See the previous example for more details.
 >>group = ones(400,1);
>>group(201:400) = 2;

Group the data points.
 >>[B,dev,stats] = mnrfit(sample,group);
>>x=[ones(1,400); sample'];

Now we use mnrfit to use logistic regression to classfy the data. This function can return $\underline{\beta}$ which is a $\,(d+1)$$\,\times$$\,(k-1)$ matrix of estimates, where each column corresponds to the estimated intercept term and predictor coefficients. In this case, $\underline{\beta}$ is a $3\times{1}$ matrix.
 >> B
B =0.1861
-5.5917
-3.0547

This is our $\underline{\beta}$. So the posterior probabilities are:
$P(Y=1 | X=x)=\frac{exp(0.1861-5.5917X_1-3.0547X_2)}{1+exp(0.1861-5.5917X_1-3.0547X_2)}$.
$P(Y=2 | X=x)=\frac{1}{1+exp(0.1861-5.5917X_1-3.0547X_2)}$
The classification rule is:
$\hat Y = 1$, if $\,0.1861-5.5917X_1-3.0547X_2\gt =0$
$\hat Y = 2$, if $\,0.1861-5.5917X_1-3.0547X_2\lt 0$
 >>f = sprintf('0 = %g+%g*x+%g*y', B(1), B(2), B(3));
>>ezplot(f,[min(sample(:,1)), max(sample(:,1)), min(sample(:,2)), max(sample(:,2))])

Plot the decision boundary by logistic regression.
This is a decision boundary by logistic regression.The line shows how the two classes split.
 >>[class, error, POSTERIOR, logp, coeff] = classify(sample, sample, group, 'linear');
>>k = coeff(1,2).const;
>>l = coeff(1,2).linear;
>>f = sprintf('0 = %g+%g*x+%g*y', k, l(1), l(2));
>>h=ezplot(f, [min(sample(:,1)), max(sample(:,1)), min(sample(:,2)), max(sample(:,2))]);

Plot the decision boundary by LDA. See the previous example for more information about LDA in matlab.
From this figure, we can see that the results of Logistic Regression and LDA are very similar.

### Lecture Summary

Traditionally, regression parameters are estimated using maximum likelihood. However, other optimization techniques may be used as well.
In the case of logistic regression, since there is no closed-form solution for finding zero of the first derivative of the log-likelihood function, the Newton-Raphson algorithm is typically used to estimate parameters. This problem is convex, so the Newton-Raphson algorithm is guaranteed to converge to a global optimum.
Logistic regression requires less parameters than LDA or QDA, which makes it favorable for high-dimensional data.

### Supplements

A detailed proof that logistic regression is convex is available here. See '1 Binary LR' for the case we discussed in lecture.

## Multi-Class Logistic Regression & Perceptron - October 19, 2010

### Multi-Class Logistic Regression

Recall that in two-class logistic regression, the class-conditional probability of one of the classes (say class 0) is modeled by a function in the form shown in figure 1.

The class-conditional probability of the second class (say class 1) is the complement of the first class (class 0).

$\displaystyle P(Y=0 | X=x) = 1 - P(Y=1 | X=x)$

This function is called sigmoid logistic function, which is the reason why this algorithm is called "logistic regression".

$Fig.1: P(Y=1 | X=x)$

$\displaystyle \sigma\,\!(a) = \frac {e^a}{1+e^a} = \frac {1}{1+e^{-a}}$

In two-class logistic regression, we compare the class-conditional probability of one class to the other using this ratio:

$\frac{P(Y=1|X=x)}{P(Y=0|X=x)}$

If we look at the natural logarithm of this ratio, we find that it is always a linear function in $\,x$:

$\log\left(\frac{P(Y=1|X=x)}{P(Y=0|X=x)}\right)=\underline{\beta}^T\underline{x} \quad \rightarrow (*)$

What if we have more than two classes?

Using (*), we can extend the notion of logistic regression for the cases where we have more than two classes.

Assume we have $\,k$ classes, where $\,k$ is greater than two. Putting an arbitrarily chosen class (which for simplicity we shall assume is class $\,k$) aside, and then looking at the logarithm of the ratio of the class-conditional probability of each of the other classes and the class-conditional probability of class $\,k$, we have:

$\log\left(\frac{P(Y=1|X=x)}{P(Y=k|X=x)}\right)=\underline{\beta}_1^T\underline{x}$
$\log\left(\frac{P(Y=2|X=x)}{P(Y=k|X=x)}\right)=\underline{\beta}_2^T\underline{x}$
$\vdots$
$\log\left(\frac{P(Y=k-1|X=x)}{P(Y=k|X=x)}\right)=\underline{\beta}_{k-1}^T\underline{x}$

Although the denominator in the above class-conditional probability ratios is chosen to be the class-conditional probability of the last class (class $\,k$), the choice of the denominator is arbitrary in that the class-conditional probability estimates are equivariant under this choice - Linear Methods for Classification.

Each of these functions is linear in $\,x$. However, we have different $\underline{\,\beta}_{i}$'s. We have to make sure that the densities assigned to all of the different classes sum to one.

In general, we can write:
$P(Y=c | X=x) = \frac{e^{\underline{\beta}_c^T \underline{x}}}{1+\sum_{l=1}^{k-1}e^{\underline{\beta}_l^T \underline{x}}},\quad c \in \{1,\dots,k-1\}$

$P(Y=k | X=x) = \frac{1}{1+\sum_{l=1}^{k-1}e^{\underline{\beta}_l^T \underline{x}}}$
These class-conditional probabilities clearly sum to one.

In the case of the two-classes problem, it is pretty simple to find the $\,\underline{\beta}$ parameter (the $\,\underline{\beta}$ in two-class logistic regression problems has dimension $\,(d+1)\times1$), as mentioned in previous lectures. In the multi-class case the iterative Newton method can be used, but here $\,\underline{\beta}$ is of dimension $(d+1)\times(k-1)$ and the weight matrix $W$ is a dense and non-diagonal matrix. This results in a computationally inefficient yet feasible-to-be-solved algorithm. A trick would be to re-parametrize the logistic regression problem. This is done by suitably expanding the following: the input vector $\,x$, the vector of parameters $\,\beta$, the vector of responses $\,y$, as well as the $\,\underline{P}$ vector and the $\,W$ matrix used in the Newton-Raphson updating rule. For interested readers, details regarding this re-parametrization can be found in Jia Li's "Logistic Regression" slides. Another major difference between the two-classes logistic regression and the general multi-classes logistic regression is that, unlike the former which uses the logistic sigmoid function, the latter uses the softmax function instead. Details regarding the softmax function can be found in Sargur N. Srihari's "Logistic Regression" slides. The Newton-Raphson updating rule however, remains the same as it is in the two-classes case, i.e. it is still $\underline{\beta}^{new} \leftarrow \underline{\beta}^{old}+(XWX^T)^{-1}X(\underline{Y}-\underline{P})$. This key point is also addressed in Jia Li's slides given above.

Note that logistic regression does not assume a distribution for the prior. whereas LDA assumes the prior to be Bernoulli.

### Neural Network Concept[12]

The concept of constructing an artificial neural network came from scientists who were interested in simulating the human neural network in their computers. They were trying to create computer programs that could learn like people. A neural network is a method in artificial intelligence, and it is a simplified model of neural processing in the brain, even though the relation between this model and brain biological architecture is not yet clear.

### Perceptron

Perceptron was invented in 1957 by Frank Rosenblatt. It is the basic building block of Feed-Forward neural networks. The perceptron quickly became very popular after it was introduced, because it was shown to be able to solve many classes of useful problems. However, in 1969, Marvin Minsky and Seymour Papert published their book Perceptrons (1969) in which the authors strongly criticized the perceptron regarding its inability of solving simple exclusive-or (XOR) problems, which are not linearly separable. Indeed, the simple perceptron and the single hidden-layer perceptron neural network are not able to solve any problem that is not linearly-separable. However, it was known to the authors of this book that the multi-layer perceptron neural network can in fact solve any type of problem, including ones that are not linearly separable such as exclusive-or problems, although no efficient learning algorithm was available at that time for this type of neural network. Because of the book Perceptrons, interest regarding perceptrons and neural networks in general greatly declined to a much lower point as compared to before this book was published and things stayed that way until 1986 when the back-propagation learning algorithm (which is discussed in detail below) for neural networks was popularized.

We know that the least-squares obtained by regression of -1/1 response variable $\displaystyle Y$ on observation $\displaystyle x$ leads to the same coefficients as LDA (recall that LDA minimizes the distance between discriminant function (decision boundary) and the data points). Least squares returns the sign of the linear combination of features as the class labels (Figure 2). This concept was called the Perceptron in Engineering literature during the 1950's.

Fig.2 Diagram of a linear perceptron

There is a cost function $\,\displaystyle D$ that the Perceptron tries to minimize:

$D(\underline{\beta},\beta_0)=-\sum_{i \in M}y_{i}(\underline{\beta}^T \underline{x_{i}}+\beta_0)$

where $\,\displaystyle M$ is the set of misclassified points.

By minimizing D, we minimize the sum of the distances between the misclassified points and the decision boundary.

Derivation: The distances between the misclassified points and the decision boundary.

Consider points $\underline{x_1}$, $\underline{x_2}$ and a decision boundary defined as $\underline{\beta}^T\underline{x}+\beta_0$ as shown in Figure 3.

Fig.3 Distance from the decision boundary

Both $\underline{x_1}$ and $\underline{x_2}$ lie on the decision boundary, thus:
$\underline{\beta}^T\underline{x_1}+\beta_0=0 \rightarrow (1)$
$\underline{\beta}^T\underline{x_2}+\beta_0=0 \rightarrow (2)$

Consider (2) - (1):
$\underline{\beta}^T(\underline{x_2}-\underline{x_1})=0$

We see that $\,\displaystyle \underline{\beta}$ is orthogonal to $\underline{x_2}-\underline{x_1}$, which is in the same direction with the decision boundary, which means that $\,\displaystyle \underline{\beta}$ is orthogonal to the decision boundary.

Then the distance of a point $\,\underline{x_0}$ from the decision boundary is:

$\underline{\beta}^T(\underline{x_0}-\underline{x_2})$

From (2):

$\underline{\beta}^T\underline{x_2}= -\beta_0$.
$\underline{\beta}^T(\underline{x_0}-\underline{x_2})=\underline{\beta}^T\underline{x_0}-\underline{\beta}^T\underline{x_2}=\underline{\beta}^T\underline{x_0}+\beta_0$

Therefore, distance between any point $\underline{x_{i}}$ to the discriminant hyperplane is defined by $\underline{\beta}^T\underline{x_{i}}+\beta_0$.

However, this quantity is not always positive. Consider $\,y_{i}(\underline{\beta}^T \underline{x_{i}}+\beta_0)$. If $\underline{x_{i}}$ is classified correctly then this product is positive, since both ($\underline{\beta}^T\underline{x_{i}}+\beta_0)$ and $\displaystyle y_{i}$ are positive or both are negative. However, if $\underline{x_{i}}$ is classified incorrectly, then one of $(\underline{\beta}^T\underline{x_{i}}+\beta_0)$ and $\displaystyle y_{i}$ is positive and the other one is negative; hence, the product $y_{i}(\underline{\beta}^T \underline{x_{i}}+\beta_0)$ will be negative for a misclassified point. The "-" sign in $D(\underline{\beta},\beta_0)$ makes this cost function always positive (since only misclassified points are passed to D).

## Perceptron Learning Algorithm and Feed Forward Neural Networks - October 21, 2010

### Lecture Summary

In this lecture, we finalize our discussion of the Perceptron by reviewing its learning algorithm, which is based on gradient descent. We then begin the next topic, Neural Networks (NN), and we focus on a NN that is useful for classification: the Feed Forward Neural Network (FFNN). The mathematical model for the FFNN is shown, and we review one of its most popular learning algorithms: Back-Propagation.

To open the Neural Network discussion, we present a formulation of the universal function approximator. The mathematical model for Neural Networks is then built upon this formulation. We also discuss the trade-off between training error and testing error -- known as the generalization problem -- under the universal function approximator section.

### Perceptron

The last lecture introduced the Perceptron and showed how it can suggest a solution for the 2-class classification problem. We saw that the solution requires minimization of a cost function, which is basically a summation of the distances of the misclassified data points to the separating hyperplane. This cost function is

$D(\underline{\beta},\beta_0)=-\sum_{i \in M}y_{i}(\underline{\beta}^T \underline{x}_i+\beta_0),$

in which, $\,M$ is the set of misclassified points. Thus, the objective is to find $\arg\min_{\underline{\beta},\beta_0} D(\underline{\beta},\beta_0)$.

#### Perceptron Learning Algorithm

To minimize $D(\underline{\beta},\beta_0)$, an algorithm that uses gradient-descent has been suggested. Gradient descent, also known as steepest descent, is a numerical optimization technique that starts from an initial value for $(\underline{\beta},\beta_0)$ and recursively approaches an optimal solution. Each step of recursion updates $(\underline{\beta},\beta_0)$ by subtracting from it a factor of the gradient of $D(\underline{\beta},\beta_0)$. Mathematically, this gradient is

$\nabla D(\underline{\beta},\beta_0) = \left( \begin{array}{c}\cfrac{\partial D}{\partial \underline{\beta}} \\ \\ \cfrac{\partial D}{\partial \beta_0} \end{array} \right) = \left( \begin{array}{c} -\displaystyle\sum_{i \in M}y_{i}\underline{x}_i^T \\ -\displaystyle\sum_{i \in M}y_{i} \end{array} \right)$

However, the perceptron learning algorithm does not use the sum of the contributions from all of the observations to calculate the gradient in each step. Instead, each step uses the gradient contribution from only a single observation, and each successive step uses a different observation. This slight modification is called stochastic gradient descent. That is, instead of subtracting some factor of $\nabla D(\underline{\beta},\beta_0)$ at each step, we subtract a factor of

$\left( \begin{array}{c} y_{i}\underline{x}_i^T \\ y_{i} \end{array} \right)$

As a result, the pseudo code for the Perceptron Learning Algorithm is as follows:

1) Choose a random initial value $\begin{pmatrix} \underline{\beta}^0\\ \beta_0^0 \end{pmatrix}$ for $(\underline{\beta},\beta_0)$.
2) $\begin{pmatrix} \underline{\beta}^{\mathrm{old}}\\ \beta_0^{\mathrm{old}} \end{pmatrix} \leftarrow \begin{pmatrix} \underline{\beta}^0\\ \beta_0^0 \end{pmatrix}$
3) $\begin{pmatrix} \underline{\beta}^{\mathrm{new}}\\ \underline{\beta_0}^{\mathrm{new}} \end{pmatrix} \leftarrow \begin{pmatrix} \underline{\beta}^{\mathrm{old}}\\ \underline{\beta_0}^{\mathrm{old}} \end{pmatrix} +\rho \begin{pmatrix} y_i \underline{x_i^T}\\ y_i \end{pmatrix}$ for some $\,i \in M$.
4) If the termination criterion has not been met, go back to step 3 and use a different observation datapoint (i.e. a different $\,i$).

The learning rate $\,\rho$ controls the step size of convergence toward $\min_{\underline{\beta},\beta_0} D(\underline{\beta},\beta_0)$. A larger value for $\,\rho$ causes the steps to be larger. If $\,\rho$ is set to be too large, however, then the minimum could be missed (over-stepped). In practice, $\,\rho$ can be adaptive and not fixed, it means that, in the first steps $\,\rho$ could be larger than the last steps, with $\,\rho$ gradually declining in size as the steps progress towards convergence. At the beginning, larger $\,\rho$ helps to find the approximate answer sooner. And smaller $\,\rho$ towards the last steps help to tune the final answer more accurately. Many works have been done relating to adaptive learning rates. For interested readers, an example of these works is this paper by Plagianakos et al. and this paper by Schraudolph.

As mentioned earlier, the learning algorithm uses just one of the data points at each iteration; this is the common practice when dealing with online applications. In an online application, datapoints are accessed one-at-a-time because training data is not available in batch form. The learning algorithm does not require the derivative of the cost function with respect to the previously seen points; instead, we just have to take into consideration the effect of each new point.

One way that the algorithm could terminate is if there are no more mis-classified points (i.e. if set $\,M$ is empty). Another way that the algorithm could terminate is continuing until some other termination criterion is reached even if there are still points in $\,M$. The termination criterion for an optimization algorithm is usually convergence, but for numerical methods this is not well-defined. In theory, convergence is realized when the gradient of the cost function is zero; in numerical methods an answer close to zero within some margin of error is taken instead.

Since the data is linearly-separable, the solution is theoretically guaranteed to converge in a finite number of iterations. This number of iterations depends on the

• learning rate $\,\rho$
• initial value $\begin{pmatrix} \underline{\beta}^0\\ \beta_0^0 \end{pmatrix}$
• difficulty of the problem. The problem is more difficult if the gap between the classes of data is very small.

Note that we consider the offset term $\,\beta_0$ separately from $\underline{\beta}$ to distinguish this formulation from those in which the direction of the hyperplane ($\underline{\beta}$) has been considered.

A major concern about gradient descent is that it may get trapped in local optimal solutions. Many works such as this paper by Cetin et al. and this paper by Atakulreka et al. have been done to tackle this issue.

#### Some notes on the Perceptron Learning Algorithm

• If there is access to the training data points in a batch form, we should better take advantage of a closed optimization technique like least-squares or maximum-likelihood estimation for linear classifiers. (These closed solutions has been around many years before invention of the Perceptron).
• Just like the linear classifier, a Perceptron can discriminate between only two classes at a time, and one can generalize its performance for multi-class problems by using one of the $k-1$, $k$, or $k(k-1)/2$-hyperplane methods.
• If the two classes are linearly separable, the algorithm will converge in a finite number of iterations to a hyperplane, which makes the error of training data zero. The convergence is guaranteed if the learning rate is set adequately.
• If the two classes are not linearly separable, the algorithm will never converge. So, one may think of a termination criterion in these cases. (e.g. a maximum number of iterations in which convergence is expected, or the rate of changes in both a cost function and its derivative).
• In the case of linearly separable classes, the final solution and number of iterations will be dependent on the initial values (which are arbitrarily chosen), the learning rate (for example, fixed or adaptive), and the gap between the two classes. In general, a smaller gap between classes requires a greater number of iterations for the algorithm to converge.
• Learning rate --or updating step-- has a direct impact on both the number of iterations and the accuracy of the solution for the optimization problem. Smaller quantities of this factor make convergence slower, even though we will end up with a more accurate solution. In the opposite way, larger values of the learning rate make the process faster, even though we may lose some precision. So, one may make a balance for this trade-off in order to get fast enough to an accurate enough solution (exploration vs. exploitation). In addition, an adaptive learning rate that starts off with a large value and then gradually decreases to a small value over the steps toward convergence can be used in place of a fixed learning rate.

In the upcoming lectures, we introduce the Support Vector Machines (SVM), which uses a method similar to the iteration optimization scheme to what the Perceptron suggests, but have a different definition for the cost function.

### Universal Function Approximator

In mathematics, the Universal Approximation Theorem states that the standard multilayer feed-forward neural network with a single hidden layer that contains a finite and sufficient number of hidden neurons and having an arbitrary activation function for each neuron is a universal approximator on a compact subset of $\mathbb{R}^n$ under the assumption that the output units are always linear. George Cybenko first proved this theorem in 1989 for a sigmoid activation function, and thus the Universal Approximation Theorem is also called Cybenko's Theorem. For interested readers, a detailed proof of Cybenko's Theorem is given in this presentation by Yousef Shajrawi and Fadi Abboud. In 1991, Kurt Hornik showed that the potential of a particular neural network of being a universal approximator does not depend on the specific choice of the activation function used by the neurons, rather it depends on the multilayer feedforward architecture itself that is used by that neural network.

The universal function approximator is a mathematical formulation for a group of estimation techniques. The usual formulation for it is

$\hat{Y}(x)=\sum\limits_{i=1}^{n}\alpha_i\sigma(\omega_i^Tx+b_i),$

where $\hat{Y}(x)$ is an estimation for a function $\,Y(x)$. According to the universal approximation theorem we have

$|\hat{Y}(x) - Y(x)|\lt \epsilon,$

which means that $\hat{Y}(x)$ can get as close to $\,Y(x)$ as necessary.

This formulation assumes that the output, $\,Y(x)$, is a linear combination of a set of functions like $\,\sigma(.)$ where $\,\sigma(.)$ is a nonlinear function of the inputs or $\,x_i$'s.

#### Generalization Factors

Even though this formulation represents a universal function approximator, which means that it can be fitted to a set of data as closely as demanded, the closeness of fit must be carefully decided upon. In many cases, the purpose of the model is to target unseen data. However, the fit to this unseen data is impossible to determine before it arrives.

To overcome this dilemma, a common practice is to divide the set of available data points into two sets: training data and validation (test) data. We use the training data to estimate the fixed parameters for the model, and then use the validation data to find values for the construction-dependent parameters. How these construction-dependent parameters vary depends on the model. In the case of a polynomial, the construction-dependent parameter would be its highest degree, and for a neural network, the construction-dependent parameter could be the number of hidden layers and the number of neurons in each layer.

These matters on model generalization vs. complexity matters will be discussed with more detail in the lectures to follow.

### Feed-Forward Neural Network

Neural Network (NN) is one instance for the universal function approximator. It can be thought of as a system of Perceptrons linked together as units of a network. One particular NN useful for classification is the Feed-Forward Neural Network (FFNN), which consists of multiple "hidden layers" of Perceptron units (also known as neurons). Our discussion here is based around the FFNN, which has a toplogy shown in Figure 1. The neurons in the first hidden layer take their inputs, the original features (the $\,x_i$'s), and pass their inputs unchanged as their outputs to the first hidden layer. From the first layer (the input layer) to the last hidden layer, connections from each neuron are always directed to the neurons in the next adjacent layer. In the output layer, which receives input only from the last hidden layer, each neuron produces a target measurement for a distinct class. $\,K$ classes typically require $\,K$ output neurons in the output layer. In the case where the target variable has two values, it suffices to have one output node in the output layer, although it is generally necessary for the single output node to have a sigmoid activation function so as to restrict the output of the neural network to be a value between 0 and 1. As shown in Figure 1, the neurons in a single layer are typically distributed vertically, and the inputs and outputs of the network are shown as the far left layer and the far right layer, respectively. Furthermore, as shown in Figure 1, it is often useful to add an extra hidden node to each hidden layer that represents the bias term (or the intercept term) of that hidden layer's hyperplane. Each bias node usually outputs a constant value of -1. The purpose of adding a bias node to each hidden layer is to ensure that the hyperplane of that hidden layer does not necessarily have to pass through the origin. In Figure 1, the bias node in the single hidden layer is the topmost hidden node in that layer.

Fig.1 A common architecture for the FFNN

#### Mathematical Model of the FFNN with One Hidden Layer

The FFNN with one hidden layer for a $\,K$-class problem is defined as follows:
Let $\,d$ be the number of input features, $\,p$ be the number of neurons in the hidden layer, and $\,K$ be the number of classes which is also typically the number of neurons in the output layer in the case where $\,K$ is greater than 2.

Each neuron calculates its derived feature (i.e. output) using a linear combination of its inputs. Suppose $\,\underline{x}$ is the $\,d$-dimensional vector of input features. Then, each hidden neuron uses a $\,d$-dimensional vector of weights to combine these input features. For the $\,i$th hidden neuron, let $\underline{u}_i$ be this neuron's vector of weights. The linear combination calculated by the $\,i$th hidden neuron is then given by

$a_i = \sum_{j=1}^{d}\underline{u}_{ij}^T\underline{x}_j, i={1,...,p}$

However, we want the derived feature of each hidden neuron and each output neuron to lie between 0 and 1, so we apply an activation function $\,\sigma(a)$ to each hidden or output neuron. The derived feature of each hidden or output neuron $\,i$ is then given by

$\,z_i = \sigma(a_i)$ where $\,\sigma$ is typically the logistic sigmoid function $\sigma(a) = \cfrac{1}{1+e^{-a}}$.

Now, we place each of the derived features $\,z_i$ from the hidden layer into a $\,p$-dimensional vector:

$\underline{z} = \left[ \begin{array}{c} z_1 \\ z_2 \\ \vdots \\ z_p \end{array}\right]$

As in the hidden layer, each neuron in the output layer calculates its derived feature using a linear combination of its inputs which are the elements of $\underline{z}$. Each output neuron uses a $\,p$-dimensional vector of weights to combine its inputs derived from the hidden layer. Let $\,\underline{w}_k$ be the vector of weights used by the $\,k$th output neuron. The linear combination calculated by the $\,k$th output neuron is then given by $\hat{y}_k = \sum_{j=1}^{p}\underline{w}_{kj}^T\underline{z}_j, k={1,...,K}$.

$\,\hat y_k$ is thus the target measurement for the $\,k$th class. It is not necessary to use an activation function $\,\sigma$ for each of the hidden and output neurons in the case of regression since the outputs are continuous, though it is necessary to use an activation function $\,\sigma$ for each of the hidden and output neurons in the case of classification so as to ensure that the outputs are discrete.

Notice that in each neuron, two operations take place one after the other:

• a linear combination of the neuron's inputs is calculated using corresponding weights
• a nonlinear operation on the linear combination is performed.

These two calculations are shown in Figure 2.

The nonlinear function $\,\sigma(.)$ is called the activation function. Activation functions, like the logistic function shown earlier, are usually continuous and usually have a finite range with regard to their outputs. Another common activation function used in neural networks is the hyperbolic tangent function $\,\sigma(a) = tanh(a)$ (Figure 3). The logistic sigmoid activation function $\sigma(a) = \cfrac{1}{1+e^{-a}}$ and the hyperbolic tangent activation function are very similar to each other. One major difference between them is that, as shown in their illustrations, the output range of the the logistic sigmoid activation function is $\,[0,1]$ while that of the hyperbolic tangent activation function is $\,[-1,1]$. Typically, in a neural network used for classification tasks, the logistic sigmoid activation function is used rather than any other type of activation function. The reason is that, as explained in detail in this paper by Helmbold et al., the logistic sigmoid activation function results in the least matching loss as compared to other types of activation functions.

File:neuron2.png
Fig.2 A general construction for a single neuron
File:actfcn.png
Fig.3 $tanh$ as activation function

The NN can be applied as a regression method or as a classifier, and the output layer differs depending on the application. The major difference between regression and classification is in the output space of the model, which is continuous in the case of regression and discrete in the case of classification. For a regression task, no consideration is needed beyond what has already been mentioned earlier, since the outputs of the network would already be continuous. However, to use the neural network as a classifier, as mentioned above, it is necessary to have a threshold stage for each of the hidden and output neurons using an activation function.

#### Mathematical Model of the FFNN with Multiple Hidden Layers

In the FFNN model with a single hidden layer, the derived features were represented as elements of the vector $\underline{z}$, and the original features were represented as elements of the vector $\underline{x}$. In the FFNN model with more than one hidden layer, $\underline{z}$ is processed by the second hidden layer in the same way that $\underline{x}$ was processed by the first hidden layer. Perceptrons in the second hidden layer each use their own combination of weights to calculate a new set of derived features. These new derived features are processed by the third hidden layer in a similar way, and the cycle repeats for each additional hidden layer. This progression of processing is depicted in Figure 4.

#### Back-Propagation Learning Algorithm

File:bpl.png
Fig.4 Labels for weights and derived features in the FFNN.

Every linear-combination calculation in the FFNN involves weights that need to be updated after they are initialized to be small random values, and these weights are updated using an algorithm called Back-Propagation when each data point in the training data-set is fed into the neural network. This algorithm is similar to the gradient-descent algorithm introduced in the discussion of the Perceptron. The primary difference is that the gradient used in Back-Propagation is calculated in a more complicated way.

First of all, we want to minimize the error between the estimated target measurement and the true target measurement of each input from the training data-set. That is, if $\,U$ is the set of all weights in the FFNN, then we want to determine

$\arg\min_U \left|y - \hat{y}\right|^2$ for each data point in the training data-set.

Now, suppose the hidden layers of the FFNN are labelled as in Figure 4. Then, we want to determine the derivative of $\left|y - \hat{y}\right|^2$ with respect to each weight in the hidden layers of the FFNN. For weights $\,u_{jl}$ this means we will need to find

$\cfrac{\partial \left|y - \hat{y}\right|^2}{\partial u_{jl}} = \cfrac{\partial \left|y - \hat{y}\right|^2}{\partial a_j}\cdot \cfrac{\partial a_j}{\partial u_{jl}} = \delta_{j}z_l$

However, the closed-form solution for $\,\delta_{j}$ is unknown, so we develop a recursive definition ($\,\delta_{j}$ in terms of $\,\delta_{i}$):

$\delta_j = \cfrac{\partial \left|y - \hat{y}\right|^2}{\partial a_j} = \sum_{i=1}^p \cfrac{\partial \left|y - \hat{y}\right|^2}{\partial a_i}\cdot \cfrac{\partial a_i}{\partial a_j} = \sum_{i=1}^p \delta_i\cdot u_{ij} \cdot \sigma'(a_j) = \sigma'(a_j)\sum_{i=1}^p \delta_i \cdot u_{ij}$

We also need to determine the derivative of $\left|y - \hat{y}\right|^2$ with respect to each weight in the output layer $\,k$ of the FFNN (this layer is not shown in Figure 4, but it would be the next layer to the right of the rightmost layer shown). For weights $\,u_{ki}$ this means

$\cfrac{\partial \left|y - \hat{y}\right|^2}{\partial u_{ki}} = \cfrac{\partial \left|y - \sum_i u_{ki}z_i\right|^2}{\partial u_{ki}} = -2(y - \sum_i u_{ki}z_i)z_i = -2(y - \hat{y})z_i$

With similarity to our computation of $\,\delta_j$, we define

$\delta_k = \cfrac{\partial \left|y - \hat{y}\right|^2}{\partial a_k}$

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However, $\,a_k = \hat{y}$ because an activation function is not applied in the output layer. So, our calculation becomes

$\delta_k = \cfrac{\partial \left|y - \hat{y}\right|^2}{\partial \hat{y}} = -2(y - \hat{y})$

Now that we have $\,\delta_k$ and a recursive definition for $\,\delta_j$, it is clear that our weights can be deduced by starting from the output layer and working leftwards through the hidden layers one layer at a time towards the input layer.

Based on the above derivation, our algorithm for determining weights in the FFNN is as follows:

First, choose small random values to initialize the network weights. Then, during each epoch (a single pass through all of the training data points), all of the training data points are sequentially fed into the FFNN one at a time. The network weights are updated using the back-propagation algorithm when each training data point $\underline{x}$is fed into the FFNN. This update procedure is done using the following steps:

• Apply $\underline{x}$ to the FFNN's input layer, and calculate the outputs of all input neurons.

• Propagate $\underline{x}$ forward through the hidden layers one layer at a time, and calculate the outputs of all hidden neurons.

• Once $\underline{x}$ reaches the output layer, calculate the output(s) of all output neuron(s).

• At the output layer, compute $\,\delta_k = -2(y_k - \hat{y}_k)$ for each output neuron(s), then compute $\cfrac{\partial \left|y - \hat{y}\right|^2}{\partial u_{jl}} = \delta_{j}z_l$ for all weights $\,u_{jl}$, and then update $u_{jl}^{\mathrm{new}} \leftarrow u_{jl}^{\mathrm{old}} - \rho \cdot \cfrac{\partial \left|y - \hat{y}\right|^2}{\partial u_{jl}}$ for all weights $\,u_{jl}$. Here, $\,\rho$ is the learning rate.

• Starting from the last hidden layer, back-propagate layer-by-layer to the first hidden layer. At each hidden layer, compute $\delta_j = \sigma'(a_j)\sum_{i=1}^p \delta_i \cdot u_{ij}$ for all hidden neurons in that layer, then compute $\cfrac{\partial \left|y - \hat{y}\right|^2}{\partial u_{jl}} = \delta_{j}z_l$ for all weights $\,u_{jl}$, and then update $u_{jl}^{\mathrm{new}} \leftarrow u_{jl}^{\mathrm{old}} - \rho \cdot \cfrac{\partial \left|y - \hat{y}\right|^2}{\partial u_{jl}}$ for all weights $\,u_{jl}$. Here, $\,\rho$ is the learning rate.

Usually, a fairly large number of epochs is necessary for training the FFNN so that the network weights would be close to being their optimal values. The learning rate $\,\rho$ should be chosen carefully. Usually, $\,\rho$ should satisfy $\,\rho \rightarrow 0$ as the iteration times $i \rightarrow \infty$. This is an interesting video depicting the training procedure of the weights of an FFNN using the back-propagation algorithm.

#### Alternative Description of the Back-Propagation Algorithm

Label the inputs and outputs of the $\,i$th hidden layer $\underline{x}_i$ and $\underline{y}_i$ respectively, and let $\,\sigma(.)$ be the activation function for all neurons. We now have

\begin{align} \begin{cases} \underline{y}_1=\sigma(W_1.\underline{x}_1),\\ \underline{y}_2=\sigma(W_2.\underline{x}_2),\\ \underline{y}_3=\sigma(W_3.\underline{x}_3), \end{cases} \end{align}

Where $\,W_i$ is a matrix of the connection's weights, between two layers of $\,i$ and $\,i+1$, and has $\,n_i$ columns and $\,n_i+1$ rows, where $\,n_i$ is the number of neurons of the $\,i^{th}$ layer.

Considering this matrix equations, one can imagine a closed form for the derivative of the error with respect to the weights of the network. For a neural network with two hidden layers, the equations are as follows:

\begin{align} \frac{\partial E}{\partial W_3}=&diag(e).\sigma'(W_3.\underline{x}_3).(\underline{x}_3)^T,\\ \frac{\partial E}{\partial W_2}=&\sigma'(W_2.\underline{x}_2).(\underline{x}_2)^T.diag\{\sum rows\{diag(e).diag(\sigma'(W_3.\underline{x}_3)).W_3\}\},\\ \frac{\partial E}{\partial W_1}=&\sigma'(W_1.\underline{x}_1).(\underline{x}_1)^T.diag\{\sum rows\{diag(e).diag(\sigma'(W_3.\underline{x}_3)).W_3.diag(\sigma'(W_2.\underline{x}_2)).W_2\}\}, \end{align}

where $\,\sigma'(.)$ is the derivative of the activation function $\,\sigma(.)$.

Using this closed form derivative, it is possible to code the procedure for any number of layers and neurons. Given below is the Matlab code for the back-propagation algorithm ($\,tanh$ is utilized as the activation function).

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% This code might be used to train a neural network, using backpropagation algorithm
% ep: maximum number of epochs
% io: matrix of all the inputs and outputs of the network's layers, given the weights matrix, w.
% w: w is the weights matrix
% gp: is the derivatives matrix
% shuffle: a function for changing the permutation of the data
%
while i < ep
i = i + 1;
data = shuffle(data,2);
for j = 1:Q
io = zeros(max(n)+1,length(n));
gp = io;
io(1:n(1)+1,1) = [1;data(1:f,j)];
for k = 1:l
io(2:n(k+1)+1,k+1) = w(2:n(k+1)+1,1:n(k)+1,k)*io(1:n(k)+1,k);
gp(1:n(k+1)+1,k) = [0;1./(cosh(io(2:n(k+1)+1,k+1))).^2];
io(1:n(k+1)+1,k+1) = [1;tanh(io(2:n(k+1)+1,k+1))];
wg(1:n(k+1)+1,1:n(k)+1,k) = diag(gp(1:n(k+1)+1,k))*w(1:n(k+1)+1,1:n(k)+1,k);
end
e = [0;io(2:n(l+1)+1,l+1) - data(f+1:dd,j)];
wg(1:n(l+1)+1,1:n(l)+1,l) = diag(e)*wg(1:n(l+1)+1,1:n(l)+1,l);
gp(1:n(l+1)+1,l) = diag(e)*gp(1:n(l+1)+1,l);
d = eye(n(l+1)+1);
E(i) = E(i) + 0.5*norm(e)^2;
for k = l:-1:1
w(1:n(k+1)+1,1:n(k)+1,k) = w(1:n(k+1)+1,1:n(k)+1,k) - ro*diag(sum(d,1))*gp(1:n(k+1)+1,k)*(io(1:n(k)+1,k)');
d = d*wg(1:n(k+1)+1,1:n(k)+1,k);
end
end
end


#### Some notes on the neural network and its learning algorithm

• The activation functions are usually linear around the origin. If this is the case, choosing random weights between the $\,-0.5$ and $\,0.5$, and normalizing the data may boost up the algorithm in the very first steps of the procedure, as the linear combination of the inputs and weights falls within the linear area of the activation function.
• Learning of the neural network using backpropagation algorithm takes place in epochs. An Epoch is a single pass through the entire training set.
• It is a common practice to randomly change the permutation of the training data in each one of the epochs, to make the learning independent of the data permutation.
• Given a set of data for training a neural network, one should keep aside a ratio of it as the validation dataset, to obtain a sufficient number of layers and number of neurons in each of the layers. The best construction may be the one which leads to the least error for the validation dataset. Validation data may not be used as the training data of the network (refer to cross-validation and k-fold validation explained in the next lecture).
• We can also use the validation-training scheme to estimate how many epochs is enough for training the network.
• It is also common to use other optimization algorithms as steepest descent and conjugate gradient in a batch form.

### Deep Neural Network

Back-propagation in practice may not work well when there are too many hidden layers, since the $\,\delta$ may become negligible and the errors vanish. This is a numerical problem, where it is difficult to estimate the errors. So in practice configuring a Neural Network with Back-propagation faces some subtleties.

Deep Neural Networks became popular two or three years ago, when introduced by Dr. Geoffrey E. Hinton. Deep Neural Network training algorithm deals with the training of a Neural Network with a large number of layers.

The approach of training the deep network is to assume the network has only two layers first and train these two layers. After that we train the next two layers, so on and so forth.

Although we know the input and we expect a particular output, we do not know the correct output of the hidden layers, and this will be the issue that the algorithm mainly deals with. There are two major techniques to resolve this problem: using Boltzman machine to minimize the energy function, which is inspired from the theory in atom physics concerning the most stable condition; or somehow finding out what output of the second layer is most likely to lead us to the expected output at the output layer.

#### Difficulties of training deep architecture <ref>H. Larochelle, Y. Bengio, J. Louradour, P. Lamblin, Exploring Strategies for Training Deep Neural Networks [13], year = 2009, Journal of Machine Learning Research, vol. 10, pp 1-40. </ref>

Given a particular task, a natural way to train a deep network is to frame it as an optimization problem by specifying a supervised cost function on the output layer with respect to the desired target and use a gradient-based optimization algorithm in order to adjust the weights and biases of the network so that its output has low cost on samples in the training set. Unfortunately, deep networks trained in that manner have generally been found to perform worse than neural networks with one or two hidden layers.

We discuss two hypotheses that may explain this difficulty. The first one is that gradient descent can easily get stuck in poor local minima (Auer et al., 1996) or plateaus of the non-convex training criterion. The number and quality of these local minima and plateaus (Fukumizu and Amari, 2000) clearly also influence the chances for random initialization to be in the basin of attraction (via gradient descent) of a poor solution. It may be that with more layers, the number or the width of such poor basins increases. To reduce the difficulty, it has been suggested to train a neural network in a constructive manner in order to divide the hard optimization problem into several greedy but simpler ones, either by adding one neuron (e.g., see Fahlman and Lebiere, 1990) or one layer (e.g., see Lengell´e and Denoeux, 1996) at a time. These two approaches have demonstrated to be very effective for learning particularly complex functions, such as a very non-linear classification problem in 2 dimensions. However, these are exceptionally hard problems, and for learning tasks usually found in practice, this approach commonly overfits.

This observation leads to a second hypothesis. For high capacity and highly flexible deep networks, there actually exists many basins of attraction in its parameter space (i.e., yielding different solutions with gradient descent) that can give low training error but that can have very different generalization errors. So even when gradient descent is able to find a (possibly local) good minimum in terms of training error, there are no guarantees that the associated parameter configuration will provide good generalization. Of course, model selection (e.g., by cross-validation) will partly correct this issue, but if the number of good generalization configurations is very small in comparison to good training configurations, as seems to be the case in practice, then it is likely that the training procedure will not find any of them. But, as we will see in this paper, it appears that the type of unsupervised initialization discussed here can help to select basins of attraction (for the supervised fine-tuning optimization phase) from which learning good solutions is easier both from the point of view of the training set and of a test set.

### Neural Networks in Practice

Now that we know so much about Neural Networks, what are suitable real world applications? Neural Networks have already been successfully applied in many industries.

Since neural networks are good at identifying patterns or trends in data, they are well suited for prediction or forecasting needs, such as customer research, sales forecasting, risk management and so on.

Take a specific marketing case for example. A feedforward neural network was trained using back-propagation to assist the marketing control of airline seat allocations. The neural approach was adaptive to the rule. The system is used to monitor and recommend booking advice for each departure.

Neural networks have been applied to almost every field that one can think of. For the interested reader, a detailed description with links that discusses some of the many application of neural networks is available here.

### Issues with Neural Network

When Neural Networks was first introduced they were thought to be modeling human brains, hence they were given the fancy name "Neural Network". But now we know that they are just logistic regression layers on top of each other but have nothing to do with the real function principle in the brain.

We do not know why deep networks turn out to work quite well in practice. Some people claim that they mimic the human brains, but this is unfounded. As a result of these kinds of claims it is important to keep the right perspective on what this field of study is trying to accomplish. For example, the goal of machine learning may be to mimic the 'learning' function of the brain, but not necessarily the processes that the brain uses to learn.

As for the algorithm, as discussed above, since it does not have a convex form, it still faces the problem of getting trapped in local minima, although people have devised techniques to help it avoid this problem.

In sum, Neural Network lacks a strong learning theory to back up its "success", thus it's hard for people to wisely apply and adjust it. Having said that, it is still an active research area in machine learning. NN still has wide applications in the engineering field such as in control.

### Business Applications of Neural Networks

Neural networks are increasingly being used in real-world business applications and, in some cases, such as fraud detection, they have already become the method of choice. Their use for risk assessment is also growing and they have been employed to visualize complex databases for marketing segmentation. This method covers a wide range of business interests — from finance management, through forecasting, to production. The combination of statistical, neural and fuzzy methods now enables direct quantitative studies to be carried out without the need for rocket-science expertise.

• On the Use of Neural Networks for Analysis Travel Preference Data
• Extracting Rules Concerning Market Segmentation from Artificial Neural Networks
• Characterization and Segmenting the Business-to-Consumer E-Commerce Market Using Neural Networks
• A Neurofuzzy Model for Predicting Business Bankruptcy
• Neural Networks for Analysis of Financial Statements
• Developments in Accurate Consumer Risk Assessment Technology
• Strategies for Exploiting Neural Networks in Retail Finance
• Novel Techniques for Profiling and Fraud Detection in Mobile Telecommunications
• Detecting Payment Card Fraud with Neural Networks
• Money Laundering Detection with a Neural-Network

<references/>

## Complexity Control - October 26, 2010

### Lecture Summary

Selecting the model structure with an appropriate complexity is a standard problem in pattern recognition and machine learning. Systems with the optimal complexity have a good generalization to unseen data.

A wide range of techniques may be used which alter the system complexity. In this lecture, we present the concepts of over-fitting & under-fitting with an example to illustrate how we choose a good classifier and how to avoid over-fitting.

Moreover, cross-validation has been introduced during the lecture which is a method for estimating generalization error based on "resampling" (Weiss and Kulikowski 1991; Plutowski, Sakata, and White 1994; Shao and Tu 1995)[1],[2],[3]. The resulting estimates of generalization error are often used for choosing among various models. Also, it can be used for model selection by choosing one of several models that has the smallest estimated generalization error. Finally, the common types of cross-validation have been addressed.

Before starting of next section a short description of model complexity is necessary. As the words show model complexity somehow describes complication of our model. Suppose we have a feed forward neural network if we increase the number of the hidden layers or the number of nodes in a specific layer it makes sense that our model is becoming more complex. Or suppose we want to fit a polynomial function on some data points if we add degree of this polynomial it seems that we are choosing a more complex model. It seems that when we choose a complex model it would be better since we have more degree of freedom and we can get more exact answer. In next section will explain why the case is not like this and why there is a trade-off between model complexity and optimal result. This make it necessary to find methods for controlling complexity in model selection. We will see this procedure in an example.

### Over-fitting and Under-fitting

File:overfitting-model.png
Figure 1. The overfitting model that uses kernel regression and smoothing splines passes through all of the points of the training set, but has poor predictive power for new data points that are not in the training set. On the other hand, the line model makes more errors on the training points but it is better at extracting the main characteristic of the training points, i.e. the underlying function. Consequently, it has better predictive power for new data points that are not in the training set.

There are two issues that we have to avoid in Machine Learning:

1. Overfitting
2. Underfitting

Suppose there is no noise in the training data, then we would face no problem with over-fitting, because in this case every training data point lies on the underlying function and the only goal would be to build a model that is as complex as needed to pass through every training data point.

However, in the real-world, the training data are noisy, i.e. they tend to not lie on the underlying function and instead they tend to be shifted away from the underlying function to unpredictable locations by random noise. If the model is more complex than what it needs to be in order to accurately fit the underlying function, then it would end up fitting through most or all of the training data. Consequently, it would be a poor approximation of the underlying function and it would have poor prediction ability on new unseen data.

The dangers of overfitting is that it can easily lead the predictions to the range that is far beyond that of the training data, and produce wild predictions in multilayer perceptrons even with noise-free data. The best way to avoid overfitting is to use lots of training data. But unfortunately it is not always useful. Increasing the training data alone does not guarantee that it will avoid over-fitting. In fact, it is the combination of a good number of training examples and the complexity of the model. The training set should have a sufficient number of data points, so that it is representative of the whole data space which needs to be sampled appropriately.

In a Neural Network if the depth is too much, the network will have many degrees of freedom and will learn every characteristic of the training data set. That means it will show a very precise outcome of the training set but will not be able to generalize the commonality of the training set to predict the outcome of new cases.

Underfitting occurs when the model we picked to describe the data is not complex enough, and has high error rate on the training set. There is always a trade-off. If our model is too simple, underfitting could occur and if it is too complex, overfitting can occur.

Example

1. Consider the example showed in the figure. We have a training set and we want to find a model which fits it the best. We can find a polynomial of high degree which almost passes through all points in the training set. But, in fact the training set is coming from a line model. Although the complex model has less error on the training set it diverges from the line in other ranges which we have no training point. As a result the high degree polynomial has very poor prediction on the test cases. This is an example of overfitting model.
2. Now consider a training set which comes from a polynomial of degree two model. If we model this training set with a polynomial of degree one, our model will have high error rate on the training set, and is not complex enough to describe the problem.
3. Consider a simple classification example. If our classification rule takes as input only the colour of a fruit and concludes that it is a banana, then it is not a good classifier. The reason is that just because a fruit is a yellow, does not mean that it is a banana. We can add complexity to our model to make it a better classifier by considering more features typical of bananas, such as size and shape. If we continue to make our model more and more complex in order to improve our classifier, we will eventually reach a point where the quality of our classifier no longer improves, ie., we have overfit the data. This occurs when we have considered so many features that we have perfectly described the existing bananas, but if presented with a new banana of slightly different shape than the existing bananas, for example, it cannot be detected. This is the tradeoff; what is the right level of complexity?

Overfitting occurs when the model becomes too complex and underfitting occurs when it is not complex enough, both of which are not desirable. To control complexity, it is necessary to make assumptions for the model before fitting the data. Assumptions that we can make for a model are with polynomials or a neural network. There are other ways as well.

Figure 2: An example of a model with a family of polynomials

We do not want a model to get too complex, so we control it by making an assumption on the model. With complexity control, we want a model or a classifier with a low error rate. The lecture will explain the tradeoff between Bias and variance for model complexity control.

Overfitted model and Underfitted model:

After the construction of model is determined, the next problem we meet is do the model selection, that is, how to estimate the parameters effectively, especially when we use iteration method to do the estimation. In the iteration method, the key point is to determine the best time to stop update parameters. Let us see a very simple example; assume the dotted line on the graph can be expressed as a function, and the data points, the circles, are generated by the function with added noise.

Model 1(as shown on the left of Figure 3) A line can be used to describe the data points, where two parameter are needed to construct the estimate of the function. However, it is clear that it performs badly. This model is a typical example of underfitted model. In this case, the model will perform well in prediction, but a large bias could be generated.

Model 2 (as shown on the right of Figure 3) in this model, lots of parameter are used to fit the data points. Although it looks pretty good on fitting, the performance on prediction could be very bad, which means this model will generate a large variance when we use it on the data points which are not in the training data. The models above are the extreme case in the model selection, we do not want to choose any of them in our classification work. So the key thing is to stop our training work at the optimal time such that the balance of bias and variance would be obtained, that is, the time t in the following graph.

To achieve the aim, one approach we can use is to divide our data points into two groups and make them independently; one (training set) is used in the training test to obtain parameters, the other one (validation set) is used for determining the optimal time. After every updated parameter, the test in the validation set is implement and plot the curve of error in the two test in order to find the optimal point t. Here, the validation test is a good measure of generalization. Remember that do not update the parameters in the validation test. If one more independent test we need to follow, three independent groups should be determined at the beginning. In addition, this approach is suitable for the case of more data points, especially a finite data set, since the effect on noise could be decreased to the lowest level.

So far, we have learn two most popular ways to estimate the expected level of fit of a model to a data set that is independent of the data that were used to train the model:

1. Cross validation
2. Regularization: refers to a series of techniques we can use to suppress overfitting,that is, making our function not so curved such that it performance badly in the prediction. The specific way is to add a new penalty term into the error function, and it tends to limit the over-increasing the weight when the weight update by iteration.

Indeed, there are many techniques could be used, such as:

1.Akaike information criterion
2.Bayesian information criterion
3.Mallows' Cp]

#### Note

When the model is linear, the true error form AIC approach is identical to that from Cp approach; When the model is nonlinear, they are different.

### How do we choose a good classifier?

Our goal is to find a classifier that minimizes the true error rate$\ L(h)$.

$\ L(h)=Pr\{h(x)\neq y\}$

Recall the empirical error rate

$\ \hat L(h)= \frac{1}{n} \sum_{i=1}^{n} I(h(x_{i}) \neq y_{i})$

Figure 3

There is a downward bias to training error estimate that it is always less than the true error rate.

If there is a change in our complexity from low to high, our training (empirical) error rate is always decreasing. When we apply our model to the test data, our error rate will start to decrease to a point, but then it will increase since the model hasn't seen the datapoints in the test set before. This results in a convex test error curve as a function of learning model complexity. This can be explained since training error will decrease when we fit the model better by increasing its complexity, but as we have seen, this complex model will not generalize well, resulting in a larger test error.

We use our test data (from the test sample line shown on Figure 2) to get our true error rate. Right complexity is defined as where true error rate ( the error rate associated with the test data) is minimum; and this is one idea behind complexity control.

Figure 4

We assume that we have samples $\,x_1, . . . ,x_n$ that follow some (possibly unknown) distribution. We want to estimate a parameter $\,f$ of the unknown distribution. This parameter may be the mean $\,E(x_i)$, the variance $\,var(x_i)$ or some other quantity.

The unknown parameter $\,f$ is a ﬁxed real number $f\in R$. To estimate it, we use an estimator which is a function of our observations, $\hat{f}(x_1,...,x_n)$.

$Bias (\hat{f}) = E(\hat{f}) - f$

$MSE (\hat{f}) = E[(\hat{f} - f)^2]=Varince (\hat f)+Bias^2(\hat f )$

$Variance (\hat{f}) = E[(\hat{f} - E(\hat{f}))^2]$

One desired property of the estimator is that it is correct on average, that is, it is unbiased. $Bias (\hat{f}) = E(\hat{f}) - f=0$. However, there is a more important property for an estimator than just being unbiased: the mean squared error. In statistics, there are problems for which it may be good to use an estimator with a small bias. In some cases, an estimator with a small bias may have lesser mean squared error or be median-unbiased (rather than mean-unbiased, the standard unbiasedness property). The property of median-unbiasedness is invariant under transformations while the property of mean-unbiasedness may be lost under nonlinear transformations. For example, while using an unbiased estimator with large mean square error to estimate the parameter, we highly risk a big error. In contrast, a biased estimator with small mean square error will improve the precision of our prediction.

Hence, our goal is to minimize $MSE (\hat{f})$.

From figure 4, we can see that the relationship of the three parameters is: $MSE (\hat{f})=Variance (\hat{f})+Bias ^2(\hat{f})$. Thus given the Mean Squared Error (MSE), if we have a low bias, then we will have a high variance and vice versa.

A test error is a good estimation on MSE. We want to have a somewhat balanced bias and variance (not high on bias or variance), although it will have some bias.

Referring to Figure 3, overfitting happens after the point where training error starts to decrease and test error starts to increase.

### Avoid Overfitting

There are 2 main approaches to avoid overfitting:

1. Estimating error rate

$\hookrightarrow$ Empirical training error is not a good estimation

$\hookrightarrow$ Empirical test error is a better estimation

$\hookrightarrow$ Cross-Validation is fast

$\hookrightarrow$ Computing error bound (analytically) using some probability inequality.

We will not discuss computing the error bound in class; however, a popular method for doing this computation is called VC Dimension (short for Vapnik–Chervonenkis Dimension). Information can be found from Andrew Moore and Steve Gunn.

2. Regularization

$\hookrightarrow$ Use of shrinkage method

$\hookrightarrow$ Decrease the chance of overfitting by controlling the weights

$\hookrightarrow$ Weight Decay: bound the complexity and non-linearity of the output by a new regularized cost function.

### Cross-Validation

Figure 1: Illustration of Cross-Validation

Cross-Validation is the simplest and most widely used method to estimate the true error.

Here is a general description of cross-validation:

Given a set of collected data for which we know the proper labels,

1) Randomly divide the data into two parts, Training data (T) and Validation data (V)
2) Train the classifier using only data in T
3) Estimate the true error rate, \begin{align}\hat L(h)\end{align}, using only data in V
$\hat L(h) = \frac{1}{|\mathrm{V}|}\sum_{x_i \in \mathrm{V}}I(h(x_i) \neq y_i)$, where \begin{align}\,|\mathrm{V}|\end{align} is the cardinality of the validation set and
$\, I(h(x_i) \neq y_i)= \left\{\begin{matrix} 1 & h(x_i) \neq y_i \\ 0 & \mathrm{otherwise} \end{matrix}\right.$

Note that the validation set will be totally unknown to the trained model but the proper label of all elements in this set are known. Therefore, it is easy to count the number of misclassified points in V.

The best classifier is the model with minimum true error, \begin{align}\hat L(h)\end{align}.

### K-Fold Cross-Validation

File:k-fold.png
Figure 2: K-fold cross-validation

The results from the method above may differ significantly based on the initial choice of T and V. Therefore, we improve simple cross-validation by introducing K-fold cross-validation. The advantage of K-fold cross validation is that all the values in the dataset are eventually used for both training and testing.

In this case, the algorithm is:

Given a set of collected data for which we know the proper labels,

1) Randomly divide the data into K parts with approximately equal size

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2) For k = 1,...,K
3) Remove part k and train the classifier using data from all classes except part k
4) Compute the error rate, \begin{align}\hat L_k(h)\end{align}, using only data in part k
$\hat L_k(h) = \frac{1}{m} \sum_{i=1}^{m} I(h(x_{i}) \neq y_{i})$, where $m$ is the number of data points in part k
5) End loop
6) Compute the average error $\hat L(h) = \frac{1}{K} \sum_{k=1}^{K} \hat L_k(h)$

Once again, the best classifier is the model with minimum average error, \begin{align}\hat L(h)\end{align}.

In class we mentioned that \begin{align}\hat L(h)\end{align} is a high variance estimator of the error rate, but it is unbiased.

Figure 4 is an illustration of data that is divided into four roughly equal parts.

### Leave-One-Out Cross-Validation - October 28, 2010

Leave-one-out cross validation is used to determine how accurately a learning algorithm will be able to predict data that it was not trained on. When using the leave-one-out method, the learning algorithm is trained multiple times, using all but one of the training set data points. The form of the algorithm is as follows:

For k = 1 to n (where n is the number of points in our dataset)

•Temporarily remove the kth data point.

•Train the learning algorithm on the remaining n - 1 points.

•Test the removed data point and note your error.

Calculate the mean error over all n data points.

Leave-one-out cross validation is useful because it does not waste data. When training, all but one of the points are used, so the resulting regression or classification rules are essentially the same as if they had been trained on all the data points. The main drawback to the leave-one-out method is that it is expensive - the computation must be repeated as many times as there are training set data points.

Leave-one-out cross-validation is similar to k-fold validation by selecting sets of equal size for error estimation. Leave-one-out cross-validation instead removes a single data point, with n-partitions. Each partition is used systematically for testing exactly once whereas the remaining partitions are used for training. For example, we estimate the $\,n-1$ data points with $\,m$ linear models over the $\,n$ sets, and compare the average error rates.The leave-one-out error is the average error over all partitions.
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In Leave-one-out cross-validation, in each iteration, we leave a point out (the kth point) and fit a model to the remaining $\,n-1$ datapoints, and then test the removed datapoint and compute the error. The overall error rate will be the average error computed over all individual errors.
The above procedure should be repeated for all the model options and the model with the least error rate will be selected.
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In the above example, we can see that k-fold cross-validation can be computationally expensive: for every possible value of the parameter, we must train the model $\,K$ times. This deficiency is even more obvious in leave-one-out cross-validation, where we must train the model $\,n$ times, where $\,n$ is the number of data points in the data set.

But an expensive computational load does not tell the whole story. Why do we need this validation? The key factor is not having enough data points! In some real world problems gathering data points can be very expensive or time consuming. Suppose we want to study the effect of a new drug on the human body. To do this, we must test the drug on some patients. However, it is very hard to convince a person to take part in this procedure since there may be risks and side effects with testing the new drug on him/her. As well, a long-term study needs to be done to observe any long-term effects. In a similar manner we lack data points or observations in some problems. But if we use K-fold cross-validation and divide the data points into a training and test data set then we may not have enough data to train the neural network or fit any other model, and under fitting may occur. To avoid this the best thing that can be done is to do leave-one-out cross-validation. In this way we will take advantage of the data points we have and yet still be able to test the model.

However, in the linear model, we can save complexity analytically. A model is correct if the mean response is the linear combination of subsets of a vector and the columns of $X_n$. Let $A_n$ be a finite set of proposed models. Let $a_n^L$ be the model minimizing average squared error, then the selection procedure is consistent if the probability of the model selected being $a_n^L$ approaches 1. Leave-one-out is correct, can be inconsistent, and given

• $\max_{i \lt = n} x_i^t (X_n^tX_n)^{-1} x_i \to 0$

is asymptotically equivalent to AIC, which performs slightly worse than k-fold <ref>Shao, J. An asymptotic theory for linear model selection, Statistica Sineca, 7, 221-264 (1997).</ref>
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Leave-one-out cross-validation can perform poorly in comparison to k-fold validation. A paper by Breiman compares k-fold (leave-many-out) cross-validation to leave-one-out cross-validation, noting that average prediction loss and downward bias increase from k-fold to leave-one-out <ref>Breiman, L. Heuristics of instability and stabilization in model selection, Annals of Statistics, 24, 2350-2383 (1996).</ref>. This can be explained by the lower bias of leave-one-out validation, causing an increase in variance. The bias is relative to the size of the sample set compared to the training set [14]. As such, as k becomes larger, it becomes more biased and has less variance. Similarly, larger data sets will direct the bias toward zero.

### References

1. Sholom M. Weiss and Casimir A. Kulikowski, Computer Systems That Learn: Classification and Prediction Methods from Statistics, Neural Nets, Machine Learning and Expert Systems. Morgan Kaufmann, 1991.

2. M. Plutowski, S. Sakata and H. White: "Cross-Validation Estimates Integrated Mean Squared Error," in J. Cowan, G. Tesauro, and J. Alspector, eds., Advances in Neural Information Processing Systems 6. San Francisco: Morgan Kaufmann, 391-398 (1994).

3. Shao, J. and Tu D. (1995). The Jackknife and Bootstrap. Springer, New York.

## Radial Basis Function (RBF) Network - October 28, 2010

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| type = style | image = | imageright = | style = | textstyle = | text = This article may require cleanup to meet Wikicoursenote's quality standards. The specific problem is: It might be interesting to explain the reason for over fitting in MLP. Based on what we learnt on the class we might think that over-fitting occurs due to complexity in structure but the most common way that occurs in MLP has to do nothing with the structure. In MLP over-fitting occurs to do iterations more than what is needed. Since if you this then weights of neural network get very adapted to training data and the training error rate will decrease, this will cause over-fitting. To avoid this problem training of neural network is stopped heuristically. Making this automatic for MLP's can be a good topic for research!. Please improve this article if you can. (October 2010) | small = | smallimage = | smallimageright = | smalltext = }} {{

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| type = style | image = | imageright = | style = | textstyle = | text = This article may require cleanup to meet Wikicoursenote's quality standards. The specific problem is: Could you please clarify what you mean?. Please improve this article if you can. (November 2010) | small = | smallimage = | smallimageright = | smalltext = }} {{

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| type = style | image = | imageright = | style = | textstyle = | text = This article may require cleanup to meet Wikicoursenote's quality standards. The specific problem is: I meant that generally our sense is that model complexity is related to structure. It means that we think that if we choose a more complex model it will result in overfitting but sometimes the structure is not bad and appropriate for our problem like a MLP. But we encounter overfitting due to over-training. Which means that we do more than what is needed iterations and this will lead to fitting the training data more than what is needed.. Please improve this article if you can. (November 2010) | small = | smallimage = | smallimageright = | smalltext = }} {{

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Figure 1: Radial Basis Function Network

### Introduction

A Radial Basis Function (RBF) network is a type of artificial neural network with:

• an output layer,
• a single hidden layer,
• weights from the hidden layer to the output layer,
• and no weights from the first layer to the hidden layer.

An RBF network can be trained without back propagation since it has a closed-form solution. The neurons in the hidden layer contain basis functions. A common basis function is RBF function which is a kind of Gaussian function without scaling factor.

• Note: Spline, RBF, Fourier, and similar methods differ only in the basis function.

RBF networks were first used in solving multivariate interpolation problems and numerical analysis. Their prospect is similar in neural network applications, where the training and query targets are rather continuous. RBF networks are artificial neural networks and they can be applied to Regression, Classification and Time series prediction.

For example, if we consider $\,n$ data points along a one dimensional line and $\,m$ clusters. An RBF network with radial basis (Gaussian) functions will cluster points around the $\,m$ means, $\displaystyle\mu_{j}$ for $j= 1, ..., m$. The other data points will be distributed normally around these centers.

• Note: The hidden layer can have a variable number of basis functions (the optimal number of basis function can be determined using the complexity control techniques discussed in the previous section). As usual the more basis functions in the hidden layer, the higher the model complexity will be.

RBF networks, K-Means clustering ,Probabilistic Neural Networks(PNN) and General Regression Neural Networks (GRNN) are almost the same. The main difference is that PNN/GRNN networks have one neuron for each point in the training file, whereas the number of RBF networks neurons (basis functions) is not distinct and it is usually much less than the number of training points. When the training size sets is not very large PNN/GRNN perform well. But for large size data sets RBF will be more useful since PNN/GRNN are impractical.

#### Reference for the above paragraph

1. Improving the performance of k-means clustering algorithm to position the centers of RBF network [15]

2. GA-RBF: A selfoptimising RBF network [16]

3. A closer look at the radial basis function (RBF) networks [17]

4. Probabilistic neural networks [18]

5. A general regression neural network [19]

6. Experience with adaptive probabilistic neural networks and adaptive general regression neural networks [20]

7. Using General Regression and Probabilistic Neural Networks To Predict Human Intestinal Absorption with Topological Descriptors Derived from Two-Dimensional Chemical Structures [21]

### Model Detail

#### RBF Network's Hidden Layer

The hidden layer has $\, m$ neurons, where the optimal number can be determined using the cross validation techniques discussed in the previous section. For example, if the data is generated from mixture of Gaussian distribution, you can cluster the data and estimate each Gaussian distribution mean and variance by EM algorithm. Their mean and variance can be used for constructing the basis functions. Each neuron consists of a basis function of an input layer point $\underline x_{i}$ referred to as $\,\Phi_{j}(\underline x_{i})$ where $\, j \in \{1 ... m\}$ and $\, i \in \{1 ... n\}$.

• Note: In the following section, $k$ is the number of outputs, $n$ is the number of data points, and $m$ is the number of hidden units. If $\,k = 1$, $\,\hat Y$ and $\,W$ are column vectors.

A common basis function is the radial basis Gaussian function:
$\Phi_{j}(\underline x_i) = e^{\frac{-\Vert\underline x_i - \mu_{j}\Vert ^2}{2\gamma_{j}}}$

• Note: An RBF function $\Phi$ is a real-valued function whose value depends only on the distance from a centre $\underline c$, such that $\Phi(\underline x,\underline c) = \Phi(\|\underline x - \underline c \|)$. Other commonly used radial basis functions are Multiquadric, Polyharmonic spline, and Thin plate spline.

The radii of the RBF functions may be different. The centers and radii can be determined by clustering the data into $\, m$ groups and taking each group mean $\,\mu_{j}$.

$\Phi_{n,m} = \left[ \begin{matrix} \Phi_{1}(\underline x_{1}) & \Phi_{2}(\underline x_{1}) & \cdots & \Phi_{m}(\underline x_{1}) \\ \Phi_{1}(\underline x_{2}) & \Phi_{2}(\underline x_{2}) & \cdots & \Phi_{m}(\underline x_{2}) \\ \vdots & \vdots & \ddots & \vdots \\ \Phi_{1}(\underline x_{n}) & \Phi_{2}(\underline x_{n}) & \cdots & \Phi_{m}(\underline x_{n}) \end{matrix}\right]$ is the matrix of Radial Basis Functions.

#### Weights

The weights $\, w_k$ used in calculating the output layer can be optimally calculated. Let

$W_{m,k} = \left[ \begin{matrix} w_{1,1} & w_{1,2} & \cdots & w_{1,k} \\ w_{2,1} & w_{2,2} & \cdots & w_{2,k} \\ \vdots & \vdots & \ddots & \vdots \\ w_{m,1} & w_{m,2} & \cdots & w_{m,k} \end{matrix}\right]$ be the matrix of weights.

#### Output Layer

The output layer can be multi-dimensional.

$Y_{n,k} = \left[ \begin{matrix} y_{1,1} & y_{1,2} & \cdots & y_{1,k} \\ y_{2,1} & y_{2,2} & \cdots & y_{2,k} \\ \vdots &\vdots & \ddots & \vdots \\ y_{n,1} & y_{n,2} & \cdots & y_{n,k} \end{matrix}\right]$ is the matrix of output variables, and the fitted output $\, \hat{Y}$ can be expressed in matrix form as:

$\hat Y = \Phi W$

Since this is a linear combination of $\, \Phi_{j}(\underline x_{i})$s, we can apply least-squares to find the optimal $\, w_j$:
$min_W \vert Y - \Phi W \vert^2 \ \Rightarrow W = (\Phi^T \Phi)^{-1}\Phi^T Y$

Model selection implies choosing the following:

• the number of basis functions (hidden nodes), and thus, the complexity of the model
• the basis function to be used (for the time being assumed to be the Gaussian function above)
• the function parameters ($\, \mu_{j}, \gamma_{j}$)

Let

• $\, \hat f$ denote the prediction model which is estimated from a training set (model estimate)
• $\, f$ denote the true model (the model which when applied to input data $\, X$ will result in $\, Y$)
• $\, err$ be the training error
• $\, Err$ be the generalized error (true error)

Assume that given data $\, D=\{x_i, y_i\}$ for $\, i \in \{1 ... n\}$,
$\, y_i = f(x_i) + \epsilon_i$
$\, \epsilon$ is what essentially contributes to the complexity of the model. If there were no noise then model selection would be trivial since there would exist many functions of various degrees of complexity that would perfectly fit the data. We assume that $\, \epsilon$ is an additive Gaussian noise: $\, \epsilon_i \sim N(0, \sigma^2)$.
$\, err = E[(y - \hat y)^2]$
$\,= E[(f(x) + \epsilon - \hat f(x))^2]$
$\,= E[(f(x) - \hat f(x))^2 + \epsilon^2 - 2\epsilon(f(x) - \hat f(x))]$
The part of the error term we want to approximate is $\, E[(f(x) - \hat f(x))^2]$. We will try to estimate this by finding the other terms of the above expression.

#### Conceptualizing RBF Networks

In the past, we have classified data using models that were explicitly linear, quadratic, or otherwise definite. In RBF networks, like in Neural Networks, we can fit an arbitrary model. How can we do this without changing the equations being used? Recall the trick we discussed at the beginning of the term: if we add new features to our original data set, we can project our input data into higher dimensions, and then use a linear algorithm to solve. Thinking of $\,\Phi$ as a feature space of the input, each hidden unit can then represent a feature; we can see that, if there are more hidden units than input units, we can essentially project to a higher-dimensional space, as we did in our earlier trick. This does not mean that an RBF network will always do this, it is merely a way to convince yourself that RBF networks (and neural networks) can fit arbitrary models.

Introduction of the Radial Basis Function (RBF) Networks [22]

Radial Basis Function (RBF) Networks [24] [25] [26]

An Example of RBF Networks [27]

## Model Selection for RBF Network (Stein's Unbiased Risk Estimator) - November 2nd, 2010

### Model Selection

Model selection is the task of selecting a model of optimal complexity for a given set of data. Learning a radial basis function network from data is a parameter estimation problem. One difficulty with this problem is selecting parameters that perform well for both the training data and the testing data. In principle, a model is selected that has parameters associated with the best observed performance on the training data, although our goal is really to achieve good performance on the unseen (to the model) testing data. Not surprisingly, a model selected on the basis of the training data set does not necessarily exhibit comparable performance on the testing data set. When squared error is used as the performance index, a zero-error model on the training data can always be achieved by using a sufficient number of basis functions.

However, training error and testing error do not demonstrate a linear relationship. In particular, a smaller training error does not necessarily result in a smaller testing error. In practice, one often observes that up to a certain point the model error on testing data tends to decrease as the training error decreases. However, if one attempts to decrease the training error too much by increasing the model complexity, the testing error often can take a dramatic turn and begin to increase. This was explained and a graphic provided in the lecture on October 26th concerning complexity control.

File:data noise.jpg
Figure 1. Data sampled from a smooth function (in black) cannot be over-fit. Data sampled from a smooth function with noise (in red) can be over-fit when the noise is modelled along with the smooth function.

The basic reason behind this phenomenon of the training and testing errors is that in the process of minimizing training error, after a certain point, the model begins to over-fit the training set. Over-fitting in this context means fitting the model to the training data at the expense of losing generality. As seen in Figure 1, the red data points have been over-fit as the general form of the underlying smooth function has been lost in the red-curve model. In the extreme case, a set of $\displaystyle N$ training data points can be modeled exactly with $\displaystyle N$ radial basis functions. Such a model will fit the training data set perfectly. However, the perfectly-fit model fails to be as accurate or perform as well on the training data set because it has modelled not only the true function $\displaystyle f(X)$ but the random noise as well, and thus has over-fit the data (as the red curve in Figure 1 has done). It is interesting to note that in the case of no noise, over-fitting will not occur and hence the complexity of the model can be increased without bound. However, this is not realistic in practice as random noise is almost always present in the data.

In general, the training error rate will be less than the testing error on the new data. A model typically adapts to the training data, and hence the training error will be an overly optimistic estimate of the testing error. An obvious way to estimate testing error is to add a penalty term to the training error to compensate for the difference. SURE is developed based on this idea.

### Stein's unbiased risk estimate (SURE)

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Stein's unbiased risk estimate (SURE) is an unbiased estimator of the mean-squared error of a given estimator in a deterministic estimation scenario. In other words, it provides an indication of the accuracy of a given estimator. This is important since, in deterministic estimation, the true mean-squared error of an estimator generally depends on the value of the unknown parameter, and thus cannot be determined completely. A standard application of SURE is to choose a parametric form for an estimator, and then optimize the values of the parameters to minimize the risk estimate. This technique has been applied in several settings. For example, a variant of the James-Stein estimator can be derived by finding the optimal shrinkage estimator. The technique has also been used by Donoho and Johnstone to determine the optimal shrinkage factor in a wavelet denoising setting [28].

For more information about the relation between Stein's unbiased risk estimator and Stein's lemma refer to[ http://www.cc.gatech.edu/~lebanon/notes/sure.pdf]

Note that the material presented here is applicable to model selection in general, and is not specific to RBF networks.

#### Important Notation [29]

Let:

• $\displaystyle f(X)$ denote the true model.
• $\hat f(X)$ denote the prediction/estimated model, which is generated from a training data set $\displaystyle D = \{(x_i, y_i)\}^n_{i=1}$.
• $\displaystyle err$ denote the training error or empirical error.
• $\displaystyle Err$ denote the true error or generalization error, and is what we are trying to minimize.
• $\displaystyle MSE=E[(\hat f-f)^2]$ denote the mean squared error, where $\hat f(X)$ is the estimated model and $\displaystyle f(X)$ is the true model.

For a single data point, we have the following two values:

• the observations $\displaystyle y_i = f(\underline x_i) + \epsilon_i$ where $\displaystyle \epsilon$ is noise
• the fitted values $\displaystyle \hat y_i = \hat f(\underline x_i)$

We will make two assumptions about the observations: 1) $\displaystyle \epsilon$ is additive Gaussian noise, and 2) $\displaystyle \epsilon_i$ ~ $\displaystyle N(0,\sigma^2)$.

We need to estimate $\hat f$ from the training data set $D=\{(x_i,y_i)\}^n_{i=1}$. Let $\hat f_i=\hat f(x_i)$ and $\displaystyle f_i= f(x_i)$， then:

$\displaystyle E[(\hat y_i-y_i)^2 ]=E[(\hat f_i-f_i-\epsilon_i)^2]$$=E[(\hat f_i-f_i)^2]+E[\epsilon_i^2]-2E[\epsilon_i(\hat f_i-f_i)]$

Let $\displaystyle E[(\hat y_i-y_i)^2 ]=E[(\hat f_i-f_i)^2]+\sigma^2-2E[\epsilon_i(\hat f_i-f_i)]$ be referred to as equation $\displaystyle (1)$.

The last term of equation (1) can be written as:

$\displaystyle E[\epsilon_i(\hat f_i-f_i)]=E[(y_i-f_i)(\hat f_i-f_i)]=cov(y_i,\hat f)$, where$\displaystyle y_i$ and $\hat f_i$ both have same mean $\displaystyle f_i$.

Note that we can compute the left-hand side of equation (1), and what we are interested in calculating is the term $\displaystyle E[(\hat f_i-f_i)^2]$. Thus, if we can somehow calculate the last term of equation (1) we will have achieved our goal.

For reference, we make note of the bias-variance decomposition:

\begin{align} \displaystyle MSE = E[(\hat f-f)^2] &= E[(\hat f-E(\hat f))+(E(\hat f)-f)]^2\\ &= E[(\hat f-E(\hat f))^2+2*(\hat f-E(\hat f))*(E(\hat f)-f)+(E(\hat f)-f)^2]\\ &= E[(\hat f-E(\hat f))^2]+E[2*(\hat f-E(\hat f))*(E(\hat f)-f)]+E[(E(\hat f)-f)^2]\\ &= Var(\hat f)+Bias^2(\hat f) \end{align}

Since, $\displaystyle E[2*(\hat f-E(\hat f))*(E(\hat f)-f)]=2*Cov[E(\hat f)-f, \hat f-E(\hat f)]$, which is equal to zero.

#### Stein's Lemma

If $\,Z$ is $\,N(\theta,\sigma^2)$ and if $\displaystyle g(Z)$ is weakly differentiable, such that $\displaystyle E[\vert g'(Z)\vert]\lt \infty$, then $\displaystyle E[g(Z)(Z-\theta)]=\sigma^2E(g'(Z))$.

According to Stein's Lemma, the last cross term of equation $\displaystyle (1)$, $\displaystyle E[\epsilon_i(\hat f_i-f_i)]$, can be written as $\sigma^2 E\left[\frac {\partial \hat f}{\partial y_i}\right]$. The derivation is as follows.

$\displaystyle Proof$: Let $\,Z = \epsilon$. Then $g(Z) = \hat f-f$, since $\hat y = f + \epsilon$, and $\,f$ is a constant. So $\,\theta = 0$ and $\,\sigma^2$ is the variance in $\,\epsilon$.

$\displaystyle E[g(Z)(Z-\theta)]=E[(\hat f-f)\epsilon]=\sigma^2E(g'(Z))=\sigma^2 E\left[\frac {\partial (\hat f-f)}{\partial y_i}\right]=\sigma^2 E\left[\frac {\partial \hat f}{\partial y_i}-\frac {\partial f}{\partial y_i}\right]$

Since $\displaystyle f$ is the true function and not a function of the observations $\displaystyle y_i$, then $\frac {\partial f}{\partial y_i}=0$.

So, $\displaystyle E[\epsilon_i(\hat f_i-f_i)]=\sigma^2 E\left[\frac {\partial \hat f}{\partial y_i}\right]$. Call this equation $\displaystyle (2)$.

#### Two Different Cases

##### Case 1

Consider the case in which a new data point has been introduced to the estimated model, i.e. $(x_i,y_i)\not\in D$; this new point belongs to the testing/validation data set $V=\{(x_i,y_i)\}^m_{i=1}$. Since $\displaystyle y_i$ is a new point, $\hat f$ and $\displaystyle y_i$ are independent. Therefore $\displaystyle cov(y_i,\hat f)=0$. Alternatively, this can be thought of when considering $\frac{\partial \hat f}{\partial y_i}$: when $\,y_i$is a new point the partial derivative has no relation with $\hat f$ because the estimation of $\hat f$ was based on the training data of which $\displaystyle y_i$ was not a part of. Thus, $\frac{\partial \hat f}{\partial y_i}=0$. In this case, equation $\displaystyle (1)$ can be written as:

$\displaystyle E[(\hat y_i-y_i)^2 ]=E[(\hat f_i-f_i)^2]+\sigma^2$ for one data point.

Summing over all m data points in the testing/validation dataset gives the following expression:

$\sum_{i=1}^m (\hat y_i-y_i)^2 = \sum_{i=1}^m (\hat f_i-f_i)^2+ m\sigma^2$

Based on the notation we defined above, we then have: $\displaystyle err=Err+m\sigma^2$.

The empirical error is a good estimator of the true error, up to a constant additive value. Since $\displaystyle m \sigma^2$ is constant, minimizing $\displaystyle err$ is equal to minimizing the true error $\displaystyle Err$. This is the justification behind the technique of cross-validation. To avoid over-fitting or under-fitting using cross-validation, a validation data set selected so that it is independent from the estimated model.

##### Case 2

A more interesting case is the case in which we do not use new data points to assess the performance of the estimated model, and the training data set is used for both estimating and assessing the model $\hat f_i$. In this case the cross-term in equation $\displaystyle (1)$ cannot be ignored because $\hat f_i$ and $\displaystyle y_i$ are not independent. Instead, the cross-term can be estimated by Stein's Lemma, which was originally proposed to estimate the mean of a Gaussian distribution.

Suppose $(x_i,y_i)\in D$. Then by applying Stein's Lemma, we obtain equation $\displaystyle (2)$ that was proven above.

This means that equation $\displaystyle (1)$ now becomes, for one data point: $\displaystyle E[(\hat y_i-y_i)^2 ]=E[(\hat f_i-f_i)^2]+\sigma^2-2\sigma^2E\left[\frac {\partial \hat f}{\partial y_i}\right]$.

Summing over all n data points in the training (and testing, since it is the same) dataset gives the following expression:

$\sum_{i=1}^n (\hat y_i-y_i)^2 = \sum_{i=1}^n (\hat f_i-f_i)^2+ n\sigma^2-2\sigma^2\sum_{i=1}^n \frac {\partial \hat f}{\partial y_i}$.

Based on the notation we defined above, we then have: $\displaystyle err=Err+n\sigma^2-2\sigma^2\sum_{i=1}^n \frac {\partial \hat f}{\partial y_i}$ or equivalently $\displaystyle Err=err-n\sigma^2+2\sigma^2\sum_{i=1}^n \frac {\partial \hat f}{\partial y_i}$. Denote this last expression as equation $\displaystyle (3)$.

In statistics, this is known as Stein's unbiased risk estimate (SURE). It is an unbiased estimator of the mean-squared error of a given estimator, in a deterministic estimation scenario. In other words, it provides an indication of the accuracy of a given estimator. This is important since, in deterministic estimation, the true mean-squared error of an estimator generally depends on the value of the unknown parameter and thus cannot be determined completely.

### SURE for RBF Network

We now consider applying SURE to Radial Basis Function networks specifically. Based on SURE, the optimum number of basis functions should be assigned so that the generalization error $\displaystyle err$ is minimized. Based on the RBF Network, by setting $\frac{\partial err}{\partial W}$ equal to zero we obtain the least squares solution of $\ W = (\Phi^{T}\Phi)^{-1}\Phi^{T}Y$. Then the fitted values are $\hat{Y} = \hat{f} = \Phi W = \Phi(\Phi^{T}\Phi)^{-1}\Phi^{T}Y = HY$, where $\ H = \Phi(\Phi^{T}\Phi)^{-1}\Phi^{T}$ is the hat matrix for this model.

Consider only one node of the network. In this case we can write: $\hat f_i=\,H_{i1}y_1+\,H_{i2}y_2+\cdots+\,H_{ii}y_i+\cdots+\,H_{in}y_n$. Denote this as equation $\displaystyle (4)$.

Note here that $\,H$ depends on the input vector $\displaystyle x_i$ but not on the observation $\displaystyle y_i$.

By taking the derivative of $\hat f_i$ with respect to $\displaystyle y_i$, we can readily obtain:

$\sum_{i=1}^n \frac {\partial \hat f}{\partial y_i}=\sum_{i=1}^n \,H_{ii}$

Here we recall that $\sum_{i=1}^n \,H_{ii}= \,Trace(H)$, the sum of the diagonal elements of $\,H$. Using the permutation property of the trace function we can further simplify the expression as follows: $\,Trace(H)= Trace(\Phi(\Phi^{T}\Phi)^{-1}\Phi^{T})= Trace(\Phi^{T}\Phi(\Phi^{T}\Phi)^{-1})=m$, by the trace cyclical permutation property, where $\displaystyle m$ is the number of basis functions in the RBF network (and hence $\displaystyle \Phi$ has dimension $\displaystyle n \times m$).

#### Sketch of trace cyclical property proof:

For $\, A_{mn}, B_{nm}, Tr(AB) = \sum_{i=1}^{n}\sum_{j=1}^{m}A_{ij}B_{ji} = \sum_{j=1}^{m}\sum_{i=1}^{n}B_{ji}A_{ij} = Tr(BA)$.
With that in mind, for $\, A_{nn}, B_{nn} = CD, Tr(AB) = Tr(ACD) = Tr(BA)$ (from above) $\, = Tr(CDA)$.

Note that since $\displaystyle \Phi$ is a projection of the input matrix $\,X$ onto a basis set spanned by $\,m$, the number of basis functions, that sometimes an extra $\displaystyle \Phi_0$ term is included without any input to represent the intercept of a fitted model. In this case, if considering an intercept, then $\,Trace(H)= m+1$.

Substituing $\sum_{i=1}^n \,H_{ii} = m+1$ into equation $\displaystyle (3)$ gives the following: $\displaystyle Err=err-n\sigma^2+2\sigma^2(m+1)$.

Computationally, to obtain an estimate for the true error $\displaystyle Err$ the above expression is repeatedly evaluated beginning at $\displaystyle m = 1$, then at $\displaystyle m = 2$, then $\displaystyle m = 3$, and so on until the minimum value for $\displaystyle Err$ is determined. The value of m that gives the minimum true error estimate is the optimal number of basis functions to be implemented in the RBF network, and hence is also the optimal degree of complexity of the model.

### Lecture Summary

Stein's unbiased risk estimate (SURE) is an unbiased estimator of the mean-squared error of a given estimator, in a deterministic estimation scenario. It provides an indication of the accuracy of a given estimator.

In RBF network, the problem of selecting the appropriate number of basis functions is a critical issue. An RBF network with an overly restricted basis gives poor predictions on new data. But if an RBF network with too many basis functions, it also gives poor generalization performance.

This lecture introduce a criterion for selecting the number of radial basis functions in an RBF network, using the generalization of Stein’s unbiased risk estimator (SURE).

#### Reference:

Automatic basis selection for RBF networks using Stein’s unbiased risk estimator [30]

From Stein's unbiased risk estimates to the method of generalized cross validation [31]

(This paper concerns the method of generalized cross validation (GCV), based on Stein estimates and the associated unbiased risk estimates.)

Adaptive denoising based on SURE risk [32]

(In this paper, a new adaptive denoising method is presented based on Stein's (1981) unbiased risk estimate (SURE) and on a new class of thresholding functions.)

Wavelet shrinkage denoising using the non-negative garrote [33]

Estimation of the Mean of a Multivariate Normal Distribution [34]

## Regularization for Neural Network - November 4, 2010

### Weight decay

Weight decay is a subset of regularization methods. The penalty term in weight decay, by definition, penalizes large weights. Other regularization methods may involve not only the weights but various derivatives of the output function [35].

Figure 3: activation function

Weight decay training is suggested as a method useful in achieving a robust neural network which is insensitive to noise. Since the number of hidden layers in a NN is usually decided by certain domain knowledge, it may easily get into the problem of overfitting.

The weight–decay method is an effective way to improve the generalization ability of neural networks. In general, the trained weights are constrained to be small when the weight-decay method is applied. Large weights leading to output units can cause outputs that are far beyond the range of the data (when test data is used); in other words, large weights can result in high output variance.

It can be seen from Figure 3 that when the weight is in the vicinity of zero, the operative part of the activation function shows linear behavior. That is, the operative part of a sigmoid function is almost linear for small weights. The NN then collapses to an approximately linear model. Note that a linear model is the simplest model, and we can avoid overfitting by constraining the weights to be small. This gives us a hint on why we initialize the random weights to be close to zero. If the weights are large, the model is more complex and the activation function tends to be nonlinear.

Note that it is not necessarily bad to go to the nonlinear section of the activation function. In fact we use nonlinear activation functions to increase the ability of neural networks and make it possible to estimate nonlinear functions. What we must avoid is using the nonlinear section more than required, which would result in overfitting the training data. For this reason we add a penalty term to the error function.

Our goal is keeping the weights small. Formally, we penalize nonlinear weights by adding a penalty term to the error function; the penalty term in weight decay, by definition, penalizes large weights. The usual penalty is the sum of squared weights times a decay constant. In a linear model, this form of weight decay is equivalent to ridge regression. Now the regularized error function becomes:

$\,REG = err + \lambda( \sum_{ij}|u_{ij}|^2)$, where $\,err$ is the original error in back-propagation;and it decreases all the time; $\,u_{ij}$ is the weights of the hidden layers.

Usually, we use $\,\lambda( \sum_{ij}|u_{ij}|^2)$ to control the value of the weights. We can use cross validation to estimate $\,\lambda$.Another approach to choosing the $\,\lambda$ is to train several networks with different amounts of decay and estimates the generalization error for each; then choose the $\,\lambda$ that minimizes the estimated generalization error.

A similar penalty, weight elimination, is given by,

$\,REG = err + \lambda(\sum_{jk}\frac{|u_{jk}|^2}{1+|u_{jk}|^2})$.

As in back-propagation, we take partial derivative with respect to the weights:

$\frac{\partial REG}{\partial u_{ij}} = \frac{\partial err}{\partial u_{ij}} + 2\lambda u_{ij}$

$u^{new} \leftarrow u^{old} - \rho\left(\frac{\partial err}{\partial u} + 2\lambda u^{old}\right)$

To conclude, the weight decay penalty term lead the weights to converge to smaller absolute values than they otherwise would. Large weights can effect generalization negatively in two different ways. Excessively large weights leading to hidden units can cause the output function to be too rough, possibly with near discontinuities. Excessively large weights leading to output units can cause wild outputs far beyond the range of the data if the output activation function is not bounded to the same range as the data. In another words, large weights can cause large variance of the output [36]. According to [37], the size (L_1 norm) of the weights is more important than the number of weights in determining generalization.

Note:
here $\,\lambda$ serves as a trade-off parameter, tuning between the error rate and the linearity. Actually, we may also set $\,\lambda$ by cross-validation. The tuning parameter is important since weights of zero will lead to zero derivatives and the algorithm will not change. On the other hand, starting with weights that are too large means starting with a nonlinear model which can often lead to poor solutions. <ref>Trevor Hastie, Robert Tibshirani, Jerome Friedman, Elements of Statistical Learning (Springer 2009) pp.398</ref>
We can standardize or normalize the inputs and targets, or adjust the penalty term for the standard deviations of all the inputs and targets in order to omit the biases and get good result from weight decay.
$\,\lambda$is different for different types of weights in the NN. We can have different $\,\lambda$ for input-to-hidden, hidden-to-hidden, and hidden-to-output weights.

For more reading about the effect of weight decay training for backpropagation on noisy data sets please refer to [38] and how weight decay can improve generalization in feed forward network refer to [39]

#### Methods to estimate the weight decay parameter

One of the biggest problems in weight decay for neural networks is how to estimate its parameter. There are many ways proposed in the literature to estimate the weight decay parameter. The following papers are good start for some one who is looking for further reading.

1- On the selection of weight decay parameter for faulty networks here

2- A Simple Trick for Estimating the Weight Decay Parameter here

The generalization ability of the network can depend crucially on the decay constant, especially with small training sets. One approach to choosing the decay constant is to train several networks with different amounts of decay and estimate the generalization error for each; then choose the decay constant that minimizes the estimated generalization error.

There are other important considerations for getting good results from weight decay. You must either standardize the inputs and targets, or adjust the penalty term for the standard deviations of all the inputs and targets. It is usually a good idea to omit the biases from the penalty term.

A fundamental problem with weight decay is that different types of weights in the network will usually require different decay constants for good generalization. At the very least, you need three different decay constants for input-to-hidden, hidden-to-hidden, and hidden-to-output weights. Adjusting all these decay constants to produce the best estimated generalization error often requires vast amounts of computation.

Fortunately, there is a superior alternative to weight decay: hierarchical Bayesian learning. Bayesian learning makes it possible to estimate efficiently numerous decay constants.For information about bayesian learning, please refer to Bayesian inference

## Support Vector Machine - November 09, 2010

### Introduction

Through the course we have seen different methods for solving linearly separable problems, e.g.: Linear regression, LDA, Neural Networks. In most cases we can find a lot of linear boundaries for a problem that separates classes (see figure 1) and all having the same training error. A question arises, which of these boundaries is optimal and have minimum true eror? Answer to this question lead to a new type of classifiers called Support Vector Machines (SVM). SVMs are a set of supervised learning methods. The original algorithm was proposed by Vladmier Vapnik and later formulated to what is in current literature by Corinna Cortes and Vapnik. Though the idea propsed by Vapnik in 1960s, since it was published in Russian language, SVM did not get popular until early 1990s when Vapnik, after immigrating to US, published his work in English. SVM was introduced after neural networks but get popular due to outperformance compared to neural networks in many applications e.g. bioinformatics, text, image recognition, … until recently when notion of deep network was introduced by Hinton which outperforms SVM in some applications. A support vector machine constructs a hyperplane which can be used as classification boundary. These linear decision boundaries explicitly try to separate the data into different classes while maximizing the margin of separation. Intuitively, a good separation is achieved by the hyperplane that has the largest distance to the nearest training data points of any class since in general the larger the margin the lower is the generalization error of the classifier. The techniques that make the extensions to the non-separable case, where the classes overlap, are generalized to what is known as the support vector machine. It produces nonlinear boundaries by constructing a linear boundary in a high-dimensional, transformed version of the feature space. It is also calculated based on only a fraction of the data rather than a calculation on every point in the data, much like the difference between the median and the mean. SVM can also be considered a special case of Tikhonov regularization. A special property is that they simultaneously minimize the empirical classification error and maximize the geometric margin; hence they are also known as maximum margin classifiers. The key features of SVMs are the use of kernels, the absence of local minima, the sparseness of the solution and the capacity control obtained by optimising the margin."(Shawe-Taylor and Cristianini (2004)).SVM has been used successfully in many real-world problems

- Pattern Recognition (Face Detection [17], Face Verification [18], Object Recognition [19], Handwritten Character/Digit Recognition [20], Speaker/Speech Recognition [21], Image Retrieval [22], Prediction [23])

- Text (and hypertext) categorization

- Image classification

- Bioinformatics (Protein classification, Cancer classification)

For a complete list of SVM application please refer to [40].

### Optimal Seperating Hyperplane

As can be seen in figure 2 there may exist many linear hyperplane between the calsses. The data point which are indicated in figure 3 are the datapoints that the Margin is pushed up against and are called Support Vectors.

Fig. 1 Linear Classifiers
Fig. 2 Maximum Margin
Fig. 3 The linear algebra of a hyperplane

### Some facts about the geometry of hyperplane

Figure 3 shows the linear algebra of the hyperplane, where $\,d_i$ is the distnace between the origion and a point $\,x_i$ on the hyperplane.

Suppose hyperplane is defined as $\displaystyle \beta^{T}x_1+\beta_0=0$ as shown in figure1 and suppose that data is linearly separable and $\displaystyle y_i \in \{-1,1 \}$.

$\displaystyle \beta_0$ is distance of the hyperplane to the origin

1)

$\displaystyle \beta$ is orthogonal to the hyperplane

Suppose that $\displaystyle x_1,x_2$ are lying on the hyperplane.

$\displaystyle \beta^{T}x_1+\beta_0=0$

$\displaystyle \beta^{T}x_2+\beta_0=0$

therefore : $\displaystyle \beta^{T}x_1+\beta_0 - (\beta^{T}x_2+\beta_0)=0$

$\displaystyle \beta^{T}(x_1-x_2)=0$

Hence,$\displaystyle \beta \bot \displaystyle (x_1 - x_2)$. But $\displaystyle x_1-x_2$ is a vector lying in the hyperplane and since the two points were arbitrary so $\displaystyle \beta$ is orthogonal to every vector lying in the hyperplane and by definition orthogonal to hyperplane

2)

For any point $\displaystyle x_0$ on the hyperplane, we can say that

$\displaystyle \beta^{T}x_0+\beta_0=0$

$\displaystyle \beta^{T}x_0=-\beta_0$ For any point on the hyperplane, multiplying by $\displaystyle \beta^{T}$ gives negative value of the intercept of the hyperplane.

3)

For any point $\displaystyle x_i$ the distance of the point tp the hyperplane is denoted by $\displaystyle d_i$, which is the projection of ($\displaystyle x_i - x_0$) on $\displaystyle\beta$ The signed distance for any point $\displaystyle x_i$ to the hyperplane is $\displaystyle d_i = \beta^{T}(x_i - x_0)$.
Since the length of $\displaystyle \beta$ changes the value of distance , normalize it by diving it into its length.

$\displaystyle d_i=\frac{\beta^{T}(x_i-x_0)}{\|\beta\|}$ $\displaystyle i=1,2,....,N$

$\displaystyle d_i=\frac{\beta^{T}x_i-\beta^{T}x_0}{\|\beta\|}$

by property 2

$\displaystyle d_i=\frac{\beta^{T}x_i+\beta_0}{\|\beta\|}$

Therefore, for any point if we want to find its dictance to the hyperplane we simply put it in this equation and normalize it.

4)

We use lables to make the distnace positive, therefore: $\displaystyle Margin=(y_id_i)$ and since we would like to minimize the Margin:

$\displaystyle Margin=min(y_id_i)$ $\displaystyle i=1,2,....,N$, and since we now know how to compute $\displaystyle d_i \Rightarrow$

$\displaystyle Margin=min\{y_i\frac{\beta^{T}x_i+\beta_0}{\|\beta\|}\}$

$\displaystyle y_i(\beta^{T}x_i+\beta_0)\geq1 \gt 0$

Since it is distance it is always positive if the point is on the hyperplane it is zero if not on the hyperplane it is greater than zero. For all points that are not on the hyperplane: $\displaystyle x_i$

$\displaystyle y_i(\beta^{T}x_i+\beta_0)\gt 0$

So we can conclude there exists $\displaystyle c \gt 0$ such that $\displaystyle y_i(\beta^{T}x_i+\beta_0)\ c$ for all $\displaystyle i$ where $\displaystyle c$ is the minimum distance to the hyperplane

$\displaystyle y_i(\frac{\beta^{T}x_i}{c}+\frac{\beta_0}{c})\geq1$

This is known as the canonical representation of the decision hyperplane.

For $\displaystyle \beta^{T}$ only the direction is important, so $\displaystyle \frac{\beta^{T}}{c}$ does not change its direction, the hyperplane will be the same.

$\displaystyle y_i(\beta^{T}x_i+\beta_0)\geq1$

$\displaystyle y_i\frac{\beta^{T}x_i+\beta_0}{\|\beta\|}\geq1 \frac{1}{\|\beta\|}$

$\displaystyle y_i \beta^{T}x_i+\beta_0 \geq 1$

Therefore, in order to maximize the Margine we have to minimize the $\beta$

minimize $\displaystyle\|\beta\|^2$

minimize $\displaystyle\frac{1}{2}\|\beta\|^2$ s.t $\displaystyle y_i(\beta^T x_i + \beta_0) \geq 1 \forall$ i

for the $\displaystyle\beta$ s which have distances greater or equal to one. We have used $\displaystyle\frac{1}{2}$ factor only for convenience in derivation of derivative.

#### Writing Lagrangian Form of Support Vector Machine

The Lagrangian form is introduced to ensure that the optimization conditions are satisfied, as well as finding an optimal solution (the optimal saddle point of the Lagrangian for the classic quadratic optimization). The problem will be solved in dual space by introducing $\,\alpha_i$ as dual constraints, this is in contrast to solving the problem in primal space as function of the betas. A simple algorithm for iteratively solving the Lagrangian has been found to run well on very large data sets, making SVM more usable. Note that this algorithm is intended to solve Support Vector Machines with some tolerance for errors - not all points are necessarily classified correctly. Several papers by Mangasarian explore different algorithms for solving SVM.

Dual form of the optimization problem:

$\,L(\beta,\beta_0,\alpha) = \frac{1}{2}\|\beta\|^2 - \sum_{i=1}^n{\alpha_i\left(y_i(\beta^Tx_i+\beta_0)-1\right)}$.

To find the optimal value, set the derivative equal to zero.

$\,\frac{\partial L}{\partial \beta} = 0$, $\,\frac{\partial L}{\partial \beta_0} = 0$. Note that $\,\frac{\partial L}{\partial \alpha_i}$ is equivalent to the constraints $\left(y_i(\beta^Tx_i+\beta_0)-1\right) \geq 0, \,\forall\, i$

First, $\,\frac{\partial L}{\partial \beta} = 0$

$\,\frac{\partial L}{\partial \beta} = \frac{\partial}{\partial \beta}\frac{1}{2}\|\beta\|^2 - \sum_{i=1}^n{\left\{\frac{\partial}{\partial \beta}(\alpha_iy_i\beta^Tx_i)+\frac{\partial}{\partial \beta}\alpha_iy_i\beta_0-\frac{\partial}{\partial \beta}\alpha_iy_i\right\}}$

$\frac{\partial}{\partial \beta}\frac{1}{2}\|\beta\|^2 = \beta$.
$\,\frac{\partial}{\partial \beta}(\alpha_iy_i\beta^Tx_i) = \alpha_iy_ix_i$
$\,\frac{\partial}{\partial \beta}\alpha_iy_i\beta_0 = 0$.
$\,\frac{\partial}{\partial \beta}\alpha_iy_i = 0$.

So this simplifies to $\,\frac{\partial L}{\partial \beta} = \beta - \sum_{i=1}^n{\alpha_iy_ix_i} = 0$. In other words,

$\,\beta = \sum_{i=1}^n{\alpha_iy_ix_i}$, $\,\beta^T = \sum_{i=1}^n{\alpha_iy_ix_i^T}$

Similarly, $\,\frac{\partial L}{\partial \beta_0} = \sum_{i=1}^n{\alpha_iy_i} = 0$.

$\,\frac{1}{2}\sum_{i=1}^n{\sum_{j=1}^n{\alpha_i\alpha_jy_iy_jx_i^Tx_j}} - \sum_{i=1}^n{\sum_{j=1}^n{\alpha_i\alpha_jy_iy_jx_i^Tx_j}} + \sum_{i=1}^n{\alpha_i}$

$\underset{\alpha}{\min} \sum_{i=1}^n{\alpha_i}- \,\frac{1}{2}\sum_{i=1}^n{\sum_{j=1}^n{\alpha_i\alpha_jy_iy_jx_i^Tx_j}}$

A dual representation of the maximum margin.

Because $\,\alpha_i$ is the Lagrange multiplier, $\,\alpha_i \geq 0 \forall i$.

Therefore, we have a new optimization problem:

$\underset{\alpha}{\min} \sum_{i=1}^n{\alpha_i}- \,\frac{1}{2}\sum_{i=1}^n{\sum_{j=1}^n{\alpha_i\alpha_jy_iy_jx_i^Tx_j}}$

1) $\,\alpha_i$ > 0

2) $\sum{\alpha_i y_i}$ = $0$

This is a much simpler optimization problem and we can solve it by [41]. Quadratic programming (QP) is a special type of mathematical optimization problem. It is the problem of optimizing (minimizing or maximizing) a quadratic function of several variables subject to linear constraints on these variables.

#### Quadratic Programming Problem of SVMs and Dual Problem

We have to find $\,\beta$ and $\,\beta_0$ such that $\,\frac{1}{2}\|\beta\|^2$ is minimized subject to $\,y_i (\beta^T x_i + \beta_0) \geq 1 \forall i$.

Therefore, we need to optimize a quadratic function subject to linear constraints. Quadratic optimization problems are a well-known class of mathematical programming problems, and many (rather intricate) algorithms exist for solving them. The solution involves constructing a dual problem where a Lagrange multiplier $\,\alpha_i$ is associated with every constraint in the primary problem.

The optimization goal is quadratic and the constraints are linear, it is a typical QP. Given such a constrained optimization problem, it is possible to construct another problem called the dual problem. We may now state the dual problem: given the training sample, find the Lagrange multipliers that maximize the objective function

$\,Q(\alpha)= \underset{\alpha}{\min} \sum_{i=1}^n{\alpha_i}- \,\frac{1}{2}\sum_{i=1}^n{\sum_{j=1}^n{\alpha_i\alpha_jy_iy_jx_i^Tx_j}}$

subject to the constraints

1) $\,\alpha_i$ > 0

2) $\sum{\alpha_i y_i}$ = $0$

### Implementation

The parameters of the maximum-margin hyperplane are derived by solving the optimization. There exist several specialized algorithms for quickly solving the QP problem that arises from SVMs, mostly reliant on heuristics for breaking the problem down into smaller, more-manageable chunks. A common method for solving the QP problem is Platt's Sequential Minimal Optimization (SMO) algorithm, which breaks the problem down into 2-dimensional sub-problems that may be solved analytically, eliminating the need for a numerical optimization algorithm. Another approach is to use an interior point method that uses Newton-like iterations to find a solution of the Karush-Kuhn-Tucker conditions of the primal and dual problems.[10] Instead of solving a sequence of broken down problems, this approach directly solves the problem as a whole. To avoid solving a linear system involving the large kernel matrix, a low rank approximation to the matrix is often used to use the kernel trick.Please refer to [42] for code implementation of SVM.

### Multiclass SVM

SVM is only directly applicable for two-class case. We want to generalize this algorithm to multi-class tasks. Multiclass SVM aims to assign labels to instances by using support vector machines, where the labels are drawn from a finite set of several elements. The dominating approach for doing so is to reduce the single multiclass problem into multiple binary problems. Each of the problems yields a binary classifier, which is assumed to produce an output function that gives relatively large values for examples from the positive class and relatively small values for examples belonging to the negative class. Two common methods to build such binary classifiers are where each classifier distinguishes between (i) one of the labels to the rest (one-versus-all) or (ii) between every pair of classes (one-versus-one). Classification of new instances for one-versus-all case is done by a winner-takes-all strategy, in which the classifier with the highest output function assigns the class (it is important that the output functions be calibrated to produce comparable scores). For the one-versus-one approach, classification is done by a max-wins voting strategy, in which every classifier assigns the instance to one of the two classes, then the vote for the assigned class is increased by one vote, and finally the class with most votes determines the instance classification.

### Support Vector Machines vs Artificial Neural Networks

The development of ANNs followed a heuristic path, with applications and extensive experimentation preceding theory. In contrast, the development of SVMs involved sound theory first, then implementation and experiments. A significant advantage of SVMs is that whilst ANNs can suffer from multiple local minima, the solution to an SVM is global and unique. Two more advantages of SVMs are that that have a simple geometric interpretation and give a sparse solution. Unlike ANNs, the computational complexity of SVMs does not depend on the dimensionality of the input space. ANNs use empirical risk minimization, whilst SVMs use structural risk minimization. The reason that SVMs often outperform ANNs in practice is that they deal with the biggest problem with ANNs, SVMs are less prone to overfitting. In contrast to neural networks SVMs automatically select their model size (by selecting the Support vectors)(Rychetsky (2001)).While the weight decay term is an important aspect for obtaining good generalization in the context of neural networks for regression, the margin plays a somewhat similar role in classification problems.

### Advantages of Support Vector Machines

• SVMs provide a good out-of-sample generalization. This means that, by choosing an appropriate generalization grade,

SVMs can be robust, even when the training sample has some bias. This is mainly due to selection of optimal hyperplane.

• SVMs deliver a unique solution, since the optimality problem is convex. This is an advantage compared

to Neural Networks, which have multiple solutions associated with local minima and for this reason may not be robust over different samples.

### Disadvantages of Support Vector Machines [43]

• Perhaps the biggest limitation of the support vector approach lies in choice of the kernel (Which we will study about in future).
• A second limitation is speed and size, both in training and testing (mostly in training - for large training sets, it typically selects a small number of support vectors, thereby minimizing the computational requirements during testing).
• Discete data presents another problem.
• The optimal design for multiclass SVM classifiers is a further area for research.
• Although SVMs have good generalization performance, they can be abysmally slow in test phase.
• Besides the advantages of SVMs - from a practical point of view - they have some drawbacks. An important practical question that is not entirely solved, is the selection of the kernel function parameters - for Gaussian kernels the width parameter [sigma] - and the value of [epsilon] in the [epsilon]-insensitive loss function.
• However, from a practical point of view perhaps the most serious problem with SVMs is the high algorithmic complexity and extensive memory requirements of the required quadratic programming in large-scale tasks.

### Applications of Support Vector Machines

The following papers describe some of the possible applications of support vector machines:

1- Training support vector machines: an application to face detection here

2- Application of support vector machines in financial time series forecasting here

3- Support vector machine active learning with applications to text classification here

4- Support Vector Machines: Training and Applications here

5- Application of support vector machines for T-cell epitopes prediction here

### References

1. V. Vapnik and A. Chervonenkis, Theory of Pattern Recognition, Nauka, Moscow, 1974.

2. V. Vapnik, Estimation of Dependencies Based on Empirical Data, Nauka, Moscow, 1979.

3. V. Vapnik, The Nature of Statistical Learning Theory, Springer, 1995.

4. V. Vapnik, Statistical Learning Theory, Wiley-Interscience, New York, 1998.

5. P. H. Chen, C. J. Lin, and B. Schölkopf, A tutorial on ν-support vector machines, Appl. Stoch. Models. Bus. Ind. 21, 111-136, 2005.

6. S.-W. Lee and A. Verri (Eds.),Applications of Support Vector Machines for Pattern Recognition: SVM 2002, LNCS 2388, pp. 213-236, 2002.

7. V. D. Sanchez, Advanced support vector machines and kernel methods, Neurocomputing 55, 5-20, 2003.

8. C. Campbell, Kernel methods: a survey of current techniques, Neurocomputing, 48, 63-84, 2002.

9. K. R. Müller, S. Mika, G. Rätsch, K. Tsuda, and B. Schölkopf, An introduction to kernel-based learning algorithms, IEEE Trans. Neural Netw., 12, 181-201, 2001.

10. J. A. K. Suykens, Support vector machines: A nonlinear modelling and control perspective, Eur. J. Control, 7, 311-327, 2001.

11. V. N. Vapnik, An overview of statistical learning theory, IEEE Trans. Neural Netw., 10, 988-999, 1999.

12. B. Schölkopf, S. Mika, C. J. C. Burges, P. Knirsch, K. R. Muller, G. Ratsch, and A. J. Smola, Input space versus feature space in kernel-based methods, IEEE Trans. Neural Netw., 10, 1000-1017, 1999.

13. C. J. C. Burges, A tutorial on Support Vector Machines for pattern recognition, Data Min. Knowl. Discov., 2, 121-167, 1998.

14. A. J. Smola and B. Schölkopf, On a kernel-based method for pattern recognition, regression, approximation, and operator inversion, Algorithmica, 22, 211-231, 1998.

15. K. Jonsson, J. Kittler, and Y.P. Matas, Support vector machines for face authentication, Journal of Image and Vision Computing, vol. 20. pp. 369-375, 2002.

16. A. Tefas, C. Kotropoulos, and I. Pitas, Using support vector machines to enhance the performance of elastic graph matching for frontal face authentication, IEEE Transaction on Pattern Analysis and Machine Intelligence, vol. 23. No. 7, pp. 735-746, 2001.

17. E. Osuna, R. Freund, and F. Girosi, Training support machines: An application to face detection. In Proceedings of IEEE Conference on Computer Vision and Pattern Recognition, pp. 130-136, 1997.

18. Y. Wang, C.S. Chua, and Y.K, Ho. Facial feature detection and face recognition from 2D and 3D images, Pattern Recognition Letters, Feb., 2002.

19. Q. Tian, P. Hong, and T.S. Huang, Update relevant image weights for contentbased image retrieval using support vector machines, In Proceedings of IEEE Int. Conference on Multimedia and Expo, vol.2, pp. 1199-1202, 2000.

20. D. Gorgevik, D. Cakmakov, and V. Radevski, Handwritten digit recognition by combining support vector machines using rule-based reasoning, In Proceedings of 23rd Int. Conference on Information Technology Interfaces, pp. 139-144, 2001.

21. V. Wan and W.M. Campbell, Support vector machines for speaker verification and identification, In Proceedings of IEEE Workshop on Neural Networks for Signal Processing X, vol. 2, 2000.

22. A. Fan and M. Palaniswami, Selecting bankruptcy predictors using a support vector machine approach, vol. 6, pp. 354-359, 2000.

## Support Vector Machine Cont., Kernel Trick - November 11, 2010

Recall in the previous lecture that instead of solving the primal problem of maximizing the margin, we can solve the dual problem without changing the solution as long as it subjects to the Karush-Kuhn-Tucker (KKT) conditions. Leading to the following:

$\min_{\alpha} L(\alpha) = \sum_{i=1}^n{\alpha_i} - \frac{1}{2}\sum_{i=1}^n{\sum_{j=1}^n{\alpha_i\alpha_jy_iy_jx_i^Tx_j}}$

such that $\alpha_i \geq 0$
and $\sum_{i=1}^n{\alpha_i y_i} = 0$

We are looking to minimize $\,\alpha$, which is our only unknown. Once we know $\,\alpha$, we can easily find $\,\beta$ and $\,\beta_0$ (see the Support Vector algorithm below for complete details).

If we examine the Lagrangian equation, we can see that $\,\alpha$ is multiplied by itself; that is, the Lagrangian is quadratic with respect to $\,\alpha$. Our constraints are linear. This is therefore a problem that can be solved through quadratic programming techniques.

We can write the Lagrangian equation in matrix form:

$L(\alpha) = \underline{\alpha}^T\underline{1} - \frac{1}{2}\underline{\alpha}^TS\underline{\alpha}$

such that $\underline{\alpha} \geq \underline{0}$
and $\underline{\alpha}^T\underline{y} = 0$

Where:

• $\underline{\alpha}$ denotes an $\,n \times 1$ vector; $\underline{\alpha}^T = [\alpha_1, ..., \alpha_n]$
• Matrix $S = y_iy_jx_i^Tx_j = (y_ix_i)^T(y_ix_i)$
• $\,\underline{0}$ and $\,\underline{1}$ are vectors containing all 0s or all 1s respectively

Using this matrix notation, we can use Matlab's built in quadratic programming routine, quadprog.

Note

Matlab's quadprog function minimizes an equation of the following form:

$\min_x\frac{1}{2}x^THx+f^Tx$
such that: $\,A \cdot x \leq b$, $\,A_{eq} \cdot x = b_{eq}$ and $\,lb \leq x \leq ub$

The function is called as such: x = quadprog(H,f,A,b,Aeq,beq,lb,ub). The variables correspond to values in the equation above.

An example of running "quadprog" method in matla:

$\min_x\frac{1}{2}x_1^2+x_2^2-x_1x_2-2x_1-6x_2$
$\,s.t. x_1+x_2 \leq 2$, $\,-x_1+2x_2 \leq 2$, $\,2x_1+x_2 \leq 3$, $\,0 \leq x_1$,$\,x_2$

### Examining K.K.T. conditions

Karush-Kuhn-Tucker conditions (more info) give us a closer look into the Lagrangian equation and the associated conditions.

Suppose we are looking to minimize $\,f(x)$ such that $\,g_i(x) \geq 0, \forall{x}$. If $\,f$ and $\,g$ are differentiable, then the necessary conditions for $\hat{x}$ to be a local minimum are:

1. At the optimal point, $\frac{\partial L}{\partial \hat{x}} = 0$; i.e. $f'(\hat{x}) - \sum{\alpha_ig'(\hat{x})}=0$
2. $\alpha_i \geq 0$. (Dual Feasibility)
3. $\alpha_ig_i({x}) = 0, \forall{i}$ (Complementary Slackness)
4. $g_i(\hat{x}) \geq 0$ (Primal Feasibility)

If any of these conditions are violated, then the problem is deemed not feasible.

These are all trivial except for condition 3. Let's examine it further in our support vector machine problem.

### Support Vectors

Support vectors are the training points that determine the optimal separating hyperplane that we seek. Also, they are the most difficult points to classify and at the same time the most informative for classification.

In our case, the $g_i({x})$ function is:

$\,g_i(x) = y_i(\beta^Tx_i+\beta_0)-1$

Substituting $\,g_i$ into KKT condition 3, we get $\,\alpha_i[y_i(\beta^Tx_i+\beta_0)-1] = 0$. <br\>In order for this condition to be satisfied either
$\,\alpha_i= 0$ or
$\,y_i(\beta^Tx_i+\beta_0)=1$

All points $\,x_i$ will be either 1 or greater than 1 distance away from the hyperplane.

Case 1: a point away from the margin

If $\,y_i(\beta^Tx_i+\beta_0) \gt 1 \Rightarrow \alpha_i = 0$.

If point $\, x_i$ is not on the margin, then the corresponding $\,\alpha_i=0$.

Case 2: a point on the margin

If $\,y_i(\beta^Tx_i+\beta_0) = 1 \Rightarrow \alpha_i \gt 0$. <br\>If point $\, x_i$ is on the margin, then the corresponding $\,\alpha_i\gt 0$.

Points on the margin, with corresponding $\,\alpha_i \gt 0$, are called support vectors. {{

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### The support vector machine algorithm

1. Solve the quadratic programming problem:$\min_{\alpha} L(\alpha) = \sum_{i=1}^n{\alpha_i} - \frac{1}{2}\sum_{i=1}^n{\sum_{j=1}^n{\alpha_i\alpha_jy_iy_jx_i^Tx_j}}$ such that $\alpha_i \geq 0$ and $\sum_{i=1}^n{\alpha_i y_i} = 0$
(Use Matlab's quadprog to find the optimal $\,\underline{\alpha}$)
2. Find $\beta = \sum_{i=1}^n{\alpha_iy_i\underline{x_i}}$
3. Find $\,\beta_0$ by choosing a support vector (a point with $\,\alpha_i \gt 0$) and solving $\,y_i(\beta^Tx_i+\beta_0) = 1$

The support vector machine algorithm is insensitive to outliers. If $\,\alpha = 0$, then the cost function is also 0, and won't contribute to the solution of the SVM problem; only points on the margin — support vectors — contribute. Hence the model given by SVM in entirely defined by the set of support vectors, a subset of the entire training set. In this case we have a data-driven or 'nonparametric' model in which is the training set and algorithm will determine the support vectors, instead of fitting a set of parameters.

References: Wang, L, 2005. Support Vector Machines: Theory and Applications, Springer, 3

### Kernel Trick

We talked about the curse of dimensionality at the beginning of this course, however, we now turn to the power of high dimensions in order to find a linearly separable hyperplane between two classes of data points. To understand this, imagine a two dimensional prison where a two dimensional person is constrained. Suppose magically we give the person a third dimension, then he can escape from the prison. In other words, the prison and the person are linearly separable now with respect to the third dimension. The intuition behind the "kernel trick" is basically to map data to a higher dimension so that they are linearly separable by a hyperplane.

Imagine the point is a person. They're stuck.
Escape through the third dimension!
It's not possible to put a hyperplane through these points.
After a simple transformation, a perfect classification plane can be found.

The original optimal hyperplane algorithm proposed by Vladimir Vapnik in 1963 was a linear classifier. However, in 1992, Bernhard Boser, Isabelle Guyon and Vapnik suggested a way to create non-linear classifiers by applying the kernel trick to maximum-margin hyperplanes. The algorithm is very similar, except that every dot product is replaced by a non-linear kernel function as below. This allows the algorithm to fit the maximum-margin hyperplane in a transformed feature space. We have seen SVM as a linear classification problem that finds the maximum margin hyperplane in the given input space. However, for many real world problems a more complex decision boundary is required. The following simple method was devised in order to solve the same linear classification problem but in a higher dimensional space, a 'feature space', under which the maximum margin hyperplane is better suited.

Let $\,\phi$ be a mapping,

$\phi:\mathbb{R}^d \rightarrow \mathbb{R}^D$

We wish to find a $\,\phi$ such that our data will be suited for separation by a hyperplane. Given this function, we are lead to solve the previous constrained quadratic optimization on the transformed dataset,

$\min_{\alpha} L(\alpha) = \sum_{i=1}^n{\alpha_i} - \frac{1}{2}\sum_{i=1}^n{\sum_{j=1}^n{\alpha_i\alpha_jy_iy_j\phi(x_i)^T\phi(x_j)}}$ such that $\alpha_i \geq 0$ and $\sum_{i=1}^n{\alpha_i y_i} = 0$

The solution to this optimization problem is now well known; however a workable $\,\phi$ must be determined. Possibly the largest drawback in this method is that we must compute the inner product of two vectors in the high dimensional space. As the number of dimensions in the initial data set increases, the inner product becomes computationally intensive or impossible.

However, we have a very useful result that says that there exists a class of functions, $\,\Phi$, which satisfy the above requirements and that for any function $\,\phi \in \Phi$,

$\,\phi(x_i)^T\phi(x_j) = K(x_i,x_j)$

Where K is the kernel function in the input space satisfying Mercer's condition (to guarantee that it indeed corresponds to certain mapping function $\,\phi$). As a result, if the objective function depends on inner products but not on coordinates, we can always use the kernel function to implicitly calculate in the feature space without storing the huge data. Not only does this solve the computation problems but it no longer requires us to explicitly determine a specific mapping function in order to use this method. In fact, it is now possible to use an infinite dimensional feature space in SVM without even knowing the function $\,\phi$.

• one may look at $\,x_i^T x_j$ as way of measuring similarity, where $\,K(\underline{x}_i,\underline{x}_j)$ is another way of measuring similarity between $\,x_i$ and $\,x_j$

#### Popular kernel choices for SVM

There are many types of kernels that can be used in Support Vector Machines models. These include linear, polynomial and radial basis function (RBF).

Linear: $\ K(\underline{x}_{i},\underline{x}_{j})= \underline{x}_{i}^T\underline{x}_{j}$,

Polynomial: $\ K(\underline{x}_{i},\underline{x}_{j})= (\gamma\underline{x}_{i}^T\underline{x}_{j}+r)^{d}, \gamma \gt 0$,

Radial Basis: $\ K(\underline{x}_{i},\underline{x}_{j})= exp(-\gamma \|\underline{x}_i - \underline{x}_j\|^{2}), \gamma \gt 0$,

Gaussian kernel: $\ K(x_i,x_j)=exp(\frac{-||x_i-x_j||^2}{2\sigma^2 })$,

The RBF is by far the most popular choice of kernel types used in Support Vector Machines. This is mainly because of their localized and finite responses across the entire range of the real x-axis.The art of flexible modeling using basis expansions consists of picking an appropriate family of basis functions, and then controlling the complexity of the representation by selection, regularization, or both. Some of the families of basis functions have elements that are defined locally; for example, $\displaystyle B$-splines are defined locally in $\displaystyle R$. If more flexibility is desired in a particular region, then that region needs to be represented by more basis functions(which in the case of $\displaystyle B$-splines translates to more knots).Kernel methods achieve flexibility by fitting simple models in a region local to the target point $\displaystyle x_0$. Localization is achieved via a weighting kernel $\displaystyle K$ and individual observations receive weights $\displaystyle K(x_0,x_i)$. RBF combine these ideas, by treating the kernel functions as basis functions.

Once we have chosen the Kernel function, we don't need to figure out what $\,\phi$ is, just use $\,\phi(\underline{x}_i)^T\phi(\underline{x}_j) = K(\underline{x}_i,\underline{x}_j)$ to replace $\,\underline{x}_i^T\underline{x}_j$

Since the transformation chosen is dependent on the shape of the data, the only automated way to choose an appropriate kernel is by trial and error. Otherwise it is chosen manually.

### Example in Matlab

The following code, taken verbatim from the lecture, shows how to use Matlab built-in SVM routines (found in the Bioinformatics toolkit) to do classification through support vector machines.

load 2_3;
[U,Y] = princomp(X');
data = Y(:,1:2);
l = [-ones(1,200) ones(1,200)];
[train,test] = crossvalind('holdOut',400);
% Gives indices of train and test; so, train is a matrix of 0 or 1, 1 where the point should be used as part of the training set
svmStruct = svmtrain(data(train,:), l(train), 'showPlot', true);

The plot produced by training on some of the 2_3 data's first two features.
yh = svmclassify(svmStruct, data(test,:), 'showPlot', true);

The plot produced by testing some of the 2_3 data.
% SVM kernel trick using rbf as the kernel
yh = svmclassify(svmStruct, data(test,:), 'showPlot', true, 'Kernel_Function','rbf');


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#### Extention:Support Vector Machines for Pattern Recognition

[44] This paper talks about linear Support Vector Machines for separable and non-separable data by working through a non-trivial example in detail, and also it describes a mechanical analog and when SVM solutions are unique and when they are global. From this paper we can know support vector training can be practically implemented, and the kernel mapping technique which is used to construct SVM solutions which are nonlinear in the data.

Results of some experiments which were inspired by these arguments are also presented. The writer gives numerous examples and proofs of most of the key theorems, he hopes the people can find old material is cast in a fresh light since the paper includes some new material.