# Foundations of Machine Learning, part 2

This post is the sixth one of our series on the history and foundations of econometric and machine learning models. The first fours were on econometrics techniques. Part 5 is online here.

## The probabilistic formalism in the 80’s

We have a training sample, with observations $(\mathbf{x}_i,y_i)$ where the variables $y$ are in a set $\mathcal{Y}$. In the case of classification, $\mathcal{Y}=\{-1,+1\}$, but a relatively general set can be considered (note that if econometricians prefer $\mathcal{Y}=\{0,1\}$ – because of the Bernoulli distribution and because $0$ and $1$ are lower and upper bounds of probabilities, people in the “machine learning” community prefer $\mathcal{Y}=\{-1,+1\}$). A predictor $m$ is an function taking values in $\mathcal{Y}$, used to label (or classify) future new observations, using some features that lie in a set $\mathcal{X}$. It is assumed that the labels are produced by an (unknown) classifier $f$ called target. For a statistician, this function would be the real model. Naturally, we want to build $m$ as close as possible to $f$. Let $\mathbb{P}$ be a (unknown) distribution on $\mathcal{X}$. The error of $m$ with respect to target $f$ is defined by $$\mathcal{R}_{\mathbb{P},f}(m)=\mathbb{P}[m(\boldsymbol{X})\neq f(\boldsymbol{X})]\text{ where }\boldsymbol{X}\sim\mathbb{P}$$or equivalently,$$\mathcal{R}_{\mathbb{P},f}(m)=\mathbb{P}\big[\{\boldsymbol{x}\in\mathcal{X}:m(\boldsymbol{x})\neq f(\boldsymbol{x})\}\big]$$To obtain our “optimal” classifier, it becomes necessary to assume that there is a link between the data in our sample and the pair $(\mathbb{P},f)$, i.e. a data generation model. We will then assume that the $\mathbf{x}_i$ are obtained by independent draws according to $\mathbb{P}$, and that then $y_i=f(\mathbf{x}_i)$ . We can define the empirical risk of a classifier $m$, as $$\widehat{{R}}(m)=\frac{1}{n}\sum_{i=1}^n \boldsymbol{1}(m(\boldsymbol{x}_i)\neq y_i)$$

It is important to recognize that a perfect model cannot be found, in the sense that $R_{\mathbb{P},f} (m)=0$. Indeed, if we consider the simplest case, with $\mathcal{X}=\{x_1,x_2\}$ and $\mathbb{P}$ is such that $\mathbb{P}(\{x_1\})=p$ and $\mathbb{P}(\{x_2\})=1-p$. The probability of never observing $\{x_2\}$ among the $n$ observations is $(1-p)^n$, and if $p<1/n$, it is quite likely never to observe $\{x_2\}$ so it can never be predicted. We cannot therefore hope to have a zero risk whatever $\mathbb{P}$. And more generally, it is also possible to observe $\{x_1\}$ and $\{x_2\}$, and despite everything, to make mistakes on the labels. Also, instead of looking for a perfect model, we can try to have an “approximately correct” model. We will then try to find $m$ such that $R_{\mathbb{P},f} (m)\leq\varepsilon$, where $\varepsilon$ is an a priori specified threshold. But even this condition is too strong, and cannot be fulfilled. Thus, we will usually as to have $R_{\mathbb{P},f} (m)\leq\varepsilon$ with some probability $1-\delta$. Hence, we will try to be “probably approximately correct” (PAC), allowing to make a mistake with a probability $\delta$, again fixed a priori.

Also, when we build a classifier, we do not know either $\mathbb{P}$ or $f$, but we give ourselves a precision criterion $\varepsilon$, and a confidence parameter $\delta$, and we have $n$ observations. Note that $n$, $\varepsilon$ and $\delta$ can be linked. We then look for a model $m$ such that $R_{\mathbb{P},f} (m)\leq\varepsilon$ with probability (at least) $1-\delta$, so that we are probably approximately correct. Wolpert (1996) has shown (see details in Wolpert & Macready (1997)) that there is no universal learning algorithm. In particular, it can be shown that there is $\mathbb{P}$ such that $R_{\mathbb{P},f} (m)$ is relatively high, with a relatively high probability (also).

The interpretation is that since we cannot learn (in the PAC sense) about all the functions $m$, we will then force $m$ to belong to a particular class, noted $\mathcal{M}$. Let us suppose, to start with, that $\mathcal{M}$ contains a finite number of possible models. We can then show that for all $\varepsilon$ and $\delta$, that for all $\mathbb{P}$ and $f$, if we have enough observations (more precisely $n\geq \varepsilon^{-1} \log[\delta^{-1} |\mathcal{M}|]$, then with a greater probability than $1-\delta$, $R_{\mathbb{P},f} (m^\star)\leq\varepsilon$ where$$m^\star \in \underset{m\in\mathcal{M}}{\text{argmin}}\Big\lbrace\frac{1}{n}\sum_{i=1}^n \boldsymbol{1}(m(\boldsymbol{x}_i)\neq y_i)\Big\rbrace$$in other words $m^\star$ is a model in $\mathcal{M}$ that minimizes empirical risk.

We can go a little further, staying in the case where $\mathcal{Y}=\{-1,+1\}$. An $\mathcal{M}$ class of classifiers will be called PAC-learnable if there is $n_M:[0,1]^2\rightarrow \mathbb{N}$ such that, for all $\varepsilon$, $\delta$, $\mathbb{P}$ and if it is assumed that the target $f$ belongs to $\mathcal{M}$, then using $n>n_M (\varepsilon,\delta)$ observations $\mathbf{x}_i$ drawn from $\mathbb{P}$, labelled $y_i$ by $f$, then there is $m\in\mathcal{M}$ such that, with probability $1-\delta$, $R_{\mathbb{P},f} (m)\leq\varepsilon$. The $n_M$ function is then called “sample complexity to learn”. In particular, we have seen that if $M$ contains a finite number of classifiers, then $\mathcal{M}$ is PAC-learnable with complexity $n_M (\varepsilon,\delta)=\varepsilon^{-1} \log[\delta^{-1} |M|]$.

Naturally, we would like to have a more general result, especially if $\mathcal{M}$ is not finite. To do this, the $VC$ dimension of Vapnik-Chervonenkis must be used, which is based on the idea of shattering points (for a binary classification). Consider $k$ points $\{x_1,\cdot,x_k\}$, and consider the set $${E}_k=\big\lbrace(m(\boldsymbol{x}_1),\cdots,m(\boldsymbol{x}_k))\text{ for }m\in\mathcal{M})\big\rbrace$$ Note that the elements of $E_k$ belong to $\{-1,+1\}^k$. In other words, $|E_k |\leq 2^k$. We will say that M shatter all the points if all the combinations are possible, i. e. $|E_k |=2^k$. Intuitively, the labels of the set of points do not provide enough information on target $f$, because anything is possible. The $VC$ dimension of $\mathcal{M}$ is then$$VC(\mathcal{M})=\sup\big\lbrace k\text{ such that }\mathcal{M}\text{ shatters }\{\boldsymbol{x}_1,\cdots\boldsymbol{x}_k\}\big\rbrace$$

For example, if $\mathcal{X}=\mathbb{R}$ and all (simple) models of the form [1] $m_{a,b}=\mathbf{1}_{\pm}(x\in[a,b])$ are considered. No set of $\{x_1,x_2,x_2,x_3\}$ ordered points can be shattered because it is sufficient to assign respectively +1, -1 and +1 to $x_1$, $x_2$ and $x_3$ respectively, therefore $VC<3$. On the other hand $\{0,1\}$ is shattered, so $VC\geq 2$. The dimension of this predictor set is $2$: If we increase by one dimension, $\mathcal{X}=\mathbb{R}^2$ and consider all (simple) models of the form $m_{a,b}=\mathbf{1}_{\pm} (x\in[a,b])$ (where $[a,b]$ refers to the rectangle), then the dimension of $\mathcal{M}$ is here $4$.

To introduce SVMs, let’s place ourselves in the case where $\mathcal{X}=\mathbb{R}^k$, and consider separations by hyperplanes passing through the origin (we will say homogeneous), in the sense that $m_{\mathbf{w}} (\mathbf{x})=\mathbf{1}_{\pm}(\mathbf{w}^T \mathbf{x}\geq 0)$. It can be shown that no set of $k+1$ points can be shattered by these two homogeneous spaces in $\mathbb{R}^k$, and therefore $VC(M)=k$. If we add a constant, in the sense that $m_{\mathbf{w},b} (\mathbf{x})=\mathbf{1}_{\pm}(\mathbf{w}^T \mathbf{x}+b\geq 0)$, we can show that no set of $k+2$ points can be sprayed by these two (non-homogeneous) spaces in $\mathbb{R}^k$, and therefore $VC(M)=k+1$. This dimension reminds us of the dimension of the model we’ve seen in the econometric context.

From this dimension $VC$, we deduce the so-called fundamental theorem of learning: if $\mathcal{M}$ is a class of dimension $d=VC(M)$, then there are positive constants $\underline{C}$ and $\overline{C}$ such as the sample complexity for M to be PAC-learnable satisfies$$\underline{C}\epsilon^{-1}\big(d+\log[\delta^{-1}]\big)\leq n_{\mathcal{M}}(\epsilon,\delta) \leq \overline{C}\epsilon^{-1}\big(d\log[\epsilon^{-1}]+\log[\delta^{-1}]\big)$$The link between the notion of learning (as defined in Vailiant (1984)) and the $VC$ dimension was clearly established in Blumer et al (1989).

Nevertheless, while the work of Vapnik and Chervonenkis is considered to be the foundation of statistical learning, Thomas Cover’s work in the 1960s and 1970s should also be mentioned, in particular Cover (1965) on the capacities of linear models, and Cover & Hart (1967) on learning in the context of the algorithm of the $k$-nearest neighbors. These studies have linked learning, information theory (with the textbook Cover & Thomas (1991)), complexity and statistics. Other authors have subsequently brought the two communities closer together, in terms of learning and statistics. For example, Halbert White proposed to see neural networks in a statistical context in White (1989), going so far as to state that « learning procedures used to train artificial neural networks are inherently statistical techniques. It follows that statistical theory can provide considerable insight into the properties, advantages, and disadvantages of different network learning methods ». This turning point in the late 1980s will anchor learning theory in a probabilistic context.

## Objective and loss function

These choices (of objective and loss function) are essential, and very dependent on the problem under consideration. Let us begin by describing a historically important model, Rosenblatt’s (1958) “perceptron”, introduced into classification problems, where $y\in\{-1,+1\}$, inspired by McCulloch & Pitts (1943). We have data $\{(y_i,\mathbf{x}_i)\}$, and we will iteratively build a set of $m_k[\mathbf{x}$ models, where at each step, we will learn from the errors of the previous model. In the perceptron, a linear model is considered so that :$$m(\mathbf{x})=\boldsymbol{1}_{\pm}(\beta_0+\mathbf{x}^T \boldsymbol{\beta}\geq 0)=\left\lbrace\begin{array}{l}+1\text{ si }\beta_0+\mathbf{x}^T \boldsymbol{\beta}\geq 0\\-1\text{ si }\beta_0+\mathbf{x}^T \boldsymbol{\beta}< 0\end{array}\right.$$where $\beta$ coefficients are often interpreted as “weights” assigned to each of the explanatory variables. We give ourselves initial weights $(\beta_0^{(0)},\beta^{(0)}$, which we will update taking into account the prediction error made, between $y_i$ and the prediction $\widehat{y}_i^{(k)}$ :$$\widehat{y}_i^{(k)}=m^{(k)}(\mathbf{x}_i)=\boldsymbol{1}_{\pm}(\beta_0^{(k)}+\mathbf{x}^T \boldsymbol{\beta}^{(k)}\geq 0),$$with, in the case of the perceptron:$$\beta_j^{(k+1)}={\beta}_j^{(k)}+\eta\underbrace{(\mathbf{y}-\widehat{\mathbf{y}}^{(k)})^T}_{=\ell({\mathbf{y}},\widehat{\mathbf{y}}^{(k)})}\mathbf{x}_j$$Here $\ell(y,y')=\mathbf{1}(y\neq y')$ is a loss function, which will allow to give a price to an error made, by predicting $\widehat{y}=m(\mathbf{x})$ and observing $y$. For a regression problem, we can consider a quadratic error $\ell_2$, such that $\ell(y,m(\mathbf{x}))=(y-m(\mathbf{x}))^2$ or in absolute value $\ell_1$, with $\ell(y,m(\mathbf{x}))=|y-m(\mathbf{x})|$. Here, for our classification problem, we used a mis-qualification indicator (we could discuss the symmetry of this loss function, suggesting that a false positive costs as much as a false negative). Once this loss function has been specified, we recognize in the problem previously described a gradient descent, and we see that we are trying to solve:$$m^\star(\mathbf{x})=\underset{m\in\mathcal{M}}{\text{argmin}}\left\lbrace\sum_{i=1}^n \ell(y_i,m(\mathbf{x}_i))\right\rbrace~~~(6)$$for a predefined set of predictors $\mathcal{M}$. Any machine learning problem is mathematically formulated as an optimization problem, whose solution determines a set of model parameters (if the $\mathcal{M}$ family is described by a set of parameters – which can be coordinates in a functional database). We can note $\mathcal{M}_0$ the space of the hyperplanes of $\mathbb{R}^p$ in the sense that$$m\in\mathcal{M}_0 \text{\quad means \quad}m(\mathbf{x})=\beta_0+\beta^T\mathbf{x}\text{ where }\beta\in\mathbb{R}^p$$ generating the class of linear predictors. We will then have the estimator that minimizes the empirical risk. Some of the recent work in statistical learning aims to study the properties of the estimator $\widehat{m}^\star$, known as “oracle”, in a family of $\mathcal{M}$ estimators, $$\widehat{m}^{\star} =\underset{\widehat{m}\in\mathcal{M}}{\text{argmin}}\big\lbrace\mathcal{R}(\widehat{m},m)\big\rbrace$$This estimator is, of course, impossible to define because it depends on $m$, the real model, unknown.

But let’s come back a little more to these loss functions. A loss function $\ell$ is a function $\mathbb{R}^d\times\mathbb{R}^d\rightarrow\mathbb{R}_+$, symmetric, which checks the triangular inequality, and such that $\ell(x,y)=0$ if and only if $x=y$. The associated norm is $\|\cdot\|$, such that $\ell(x,y)=\|x-y\|=\ell(x-y,0)$ (using the fact that $\ell(x,y+z)=\ell(x-y,z)$ – we will review this fundamental property later).

For a quadratic loss function, it should be noted that we can have a particular interpretation of this problem, since:$$\overline{y}=\underset{m\in\mathbb{R}}{\text{argmin}} \left\lbrace\sum_{i=1}^n\frac{1}{n} [y_i-m]^2\right\rbrace=\underset{m\in\mathbb{R}}{\text{argmin}} \left\lbrace \sum_{i=1}^n \ell_2(y_i,m)\right\rbrace$$ where $\ell_2$ is the usual quadratic distance If we assume – as we did in econometrics – that there is an underlying probabilistic model, and observe that : $$\displaystyle{\mathbb{E}(Y)=\underset{m\in\mathbb{R}}{\text{argmin}}\left\lbrace\mathbb{E}\left([Y-m]^2\right)\right\rbrace=\underset{m\in\mathbb{R}}{\text{argmin}}\left\lbrace\mathbb{E}\big[\ell_2(Y,m)\big]\right\rbrace}$$it should be noted that what we are trying to obtain here, by solving the problem $(6)$ by taking the norm $\ell_2$, is an approximation (in a given functional space, $\mathcal{M}$) of the conditional expectation $x\mapsto\mathbb{E}[Y|\mathbf{X}=\mathbf{x}]$. Another particularly interesting loss function is the loss $\ell_1$,$\ell_1 (y,m)=|y-m|[\latex]. It should be recalled that [latex display="true"]\displaystyle{\text{median}(\boldsymbol{y})=\underset{m\in\mathbb{R}}{\text{argmin}}\left\lbrace\sum_{i=1}^n\ell_1(y_i,m)\right\rbrace}$The optimization problem :$$\widehat{m}^{\star}=\underset{m\in\mathcal{M}_0}{\text{argmin}}\left\lbrace\sum_{i=1}^n\vert y_i-m(\mathbf{x}_i)\vert\right\rbrace$$ is obtained in econometrics by assuming that the conditional law of $Y$ follows a Laplace law centered on $m(\mathbf{x})$, and by maximizing the likelihood (log) (the sum of the absolute values of the errors corresponds to the log-reasonableness of a Laplace law). It should also be noted that if the conditional law of $Y$ is symmetrical with respect to $0$, the median and the mean coincide If this loss function is rewritten  $$\ell_1(y,m)=\vert (y-m)(1/2-\boldsymbol{1}_{y\leq m})\vert$$ a generalization can be obtained for $\tau\in[0.1]$:$$\widehat{m}^\star_\tau=\underset{m\in\mathcal{M}_0}{\text{argmin}}\left\lbrace\sum_{i=1}^n \ell_\tau^{ q} (y_i,m(\mathbf{x}_i)) \right\rbrace$$where$$\ell_{\tau}^{q}(x,y)= (x-y)(\tau-\boldsymbol{1}_{x\leq y})$$  is then the quantile regression of level $\tau$ (Koenker, 2003; d'Haultefœuille & Givord, 2014). Another loss function, introduced by Aigner et al (1977) and analysed in Waltrup et al (2014), is the function associated with the notion of expectations: $$\displaystyle{\ell}^{\text{ e}}_{\tau}(x,y)= (x-y)^2\cdot\big\vert\tau-\boldsymbol{1}_{x\leq y}\big\vert$$with $\tau\in[0.1]$. We see the parallel with the quantile function: $$\displaystyle{\ell}^{\text{ q}}_{\tau}(x,y)= \vert x-y\vert \cdot\big\vert\tau-\boldsymbol{1}_{x\leq y}\big\vert$$Koenker & Machado (1999) and Yu & Moyeed (2001) also noted a link between this condition and the search for maximum likelihood when $Y$'s conditional law follows an asymmetric Laplace law.

In connection with this approach, Gneiting (2011) introduced the notion of "ellicable statistics" - or "ellicable measurement" in its probabilistic (or distributional) version: a statistic $T$ will be said to be "ellicitable" if there is a loss function $\ell:\mathbb{R}\times\mathbb{R}\rightarrow\mathbb{R}_+$ such that:$$T(Y)=\underset{x\in\mathbb{R}}{\text{argmin}}\left\lbrace\int_{\mathbb{R}} \ell(x,y)dF(y)\right\rbrace=\underset{x\in\mathbb{R}}{\text{argmin}}\left\lbrace\mathbb{E}\big[ \ell(x,Y)\big]\text{ where }Y\overset{\mathcal{L}}{\sim} F\right\rbrace$$ The mean (mathematical expectation) is thus ellicable by the quadratic distance, $\ell_2$, while the median is ellicable by the distance $\ell_1$. According to Gneiting (2011), this property is essential for obtain predictions and forecasts. There may then be a strong link between measures associated with probabilistic models and loss functions. Finally, Bayesian statistics provide a direct link between the form of the a priori law and the loss function, as studied by Berger (1985) and Bernardo & Smith (2000). We will come back to the use of these different norms in the section on penalization.

To be continued (keep in mind that references are online here)…

[1] Where the indicator $\mathbf{1}_{\pm}$ does not take values 0 or 1 (like the classical $\mathbf{1}$ function), but -1 and +1.

# Classification from scratch, SVM 7/8

Seventh post of our series on classification from scratch. The latest one was on the neural nets, and today, we will discuss SVM, support vector machines.

## A formal introduction

Here $y$ takes values in $\{-1,+1\}$. Our model will be $$m(\mathbf{x})=\text{sign}[\mathbf{\omega}^T\mathbf{x}+b]$$ Thus, the space is divided by a (linear) border$$\Delta:\lbrace\mathbf{x}\in\mathbb{R}^p:\mathbf{\omega}^T\mathbf{x}+b=0\rbrace$$

The distance from point $\mathbf{x}_i$ to $\Delta$ is $$d(\mathbf{x}_i,\Delta)=\frac{\mathbf{\omega}^T\mathbf{x}_i+b}{\|\mathbf{\omega}\|}$$If the space is linearly separable, the problem is ill posed (there is an infinite number of solutions). So consider
$$\max_{\mathbf{\omega},b}\left\lbrace\min_{i=1,\cdots,n}\left\lbrace\text{distance}(\mathbf{x}_i,\Delta)\right\rbrace\right\rbrace$$

The strategy is to maximize the margin. One can prove that we want to solve $$\max_{\mathbf{\omega},m}\left\lbrace\frac{m}{\|\mathbf{\omega}\|}\right\rbrace$$
subject to $y_i\cdot(\mathbf{\omega}^T\mathbf{x}_i)=m$, $\forall i=1,\cdots,n$. Again, the problem is ill posed (non identifiable), and we can consider $m=1$: $$\max_{\mathbf{\omega}}\left\lbrace\frac{1}{\|\mathbf{\omega}\|}\right\rbrace$$
subject to $y_i\cdot(\mathbf{\omega}^T\mathbf{x}_i)=1$, $\forall i=1,\cdots,n$. The optimization objective can be written$$\min_{\mathbf{\omega}}\left\lbrace\|\mathbf{\omega}\|^2\right\rbrace$$

## The primal problem

In the separable case, consider the following primal problem,$$\min_{\mathbf{w}\in\mathbb{R}^d,b\in\mathbb{R}}\left\lbrace\frac{1}{2}\|\mathbf{\omega}\|^2\right\rbrace$$subject to $y_i\cdot (\mathbf{\omega}^T\mathbf{x}_i+b)\geq 1$, $\forall i=1,\cdots,n$.

In the non-separable case, introduce slack (error) variables $\mathbf{\xi}$ : if $y_i\cdot (\mathbf{\omega}^T\mathbf{x}_i+b)\geq 1$, there is no error $\xi_i=0$.

Let $C$ denote the cost of misclassification. The optimization problem becomes$$\min_{\mathbf{w}\in\mathbb{R}^d,b\in\mathbb{R},{\color{red}{\mathbf{\xi}}}\in\mathbb{R}^n}\left\lbrace\frac{1}{2}\|\mathbf{\omega}\|^2 + C\sum_{i=1}^n\xi_i\right\rbrace$$subject to $y_i\cdot (\mathbf{\omega}^T\mathbf{x}_i+b)\geq 1-{\color{red}{\xi_i}}$, with ${\color{red}{\xi_i}}\geq 0$, $\forall i=1,\cdots,n$.

Let us try to code this optimization problem. The dataset is here

n = length(myocarde[,"PRONO"]) myocarde0 = myocarde myocarde0$PRONO = myocarde$PRONO*2-1 C = .5

and we have to set a value for the cost $C$. In the (linearly) constrained optimization function in R, we need to provide the objective function $f(\mathbf{\theta})$ and the gradient $\nabla f(\mathbf{\theta})$.

f = function(param){ w = param[1:7] b = param[8] xi = param[8+1:nrow(myocarde)] .5*sum(w^2) + C*sum(xi)} grad_f = function(param){ w = param[1:7] b = param[8] xi = param[8+1:nrow(myocarde)] c(2*w,0,rep(C,length(xi)))}

and (linear) constraints are written as $\mathbf{U}\mathbf{\theta}-\mathbf{c}\geq \mathbf{0}$

U = rbind(cbind(myocarde0[,"PRONO"]*as.matrix(myocarde[,1:7]),diag(n),myocarde0[,"PRONO"]), cbind(matrix(0,n,7),diag(n,n),matrix(0,n,1))) C = c(rep(1,n),rep(0,n))

Then we use

constrOptim(theta=p_init, f, grad_f, ui = U,ci = C)

Observe that something is missing here: we need a starting point for the algorithm, $\mathbf{\theta}_0$. Unfortunately, I could not think of a simple technique to get a valid starting point (that satisfies those linear constraints).

Let us try something else. Because those functions are quite simple: either linear or quadratic. Actually, one can recognize in the separable case, but also in the non-separable case, a classic quadratic program$$\min_{\mathbf{z}\in\mathbb{R}^d}\left\lbrace\frac{1}{2}\mathbf{z}^T\mathbf{D}\mathbf{z}-\mathbf{d}\mathbf{z}\right\rbrace$$subject to $\mathbf{A}\mathbf{z}\geq\mathbf{b}$.

library(quadprog) eps = 5e-4 y = myocarde[,&quot;PRONO&quot;]*2-1 X = as.matrix(cbind(1,myocarde[,1:7])) n = length(y) D = diag(n+7+1) diag(D)[8+0:n] = 0 d = matrix(c(rep(0,7),0,rep(C,n)), nrow=n+7+1) A = Ui b = Ci sol = solve.QP(D+eps*diag(n+7+1), d, t(A), b, meq=1, factorized=FALSE) qpsol = sol$solution (omega = qpsol[1:7]) [1] -0.106642005446 -0.002026198103 -0.022513312261 -0.018958578746 -0.023105767847 -0.018958578746 -1.080638988521 (b = qpsol[n+7+1]) [1] 997.6289927 Given an observation $\mathbf{x}$, the prediction is $$y=\text{sign}[\mathbf{\omega}^T\mathbf{x}+b]$$ y_pred = 2*((as.matrix(myocarde0[,1:7])%*%omega+b)&gt;0)-1 Observe that here, we do have a classifier, depending if the point lies on the left or on the right (above or below, etc) the separating line (or hyperplane). We do not have a probability, because there is no probabilistic model here. So far. ## The dual problem The Lagrangian of the separable problem could be written introducing Lagrange multipliers $\mathbf{\alpha}\in\mathbb{R}^n$, $\mathbf{\alpha}\geq \mathbf{0}$ as$$\mathcal{L}(\mathbf{\omega},b,\mathbf{\alpha})=\frac{1}{2}\|\mathbf{\omega}\|^2-\sum_{i=1}^n \alpha_i\big(y_i(\mathbf{\omega}^T\mathbf{x}_i+b)-1\big)$$Somehow, $\alpha_i$ represents the influence of the observation $(y_i,\mathbf{x}_i)$. Consider the Dual Problem, with $\mathbf{G}=[G_{ij}]$ and $G_{ij}=y_iy_j\mathbf{x}_j^T\mathbf{x}_i$ $$\min_{\mathbf{\alpha}\in\mathbb{R}^n}\left\lbrace\frac{1}{2}\mathbf{\alpha}^T\mathbf{G}\mathbf{\alpha}-\mathbf{1}^T\mathbf{\alpha}\right\rbrace$$ subject to $\mathbf{y}^T\mathbf{\alpha}=\mathbf{0}$ and $\mathbf{\alpha}\geq\mathbf{0}$. The Lagrangian of the non-separable problem could be written introducing Lagrange multipliers $\mathbf{\alpha},{\color{red}{\mathbf{\beta}}}\in\mathbb{R}^n$, $\mathbf{\alpha},{\color{red}{\mathbf{\beta}}}\geq \mathbf{0}$, and define the Lagrangian $\mathcal{L}(\mathbf{\omega},b,{\color{red}{\mathbf{\xi}}},\mathbf{\alpha},{\color{red}{\mathbf{\beta}}})$ as$$\frac{1}{2}\|\mathbf{\omega}\|^2+{\color{blue}{C}}\sum_{i=1}^n{\color{red}{\xi_i}}-\sum_{i=1}^n \alpha_i\big(y_i(\mathbf{\omega}^T\mathbf{x}_i+b)-1+{\color{red}{\xi_i}}\big)-\sum_{i=1}^n{\color{red}{\beta_i}}{\color{red}{\xi_i}}$$ Somehow, $\alpha_i$ represents the influence of the observation $(y_i,\mathbf{x}_i)$. The Dual Problem become with $\mathbf{G}=[G_{ij}]$ and $G_{ij}=y_iy_j\mathbf{x}_j^T\mathbf{x}_i$$$\min_{\mathbf{\alpha}\in\mathbb{R}^n}\left\lbrace\frac{1}{2}\mathbf{\alpha}^T\mathbf{G}\mathbf{\alpha}-\mathbf{1}^T\mathbf{\alpha}\right\rbrace$$ subject to $\mathbf{y}^T\mathbf{\alpha}=\mathbf{0}$, $\mathbf{\alpha}\geq\mathbf{0}$ and $\mathbf{\alpha}\leq {\color{blue}{C}}$. As previsouly, one can also use quadratic programming library(quadprog) eps = 5e-4 y = myocarde[,"PRONO"]*2-1 X = as.matrix(cbind(1,myocarde[,1:7])) n = length(y) Q = sapply(1:n, function(i) y[i]*t(X)[,i]) D = t(Q)%*%Q d = matrix(1, nrow=n) A = rbind(y,diag(n),-diag(n)) C = .5 b = c(0,rep(0,n),rep(-C,n)) sol = solve.QP(D+eps*diag(n), d, t(A), b, meq=1, factorized=FALSE) qpsol = sol$solution

The two problems are connected in the sense that for all $\mathbf{x}$$$\mathbf{\omega}^T\mathbf{x}+b = \sum_{i=1}^n \alpha_i y_i (\mathbf{x}^T\mathbf{x}_i)+b$$

To recover the solution of the primal problem,$$\mathbf{\omega}=\sum_{i=1}^n \alpha_iy_i \mathbf{x}_i$$thus

omega = apply(qpsol*y*X,2,sum) omega 1 FRCAR INCAR INSYS 0.0000000000000002439074265 0.0550138658687635215271960 -0.0920163239049630876653652 0.3609571899422952534486342 PRDIA PAPUL PVENT REPUL -0.1094017965288692356695677 -0.0485213403643276475207813 -0.0660058643191372279579454 0.0010093656567606212794835

while $b=y-\mathbf{\omega}^T\mathbf{x}$ (but actually, one can add the constant vector in the matrix of explanatory variables).

More generally, consider the following function (to make sure that $D$ is a definite-positive matrix, we use the nearPD function).

svm.fit = function(X, y, C=NULL) { n.samples = nrow(X) n.features = ncol(X) K = matrix(rep(0, n.samples*n.samples), nrow=n.samples) for (i in 1:n.samples){ for (j in 1:n.samples){ K[i,j] = X[i,] %*% X[j,] }} Dmat = outer(y,y) * K Dmat = as.matrix(nearPD(Dmat)$mat) dvec = rep(1, n.samples) Amat = rbind(y, diag(n.samples), -1*diag(n.samples)) bvec = c(0, rep(0, n.samples), rep(-C, n.samples)) res = solve.QP(Dmat,dvec,t(Amat),bvec=bvec, meq=1) a = res$solution bomega = apply(a*y*X,2,sum) return(bomega) }

On our dataset, we obtain

M = as.matrix(myocarde[,1:7]) center = function(z) (z-mean(z))/sd(z) for(j in 1:7) M[,j] = center(M[,j]) bomega = svm.fit(cbind(1,M),myocarde$PRONO*2-1,C=.5) y_pred = 2*((cbind(1,M)%*%bomega)&gt;0)-1 table(obs=myocarde0$PRONO,pred=y_pred) pred obs -1 1 -1 27 2 1 9 33

i.e. 11 misclassification, out of 71 points (which is also what we got with the logistic regression).

## Kernel Based Approach

In some cases, it might be difficult to “separate” by a linear separators the two sets of points, like below,

It might be difficult, here, because which want to find a straight line in the two dimensional space $(x_1,x_2)$. But maybe, we can distort the space, possible by adding another dimension

That’s heuristically the idea. Because on the case above, in dimension 3, the set of points is now linearly separable. And the trick to do so is to use a kernel. The difficult task is to find the good one (if any).

A positive kernel on $\mathcal{X}$ is a function $K:\mathcal{X}\times\mathcal{X}\rightarrow\mathbb{R}$ symmetric, and such that for any $n$, $\forall\alpha_1,\cdots,\alpha_n$ and $\forall\mathbf{x}_1,\cdots,\mathbf{x}_n$,$$\sum_{i=1}^n\sum_{j=1}^n\alpha_i\alpha_j k(\mathbf{x}_i,\mathbf{x}_j)\geq 0.$$
For example, the linear kernel is $k(\mathbf{x}_i,\mathbf{x}_j)=\mathbf{x}_i^T\mathbf{x}_j$. That’s what we’ve been using here, so far. One can also define the product kernel $k(\mathbf{x}_i,\mathbf{x}_j)=\kappa(\mathbf{x}_i)\cdot\kappa(\mathbf{x}_j)$ where $\kappa$ is some function $\mathcal{X}\rightarrow\mathbb{R}$.

Finally, the Gaussian kernel is $k(\mathbf{x}_i,\mathbf{x}_j)=\exp[-\|\mathbf{x}_i-\mathbf{x}_j\|^2]$.

Since it is a function of $\|\mathbf{x}_i-\mathbf{x}_j\|$, it is also called a radial kernel.

linear.kernel = function(x1, x2) { return (x1%*%x2) } svm.fit = function(X, y, FUN=linear.kernel, C=NULL) { n.samples = nrow(X) n.features = ncol(X) K = matrix(rep(0, n.samples*n.samples), nrow=n.samples) for (i in 1:n.samples){ for (j in 1:n.samples){ K[i,j] = FUN(X[i,], X[j,]) } } Dmat = outer(y,y) * K Dmat = as.matrix(nearPD(Dmat)$mat) dvec = rep(1, n.samples) Amat = rbind(y, diag(n.samples), -1*diag(n.samples)) bvec = c(0, rep(0, n.samples), rep(-C, n.samples)) res = solve.QP(Dmat,dvec,t(Amat),bvec=bvec, meq=1) a = res$solution bomega = apply(a*y*X,2,sum) return(bomega) }

To relate this duality optimization problem to OLS, recall that $y=\mathbf{x}^T\mathbf{\omega}+\varepsilon$, so that $\widehat{y}=\mathbf{x}^T\widehat{\mathbf{\omega}}$, where $\widehat{\mathbf{\omega}}=[\mathbf{X}^T\mathbf{X}]^{-1}\mathbf{X}^T\mathbf{y}$
But one can also write $$y=\mathbf{x}^T\widehat{\mathbf{\omega}}=\sum_{i=1}^n \widehat{\alpha}_i\cdot \mathbf{x}^T\mathbf{x}_i$$
where $\widehat{\mathbf{\alpha}}=\mathbf{X}[\mathbf{X}^T\mathbf{X}]^{-1}\widehat{\mathbf{\omega}}$, or conversely, $\widehat{\mathbf{\omega}}=\mathbf{X}^T\widehat{\mathbf{\alpha}}$.

## Application (on our small dataset)

One can actually use a dedicated R package to run a SVM. To get the linear kernel, use

library(kernlab) df0 = df df0$y = 2*(df$y=="1")-1 SVM1 = ksvm(y ~ x1 + x2, data = df0, C=.5, kernel = "vanilladot" , type="C-svc")

Since the dataset is not linearly separable, there will be some mistakes here

table(df0$y,predict(SVM1)) -1 1 -1 2 2 1 1 5 The problem with that function is that it cannot be used to get a prediction for other points than those in the sample (and I could neither extract $\omega$ nor $b$ from the 24 slots of that objet). But it’s possible by adding a small option in the function SVM2 = ksvm(y ~ x1 + x2, data = df0, C=.5, kernel = "vanilladot" , prob.model=TRUE, type="C-svc") With that function, we convert the distance as some sort of probability. Someday, I will try to replicate the probabilistic version of SVM, I promise, but today, the goal is just to understand what is done when running the SVM algorithm. To visualize the prediction, use pred_SVM2 = function(x,y){ return(predict(SVM2,newdata=data.frame(x1=x,x2=y), type="probabilities")[,2])} plot(df$x1,df$x2,pch=c(1,19)[1+(df$y=="1")], cex=1.5,xlab="", ylab="",xlim=c(0,1),ylim=c(0,1)) vu = seq(-.1,1.1,length=251) vv = outer(vu,vu,function(x,y) pred_SVM2(x,y)) contour(vu,vu,vv,add=TRUE,lwd=2,nlevels = .5,col="red")

Here the cost is $C$=.5, but of course, we can change it

SVM2 = ksvm(y ~ x1 + x2, data = df0, C=2, kernel = "vanilladot" , prob.model=TRUE, type="C-svc") pred_SVM2 = function(x,y){ return(predict(SVM2,newdata=data.frame(x1=x,x2=y), type="probabilities")[,2])} plot(df$x1,df$x2,pch=c(1,19)[1+(df$y=="1")], cex=1.5,xlab="", ylab="",xlim=c(0,1),ylim=c(0,1)) vu = seq(-.1,1.1,length=251) vv = outer(vu,vu,function(x,y) pred_SVM2(x,y)) contour(vu,vu,vv,add=TRUE,lwd=2,levels = .5,col="red") As expected, we have a linear separator. But slightly different. Now, let us consider the “Radial Basis Gaussian kernel” SVM3 = ksvm(y ~ x1 + x2, data = df0, C=2, kernel = "rbfdot" , prob.model=TRUE, type="C-svc") Observe that here, we’ve been able to separare the white and the black points table(df0$y,predict(SVM3))   -1 1 -1 4 0 1 0 6
plot(df$x1,df$x2,pch=c(1,19)[1+(df$y=="1")], cex=1.5,xlab="", ylab="",xlim=c(0,1),ylim=c(0,1)) vu = seq(-.1,1.1,length=251) vv = outer(vu,vu,function(x,y) pred_SVM3(x,y)) contour(vu,vu,vv,add=TRUE,lwd=2,levels = .5,col="red") Now, to be completely honest, if I understand the theory of the algorithm used to compute $\omega$ and $b$ with linear kernel (using quadratic programming), I do not feel confortable with this R function. Especially if you run it several times… you can get (with exactly the same set of parameters) or (to be continued…) # Actuariat de l’Assurance Non-Vie #2 Pour le second cours d’actuariat de l’assurance non-vie à l’ENSAE, qui aura lieu lundi après midi, les slides présentant les modèles classiques pour prédire des variables factorielles (classification) sont en ligne, # Computational Time of Predictive Models Tuesday, at the end of my 5-hour crash course on machine learning for actuaries, Pierre asked me an interesting question about computational time of different techniques. I’ve been presenting the philosophy of various algorithm, but I forgot to mention computational time. I wanted to try several classification algorithms on the dataset used to illustrate the techniques > rm(list=ls()) > myocarde=read.table( "http://freakonometrics.free.fr/myocarde.csv", head=TRUE,sep=";") > levels(myocarde$PRONO)=c("Death","Survival")

But the dataset is rather small, with 71 observations and 7 explanatory variables. So I decided to replicate the observations, and to add some covariates,

> levels(myocarde\$PRONO)=c("Death","Survival")
> idx=rep(1:nrow(myocarde),each=100)
> TPS=matrix(NA,30,10)
> myocarde_large=myocarde[idx,]
> k=23
> M=data.frame(matrix(rnorm(k*
+ nrow(myocarde_large)),nrow(myocarde_large),k))
> names(M)=paste("X",1:k,sep="")
> myocarde_large=cbind(myocarde_large,M)
> dim(myocarde_large)
[1] 7100   31
> object.size(myocarde_large)
2049.064 kbytes

The dataset is not big… but at least, it does not take 0.0001 sec. to run a regression.  Actually, to run a logistic regression, it takes 0.1 second

> system.time(fit< glm(PRONO~.,
+ data=myocarde_large, family="binomial"))
user      system     elapsed
0.114       0.016       0.134
> object.size(fit)
9,313.600 kbytes

And I was surprised that the regression object was 9Mo, which is more than four times the size of the dataset. With a large dataset, 100 times larger,

> dim(myocarde_large_2)
[1] 710000     31

it takes 20 sec.

> system.time(fit<-glm(PRONO~.,
+ data=myocarde_large_2, family="binomial"))
utilisateur     système      écoulé
16.394       2.576      19.819
> object.size(fit)
90,9025.600 kbytes

and the object is ‘only’ ten times bigger.