# Probabilistic Foundations of Econometrics, part 2

This post is the second one of our series on the history and foundations of econometric and machine learning models. Part 1 is online here.

## Geometric Properties of this Linear Model

Let’s define the scalar product in $\mathbb{R}^n$, $⟨\mathbf{a},\mathbf{b}⟩=\mathbf{a}^T\mathbf{b}$, and let’s note $\|\cdot\|$ the associated Euclidean standard, $\|\mathbf{a}\|=\sqrt{\mathbf{a}^T\mathbf{a}}$ (denoted $\|\cdot\|_{\ell_2}$ in the next post). Note $\mathcal{E}_X$ the space generated by all linear combinations of the $\mathbf{X}$ components (adding the constant). If the explanatory variables are linearly independent, $\mathbf{X}$ is a full (column) rank matrix and $\mathcal{E}_X$ is a space of dimension $p+1$. Let’s assume from now on that the variables $\mathbf{x}$  and $y$ are centered here. Note that no law hypothesis is made in this section, the geometric properties are derived from the properties of expectation and variance in the set of finite variance variables.

With this notation, it should be noted that the linear model is written $m(\mathbf{x})=⟨\mathbf{x},\beta⟩$. The space $H_z=\{\mathbf{x}\in\mathbb{R}^{p+1}:m(\mathbf{x})=z\}$ is a hyperplane (affine) that separates the space in two. Let’s define the orthogonal projection operator on $\mathcal{E}_X$, $\Pi_X =\mathbf{X}(\mathbf{X}^T\mathbf{X})^{-1} \mathbf{X}^T$. Thus, the forecast that can be made for it is: $$\widehat{\mathbf{y}}=\mathbf{X}(\mathbf{X}^T\mathbf{X})^{-1} \mathbf{X}^T\mathbf{y}=\Pi_X\mathbf{y}$$. As, $\widehat{\varepsilon}=\mathbf{y}-\widehat{\mathbf{y}}=(\mathbb{I}-\Pi_X)\mathbf{y}=\Pi_{X^\perp}\mathbf{y}$, we note that $\widehat{\varepsilon}\perp\mathbf{x}$, which will be interpreted as meaning that residuals are a term of innovation, unpredictable in the sense that $\Pi_{X }\widehat{\varepsilon}=\mathbf{0}$. The Pythagorean theorem is written here: $$\Vert \mathbf{y} \Vert^2=\Vert \Pi_{ {X}}\mathbf{y} \Vert^2+\Vert \Pi_{ {X}^\perp}\mathbf{y} \Vert^2=\Vert \Pi_{ {X}}\mathbf{y}\Vert^2+\Vert \mathbf{y}-\Pi_{ {X}}\mathbf{y}\Vert^2=\Vert\widehat{\mathbf{y}}\Vert^2+\Vert\widehat{\mathbf{\varepsilon}}\Vert^2$$which is classically translated in terms of the sum of squares: $$\underbrace{\sum_{i=1}^n y_i^2}_{n\times\text{total variance}}=\underbrace{\sum_{i=1}^n \widehat{y}_i^2}_{n\times\text{explained variance}}+\underbrace{\sum_{i=1}^n (y_i-\widehat{y}_i)^2}_{n\times\text{residual variance}}$$The coefficient of determination, $R^2$, is then interpreted as the square of the cosine of the angle $\theta$ between $\mathbf{y}$ and $\Pi_X \mathbf{y}$ : $$R^2=\frac{\Vert \Pi_{{X}} \mathbf{y}\Vert^2}{\Vert \mathbf{y}\Vert^2}=1-\frac{\Vert \Pi_{ {X}^\perp} \mathbf{y}\Vert^2}{\Vert \mathbf {y}\Vert^2}=\cos^2(\theta)$$An important application was obtained by Frish & Waugh (1933), when the explanatory variables are divided into two groups, $\mathbf{X}=[\mathbf{X}_1 |\mathbf{X}_2]$, so that the regression becomes $y=\beta_0+\mathbf{X}_1 β_1+\mathbf{X}_2 β_2+\varepsilon$. Frish & Waugh (1933) showed that two successive projections could be considered. Indeed, if $\mathbf{y}_2^\star=\Pi_{X_1^\perp} \mathbf{y}$ and $X_2^\star=\Pi_{X_1^\perp}\mathbf{X}_2$, we can show that $$\widehat{\beta} _2=[{\mathbf{X}_2^\star}^T \mathbf{X}_2^\star]^{-1}{\mathbf{X}_2^\star}^T \mathbf{y}_2^\star$$ In other words, the overall estimate is equivalent to the combination of independent estimates of the two models if $\mathbf{X}_2^\star=\mathbf{X}_2$, i.e. $\mathbf{X}_2\in \mathcal{E}_{X_1}^\perp$, which can be noted $\mathbf{x}_1\perp\mathbf{x}_2$ We obtain here the Frisch-Waugh theorem which guarantees that if the explanatory variables between the two groups are orthogonal, then the overall estimate is equivalent to two independent regressions, on each of the sets of explanatory variables. This is a theorem of double projection, on orthogonal spaces. Many results and interpretations are obtained through geometric interpretations (fundamentally related to the links between conditional expectation and the orthogonal projection in space of variables of finite variance).

This geometric interpretation might help to get a better understanding of the problem of under-identification, i.e. the case where the real model would be $y_i=\beta_0+ \mathbf{x}_1^T \beta_1+\mathbf{x}_2^T \beta_2+\varepsilon_i$, but the estimated model is $y_i=b_0+\mathbf{x}_1^T \mathbf{b}_1+\eta_i$. The maximum likelihood estimator of $\mathbf{b}_1$ is $$\widehat{\mathbf{b}}_1=\mathbf {\beta}_1 + \underbrace{ (\mathbf {X}_1^T\mathbf {X}_1)^{-1} \mathbf {X}_1^T \mathbf {X}_{2} \mathbf{\beta}_2}_{\mathbf{\beta}_{12}}+\underbrace{(\mathbf{X}_1^{T}\mathbf{X}_1)^{-1} \mathbf{X}_1^T\varepsilon}_{\nu}$$so that $\mathbb{E}[\widehat{\mathbf{b}}_1]=\beta_1+\beta_{12}$, the bias ($\beta_{12}$) being null only in the case where $\mathbf{X}_1^T \mathbf{X}_2=\mathbf{0}$ (i. e. $\mathbf{X}_1\perp \mathbf{X}_2$): we find here a consequence of the Frisch-Waugh theorem.

On the other hand, over-identification corresponds to the case where the real model would be $y_i=\beta_0+\mathbf{x}_1^T \beta_1+\varepsilon_i$, but the estimated model is $y_i=b_0+ \mathbf{x}_1^T \mathbf{b} _1+\mathbf{x}_2^T \mathbf{b}_2+\eta_i$. In this case, the estimate is unbiased, in the sense that $\mathbb{E}[\widehat{\mathbf{b}}_1]=\beta_1$ but the estimator is not efficient. Later on, we will discuss an effective method for selecting variables (and avoid over-identification).

## From parametric to non-parametric

We can rewrite equation (4) in the form $\widehat{\mathbf{y}}=\Pi_X\mathbf{y}$ which helps us to see the forecast directly as a linear transformation of the observations. More generally, a linear predictor can be obtained by considering $m(\mathbf{x})=\mathbf{s}_{\mathbf{x}}^T \mathbf{y}$, where $\mathbf{s}_{\mathbf{x}}$ is a weight vector, which depends on $\mathbf{x}$, interpreted as a smoothing vector. Using the vectors $\mathbf{s}_{\mathbf{x}_i}$, calculated from the observations $\mathbf{x}_i$, we obtain a matrix $\mathbf{S}$ of size $n\times n$, and $\widehat{\mathbf{y}}=\mathbf{S}\mathbf{y}$. In the case of the linear regression described above, $\mathbf{s}_{\mathbf{x}}=\mathbf{X}[\mathbf{X}^T\mathbf{X}]^{-1}\mathbf{x}$, and in that case $\text{trace}(\mathbf{S})$ is the number of columns in the $\mathbf{X}$ matrix (the number of explanatory variables). In this context of more general linear predictors, $\text{trace}(\mathbf{S})$ is often seen as equivalent to the number of parameters (or complexity, or dimension, of the model), and $\nu=n-\text{trace}(\mathbf{S})$ is then the number of degrees of freedom (see Ruppert et al., 2003; Simonoff, 1996). The principle of parsimony says that we should minimize this dimension (the trace of the matrix $\mathbf{S}$) as much as possible. But in the general case, this dimension is more to obtain, explicitely.

The estimator introduced by Nadaraya (1964) and Watson (1964), in the case of a simple non-parametric regression, is also written in this form since$$\widehat{m}_h(x)=\mathbf{s}_{x}^T\mathbf{y}=\sum_{i=1}^n \mathbf{s}_{x,i}y_i$$where$$\mathbf{s}_{x,i}=\frac{K_h(x-x_i)}{K_h(x-x_1)+\cdots+K_h(x-x_n)}$$ where $K(\cdot)$ is a kernel function, which assigns a value that is lower the closer $x_i$ is to $x$, and $h>0$ is the bandwidth. The introduction of this metaparameter $h$ is an important issue, as it should be chosen wisely. Using asymptotic developments, we can show that if $X$ has density $f$, $$\text{biais}[\widehat{m}_h(x)]=\mathbb{E}[\widehat{m}_h(x)]-m(x)\sim {h^2}\left(\frac{C_1 }{2}m''(x)+C_2 m'(x)\frac{f'(x)}{f(x)}\right)$$and $$\displaystyle{{\text{Var}[\widehat{m}_h(x)]\sim\frac{C_3}{{nh}}\frac{\sigma(x)}{f(x)}}}$$for some constants that can be estimated (see Simonoff (1996) for a discussion). These two functions evolve inversely with $h$, as shown in Figure 1 (where the metaparameter on the $x$-axis is here, actually, $h^{-1}$). Keep in ming that we will see a similar graph in the context of machine learning models.

Figure 1. Choice of meta-parameter and the Goldilocks problem: it must not be too large (otherwise there is too much variance), nor too small (otherwise there is too much bias).

The natural idea is then to try to minimize the mean square error, the MSE, defined as $bias[\widehat{m}_h (x)]^2+Var[\widehat{m}_h (x)]$, and them integrate over $x$, which gives an optimal value for $h$ of the form $h^\star=O(n^{-1/5})$, and reminds us of Silverman’s rule – see Silverman (1986). In larger dimensions, for continuous $\mathbf{x}$ variables, a multivariate kernel with matrix bandwidth $\mathbf{H}$ can be used, and$$\mathbb{E}[\widehat{m}_{\mathbf{H}}(\mathbf{x})]\sim m(\mathbf{x})+\frac{C_1}{2}\text{trace}\big(\mathbf{H}^Tm''(\mathbf{x})\mathbf{H}\big)+C_2\frac{m'(\boldsymbol{x})^T\mathbf{H}\mathbf{H}^T \nabla f(\mathbf{x})}{f(\mathbf{x})}$$while$$\text{Var}[\widehat{m}_{\mathbf{H}}(\mathbf{x})]\sim\frac{C_3}{n~\text{det}(\mathbf{H})}\frac{\sigma(\mathbf{x})}{f(\mathbf{x})}$$
If $\mathbf{H}$ is a diagonal matrix, with the same term $h$  on the diagonal, then $h^\star=O(n^{-1/(4+dim(\mathbf{x}))}$. However, in practice, there will be more interest in the integrated version of the quadratic error, $$MISE(\widehat{m}_{h})=\mathbb{E}[MSE(\widehat{m}_{h}(X))]=\int MSE(\widehat{m}_{h}(x))dF(x)$$and we can prove that $$MISE[\widehat{m}_h]\sim \overbrace{\frac{h^4}{4}\left(\int x^2k(x)dx\right)^2\int\big[m''(x)+2m'(x)\frac{f'(x)}{f(x)}\big]^2dx}^{\text{bias}^2} +\overbrace{\frac{\sigma^2}{nh}\int k^2(x)dx \cdot\int\frac{dx}{f(x)}}^{\text{variance}}$$as n→∞ and nh→∞. Here we find an asymptotic relationship that again recalls Silverman’s (1986) order of magnitude, $$h^\star =n^{-\frac{1}{5}}\left(\frac{C_1\int \frac{dx}{f(x)}}{C_2\int \big[m''(x)+2m'(x)\frac{f'(x)}{f(x)}\big]dx}\right)^{\frac{1}{5}}$$The main problem here, in practice, is that many of the terms in the expression above are unknown. Automatic learning offers computational techniques, when the econometrician used to searching for asymptotic (mathematical) properties.

To be continued (references mentioned above are online here)…

# 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…) # Classification from scratch, logistic with kernels 3/8 Third post of our series on classification from scratch, following the previous post introducing smoothing techniques, with (b)-splines. Consider here kernel based techniques. Note that here, we do not use the “logistic” model… it is purely non-parametric. ## kernel based estimated, from scratch I like kernels because they are somehow very intuitive. With GLMs, the goal is to estimate $\hat{m}(\mathbf{x})=\mathbb{E}(Y|\mathbf{X}=\mathbf{x})$. Heuritically, we want to compute the (conditional) expected value on the neighborhood of $\mathbf{x}$. If we consider some spatial model, where $\mathbf{x}$ is the location, we want the expected value of some variable $Y$, “on the neighborhood” of $\mathbf{x}$. A natural approach is to use some administrative region (county, departement, region, etc). This means that we have a partition of $\mathcal{X}$ (the space with the variable(s) lies). This will yield the regressogram, introduced in Tukey (1961). For convenience, assume some interval / rectangle / box type of partition. In the univariate case, consider $$\hat{m}_{\mathbf{a}}(x)=\frac{\sum_{i=1}^n \mathbf{1}(x_i\in[a_j,a_{j+1}))y_i}{\sum_{i=1}^n \mathbf{1}(x_i\in[a_j,a_{j+1}))}$$or the moving regressogram $$\hat{m}(x)=\frac{\sum_{i=1}^n \mathbf{1}(x_i\in[x\pm h])y_i}{\sum_{i=1}^n \mathbf{1}(x_i\in[x\pm h])}$$In that case, the neighborhood is defined as the interval $(x\pm h)$. That’s nice, but clearly very simplistic. If $\mathbf{x}_i=\mathbf{x}$ and $\mathbf{x}_j=\mathbf{x}-h+\varepsilon$ (with $\varepsilon>0$), both observations are used to compute the conditional expected value. But if $\mathbf{x}_{j'}=\mathbf{x}-h-\varepsilon$, only $\mathbf{x}_i$ is considered. Even if the distance between $\mathbf{x}_{j}$ and $\mathbf{x}_{j'}$ is extremely extremely small. Thus, a natural idea is to use weights that are function of the distance between $\mathbf{x}_{i}$‘s and $\mathbf{x}$.Use$$\tilde{m}(x)=\frac{\sum_{i=1}^ny_i\cdot k_h\left({x-x_i}\right)}{\sum_{i=1}^nk_h\left({x-x_i}\right)}$$where (classically)$$k_h(x)=k\left(\frac{x}{h}\right)$$for some kernel $k$ (a non-negative function that integrates to one) and some bandwidth $h$. Usually, kernels are denoted with capital letter $K$, but I prefer to use $k$, because it can be interpreted as the density of some random noise we add to all observations (independently). Actually, one can derive that estimate by using kernel-based estimators of densities. Recall that$$\tilde{f}(\mathbf{y})=\frac{1}{n|\mathbf{H}|^{1/2}}\sum_{i=1}^n k\left(\mathbf{H}^{-1/2}(\mathbf{y}-\mathbf{y}_i)\right)$$ Now, use the fact that the expected value can be defined as$$m(x)=\int yf(y|x)dy=\frac{\int y f(y,x)dy}{\int f(y,x)dy}$$Consider now a bivariate (product) kernel to estimate the joint density. The numerator is estimated by$$\frac{1}{nh}\sum_{i=1}^n\int y_i k\left(t,\frac{x-x_i}{h}\right)dt=\frac{1}{nh}\sum_{i=1}^ny_i \kappa\left(\frac{x-x_i}{h}\right)$$while the denominator is estimated by$$\frac{1}{nh^2}\sum_{i=1}^n \int k\left(\frac{y-y_i}{h},\frac{x-x_i}{h}\right)=\frac{1}{nh}\sum_{i=1}^n\kappa\left(\frac{x-x_i}{h}\right)$$In a general setting, we still use product kernels between $Y$ and $\mathbf{X}$ and write $$\widehat{m}_{\mathbf{H}}(\mathbf{x})=\displaystyle{\frac{\sum_{i=1}^ny_i\cdot k_{\mathbf{H}}(\mathbf{x}_i-\mathbf{x})}{\sum_{i=1}^n k_{\mathbf{H}}(\mathbf{x}_i-\mathbf{x})}}$$for some symmetric positive definite bandwidth matrix $\mathbf{H}$, and $$k_{\mathbf{H}}(\mathbf{x})=\det[\mathbf{H}]^{-1}k(\mathbf{H}^{-1}\mathbf{x})$$ Now that we know what kernel estimates are, let us use them. For instance, assume that $k$ is the density of the $\mathcal{N}(0,1)$ distribution. At point $x$, with a bandwidth $h$ we get the following code mean_x = function(x,bw){ w = dnorm((myocarde$INSYS-x)/bw, mean=0,sd=1) weighted.mean(myocarde$PRONO,w)} u = seq(5,55,length=201) v = Vectorize(function(x) mean_x(x,3))(u) plot(u,v,ylim=0:1,type="l",col="red") points(myocarde$INSYS,myocarde$PRONO,pch=19) and of course, we can change the bandwidth. v = Vectorize(function(x) mean_x(x,2))(u) plot(u,v,ylim=0:1,type="l",col="red") points(myocarde$INSYS,myocarde$PRONO,pch=19) We observe what we can read in any textbook : with a smaller bandwidth, we get more variance, less bias. “More variance” means here more variability (since the neighborhood is smaller, there are less points to compute the average, and the estimate is more volatile), and “less bias” in the sense that the expected value is supposed to be compute at point $x$, so the smaller the neighborhood, the better. ## Using ksmooth R function Actually, there is a function in R to compute this kernel regression. reg = ksmooth(myocarde$INSYS,myocarde$PRONO,"normal",bandwidth = 2*exp(1)) plot(reg$x,reg$y,ylim=0:1,type="l",col="red",lwd=2,xlab="INSYS",ylab="") points(myocarde$INSYS,myocarde$PRONO,pch=19) We can replicate our previous estimate. Nevertheless, the output is not a function, but two series of vectors. That’s nice to get a graph, but that’s all we get. Furthermore, as we can see, the bandwidth is not exactly the same as the one we used before. I did not find any information online, so I tried to replicate the function we wrote before g=function(bk=3){ reg = ksmooth(myocarde$INSYS,myocarde$PRONO,"normal",bandwidth = bk) f=function(bm){ v = Vectorize(function(x) mean_x(x,bm))(reg$x) z=reg$y-v sum((z[!is.na(z)])^2)} optim(bk,f)$par} x=seq(1,10,by=.1) y=Vectorize(g)(x) plot(x,y) abline(0,exp(-1),col="red") abline(0,.37,col="blue")

There is a slope of $0.37$, which is actually $e^{-1}$. Coincidence ? I don’t know to be honest…

## Application in higher dimension

Consider now our bivariate dataset, and consider some product of univariate (Gaussian) kernels

u = seq(0,1,length=101) p = function(x,y){ bw1 = .2; bw2 = .2 w = dnorm((df$x1-x)/bw1, mean=0,sd=1)* dnorm((df$x2-y)/bw2, mean=0,sd=1) weighted.mean(df$y=="1",w) } v = outer(u,u,Vectorize(p)) image(u,u,v,col=clr10,breaks=(0:10)/10) points(df$x1,df$x2,pch=19,cex=1.5,col="white") points(df$x1,df$x2,pch=c(1,19)[1+(df$y=="1")],cex=1.5) contour(u,u,v,levels = .5,add=TRUE)

We get the following prediction

Here, the different colors are probabilities.

## k-nearest neighbors

An alternative is to consider a neighborhood not defined using a distance to point $\mathbf{x}$ but the $k$-neighbors, with the $n$ observations we got.$$\tilde{m}_k(\mathbf{x})=\frac{1}{n}\sum_{i=1}^n\omega_{i,k}(\mathbf{x})y_i$$
where $\omega_{i,k}(\mathbf{x})=n/k$ if $i\in\mathcal{I}_{\mathbf{x}}^k$ with
$$\mathcal{I}_{\mathbf{x}}^k=\{i:\mathbf{x}_i\text{ one of the }k\text{ nearest observations to }\mathbf{x}\}$$
The difficult part here is that we need a valid distance. If units are very different on each component, using the Euclidean distance will be meaningless. So, quite naturally, let us consider here the Mahalanobis distance

Sigma = var(myocarde[,1:7]) Sigma_Inv = solve(Sigma) d2_mahalanobis = function(x,y,Sinv){as.numeric(x-y)%*%Sinv%*%t(x-y)} k_closest = function(i,k){ vect_dist = function(j) d2_mahalanobis(myocarde[i,1:7],myocarde[j,1:7],Sigma_Inv) vect = Vectorize(vect_dist)((1:nrow(myocarde))) which((rank(vect)))}

Here we have a function to find the $k$ closest neighbor for some observation. Then two things can be done to get a prediction. The goal is to predict a class, so we can think of using a majority rule : the prediction for $y_i$ is the same as the one the majority of the neighbors.

k_majority = function(k){ Y=rep(NA,nrow(myocarde)) for(i in 1:length(Y)) Y[i] = sort(myocarde$PRONO[k_closest(i,k)])[(k+1)/2] return(Y)} But we can also compute the proportion of black points among the closest neighbors. It can actually be interpreted as the probability to be black (that’s actually what was said at the beginning of this post, with kernels), k_mean = function(k){ Y=rep(NA,nrow(myocarde)) for(i in 1:length(Y)) Y[i] = mean(myocarde$PRONO[k_closest(i,k)]) return(Y)}

We can see on our dataset the observation, the prediction based on the majority rule, and the proportion of dead individuals among the 7 closest neighbors

cbind(OBSERVED=myocarde$PRONO, MAJORITY=k_majority(7),PROPORTION=k_mean(7)) OBSERVED MAJORITY PROPORTION [1,] 1 1 0.7142857 [2,] 0 1 0.5714286 [3,] 0 0 0.1428571 [4,] 1 1 0.5714286 [5,] 0 1 0.7142857 [6,] 0 0 0.2857143 [7,] 1 1 0.7142857 [8,] 1 0 0.4285714 [9,] 1 1 0.7142857 [10,] 1 1 0.8571429 [11,] 1 1 1.0000000 [12,] 1 1 1.0000000 Here, we got a prediction for an observed point, located at $\boldsymbol{x}_i$, but actually, it is possible to seek the $k$ closest neighbors of any point $\boldsymbol{x}$. Back on our univariate example (to get a graph), we have mean_x = function(x,k=9){ w = rank(abs(myocarde$INSYS-x),ties.method ="random") mean(myocarde$PRONO[which(w&lt;=9)])} u=seq(5,55,length=201) v=Vectorize(function(x) mean_x(x,3))(u) plot(u,v,ylim=0:1,type="l",col="red",lwd=2,xlab="INSYS",ylab="") points(myocarde$INSYS,myocarde$PRONO,pch=19) That’s not very smooth, but we do not have a lot of points either. If we use that technique on our two-dimensional dataset, we obtain the following Sigma_Inv = solve(var(df[,c("x1","x2")])) u = seq(0,1,length=51) p = function(x,y){ k = 6 vect_dist = function(j) d2_mahalanobis(c(x,y),df[j,c("x1","x2")],Sigma_Inv) vect = Vectorize(vect_dist)(1:nrow(df)) idx = which(rank(vect)&lt;=k) return(mean((df$y==1)[idx]))} v = outer(u,u,Vectorize(p)) image(u,u,v,xlab="Variable 1",ylab="Variable 2",col=clr10,breaks=(0:10)/10) points(df$x1,df$x2,pch=19,cex=1.5,col="white") points(df$x1,df$x2,pch=c(1,19)[1+z],cex=1.5) contour(u,u,v,levels = .5,add=TRUE)

This is the idea of local inference, using either kernel on a neighborhood of $\mathbf{x}$ or simply using the $k$ nearest neighbors. Next time, we will investigate penalized logistic regressions, to be continued

On Thursday, March 2nd, I will give the first lecture of the PhD course on advanced tools for econometrics, on nonlinearities. Slides are available online.

# Heuristics on bias and variance for kernel density estimators

Consider the simple case of a moving histogram (which is a very simple kernel). The idea is to recall that

$f(x)=\lim_{h\downarrow 0} f_h(x)$

where

$f_h(x)=\frac{1}{h}\left[F\left(x+\frac{h}{2}\right)-F\left(x-\frac{h}{2}\right)\right]$

is the slope close to point $x$.

Then we use the empirical cumulative density to approximate the slope, i.e.

$\hat f_h(x)=\frac{1}{h}\left[\hat F_n\left(x+\frac{h}{2}\right)-\hat F_n\left(x-\frac{h}{2}\right)\right]$

which can also be writen

$\hat f_h(x)=\frac{1}{nh}\sum_{i=1}^n \boldsymbol{1}\left(x_i\in\left[x-\frac{h}{2},x+\frac{h}{2}\right]\right)$

Consider now the density seen as a random variable

$\hat f_h(x)=\frac{1}{nh}\sum_{i=1}^n \underbrace{\boldsymbol{1}\left(X_i\in\left[x-\frac{h}{2},x+\frac{h}{2}\right]\right)}_{Y_i}$

where the$Y_i$‘s are i.i.d. where $Y_i\sim\mathcal{B}(p_x)$, with

$p_x=\mathbb{P}\left(X_i\in\left[x-\frac{h}{2},x+\frac{h}{2}\right]\right)=h\cdot f_h(x)$

Thus, observe that $\mathbb{E}(\hat f_h(x))=f_h(x)$, but that’s not what we’re looking for… From Taylor’s expansion,

$f_h(x)=f(x)+\frac{h^2}{24}f''(x)+o(h^2)$

thus

$\mathbb{E}(\hat f_h(x))\sim f(x)+\frac{h^2}{24}f''(x)$

where the bias comes from the approximation of the density by some string. About the variance,

$\text{Var}(\hat f_h(x))=\frac{1}{n^2h^2}\cdot n p_x[1-p_x]$

thus, since $p_x=h\cdot f_h(x)=o(h)$,

$\text{Var}(\hat f_h(x))\sim\frac{nh}{n^2h^2}\cdot f_h(x)$

i.e.

$\text{Var}(\hat f_h(x))\sim\frac{1}{nh}\cdot f(x)$

We can observe that

$\text{bias}(\hat f_h(x))=\mathbb{E}(\hat f_h(x))-f(x)\sim\frac{h^2}{24}f''(x)$

is decreasing as $h \downarrow 0$, while the variance is increasing as $h \downarrow 0$. This is the standard bias-variance tradeoff in statistics.

# Log-transform kernel density estimation of income distribution

Our paper Log-transform kernel density estimationof income distribution, written with Emmanuel Flachaire is now available on http://papers.ssrn.com/id=2514882,

Standard kernel density estimation methods are very often used in practice to estimate density function. It works well in numerous cases. However, it is known not to work so well with skewed, multimodal and heavy-tailed distributions. Such features are usual with income distributions, defined over the positive support. We first show that a preliminary logarithmic transformation of the data, combined with standard kernel density estimation methods, can provide a much better fit of the overall density estimation. Then, we show that the fit of the bottom of the distribution may not be satisfactory, even if a better fit of the upper tail can be obtained in general.

# Kernel Density Estimation with Ripley’s Circumferential Correction

The revised version of the paper Kernel Density Estimation with Ripley’s Circumferential Correction is now online, on hal.archives-ouvertes.fr/.

In this paper, we investigate (and extend) Ripley’s circumference method to correct bias of density estimation of edges (or frontiers) of regions. The idea of the method was theoretical and difficult to implement. We provide a simple technique — based of properties of Gaussian kernels — to efficiently compute weights to correct border bias on frontiers of the region of interest, with an automatic selection of an optimal radius for the method. We illustrate the use of that technique to visualize hot spots of car accidents and campsite locations, as well as location of bike thefts.

There are new applications, and new graphs, too

Most of the codes can be found on https://github.com/ripleyCorr/Kernel_density_ripley (as well as datasets).

# Statistics, and the Goldilocks Principle

By the end of May, in Toronto, we had that great talk at the SSC by Jeff Rosenthal, on monte carlo techniques, and Jeff mention the name of “the Goldilocks principle” (it was in the contect of MCMC, and I did mention it in my talk in London on MCMC, when I discussed the value of the rejection rate of the Hastings Metropolis algorithm, which should be not to large, and not too small…). In the story, Goldilocks, there are always three alternative, one is always too much in one extreme (too hot – for the soup – or too large – for the bed, or the chaiir), one is too much in the opposite extreme (too cold, or too small), and one is “just right“.

# Càdiz, Nonparametric Statistics

Emmanuel Flachaire will be presenting some joint work in Càdiz, Spain, tomorrow evening, at the second conference of the International Society of NonParametric Statistics. Jeff invited me a few months ago, to go there, but unfortunately, I’ve already been moving a lot recently. The talk will be based on the same work that I mentioned at the SSC annual conference (Canadian Statistical Society), in Toronto, at the end of May. His talk is on quantiles and inequality indices estimation from heavy-tailed distribution. As mentioned in my previous post, we will upload the slides (and the paper) in a close future.

So, Emmanuel will go there, and enjoy the beach (and the conference, the program is truly amazing).

# Some heuristics about local regression and kernel smoothing

In a standard linear model, we assume that $\mathbb{E}(Y\vert X=x)=\beta_0+\beta_1 x$. Alternatives can be considered, when the linear assumption is too strong.

• Polynomial regression

A natural extension might be to assume some polynomial function,

$\mathbb{E}(Y\vert X=x)=\beta_0+\beta_1 x+\beta_2 x^2 +\cdots +\beta_k x^k$

Again, in the standard linear model approach (with a conditional normal distribution using the GLM terminology), parameters $\boldsymbol{\beta}=(\beta_0,\beta_1,\cdots,\beta_k)$ can be obtained using least squares, where a regression of $Y$ on $\boldsymbol{X}=(1,X,X^2,\cdots,X^k)$ is considered.

Even if this polynomial model is not the real one, it might still be a good approximation for $\mathbb{E}(Y\vert X=x)=h(x)$. Actually, from Stone-Weierstrass theorem, if $h(\cdot)$ is continuous on some interval, then there is a uniform approximation of $h(\cdot)$ by polynomial functions.

Just to illustrate, consider the following (simulated) dataset

set.seed(1)
n=10
xr = seq(0,n,by=.1)
yr = sin(xr/2)+rnorm(length(xr))/2
db = data.frame(x=xr,y=yr)
plot(db)

with the standard regression line

reg = lm(y ~ x,data=db)
abline(reg,col="red")

Consider some polynomial regression. If the degree of the polynomial function is large enough, any kind of pattern can be obtained,

reg=lm(y~poly(x,5),data=db)

But if the degree is too large, then too many ‘oscillations’ are obtained,

reg=lm(y~poly(x,25),data=db)

and the estimation might be be seen as no longer robust: if we change one point, there might be important (local) changes

plot(db)
attach(db)
lines(xr,predict(reg),col="red",lty=2)
yrm=yr;yrm[31]=yr[31]-2
regm=lm(yrm~poly(xr,25))
lines(xr,predict(regm),col="red")

• Local regression

Actually, if our interest is to have locally a good approximation of  $h(\cdot)$, why not use a local regression?

This can be done easily using a weighted regression, where, in the least square formulation, we consider

$\min\left\{ \sum_{i=1}^n \omega_i [Y_i-(\beta_0+\beta_1 X_i)]^2 \right\}$

(it is possible to consider weights in the GLM framework, but let’s keep that for another post). Two comments here:

• here I consider a linear model, but any polynomial model can be considered. Even a constant one. In that case, the optimization problem is

$\min\left\{ \sum_{i=1}^n \omega_i [Y_i-\beta_0]^2 \right\}$which can be solve explicitly, since

$\widehat{\beta}_0=\frac{\sum \omega_i Y_i}{\sum \omega_i}$

• so far, nothing was mentioned about the weights. The idea is simple, here: if you can a good prediction at point $x_0$, then $\omega_i$ should be proportional to some distance between $X_i$ and $x_0$: if $X_i$ is too far from $x_0$, then it should not have to much influence on the prediction.

For instance, if we want to have a prediction at some point $x_0$, consider $\omega_i\propto \boldsymbol{1}(\vert X_i-x_0 \vert<1)$. With this model, we remove observations too far away,

Actually, here, it is the same as

reg=lm(yr~xr,subset=which(abs(xr-x0)<1)

A more general idea is to consider some kernel function $K(\cdot)$ that gives the shape of the weight function, and some bandwidth (usually denoted h) that gives the length of the neighborhood, so that

$\omega_i = K\left(\frac{x_0-X_i}{b}\right)$

This is actually the so-called Nadaraya-Watson estimator of function $h(\cdot)$.
In the previous case, we did consider a uniform kernel $K(x)=\boldsymbol{1}(x\in[-1/2,+1/2])$, with bandwith $2$,

But using this weight function, with a strong discontinuity may not be the best idea… Why not a Gaussian kernel,

$K(x)=\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{x^2}{2}\right)$

This can be done using

fitloc0 = function(x0){
w=dnorm((xr-x0))
reg=lm(y~1,data=db,weights=w)
return(predict(reg,newdata=data.frame(x=x0)))}

On our dataset, we can plot

ul=seq(0,10,by=.01)
vl0=Vectorize(fitloc0)(ul)
u0=seq(-2,7,by=.01)
linearlocalconst=function(x0){
w=dnorm((xr-x0))
plot(db,cex=abs(w)*4)
lines(ul,vl0,col="red")
axis(3)
axis(2)
reg=lm(y~1,data=db,weights=w)
u=seq(0,10,by=.02)
v=predict(reg,newdata=data.frame(x=u))
lines(u,v,col="red",lwd=2)
abline(v=c(0,x0,10),lty=2)
}
linearlocalconst(2)

Here, we want a local regression at point 2. The horizonal line below is the regression (the size of the point is proportional to the wieght). The curve, in red, is the evolution of the local regression

Let us use an animation to visualize the construction of the curve. One can use

library(animate)

but for some reasons, I cannot install the package easily on Linux. And it is not a big deal. We can still use a loop to generate some graphs

vx0=seq(1,9,by=.1)
vx0=c(vx0,rev(vx0))
graphloc=function(i){
name=paste("local-reg-",100+i,".png",sep="")
png(name,600,400)
linearlocalconst(vx0[i])
dev.off()}

for(i in 1:length(vx0)) graphloc(i)

and then, in a terminal, I simply use

    convert -delay 25 /home/freak/local-reg-1*.png /home/freak/local-reg.gif

Of course, it is possible to consider a linear model, locally,

fitloc1 = function(x0){
w=dnorm((xr-x0))
reg=lm(y~poly(x,degree=1),data=db,weights=w)
return(predict(reg,newdata=data.frame(x=x0)))}

or even a quadratic (local) regression,

fitloc2 = function(x0){
w=dnorm((xr-x0))
reg=lm(y~poly(x,degree=2),data=db,weights=w)
return(predict(reg,newdata=data.frame(x=x0)))}

Of course, we can change the bandwidth

To conclude the technical part this post, observe that, in practise, we have to choose the shape of the weight function (the so-called kernel). But there are (simple) technique to select the “optimal” bandwidth h. The idea of cross validation is to consider

$\min\left\{ \sum_{i=1}^n [Y_i-\widehat{Y}_i(b)]^2 \right\}$

where $\widehat{Y}_i(b)$ is the prediction obtained using a local regression technique, with bandwidth $b$. And to get a more accurate (and optimal) bandwith $\widehat{Y}_i(b)$ is obtained using a model estimated on a sample where the ith observation was removed. But again, that is not the main point in this post, so let’s keep that for another one…

Perhaps we can try on some real data? Inspired from a great post on http://f.briatte.org/teaching/ida/092_smoothing.html, by François Briatte, consider the Global Episode Opinion Survey, from some TV show, http://geos.tv/index.php/index?sid=189 , like Dexter.

library(XML)
file = "geos-tww.csv"
html = htmlParse("http://www.geos.tv/index.php/list?sid=189&collection=all")
html = xpathApply(html, "//table[@id='collectionTable']")[[1]]
data = data[,-3]
names(data)=c("no",names(data)[-1])
data=data[-(61:64),]

Let us reshape the dataset,

data$no = 1:96 data$mu = as.numeric(substr(as.character(data$Mean), 0, 4)) data$se =  sd(data$mu,na.rm=TRUE)/sqrt(as.numeric(as.character(data$Count)))
data$season = 1 + (data$no - 1)%/%12
data$season = factor(data$season)
plot(data$no,data$mu,ylim=c(6,10))
segments(data$no,data$mu-1.96*data$se, data$no,data$mu+1.96*data$se,col="light blue")

As done by François, we compute some kind of standard error, just to reflect uncertainty. But we won’t really use it.

plot(data$no,data$mu,ylim=c(6,10))
abline(v=12*(0:8)+.5,lty=2)
for(s in 1:8){reg=lm(mu~no,data=db,subset=season==s)
lines((s-1)*12+1:12,predict(reg)[1:12],col="red") }

Henre, we assume that all seasons should be considered as completely independent… which might not be a great assumption.

db = data
NW = ksmooth(db$no,db$mu,kernel = "normal",bandwidth=5)
plot(data$no,data$mu)
lines(NW,col="red")

We can try to look the curve with a larger bandwidth. The problem is that there is a missing value, at the end. If we (arbitrarily) fill it, we can run a kernel regression,

db$mu[95]=7 NW = ksmooth(db$no,db$mu,kernel = "normal",bandwidth=12) plot(data$no,data$mu,ylim=c(6,10)) lines(NW,col="red") # Natura non facit saltus (see John Wilkins’ article on the – interesting – history of that phrase http://scienceblogs.com/evolvingthoughts/…). We will see, this week in class, several smoothing techniques, for insurance ratemaking. As a starting point, assume that we do not want to use segmentation techniques: everyone will pay exactly the same price. • no segmentation of the premium And that price should be related to the pure premium, which is proportional to the frequency (or the annualized frequency, as discussed previously), since $\mathbb{E}_{\mathbb{P}}\left(\sum_{i=1}^N Y_i\right)=\mathbb{E}_{\mathbb{P}}(N) \cdot \mathbb{E}_{\mathbb{P}}(Y_i)$ The probability measure is mentioned here just to recall that we can use any measure. Even $\mathbb{P}_{\boldsymbol{X}}$ (based on some covariates). Without any covariate, the expected frequency should be > regglm0=glm(nbre~1+offset(log(exposition)),data=sinistres,family=poisson) > summary(regglm0) Call: glm(formula = nbre ~ 1 + offset(log(exposition)), family = poisson, data = sinistres) Deviance Residuals: Min 1Q Median 3Q Max -0.5033 -0.3719 -0.2588 -0.1376 13.2700 Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) -2.6201 0.0228 -114.9 <2e-16 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 (Dispersion parameter for poisson family taken to be 1) Null deviance: 12680 on 49999 degrees of freedom Residual deviance: 12680 on 49999 degrees of freedom AIC: 16353 Number of Fisher Scoring iterations: 6 > exp(coefficients(regglm0)) (Intercept) 0.07279295 Thus, if we do not want to take into account potential heterogeneity, we should assume that $N\sim\mathcal{P}(\lambda)$ where $\lambda$ is closed to 7.28%. Yes, as mentioned in class, it is rather common to see $\lambda$ as a percentage, i.e. a probability, since $\mathbb{P}(N\neq 0)=1-e^{-\lambda}\approx \lambda$ i.e. $\lambda$ can be interpreted as the probability of not have a claim (see also the law of small numbers). Let us visualize this as a function of the age of the driver, > a=18:100 > yp=predict(regglm0,newdata=data.frame(ageconducteur=a,exposition=1),type="response",se.fit=TRUE) > yp0=yp$fit
> yp1=yp$fit+2*yp$se.fit
> yp2=yp$fit-2*yp$se.fit
> plot(a,yp0,type="l",ylim=c(.03,.12))
> abline(v=40,col="grey")
> lines(a,yp1,lty=2)
> lines(a,yp2,lty=2)
> k=23
> points(a[k],yp0[k],pch=3,lwd=3,col="red")
> segments(a[k],yp1[k],a[k],yp2[k],col="red",lwd=3)

We do predict the same frequency for all drivers, e.g. for some drive aged 40,

> cat("Frequency =",yp0[k]," confidence interval",yp1[k],yp2[k])
Frequency = 0.07279295  confidence interval 0.07611196 0.06947393

Let us now consider the case where we try to take into account heterogeneity, e.g. by age,

• The (standard) Poisson regression

The idea of the (log-)Poisson regression is to assume that instead of having $N\sim\mathcal{P}(\lambda)$, we should have $N|\boldsymbol{X}\sim\mathcal{P}(\lambda_{\boldsymbol{X}})$, where

$\lambda_{\boldsymbol{X}}=\exp(\beta_0+\beta_1 \boldsymbol{X}_1+\cdots+\beta_k\boldsymbol{X}_k)$

in a very general setting. Here, let us consider only one explanatory variable, i.e.

$\lambda_{X}=\exp(\beta_0+\beta_1 {X})$

Here, we have

> yp=predict(regglm1,newdata=data.frame(ageconducteur=a,exposition=1),
+ type="response",se.fit=TRUE)
> yp0=yp$fit > yp1=yp$fit+2*yp$se.fit > yp2=yp$fit-2*yp$se.fit > plot(a,yp0,type="l",ylim=c(.03,.12)) > abline(v=40,col="grey") > lines(a,yp1,lty=2) > lines(a,yp2,lty=2) > points(a[k],yp0[k],pch=3,lwd=3,col="red") > segments(a[k],yp1[k],a[k],yp2[k],col="red",lwd=3) i.e. the prediction for the annualized claim frequency for our 40 year old driver is now 7.74% (which is slightly higher than what we had before, 7.28%) > cat("Frequency =",yp0[k]," confidence interval",yp1[k],yp2[k]) Frequency = 0.07740574 confidence interval 0.08117512 0.07363636 It is possible to compute not the expected frequency , but the ratio $\mathbb{E}(N|X)/\mathbb{E}(N)$. Above the horizontal blue line, the premium will be higher than the one obtained without segmentation, and (of course) lower below. Here, drivers younger than 44 year old will pay more, while driver older than 44 year old will be less. We have discussed, in the introduction, the necessity of segmentation. If we consider two companies, one segmenting, while the other one has a flat rate, then older drivers will go to the first company (since insurance is cheaper) while younger ones will go to the second one (again, it is cheaper). The problem is that the second company implicitly hopes that older drivers will compensate the risk. But since they’re gone, insurance will be too cheap, and the company will loose money (if not goes bankrupt). So companies have to use segmentation techniques to survive. Now, the problem is that we cannot be sure that this exponential decay of the premium is the proper way the premium should evolve as a function of the age. An alternative can be to use nonparametric techniques to visualize to true influence of the age on claims frequency. • A pure nonparametric model A first model can be to consider a premium, per age. This can be done considering the age of the driver as a factor in the regression, > regglm2=glm(nbre~as.factor(ageconducteur)+offset(log(exposition)), + data=sinistres,family=poisson) > yp=predict(regglm2,newdata=data.frame(ageconducteur=a0,exposition=1), + type="response",se.fit=TRUE) > yp0=yp$fit
> yp1=yp$fit+2*yp$se.fit
> yp2=yp$fit-2*yp$se.fit
> plot(a0,yp0,type="l",ylim=c(.03,.12))
> abline(v=40,col="grey")

Here, the forecast for our 40 year old driver is slightly lower than be previous one, but the confidence interval is much larger (since we focus on a very small subclass of the portfolio: drivers aged exactly 40)

Frequency = 0.06686658  confidence interval 0.08750205 0.0462311

Here, we consider too small classes, and the premium is too erratic: the premium will decrease of 20% from age 40 to 41, and then increase of 50% from age 41 to 42,

> diff(log(yp0[23:25]))
24         25
-0.2330241  0.5223478

There is no chance that the company will keep the insured with this strategy. This discontinuity of the premium is clearly an important issue here.

• Using age classes

An alternative can be to consider age classes, from very young drivers to senior drivers.

> level1=seq(15,105,by=5)
> regglmc1=glm(nbre~cut(ageconducteur,level1)+offset(log(exposition)),
+ data=sinistres,family=poisson)
> summary(regglmc1)

Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept)                         -1.6036     0.1741  -9.212  < 2e-16 ***
cut(ageconducteur, level1)(20,25]   -0.4200     0.1948  -2.157   0.0310 *
cut(ageconducteur, level1)(25,30]   -0.9378     0.1903  -4.927 8.33e-07 ***
cut(ageconducteur, level1)(30,35]   -1.0030     0.1869  -5.367 8.02e-08 ***
cut(ageconducteur, level1)(35,40]   -1.0779     0.1866  -5.776 7.65e-09 ***
cut(ageconducteur, level1)(40,45]   -1.0264     0.1858  -5.526 3.28e-08 ***
cut(ageconducteur, level1)(45,50]   -0.9978     0.1856  -5.377 7.58e-08 ***
cut(ageconducteur, level1)(50,55]   -1.0137     0.1855  -5.464 4.65e-08 ***
cut(ageconducteur, level1)(55,60]   -1.2036     0.1939  -6.207 5.40e-10 ***
cut(ageconducteur, level1)(60,65]   -1.1411     0.2008  -5.684 1.31e-08 ***
cut(ageconducteur, level1)(65,70]   -1.2114     0.2085  -5.811 6.22e-09 ***
cut(ageconducteur, level1)(70,75]   -1.3285     0.2210  -6.012 1.83e-09 ***
cut(ageconducteur, level1)(75,80]   -0.9814     0.2271  -4.321 1.55e-05 ***
cut(ageconducteur, level1)(80,85]   -1.4782     0.3371  -4.385 1.16e-05 ***
cut(ageconducteur, level1)(85,90]   -1.2120     0.5294  -2.289   0.0221 *
cut(ageconducteur, level1)(90,95]   -0.9728     1.0150  -0.958   0.3379
cut(ageconducteur, level1)(95,100] -11.4694   144.2817  -0.079   0.9366
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

> yp=predict(regglmc1,newdata=data.frame(ageconducteur=a,exposition=1),
+ type="response",se.fit=TRUE)
> yp0=yp$fit > yp1=yp$fit+2*yp$se.fit > yp2=yp$fit-2*yp$se.fit > plot(a,yp0,ylim=c(.03,.12),type="s") > abline(v=40,col="grey") > lines(a,yp1,lty=2,type="s") > lines(a,yp2,lty=2,type="s") Here we obtain the following predictions, and for our 40 year old driver, the frequency is now 6.84%. Frequency = 0.0684573 confidence interval 0.07766717 0.05924742 But our classes were defined arbitrarily here. Perhaps should we consider other classes, to see if the prediction is sensitive to the cutting values, > level2=level1-2 > regglmc2=glm(nbre~cut(ageconducteur,level2)+offset(log(exposition)), + data=sinistres,family=poisson) which yields the following values for our 40 year old driver, Frequency = 0.07050614 confidence interval 0.07980422 0.06120807 So here, we did not remove the discontinuity problem. An idea here can be to consider moving regions: if the goal is to predict the frequency for a 40 year old driver, perhaps the class should be (somehow) centered around 40. And center the interval around 35 for drivers aged 35. Etc. • Moving average Thus, it is natural to consider some local regressions, where only drivers aged almost 40 should be considered. This almost concept is related to the bandwidth. For instance, drivers between 35 and 45 can be considered as being almost40. In practice we can either consider a subset function, or we can use weights in the regressions > value=40 > h=5 > sinistres$omega=(abs(sinistres$ageconducteur-value)<=h)*1 > regglmomega=glm(nbre~ageconducteur+offset(log(exposition)), + data=sinistres,family=poisson,weights=omega) To see what’s going on, let us consider an animated plot, where the age of interest is changing, Here, for our 40 year old drive, we get Frequency = 0.06913391 confidence interval 0.07535564 0.06291218 We do obtain a curve that can be interpreted as a local regression. But here, we do not take into account that 35 is not as close to 40 as 39 could be. An here, 34 is assumed to be very far away from 40. Clearly, we could improve that technique: kernel functions can considered, i.e. the closer to 40, the larger the weight. > value=40 > h=5 > sinistres$omega=dnorm(abs(sinistres$ageconducteur-value)/h) > regglmomega=glm(nbre~ageconducteur+offset(log(exposition)), + data=sinistres,family=poisson,weights=omega) which can be plotted below Here, our prediction for our 40 year old drive is Frequency = 0.07040464 confidence interval 0.07981521 0.06099408 This is the idea of kernel regression techniques. But as explained in the slides, other non parametric techniques can be considered, like spline functions. • Smoothing with splines In R, it is simple to use spline function (somehow much more simple than kernel smoothers) > library(splines) > regglmbs=glm(nbre~bs(ageconducteur)+offset(log(exposition)), + data=sinistres,family=poisson) The prediction for our 40 year old driver is now Frequency = 0.06928169 confidence interval 0.07397124 0.06459215 Note that this techniques is related to another class of models, the so-called Generalized Additive Models, i.e. GAMs. > library(mgcv) > reggam=gam(nbre~s(ageconducteur)+offset(log(exposition)), + data=sinistres,family=poisson) The prediction is extremely close to the one we obtained above (the main differences being observed for very old drivers) Frequency = 0.06912683 confidence interval 0.07501663 0.06323702 • Comparison of the different models Somehow, one way or another, all those models are valid. So perhaps we should compare them, On the graph above, we can visualize the upper and the lower bound of the prediction, for the 9 models. The horizontal line is the predicted value without taking into account heterogeneity. It is possible to consider relative values, with respect to this value, # (nonparametric) copula density estimation Today, we will go further on the inference of copula functions. Some codes (and references) can be found on a previous post, on nonparametric estimators of copula densities (among other related things). Consider (as before) the loss-ALAE dataset (since we’ve been working a lot on that dataset) > library(MASS) > library(evd) > X=lossalae > U=cbind(rank(X[,1])/(nrow(X)+1),rank(X[,2])/(nrow(X)+1)) The standard tool to plot nonparametric estimators of densities is to use multivariate kernels. We can look at the density using > mat1=kde2d(U[,1],U[,2],n=35) > persp(mat1$x,mat1$y,mat1$z,col="green",
+ xlab="",ylab="",zlab="",zlim=c(0,7))

or level curves (isodensity curves) with more detailed estimators (on grids with shorter steps)

> mat1=kde2d(U[,1],U[,2],n=101)
> image(mat1$x,mat1$y,mat1$z,col= + rev(heat.colors(100)),xlab="",ylab="") > contour(mat1$x,mat1$y,mat1$z,add=
+ TRUE,levels = pretty(c(0,4), 11))

Kernels are nice, but we clearly observe some border bias, extremely strong in corners (the estimator is 1/4th of what it should be, see another post for more details). Instead of working on sample $(U_i,V_i)$ on the unit square, consider some transformed sample $(Q(U_i),Q(V_i))$, where $Q:(0,1)\rightarrow\mathbb{R}$ is a given function. E.g. a quantile function of an unbounded distribution, for instance the quantile function of the $\mathcal{N}(0,1)$ distribution. Then, we can estimate the density of the transformed sample, and using the inversion technique, derive an estimator of the density of the initial sample. Since the inverse of a (general) function is not that simple to compute, the code might be a bit slow. But it does work,

> gaussian.kernel.copula.surface <- function (u,v,n) {
+   s=seq(1/(n+1), length=n, by=1/(n+1))
+   mat=matrix(NA,nrow = n, ncol = n)
+ sur=kde2d(qnorm(u),qnorm(v),n=1000,
+ lims = c(-4, 4, -4, 4))
+ su<-sur$z + for (i in 1:n) { + for (j in 1:n) { + Xi<-round((qnorm(s[i])+4)*1000/8)+1; + Yj<-round((qnorm(s[j])+4)*1000/8)+1 + mat[i,j]<-su[Xi,Yj]/(dnorm(qnorm(s[i]))* + dnorm(qnorm(s[j]))) + } + } + return(list(x=s,y=s,z=data.matrix(mat))) + } Here, we get Note that it is possible to consider another transformation, e.g. the quantile function of a Student-t distribution. > student.kernel.copula.surface = + function (u,v,n,d=4) { + s <- seq(1/(n+1), length=n, by=1/(n+1)) + mat <- matrix(NA,nrow = n, ncol = n) + sur<-kde2d(qt(u,df=d),qt(v,df=d),n=5000, + lims = c(-8, 8, -8, 8)) + su<-sur$z
+ for (i in 1:n) {
+     for (j in 1:n) {
+ 	Xi<-round((qt(s[i],df=d)+8)*5000/16)+1;
+ 	Yj<-round((qt(s[j],df=d)+8)*5000/16)+1
+ 	mat[i,j]<-su[Xi,Yj]/(dt(qt(s[i],df=d),df=d)*
+ 	dt(qt(s[j],df=d),df=d))
+     }
+ }
+ return(list(x=s,y=s,z=data.matrix(mat)))
+ }

Another strategy is to consider kernel that have precisely the unit interval as support. The idea is here to consider the product of Beta kernels, where parameters depend on the location

> beta.kernel.copula.surface=
+  function (u,v,bx=.025,by=.025,n) {
+  s <- seq(1/(n+1), length=n, by=1/(n+1))
+  mat <- matrix(0,nrow = n, ncol = n)
+ for (i in 1:n) {
+     a <- s[i]
+     for (j in 1:n) {
+     b <- s[j]
+ 	mat[i,j] <- sum(dbeta(a,u/bx,(1-u)/bx) *
+     dbeta(b,v/by,(1-v)/by)) / length(u)
+     }
+ }
+ return(list(x=s,y=s,z=data.matrix(mat)))
+ }

On those two graphs, we can clearly observe strong tail dependence in the upper (right) corner, that cannot be intuited using a standard kernel estimator…

# Border bias and weighted kernels

With Ewen (aka @3wen), not only we have been playing on Twitter this month, we have also been working on kernel estimation for densities of spatial processes. Actually, it is only a part of what he was working on, but that part on kernel estimation has been the opportunity to write a short paper, that can now be downloaded on hal.

The problem with kernels is that kernel density estimators suffer a strong bias on borders. And with geographic data, it is not uncommon to have observations very close to the border (frontier, or ocean). With standard kernels, some weight is allocated outside the area: the density does not sum to one. And we should not look for a global correction, but for a local one. So we should use weighted kernel estimators (see on hal for more details). The problem that weights can be difficult to derive, when the shape of the support is a strange polygon. The idea is to use a property of product Gaussian kernels (with identical bandwidth) i.e. with the interpretation of having noisy observation, we can use the property of circular isodensity curve. And this can be related to Ripley (1977) circumferential correction. And the good point is that, with R, it is extremely simple to get the area of the intersection of two polygons. But we need to upload some R packages first,

require(maps)
require(sp)
require(snow)
require(ellipse)
require(ks)
require(gpclib)
require(rgeos)
require(fields)

To be more clear, let us illustrate that technique on a nice example. For instance, consider some bodiliy injury car accidents in France, in 2008 (that I cannot upload but I can upload a random sample),

base_cara=read.table(
"http://freakonometrics.blog.free.fr/public/base_fin_morb.txt",
sep=";",header=TRUE)

The border of the support of our distribution of car accidents will be the contour of the Finistère departement, that can be found in standard packages

geoloc=read.csv(
"http://freakonometrics.free.fr/popfr19752010.csv",
colClasses=c(rep("character",5),rep("numeric",38)))
geoloc=geoloc[,c("dep","com","com_nom",
"long","lat","pop_2008")]
geoloc$id=paste(sprintf("%02s",geoloc$dep),
sprintf("%03s",geoloc$com),sep="") geoloc=geoloc[,c("com_nom","long","lat","pop_2008")] head(geoloc) france=map('france',namesonly=TRUE, plot=FALSE) francemap=map('france', fill=TRUE, col="transparent", plot=FALSE) detpartement_bzh=france[which(france%in% c("Finistere","Morbihan","Ille-et-Vilaine", "Cotes-Darmor"))] bretagne=map('france',regions=detpartement_bzh, fill=TRUE, col="transparent", plot=FALSE,exact=TRUE) finistere=cbind(bretagne$x[321:678],bretagne$y[321:678]) FINISTERE=map('france',regions="Finistere", fill=TRUE, col="transparent", plot=FALSE,exact=TRUE) monFINISTERE=cbind(FINISTERE$x[c(8:414)],FINISTERE$y[c(8:414)]) Now we need simple functions, cercle=function(n=200,centre=c(0,0),rayon) {theta=seq(0,2*pi,length=100) m=cbind(cos(theta),sin(theta))*rayon m[,1]=m[,1]+centre[1] m[,2]=m[,2]+centre[2] names(m)=c("x","y") return(m)} poids=function(x,h,POL) {leCercle=cercle(centre=x,rayon=5/pi*h) POLcercle=as(leCercle, "gpc.poly") return(area.poly(intersect(POL,POLcercle))/ area.poly(POLcercle))} lissage = function(U,polygone,optimal=TRUE,h=.1) {n=nrow(U) IND=which(is.na(U[,1])==FALSE) U=U[IND,] if(optimal==TRUE) {H=Hpi(U,binned=FALSE); H=matrix(c(sqrt(H[1,1]*H[2,2]),0,0, sqrt(H[1,1]*H[2,2])),2,2)} if(optimal==FALSE){H= matrix(c(h,0,0,h),2,2) before defining our weights. poidsU=function(i,U,h,POL) {x=U[i,] poids(x,h,POL)} OMEGA=parLapply(cl,1:n,poidsU,U=U,h=sqrt(H[1,1]), POL=as(polygone, "gpc.poly")) OMEGA=do.call("c",OMEGA) stopCluster(cl) }else {OMEGA=lapply(1:n,poidsU,U=U,h=sqrt(H[1,1]), POL=as(polygone, "gpc.poly")) OMEGA=do.call("c",OMEGA)} Note that it is possible to parallelize if there are a lot of observations, if(n>=500) {cl <- makeCluster(4,type="SOCK") worker.init <- function(packages) {for(p in packages){library(p, character.only=T)} NULL} clusterCall(cl, worker.init, c("gpclib","sp")) clusterExport(cl,c("cercle","poids")) Then, we can use standard bivariate kernel smoothing functions, but with the weights we just calculated, using a simple technique that can be related to one suggested in Ripley (1977), fhat=kde(U,H,w=1/OMEGA,xmin=c(min(polygone[,1]), min(polygone[,2])),xmax=c(max(polygone[,1]), max(polygone[,2]))) fhat$estimate=fhat$estimate*sum(1/OMEGA)/n vx=unlist(fhat$eval.points[1])
vy=unlist(fhat$eval.points[2]) VX = cbind(rep(vx,each=length(vy))) VY = cbind(rep(vy,length(vx))) VXY=cbind(VX,VY) Ind=matrix(point.in.polygon(VX,VY, polygone[,1], polygone[,2]),length(vy),length(vx)) f0=fhat f0$estimate[t(Ind)==0]=NA
return(list(
X=fhat$eval.points[[1]], Y=fhat$eval.points[[2]],
Z=fhat$estimate, ZNA=f0$estimate,
H=fhat$H, W=fhat$W))}
lissage_without_c = function(U,polygone,optimal=TRUE,h=.1)
{n=nrow(U)
IND=which(is.na(U[,1])==FALSE)
U=U[IND,]
if(optimal==TRUE) {H=Hpi(U,binned=FALSE);
H=matrix(c(sqrt(H[1,1]*H[2,2]),0,0,sqrt(H[1,1]*H[2,2])),2,2)}
if(optimal==FALSE){H= matrix(c(h,0,0,h),2,2)}
fhat=kde(U,H,xmin=c(min(polygone[,1]),
min(polygone[,2])),xmax=c(max(polygone[,1]),
max(polygone[,2])))
vx=unlist(fhat$eval.points[1]) vy=unlist(fhat$eval.points[2])
VX = cbind(rep(vx,each=length(vy)))
VY = cbind(rep(vy,length(vx)))
VXY=cbind(VX,VY)
Ind=matrix(point.in.polygon(VX,VY, polygone[,1],
polygone[,2]),length(vy),length(vx))
f0=fhat
f0$estimate[t(Ind)==0]=NA return(list( X=fhat$eval.points[[1]],
Y=fhat$eval.points[[2]], Z=fhat$estimate,
ZNA=f0$estimate, H=fhat$H,
W=fhat$W))} So, now we can play with those functions, base_cara_FINISTERE=base_cara[which(point.in.polygon( base_cara$long,base_cara$lat,monFINISTERE[,1], monFINISTERE[,2])==1),] coord=cbind(as.numeric(base_cara_FINISTERE$long),
as.numeric(base_cara_FINISTERE$lat)) nrow(coord) map(francemap) lissage_FIN_withoutc=lissage_without_c(coord, monFINISTERE,optimal=TRUE) lissage_FIN=lissage(coord,monFINISTERE, optimal=TRUE) lesBreaks_sans_pop=range(c( range(lissage_FIN_withoutc$Z),
range(lissage_FIN$Z))) lesBreaks_sans_pop=seq(min(lesBreaks_sans_pop)*.95, max(lesBreaks_sans_pop)*1.05,length=21) plot_article=function(lissage,breaks, polygone,coord){ par(mar=c(3,1,3,1)) image.plot(lissage$X,lissage$Y,(lissage$ZNA),
xlim=range(polygone[,1]),ylim=range(polygone[,2]),
breaks=breaks, col=rev(heat.colors(20)),xlab="",
ylab="",xaxt="n",yaxt="n",bty="n",zlim=range(breaks),
horizontal=TRUE)
contour(lissage$X,lissage$Y,lissage$ZNA,add=TRUE, col="grey") points(coord[,1],coord[,2],pch=19,cex=.1, col="dodger blue") polygon(polygone,lwd=2,)} plot_article(lissage_FIN_withoutc,breaks= lesBreaks_sans_pop,polygone=monFINISTERE, coord=coord) plot_article(lissage_FIN,breaks= lesBreaks_sans_pop,polygone=monFINISTERE, coord=coord) If we look at the graphs, we have the following densities of car accident, with a standard kernel on the left, and our proposal on the right (with local weight adjustment when the estimation is done next to the border of the region of interest), Similarly, in Morbihan, With those modified kernels, hot spots appear much more clearly. For more details, the paper is online on hal. # Beta kernel and transformed kernel This Thursday I will give a talk at Laval University, on “Beta kernel and transformed kernel : applications to copula density estimation and quantile estimation“. This time, I will talk at the department of Mathematics and Statistics (13:30 at the pavillon Adrien-Pouliot). “Because copulas have bounded support (the unit square in dimension 2), standard kernel based estimators of densities are (multiplicatively) biased on borders and in corners of the support. Two techniques can be used to avoid that underestimation: Beta kernels and Transformed kernel. We will describe and discuss those two techniques in the first part of the talk. Then, we will see that it is possible to combine those two techniques to get nice estimator of several quantities (e.g. quantiles): transform the data to get on the unit interval – using a transformed kernel – then estimate the (transformed) quantile on [0,1] using a beta kernel, then get back on the initial support. As we will see on simulations, that technique can be better than standard quantile estimators, especially when data are heavy tailed.” Slides can be downloaded here. • kernel based density estimation Kernel based estimation are a popular (and natural) technique to estimate densities. It is simply and extension of the moving histogram: so we count how many observations are a the neighborhood of the point where we want to estimate the density of the distribution. Then it is natural so consider a smoothing function, i.e. instead of a step function (either observations are close enough, or not), it is possible to give weights to observations, which will be a decreasing function of the distance, With a smooth kernel, we have a smooth estimation of the density Then it is possible to play on the bandwidth, either to get a more accurate estimation of the density, but not that smooth (small bias but large variance), or a smoother one (large bias, but small variance), In R, it is simply > X=rnorm(100) > (D=density(X)) Call: density.default(x = X) Data: X (100 obs.); Bandwidth 'bw' = 0.3548 x y Min. :-3.910799 Min. :0.0001265 1st Qu.:-1.959098 1st Qu.:0.0108900 Median :-0.007397 Median :0.0513358 Mean :-0.007397 Mean :0.1279645 3rd Qu.: 1.944303 3rd Qu.:0.2641952 Max. : 3.896004 Max. :0.3828215 > plot(D$x,D$y) • Beta kernel The idea of Beta kernel is to consider kernels having support [0,1]. In the univariate case, where is the density of a Beta distribution, i.e. For additional material, I have uploaded some R code to fit copula densities using beta kernels, library(copula) beta.kernel.copula.surface = function (u,v,bx,by,p) { s = seq(1/p, len=(p-1), by=1/p) mat = matrix(0,nrow = p-1, ncol = p-1) for (i in 1:(p-1)) { a = s[i] for (j in 1:(p-1)) { b = s[j] mat[i,j] = sum(dbeta(a,u/bx,(1-u)/bx) * dbeta(b,v/by,(1-v)/by)) / length(u) } } return(data.matrix(mat)) } Then we can used it to see what we get on a simulated sample library(copula) COPULA = frankCopula(param=5, dim = 2) X = rcopula(n=1000,COPULA) p0 = 26 Z= beta.kernel.copula.surface(X[,1],X[,2],bx=.01,by=.01,p=p0) u = seq(1/p0, len=(p0-1), by=1/p0) persp(u,u,Z,theta=30,col="green",shade=TRUE, box=FALSE,zlim=c(0,6)) (yes, the surface is changing… to illustrate the impact of the bandwidth on the estimation). • transformed kernel estimation I the talk, I will also mention the transformed Kernel estimate, as introduced in the book on L1 density estimation by Luc Devroye and Laszlo Györfi (the book can be downloaded here). I probably spend a few minutes on the original chapter, in order to provide another application of that techniques (not only to estimate copula densities, but here to estimate quantiles of heavy tailed distribution). In the univariate case, the R code is the following (here I consider two transformation, the quantile function of the Gaussian distribution, and the quantile function of the Student distribution with 3 degrees of freedom), set.seed(1) sample=rbeta(100,4,3) transfN = function(x){ Y=qnorm(sample) f=density(Y,from=-4,to=4,n=2001) ny=sum(f$x<=qnorm(x));
g=f$y[ny]/dnorm(qnorm(x)) return(g) } df0=3 transfT = function(x){ Y=qt(sample,df=df0) f=density(Y,from=-4,to=4,n=2001) ny=sum(f$x<=qt(x,3));
g=f$y[ny]/dt(qt(x,df=df0),df=df0) return(g) } tN=Vectorize(transfN) tT=Vectorize(transfT) u=seq(.01,.99,by=.01) vN=tN(u) vT=tT(u) plot(u,vN,type="l",lwd=3,col="blue") lines(u,vT,lwd=3,col="green") lines(u,dbeta(u,4,3),col="red",lty=2) The density estimation is the following, (the red dotted line is the true density, since we work on a simulated sample). Now, let us get back on the initial chapter, In the book, this is introduced as follows, The original idea we add it to use this kernel based estimator for copulas, i.e. since we can estimate densities in high dimension with unbounded support, using the idea is to transform marginal observations, and to use the fact that the associated copula density can be written to derive an intuitive estimator for the copula density An important issue is how do we choose the transformation And Luc Devroye and Laszlo Györfi mention that this can be used to deal with extremes. well, extremes are introduced through bumps (which is not the way I would have been dealing with extremes) and note that several results can be derived on those bumps, e.g. Then, there is an interesting discussion about estimating the optimal transformation and I will prove that this can be an extremely interesting idea, for instance to estimate quantiles of heavy tailed distribution, if we use also the beta kernel estimator on the unit interval. This idea was developed in a paper with Abder Oulidi, online here. Remark: actually, in the book, an additional reference is mentioned, but I have never been able to find a copy… if anyone has one, I’d be glad to read it… # Circular or spherical data, and density estimation I few years ago, while I was working on kernel based density estimation on compact support distribution (like copulas) I went through a series of papers on circular distributions. By that time, I thought it was something for mathematicians working on weird spaces…. but during the past weeks, I saw several potential applications of those estimators. • circular data density estimation Consider the density of an angle say, i.e. a function such that with a circular relationship, i.e. . It can be seen as an invariance by rotation. von Mises proposed a parametric model in 1918 (see here or there), assuming that where is Bessel modified function of order 1, (which is simply a normalization parameter). There are two parameters here, (some concentration parameter) and mu a direction. From a series of observed angles, the maximum likelihood estimator for kappa is solution of where and and where , where those functions are modified Bessel functions. Well, that estimator is biased, but it is possible to improve it (see here or there). This can be done easily in R (actually Jeff Gill – here – used that package in several applications). But I am not a big fan of that technique…. • density estimation for hours on simulated data A nice application can be on the estimation of the daily density of a temporal events (e.g. phone calls as we’ll see later on, or email arrival time). Let is the time (in hours) for the th observation (the th phone call received). Then set The time is now seen as an angle. It is possible to consider the equivalent of an histogram, set.seed(1) library(circular) X=rbeta(100,shape1=2,shape2=4)*24 Omega=2*pi*X/24 Omegat=2*pi*trunc(X)/24 H=circular(Omega,type="angle",units="radians",rotation="clock") Ht=circular(Omegat,type="angle",units="radians",rotation="clock") plot(Ht, stack=FALSE, shrink=1.3, cex=1.03, axes=FALSE,tol=0.8,zero=c(rad(90)),bins=24,ylim=c(0,1)) points(Ht, rotation = "clock", zero =c(rad(90)), col = "1", cex=1.03, stack=TRUE ) rose.diag(Ht-pi/2,bins=24,shrink=0.33,xlim=c(-2,2),ylim=c(-2,2), axes=FALSE,prop=1.5) or a kernel based estimation of the density (the gray line on the right). circ.dens = density(Ht+3*pi/2,bw=20) plot(Ht, stack=TRUE, shrink=.35, cex=0, sep=0.0, axes=FALSE,tol=.8,zero=c(0),bins=24, xlim=c(-2,2),ylim=c(-2,2), ticks=TRUE, tcl=.075) lines(circ.dens, col="darkgrey", lwd=3) text(0,0.8,"24", cex=2); text(0,-0.8,"12",cex=2); text(0.8,0,"6",cex=2); text(-0.8,0,"18",cex=2) The code looks rather simple. But I am not very comfortable using codes that I do not completely understand. So I did my own. The first step was to get a graph similar to the one we have on the right, except that I prefer my own kernel based estimator. The idea is that instead of estimating the density on , we estimate it on the sample . Then we multiply by 3 to get the density only on . For the bandwidth, I took the same as the one that we would have taken on The code is simply the following U=seq(0,1,by=1/250) O=U*2*pi U12=seq(0,1,by=1/24) O12=U12*2*pi X=rbeta(100,shape1=2,shape2=4)*24 OM=2*pi*X/24 XL=c(X-24,X,X+24) d=density(X) d=density(XL,bw=d$bw,n=1500)
I=which((d$x>=6)&(d$x<=30))
Od=d$x[I]/24*2*pi-pi/2 Dd=d$y[I]/max(d$y)+1 plot(cos(O),-sin(O),xlim=c(-2,2),ylim=c(-2,2), type="l",axes=FALSE,xlab="",ylab="") for(i in pi/12*(0:12)){ abline(a=0,b=tan(i),lty=1,col="light yellow")} segments(.9*cos(O12),.9*sin(O12),1.1*cos(O12),1.1*sin(O12)) lines(Dd*cos(Od),-Dd*sin(Od),col="red",lwd=1.5) text(.7,0,"6"); text(-.7,0,"18") text(0,-.7,"12"); text(0,.7,"24") R=1/24/max(d$y)/3+1 lines(R*cos(O),R*sin(O),lty=2)

Note that it is possible to stress more (visually) on hours having few phone calls, or a lot (compared with an homogeneous Poisson process), e.g.

plot(cos(O),-sin(O),xlim=c(-2,2),ylim=c(-2,2),
type="l",axes=FALSE,xlab="",ylab="")
for(i in pi/12*(0:12)){
abline(a=0,b=tan(i),lty=1,col="light yellow")}
segments(2*cos(O12),2*sin(O12),1.1*cos(O12),1.1*sin(O12), col="light grey")
segments(.9*cos(O12),.9*sin(O12),1.1*cos(O12),1.1*sin(O12))
text(.7,0,"6")
text(-.7,0,"18")
text(0,-.7,"12")
text(0,.7,"24")
R=1/24/max(d$y)/3+1 lines(R*cos(O),R*sin(O),lty=2) AX=R*cos(Od);AY=-R*sin(Od) BX=Dd*cos(Od);BY=-Dd*sin(Od) COUL=rep("blue",length(AX)) COUL[R<Dd]="red" CM=cm.colors(200) a=trunc(100*Dd/R) COUL=CM[a] segments(AX,AY,BX,BY,col=COUL,lwd=2) lines(Dd*cos(Od),-Dd*sin(Od),lwd=2) We get here those two graphs, To be honest, I do not really like that representation – even if it looks nice. If we compare that circular representation to a more classical one (from 0:00 till 23:59 one the graph on the left, below), I do have a problem to interpret the areas in blue and pink. density of wind direction On the left, we compare two densities, so the area in pink is the same as the area in blue. But here, it is no longer the case: the area in pink is always larger to the one in blue. So it might help so see when we have a difference, but there is a scaling issue that we cannot discuss further… But less us see if we can use that estimation technique to several problems. A standard application when studying angles is wind direction. For instance, in Montréal, it is possible to find hourly observations, starting in 1974 (we just need a R robot to pick up the information, but I’ll tell more about that in another post, someday). Here, we have directly an angle. So we can use a code rather similar to the one used above to estimate the distribution of wind direction in Montréal. density of 911 phone calls Note that our estimate is consistent with several graphs that can be found on meteorological websites (e.g. the one above on the right, that was found here). In a recent post (here) I wanted to check about the “midnight crime” myth, using hours of 911 phone calls in Montréal. That was for all phone calls. But if we look more specifically, for burglaries, we have the distribution on the left, and for conflicts the one on the right We do clearly observe that gun shots occur a bit before midnight. See also here for another study, but this time in NYC (thanks @PAC for the link).while for gun shots, we have the distribution on the left, and for “troubles” (basically people making too much noisy in parties) or “noise” the one on the right • density of earth temperatures, or earthquakes Of course it is also possible to work in higher dimension. Before, we went from densities on to densities on the unit circle . But similarly, it is possible to go from to the unit sphere . A nice application being global climate studies, The idea being that point on the left above are extremely close to the one on the right. An application can be e.g. on earthquakes occurrence. Data can be found here. library(ks) X=cbind(EQ$Longitude,EQ\$Latitude)
Hpi1 = Hpi(x = X)
DX=kde(x = X, H = Hpi1)
library(maps)
map("world")
cbind(X[,1]+360,X[,2]+180),cbind(X[,1]-360,X[,2]+180), cbind(X[,1],X[,2]-180),cbind(X[,1]+360, X[,2]-180),cbind(X[,1]-360,X[,2]-180)) DY=kde(x = Y, H = Hpi1) library(maps) plot (DY,add=TRUE,col="purple")