Tag Archives: optimization

Classification from scratch, linear discrimination 8/8

Eighth post of our series on classification from scratch. The latest one was on the SVM, and today, I want to get back on very old stuff, with here also a linear separation of the space, using Fisher’s linear discriminent analysis.

Bayes (naive) classifier

Consider the follwing naive classification rulem^\star(\mathbf{x})=\text{argmin}_y\{\mathbb{P}[Y=y\vert\mathbf{X}=\mathbf{x}]\}orm^\star(\mathbf{x})=\text{argmin}_y\left\{\frac{\mathbb{P}[\mathbf{X}=\mathbf{x}\vert Y=y]}{\mathbb{P}[\mathbf{X}=\mathbf{x}]}\right\}(where \mathbb{P}[\mathbf{X}=\mathbf{x}] is the density in the continuous case).

In the case where y takes two values, that will be standard \{0,1\} here, one can rewrite the later asm^\star(\mathbf{x})=\begin{cases}1\text{ if }\mathbb{E}(Y\vert \mathbf{X}=\mathbf{x})>\displaystyle{\frac{1}{2}}\\0\text{ otherwise}\end{cases}and the set\mathcal{D}_S =\left\{\mathbf{x},\mathbb{E}(Y\vert \mathbf{X}=\mathbf{x})=\frac{1}{2}\right\}is called the decision boundary.

Assume that\mathbf{X}\vert Y=0\sim\mathcal{N}(\mathbf{\mu}_0,\mathbf{\Sigma})and\mathbf{X}\vert Y=1\sim\mathcal{N}(\mathbf{\mu}_1,\mathbf{\Sigma})then explicit expressions can be derived.m^\star(\mathbf{x})=\begin{cases}1\text{ if }r_1^2< r_0^2+2\displaystyle{\log\frac{\mathbb{P}(Y=1)}{\mathbb{P}(Y=0)}+\log\frac{\vert\mathbf{\Sigma}_0\vert}{\vert\mathbf{\Sigma}_1\vert}}\\0\text{ otherwise}\end{cases}where r_y^2 is the Manalahobis distance, r_y^2 = [\mathbf{X}-\mathbf{\mu}_y]^{\text{{T}}}\mathbf{\Sigma}_y^{-1}[\mathbf{X}-\mathbf{\mu}_y]

Let \delta_ybe defined as\delta_y(\mathbf{x})=-\frac{1}{2}\log\vert\mathbf{\Sigma}_y\vert-\frac{1}{2}[{\color{blue}{\mathbf{x}}}-\mathbf{\mu}_y]^{\text{{T}}}\mathbf{\Sigma}_y^{-1}[{\color{blue}{\mathbf{x}}}-\mathbf{\mu}_y]+\log\mathbb{P}(Y=y)the decision boundary of this classifier is \{\mathbf{x}\text{ such that }\delta_0(\mathbf{x})=\delta_1(\mathbf{x})\}which is quadratic in {\color{blue}{\mathbf{x}}}. This is the quadratic discriminant analysis. This can be visualized bellow.

The decision boundary is here

But that can’t be the linear discriminant analysis, right? I mean, the frontier is not linear… Actually, in Fisher’s seminal paper, it was assumed that \mathbf{\Sigma}_0=\mathbf{\Sigma}_1.

In that case, actually, \delta_y(\mathbf{x})={\color{blue}{\mathbf{x}}}^{\text{T}}\mathbf{\Sigma}^{-1}\mathbf{\mu}_y-\frac{1}{2}\mathbf{\mu}_y^{\text{T}}\mathbf{\Sigma}^{-1}\mathbf{\mu}_y+\log\mathbb{P}(Y=y) and the decision frontier is now linear in {\color{blue}{\mathbf{x}}}. This is the linear discriminant analysis. This can be visualized bellow

Here the two samples have the same variance matrix and the frontier is

Link with the logistic regression

Assume as previously that\mathbf{X}\vert Y=0\sim\mathcal{N}(\mathbf{\mu}_0,\mathbf{\Sigma})and\mathbf{X}\vert Y=1\sim\mathcal{N}(\mathbf{\mu}_1,\mathbf{\Sigma})then\log\frac{\mathbb{P}(Y=1\vert \mathbf{X}=\mathbf{x})}{\mathbb{P}(Y=0\vert \mathbf{X}=\mathbf{x})}is equal to \mathbf{x}^{\text{{T}}}\mathbf{\Sigma}^{-1}[\mathbf{\mu}_y]-\frac{1}{2}[\mathbf{\mu}_1-\mathbf{\mu}_0]^{\text{{T}}}\mathbf{\Sigma}^{-1}[\mathbf{\mu}_1-\mathbf{\mu}_0]+\log\frac{\mathbb{P}(Y=1)}{\mathbb{P}(Y=0)}which is linear in \mathbf{x}\log\frac{\mathbb{P}(Y=1\vert \mathbf{X}=\mathbf{x})}{\mathbb{P}(Y=0\vert \mathbf{X}=\mathbf{x})}=\mathbf{x}^{\text{{T}}}\mathbf{\beta}Hence, when each groups have Gaussian distributions with identical variance matrix, then LDA and the logistic regression lead to the same classification rule.

Observe furthermore that the slope is proportional to \mathbf{\Sigma}^{-1}[\mathbf{\mu}_1-\mathbf{\mu}_0], as stated in Fisher’s article. But to obtain such a relationship, he observe that the ratio of between and within variances (in the two groups) was\frac{\text{variance between}}{\text{variance within}}=\frac{[\mathbf{\omega}\mathbf{\mu}_1-\mathbf{\omega}\mathbf{\mu}_0]^2}{\mathbf{\omega}^{\text{T}}\mathbf{\Sigma}_1\mathbf{\omega}+\mathbf{\omega}^{\text{T}}\mathbf{\Sigma}_0\mathbf{\omega}}which is maximal when \mathbf{\omega} is proportional to \mathbf{\Sigma}^{-1}[\mathbf{\mu}_1-\mathbf{\mu}_0], when \mathbf{\Sigma}_0=\mathbf{\Sigma}_1.

Homebrew linear discriminant analysis

To compute vector \mathbf{\omega}

m0 = apply(myocarde[myocarde$PRONO=="0",1:7],2,mean)
m1 = apply(myocarde[myocarde$PRONO=="1",1:7],2,mean)
Sigma = var(myocarde[,1:7])
omega = solve(Sigma)%*%(m1-m0)
omega
                 [,1]
FRCAR -0.012909708542
INCAR  1.088582058796
INSYS -0.019390084344
PRDIA -0.025817110020
PAPUL  0.020441287970
PVENT -0.038298291091
REPUL -0.001371677757

For the constant – in the equation \omega^T\mathbf{x}+b=0 – if we have equiprobable probabilities, use

b = (t(m1)%*%solve(Sigma)%*%m1-t(m0)%*%solve(Sigma)%*%m0)/2

Application (on the small dataset)

In order to visualize what’s going on, consider the small dataset, with only two covariates,

x = c(.4,.55,.65,.9,.1,.35,.5,.15,.2,.85)
y = c(.85,.95,.8,.87,.5,.55,.5,.2,.1,.3)
z = c(1,1,1,1,1,0,0,1,0,0)
df = data.frame(x1=x,x2=y,y=as.factor(z))
m0 = apply(df[df$y=="0",1:2],2,mean)
m1 = apply(df[df$y=="1",1:2],2,mean)
Sigma = var(df[,1:2])
omega = solve(Sigma)%*%(m1-m0)
omega
         [,1]
x1 -2.640613174
x2  4.858705676


Using R regular function, we get

library(MASS)
fit_lda = lda(y ~x1+x2 , data=df)
fit_lda
 
Coefficients of linear discriminants:
            LD1
x1 -2.588389554
x2  4.762614663

which is the same coefficient as the one we got with our own code. For the constant, use

b = (t(m1)%*%solve(Sigma)%*%m1-t(m0)%*%solve(Sigma)%*%m0)/2

If we plot it, we get the red straight line

plot(df$x1,df$x2,pch=c(1,19)[1+(df$y=="1")])
abline(a=b/omega[2],b=-omega[1]/omega[2],col="red")


As we can see (with the blue points), our red line intersects the middle of the segment of the two barycenters

points(m0["x1"],m0["x2"],pch=4)
points(m1["x1"],m1["x2"],pch=4)
segments(m0["x1"],m0["x2"],m1["x1"],m1["x2"],col="blue")
points(.5*m0["x1"]+.5*m1["x1"],.5*m0["x2"]+.5*m1["x2"],col="blue",pch=19)

Of course, we can also use R function

predlda = function(x,y) predict(fit_lda, data.frame(x1=x,x2=y))$class==1
vv=outer(vu,vu,predlda)
contour(vu,vu,vv,add=TRUE,lwd=2,levels = .5)


One can also consider the quadratic discriminent analysis since it might be difficult to argue that \mathbf{\Sigma}_0=\mathbf{\Sigma}_1

fit_qda = qda(y ~x1+x2 , data=df)

The separation curve is here

plot(df$x1,df$x2,pch=19,
col=c("blue","red")[1+(df$y=="1")])
predqda=function(x,y) predict(fit_qda, data.frame(x1=x,x2=y))$class==1
vv=outer(vu,vu,predlda)
contour(vu,vu,vv,add=TRUE,lwd=2,levels = .5)

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\rbracesubject 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\rbracesubject 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\rbracesubject 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}_ithus

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)
}

Link to the regression

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…)

Simple and heuristic optimization

This week, at the Rmetrics conference, there has been an interesting discussion about heuristic optimization. The starting point was simple: in complex optimization problems (here we mean with a lot of local maxima, for instance), we do not necessarily need extremely advanced algorithms that do converge extremly fast, if we cannot ensure that they reach the optimum. Converging extremely fast, with a great numerical precision to some point (that is not the point we’re looking for) is useless. And some algorithms might be much slower, but at least, it is much more likely to converge to the optimum. Wherever we start from.
We have experienced that with Mathieu, while we were looking for maximum likelihood of our MINAR process: genetic algorithm have performed extremely well. The idea is extremly simple, and natural. Let us consider as a starting point the following algorithm,

  1. Start from some 
  2. At step , draw a point  in a neighborhood of 
  • either  then 
  • or  then 

This is simple (if you do not enter into details about what such a neighborhood should be). But using that kind of algorithm, you might get trapped and attracted to some local optima if the neighborhood is not large enough. An alternative to this technique is the following: it might be interesting to change a bit more, and instead of changing when we have a maximum, we change if we have almost a maximum. Namely at step ,

  • either then 
  • or  then 

for some . To illustrate the idea, consider the following function

> f=function(x,y) { r <- sqrt(x^2+y^2);
+ 1.1^(x+y)*10 * sin(r)/r }
(on some bounded support). Here, by picking noise and  values arbitrary, we have obtained the following scenarios
> x0=15
> MX=matrix(NA,501,2)
> MX[1,]=runif(2,-x0,x0)
> k=.5
> for(s in 2:501){
+  bruit=rnorm(2)
+  X=MX[s-1,]+bruit*3
+  if(X[1]>x0){X[1]=x0}
+  if(X[1]<(-x0)){X[1]=-x0}
+  if(X[2]>x0){X[2]=x0}
+  if(X[2]<(-x0)){X[2]=-x0}
+  if(f(X[1],X[2])+k>f(MX[s-1,1],
+    MX[s-1,2])){MX[s,]=X}
+  if(f(X[1],X[2])+k<=f(MX[s-1,1],
+    MX[s-1,2])){MX[s,]=MX[s-1,]}
+}

It does not always converge towards the optimum,

and sometimes, we just missed it after being extremely unlucky

Note that if we run 10,000 scenarios (with different random noises and starting point), in 50% scenarios, we reach the maxima. Or at least, we are next to it, on top.

What if we compare with a standard optimization routine, like Nelder-Mead, or quasi gradient ?Since we look for the maxima on a restricted domain, we can use the following function,

> g=function(x) f(x[1],x[2])
> optim(X0, g,method="L-BFGS-B",
+ lower=-c(x0,x0),upper=c(x0,x0))$par

In that case, if we run the algorithm with 10,000 random starting point, this is where we end, below on the right (while the heuristic technique is on the left),

In only 15% of the scenarios, we have been able to reach the region where the maximum is.

So here, it looks like an heuristic method works extremelly well, if do not need to reach the maxima with a great precision. Which is usually the case actually.

EM and mixture estimation

Following my previous post on optimization and mixtures (here), Nicolas told me that my idea was probably not the most clever one (there).
So, we get back to our simple mixture model,

https://perso.univ-rennes1.fr/arthur.charpentier/latex/mixEM01.png

In order to describe how EM algorithm works, assume first that both https://perso.univ-rennes1.fr/arthur.charpentier/latex/mixEM02.png and  https://perso.univ-rennes1.fr/arthur.charpentier/latex/mixEM03.pngare perfectly known, and the mixture parameter is the only one we care about.

  • The simple model, with only one parameter that is unknown

Here, the likelihood is

https://perso.univ-rennes1.fr/arthur.charpentier/latex/mixEM04.png

so that we write the log likelihood as

https://perso.univ-rennes1.fr/arthur.charpentier/latex/mixEM05.png

which might not be simple to maximize. Recall that the mixture model can interpreted through a latent variate https://perso.univ-rennes1.fr/arthur.charpentier/latex/mixEM06.png (that cannot be observed), taking value when https://perso.univ-rennes1.fr/arthur.charpentier/latex/mixEM07.png is drawn from https://perso.univ-rennes1.fr/arthur.charpentier/latex/mixEM02.png, and 0 if it is drawn from https://perso.univ-rennes1.fr/arthur.charpentier/latex/mixEM03.png. More generally (especially in the case we want to extend our model to 3, 4, … mixtures), https://perso.univ-rennes1.fr/arthur.charpentier/latex/mixEM08.png and https://perso.univ-rennes1.fr/arthur.charpentier/latex/mixEM09.png.
With that notation, the likelihood becomes

https://perso.univ-rennes1.fr/arthur.charpentier/latex/mixEM10.png

and the log likelihood

https://perso.univ-rennes1.fr/arthur.charpentier/latex/mixEM11.png

the term on the right is useless since we only care about p, here. From here, consider the following iterative procedure,
Assume that the mixture probability https://perso.univ-rennes1.fr/arthur.charpentier/latex/mixEM13.png is known, denoted https://perso.univ-rennes1.fr/arthur.charpentier/latex/mixEM12.png. Then I can predict the value of https://perso.univ-rennes1.fr/arthur.charpentier/latex/mixEM06.png (i.e. https://perso.univ-rennes1.fr/arthur.charpentier/latex/mixEM08.png and https://perso.univ-rennes1.fr/arthur.charpentier/latex/mixEM09.png) for all observations,

https://perso.univ-rennes1.fr/arthur.charpentier/latex/mixEM14.png

So I can inject those values into my log likelihood, i.e. in

https://perso.univ-rennes1.fr/arthur.charpentier/latex/mixEM15.png

having maximum (no need to run numerical tools here)

https://perso.univ-rennes1.fr/arthur.charpentier/latex/mixEM16.png

that will be denoted https://perso.univ-rennes1.fr/arthur.charpentier/latex/mixEM17.png. And I can iterate from here.
Formally, the first step is where we calculate an expected (E) value, where https://perso.univ-rennes1.fr/arthur.charpentier/latex/mixEM18.pngis the best predictor of https://perso.univ-rennes1.fr/arthur.charpentier/latex/mixEM19.png given my observations (as well as my belief in https://perso.univ-rennes1.fr/arthur.charpentier/latex/mixEM13.png). Then comes a maximization (M) step, where using https://perso.univ-rennes1.fr/arthur.charpentier/latex/mixEM06.png, I can estimate probability https://perso.univ-rennes1.fr/arthur.charpentier/latex/mixEM13.png.

  • A more general framework, all parameters are now unkown

So far, it was simple, since we assumed that https://perso.univ-rennes1.fr/arthur.charpentier/latex/mixEM02.png and  https://perso.univ-rennes1.fr/arthur.charpentier/latex/mixEM03.png were perfectly known. Which is not reallistic. An there is not much to change to get a complete algorithm, to estimate https://perso.univ-rennes1.fr/arthur.charpentier/latex/mixEM30.png. Recall that we had https://perso.univ-rennes1.fr/arthur.charpentier/latex/mixEM18.png which was the expected value of Z_{1,i}, i.e. it is a probability that observation i has been drawn from https://perso.univ-rennes1.fr/arthur.charpentier/latex/mixEM02.png.
If https://perso.univ-rennes1.fr/arthur.charpentier/latex/mixEM18.png, instead of being in the segment https://perso.univ-rennes1.fr/arthur.charpentier/latex/mixEM31.png was in https://perso.univ-rennes1.fr/arthur.charpentier/latex/mixEM32.png, then we could have considered mean and standard deviations of observations such that https://perso.univ-rennes1.fr/arthur.charpentier/latex/mixEM18.png=0, and similarly on the subset of observations such that https://perso.univ-rennes1.fr/arthur.charpentier/latex/mixEM18.png=1.
But we can’t. So what can be done is to consider https://perso.univ-rennes1.fr/arthur.charpentier/latex/mixEM18.png as the weight we should give to observation i when estimating parameters of https://perso.univ-rennes1.fr/arthur.charpentier/latex/mixEM02.png, and similarly, 1-https://perso.univ-rennes1.fr/arthur.charpentier/latex/mixEM18.pngwould be weights given to observation i when estimating parameters of https://perso.univ-rennes1.fr/arthur.charpentier/latex/mixEM03.png.
So we set, as before

https://perso.univ-rennes1.fr/arthur.charpentier/latex/mixEM33.png

and then

https://perso.univ-rennes1.fr/arthur.charpentier/latex/mixEM34.png
https://perso.univ-rennes1.fr/arthur.charpentier/latex/mixEM35.png

and for the variance, well, it is a weighted mean again,

https://perso.univ-rennes1.fr/arthur.charpentier/latex/mixEM36.png
https://perso.univ-rennes1.fr/arthur.charpentier/latex/mixEM37.png

and this is it.

  • Let us run the code on the same data as before

Here, the code is rather simple: let us start generating a sample
> X1 = rnorm(n,0,1)
> X20 = rnorm(n,0,1)
> Z  = sample(c(1,2,2),size=n,replace=TRUE)
> X2=4+X20
> X = c(X1[Z==1],X2[Z==2])
then, given a vector of initial values (that I called https://perso.univ-rennes1.fr/arthur.charpentier/latex/mixEM12.png and then https://perso.univ-rennes1.fr/arthur.charpentier/latex/mixEM99.png before),
>  s = c(0.5, mean(X)-1, var(X), mean(X)+1, var(X))
I define my function as,
>  em = function(X0,s) {
+  Ep = s[1]*dnorm(X0, s[2], sqrt(s[4]))/(s[1]*dnorm(X0, s[2], sqrt(s[4])) +
+  (1-s[1])*dnorm(X0, s[3], sqrt(s[5])))
+  s[1] = mean(Ep)
+  s[2] = sum(Ep*X0) / sum(Ep)
+  s[3] = sum((1-Ep)*X0) / sum(1-Ep)
+  s[4] = sum(Ep*(X0-s[2])^2) / sum(Ep)
+  s[5] = sum((1-Ep)*(X0-s[3])^2) / sum(1-Ep)
+  return(s)
+  }
Then I get https://perso.univ-rennes1.fr/arthur.charpentier/latex/mixEM12.png, or https://perso.univ-rennes1.fr/arthur.charpentier/latex/mixEM99.png. So this is it ! We just need to iterate (here I stop after 200 iterations) since we can see that, actually, our algorithm converges quite fast,
> for(i in 2:200){
+ s=em(X,s)
+ }

Let us run the same procedure as before, i.e. I generate samples of size 200, where difference between means can be small (0) or large (4),

Ok, Nicolas, you were right, we’re doing much better ! Maybe we should also go for a Gibbs sampling procedure ?… next time, maybe….

Optimization and mixture estimation

Recently, one of my students asked me about optimization routines in R. He told me he that R performed well on the estimation of a time series model with different regimes, while he had trouble with a (simple) GARCH process, and he was wondering if R was good in optimization routines. Actually, I always thought that mixtures (and regimes) was something difficult to estimate, so I was a bit surprised…

Indeed, it reminded me some trouble I experienced once, while I was talking about maximum likelihooh estimation, for non standard distribution, i.e. when optimization had to be done on the log likelihood function. And even when generating nice samples, giving appropriate initial values (actually the true value used in random generation), each time I tried to optimize my log likelihood, it failed. So I decided to play a little bit with standard optimization functions, to see which one performed better when trying to estimate mixture parameter (from a mixture based sample). Here, I generate a mixture of two gaussian distributions, and I would like to see how different the mean should be to have a high probability to estimate properly the parameters of the mixture.

The density is here https://perso.univ-rennes1.fr/arthur.charpentier/latex/mix-ml-01.png proportional to

https://perso.univ-rennes1.fr/arthur.charpentier/latex/mix-ml-02.png

The true model is https://perso.univ-rennes1.fr/arthur.charpentier/latex/mix-ml-03.png, and https://perso.univ-rennes1.fr/arthur.charpentier/latex/mix-ml-04.png being a parameter that will change, from 0 to 4.
The log likelihood (actually, I add a minus since most of the optimization functions actually minimize functions) is
> logvraineg <- function(param, obs) {
+ p <- param[1]
+ m1 <- param[2]
+ sd1 <- param[3]
+ m2 <- param[4]
+  sd2 <- param[5]
+  -sum(log(p * dnorm(x = obs, mean = m1, sd = sd1) + (1 – p) *
+ dnorm(x = obs, mean = m2, sd = sd2)))
+  }
The code to generate my samples is the following,
>X1 = rnorm(n,0,1)
> X20 = rnorm(n,0,1)
> Z  = sample(c(1,2,2),size=n,replace=TRUE)
> X2=m+X20
> X = c(X1[Z==1],X2[Z==2])
Then I use two functions to optimize my log likelihood, with identical intial values,
> O1=nlm(f = logvraineg, p = c(.5, mean(X)-sd(X)/5, sd(X), mean(X)+sd(X)/5, sd(X)), obs = X)
> logvrainegX <- function(param) {logvraineg(param,X)}
> O2=optim( par = c(.5, mean(X)-sd(X)/5, sd(X), mean(X)+sd(X)/5, sd(X)),
+   fn = logvrainegX)
Actually, since I might have identification problems, I take either https://perso.univ-rennes1.fr/arthur.charpentier/latex/mix-ml-05.png or https://perso.univ-rennes1.fr/arthur.charpentier/latex/mix-ml-06.png, depending whether https://perso.univ-rennes1.fr/arthur.charpentier/latex/mix-ml-07.png or https://perso.univ-rennes1.fr/arthur.charpentier/latex/mix-ml-08.png is the smallest parameter.

On the graph above, the x-axis is the difference between means of the mixture (as on the animated grap above). Then, the red point is the median of estimated parameter I have (here https://perso.univ-rennes1.fr/arthur.charpentier/latex/mix-ml-05.png), and I have included something that can be interpreted as a confidence interval, i.e. where I have been in 90% of my scenarios: theblack vertical segments. Obviously, when the sample is not enough heterogeneous (i.e. https://perso.univ-rennes1.fr/arthur.charpentier/latex/mix-ml-09.png and https://perso.univ-rennes1.fr/arthur.charpentier/latex/mix-ml-04.png rather different), I cannot estimate properly my parameters, I might even have a probability that exceed 1 (I did not add any constraint). The blue plain horizontal line is the true value of the parameter, while the blue dotted horizontal line is the initial value of the parameter in the optimization algorithm (I started assuming that the mixture probability was around 0.2).
The graph below is based on the second optimization routine (with identical  starting values, and of course on the same generated samples),

(just to be honest, in many cases, it did not converge, so the loop stopped, and I had to run it again… so finally, my study is based on a bit less than 500 samples (times 15 since I considered several values for the mean of my second underlying distribution), with 200 generated observations from a mixture).
The graph below compares the two (empty circles are the first algorithm, while plain circles the second one),

On average, it is not so bad…. but the probability to be far away from the tru value is not small at all… except when the difference between the two means exceeds 3…
If I change starting values for the optimization algorithm (previously, I assumed that the mixture probability was 1/5, here I start from 1/2), we have the following graph

which look like the previous one, except for small differences between the two underlying distributions (just as if initial values had not impact on the optimization, but it might come from the fact that the surface is nice, and we are not trapped in regions of local minimum).
Thus, I am far from being an expert in optimization routines in R (see here for further information), but so far, it looks like R is not doing so bad… and the two algorithm perform similarly (maybe the first one being a bit closer to the trueparameter).