In the introduction of my course next week, I will (briefly) mention networks, and I wanted to provide some illustration of the Friendship Paradox. On network of thrones (discussed in Beveridge and Shan (2016)), there is a dataset with the network of characters in Game of Thrones. The word “friend” might be abusive here, but let’s continue to call connected nodes “friends”. The friendship paradox states that

People on average have fewer friends than their friends

This was discussed in Feld (1991) for instance, or Zuckerman & Jost (2001). Let’s try to see what it means here. First, let us get a copy of the dataset

download.file("https://www.macalester.edu/~abeverid/data/stormofswords.csv","got.csv") GoT=read.csv("got.csv") library(networkD3) simpleNetwork(GoT[,1:2])

Because it is difficult for me to incorporate some d3js script in the blog, I will illustrate with a more basic graph,

Consider a vertex $v\in V$ in the undirected graph $G=(V,E)$ (with classical graph notations), and let $d(v)$ denote the number of edges touching it (i.e. $v$ has $d(v)$ friends). The average number of friends of a random person in the graph is $$\mu = \frac{1}{n_V}\sum_{v\in V} d(v)=\frac{2 n_E}{n_V}$$ The average number of friends that a typical friend has is
$$\frac{1}{n_V}\sum_{v\in V} \left(\frac{1}{d(v)}\sum_{v'\in E_v} d(v')\right)$$But
$$\sum_{v\in V} \left(\frac{1}{d(v)}\sum_{v'\in E_v} d(v')\right)=\sum_{v,v' \in G} \left(
\frac{d(v')}{d(v)}+\frac{d(v)}{d(v')}\right)$$
$$=\sum_{v,v' \in G}\left(\frac{d(v')^2+d(v)^2}{d(v)d(v')}\right)=\sum_{v,v' \in G} \left(\frac{(d(v')-d(v))^2}{d(v)d(v')}+2\right){\color{red}{\succ}}\sum_{v,v' \in G} \left(2\right)=\sum_{v\in V} d(v)$$
Thus,$$\frac{1}{n_V}\sum_{v\in V} \left(\frac{1}{d(v)}\sum_{v'\in E_v} d(v')\right)\succ \frac{1}{n_V}\sum_{v\in V} d(v)$$
Note that this can be related to the variance decomposition $$\text{Var}[X]=\mathbb{E}[X^2]-\mathbb{E}[X]^2$$i.e.$$\frac{\mathbb{E}[X^2]}{\mathbb{E}[X]} =\mathbb{E}[X]+\frac{\text{Var}[X]}{\mathbb{E}[X]}\succ\mathbb{E}[X]$$(Jensen inequality). But let us get back to our network. The list of nodes is

M=(rbind(as.matrix(GoT[,1:2]),as.matrix(GoT[,2:1]))) nodes=unique(M[,1])

and we each of them, we can get the list of friends, and the number of friends

friends = function(x) as.character(M[which(M[,1]==x),2]) nb_friends = Vectorize(function(x) length(friends(x)))

as well as the number of friends friends have, and the average number of friends

friends_of_friends = function(y) (Vectorize(function(x) length(friends(x)))(friends(y))) nb_friends_of_friends = Vectorize(function(x) mean(friends_of_friends(x)))

We can look at the density of the number of friends, for a random node,

Nb = nb_friends(nodes) Nb2 = nb_friends_of_friends(nodes) hist(Nb,breaks=0:40,col=rgb(1,0,0,.2),border="white",probability = TRUE) hist(Nb2,breaks=0:40,col=rgb(0,0,1,.2),border="white",probability = TRUE,add=TRUE) lines(density(Nb),col="red",lwd=2) lines(density(Nb2),col="blue",lwd=2)

and we can also compute the averages, just to check

mean(Nb) [1] 6.579439 mean(Nb2) [1] 13.94243

So, indeed, people on average have fewer friends than their friends.

# PhD Defense in Lyon

Today, I will go to Lyon for the PhD defense of Edouard Debonneuil (that will be on Monday morning)

His thesis is on financial impacts of mortality improvements

Several models and scenarios are considered…

Probably more on that very interesting (and important) topic soon.

# Parallelizing Linear Regression or Using Multiple Sources

My previous post was explaining how mathematically it was possible to parallelize computation to estimate the parameters of a linear regression. More speficially, we have a matrix $\mathbf{X}$ which is $n\times k$ matrix and $\mathbf{y}$ a $n$-dimensional vector, and we want to compute $\widehat{\mathbf{\beta}}=[\mathbf{X}^T\mathbf{X}]^{-1}\mathbf{X}^T\mathbf{y}$ by spliting the job. Instead of using the $n$ observations, we’ve seen that it was to possible to compute “something” using the first $n_1$ rows, then the next $n_2$ rows, etc. Then, finally, we “aggregate” the $m$ objects created to get our overall estimate.

## Parallelizing on multiple cores

Let us see how it works from a computational point of view, to run each computation on a different core of the machine. Each core will see a slave, computing what we’ve seen in the previous post. Here, the data we use are

y = cars$dist X = data.frame(1,cars$speed) k = ncol(X)

On my laptop, I have three cores, so we will split it in $m=3$ chunks

library(parallel) library(pbapply) ncl = detectCores()-1 cl = makeCluster(ncl)

This is more or less what we will do: we have our dataset, and we split the jobs,

We can then create lists containing elements that will be sent to each core, as Ewen suggested,

chunk = function(x,n) split(x, cut(seq_along(x), n, labels = FALSE)) a_parcourir = chunk(seq_len(nrow(X)), ncl) for(i in 1:length(a_parcourir)) a_parcourir[[i]] = rep(i, length(a_parcourir[[i]])) Xlist = split(X, unlist(a_parcourir)) ylist = split(y, unlist(a_parcourir))

It is also possible to simplify the QR functions we will use

compute_qr = function(x){ list(Q=qr.Q(qr(as.matrix(x))),R=qr.R(qr(as.matrix(x)))) } get_Vlist = function(j){ Q3 = QR1[[j]]$Q %*% Q2list[[j]] t(Q3) %*% ylist[[j]] } clusterExport(cl, c("compute_qr", "get_Vlist"), envir=environment()) Then, we can run our functions on each core. The first one is  QR1 = parLapply(cl=cl,Xlist, compute_qr) note that it is also possible to use  QR1 = pblapply(Xlist, compute_qr, cl=cl) which will include a progress bar (that can be nice when the database is rather large). Then use  R1 = pblapply(QR1, function(x) x$R, cl=cl) %&gt;% do.call("rbind", .) Q1 = qr.Q(qr(as.matrix(R1))) R2 = qr.R(qr(as.matrix(R1))) Q2list = split.data.frame(Q1, rep(1:ncl, each=k)) clusterExport(cl, c("QR1", "Q2list", "ylist"), envir=environment()) Vlist = pblapply(1:length(QR1), get_Vlist, cl=cl) sumV = Reduce('+', Vlist)

and finally the ouput is

solve(R2) %*% sumV [,1] X1 -17.579095 X2 3.932409

which is what we were expecting…

## Using multiple sources

In practice, it might also happen that various “servers” have the data, but we cannot get a copy. But it is possible to run some functions on their server, and get some output, that we can use afterwards.

Datasets are supposed to be available somewhere. We can send a request, and get a matrix. Then we we aggregate all of them, and send another request. That’s what we will do here. Provider $j$ should run $f_1(\mathbf{X})$ on his part of the data, that function will return $R^{(1)}_j$. More precisely, to the first provider, send

function1 = function(subX){ return(qr.R(qr(as.matrix(subX))))} R1 = function1(Xlist[[1]])

and actually, send that function to all providers, and aggregate the output

for(j in 2:m) R1 = rbind(R1,function1(Xlist[[j]]))

The create on your side the following objects

Q1 = qr.Q(qr(as.matrix(R1))) R2 = qr.R(qr(as.matrix(R1))) Q2list=list() for(j in 1:m) Q2list[[j]] = Q1[(j-1)*k+1:k,]

Finally, contact one last time the providers, and send one of your objects

function2=function(subX,suby,Q){ Q1=qr.Q(qr(as.matrix(subX))) Q2=Q return(t(Q1%*%Q2) %*% suby)}

Provider $j$ should then run $f_2(\mathbf{X},\mathbf{y},Q_j^{(2)})$ on his part of the data, using also $Q_j^{(2)}$ as argument (that we obtained on own side) and that function will return $(\mathbf{Q}^{(2)}_j\mathbf{Q}^{(1)}_j)^{T}_j\mathbf{y}_j$. For instance, ask the first provider to run

sumV = function2(Xlist[[1]],ylist[[1]], Q2list[[1]])

and do the same with all providers

for(j in 2:m) sumV = sumV+ function2(Xlist[[j]],ylist[[j]], Q2list[[j]])
solve(R2) %*% sumV [,1] X1 -17.579095 X2 3.932409

which is what we were expecting…

# Linear Regression, with Map-Reduce

Sometimes, with big data, matrices are too big to handle, and it is possible to use tricks to numerically still do the map. Map-Reduce is one of those. With several cores, it is possible to split the problem, to map on each machine, and then to agregate it back at the end.

Consider the case of the linear regression, $\mathbf{y}=\mathbf{X}\mathbf{\beta}+\mathbf{\varepsilon}$ (with classical matrix notations). The OLS estimate of $\mathbf{\beta}$ is $\widehat{\mathbf{\beta}}=[\mathbf{X}^T\mathbf{X}]^{-1}\mathbf{X}^T\mathbf{y}$. To illustrate, consider a not too big dataset, and run some regression.

lm(dist~speed,data=cars)$coefficients (Intercept) speed -17.579095 3.932409 y=cars$dist X=cbind(1,cars$speed) solve(crossprod(X,X))%*%crossprod(X,y) [,1] [1,] -17.579095 [2,] 3.932409 How is this computed in R? Actually, it is based on the QR decomposition of $\mathbf{X}$, $\mathbf{X}=\mathbf{Q}\mathbf{R}$, where $\mathbf{Q}$ is an orthogonal matrix (ie $\mathbf{Q}^T\mathbf{Q}=\mathbb{I}$). Then $\widehat{\mathbf{\beta}}=[\mathbf{X}^T\mathbf{X}]^{-1}\mathbf{X}^T\mathbf{y}=\mathbf{R}^{-1}\mathbf{Q}^T\mathbf{y}$ solve(qr.R(qr(as.matrix(X)))) %*% t(qr.Q(qr(as.matrix(X)))) %*% y [,1] [1,] -17.579095 [2,] 3.932409 So far, so good, we get the same output. Now, what if we want to parallelise computations. Actually, it is possible. Consider $m$ blocks m = 5 and split vectors and matrices $$\mathbf{y}=\left[\begin{matrix}\mathbf{y}_1\\\mathbf{y}_2\\\vdots \\\mathbf{y}_m\end{matrix}\right]$$ and $$\mathbf{X}=\left[\begin{matrix}\mathbf{X}_1\\\mathbf{X}_2\\\vdots\\\mathbf{X}_m\end{matrix}\right]=\left[\begin{matrix}\mathbf{Q}_1^{(1)}\mathbf{R}_1^{(1)}\\\mathbf{Q}_2^{(1)}\mathbf{R}_2^{(1)}\\\vdots \\\mathbf{Q}_m^{(1)}\mathbf{R}_m^{(1)}\end{matrix}\right]$$ To split vectors and matrices, use (eg) Xlist = list() for(j in 1:m) Xlist[[j]] = X[(j-1)*10+1:10,] ylist = list() for(j in 1:m) ylist[[j]] = y[(j-1)*10+1:10] and get small QR recomposition (per subset) QR1 = list() for(j in 1:m) QR1[[j]] = list(Q=qr.Q(qr(as.matrix(Xlist[[j]]))),R=qr.R(qr(as.matrix(Xlist[[j]])))) Consider the QR decomposition of $\mathbf{R}^{(1)}$ which is the first step of the reduce part$$\mathbf{R}^{(1)}=\left[\begin{matrix}\mathbf{R}_1^{(1)}\\\mathbf{R}_2^{(1)}\\\vdots \\\mathbf{R}_m^{(1)}\end{matrix}\right]=\mathbf{Q}^{(2)}\mathbf{R}^{(2)}$$where$$\mathbf{Q}^{(2)}=\left[\begin{matrix}\mathbf{Q}^{(2)}_1\\\mathbf{Q}^{(2)}_2\\\vdots\\\mathbf{Q}^{(2)}_m\end{matrix}\right]$$ R1 = QR1[[1]]$R for(j in 2:m) R1 = rbind(R1,QR1[[j]]$R) Q1 = qr.Q(qr(as.matrix(R1))) R2 = qr.R(qr(as.matrix(R1))) Q2list=list() for(j in 1:m) Q2list[[j]] = Q1[(j-1)*2+1:2,] Define – as step 2 of the reduce part$$\mathbf{Q}^{(3)}_j=\mathbf{Q}^{(2)}_j\mathbf{Q}^{(1)}_j$$ and$$\mathbf{V}_j=\mathbf{Q}^{(3)T}_j\mathbf{y}_j$$ Q3list = list() for(j in 1:m) Q3list[[j]] = QR1[[j]]$Q %*% Q2list[[j]] Vlist = list() for(j in 1:m) Vlist[[j]] = t(Q3list[[j]]) %*% ylist[[j]]

and finally set – as the step 3 of the reduce part$$\widehat{\mathbf{\beta}}=[\mathbf{R}^{(2)}]^{-1}\sum_{j=1}^m\mathbf{V}_j$$

sumV = Vlist[[1]] for(j in 2:m) sumV = sumV+Vlist[[j]] solve(R2) %*% sumV [,1] [1,] -17.579095 [2,] 3.932409

It looks like we’ve been able to parallelise our linear regression…

# Workshop on Inequalities

Next week, we organize in Rennes a two day workshop on inequalities, with Olivier l’Haridon and Benoit Tarroux.

Stéphane Zuber, Frank Cowell, Emmanuel Flachaire, Kirsten Rohde, Nicolas Gravel, Brice Magdalou, Alain Chateauneuf and many other will be around. The program is now online

[an updated version is now online here]

After my series of post on classification algorithms, it’s time to get back to R codes, this time for quantile regression. Yes, I still want to get a better understanding of optimization routines, in R. Before looking at the quantile regression, let us compute the median, or the quantile, from a sample.

## Median

Consider a sample $\{y_1,\cdots,y_n\}$. To compute the median, solve$$\min_\mu \left\lbrace\sum_{i=1}^n|y_i-\mu|\right\rbrace$$which can be solved using linear programming techniques. More precisely, this problem is equivalent to$$\min_{\mu,\mathbf{a},\mathbf{b}}\left\lbrace\sum_{i=1}^na_i+b_i\right\rbrace$$with $a_i,b_i\geq 0$ and $y_i-\mu=a_i-b_i$, $\forall i=1,\cdots,n$.
To illustrate, consider a sample from a lognormal distribution,

n = 101 set.seed(1) y = rlnorm(n) median(y) [1] 1.077415

For the optimization problem, use the matrix form, with $3n$ constraints, and $2n+1$ parameters,

So far so good…

## Quantile Regression (simple)

Consider the following dataset, with rents of flat, in a major German city, as function of the surface, the year of construction, etc.

base=read.table("http://freakonometrics.free.fr/rent98_00.txt",header=TRUE)

The linear program for the quantile regression is now$$\min_{\mu,\mathbf{a},\mathbf{b}}\left\lbrace\sum_{i=1}^n\tau a_i+(1-\tau)b_i\right\rbrace$$with $a_i,b_i\geq 0$ and $$y_i-[\beta_0^\tau+\beta_1^\tau x_i]=a_i-b_i$$$\forall i=1,\cdots,n$. So use here

require(lpSolve) tau = .3 n=nrow(base) X = cbind( 1, base$area) y = base$rent_euro A1 = cbind(diag(2*n), 0,0) A2 = cbind(diag(n), -diag(n), X) r = lp("min", c(rep(tau,n), rep(1-tau,n),0,0), rbind(A1, A2), c(rep("&gt;=", 2*n), rep("=", n)), c(rep(0,2*n), y)) tail(r$solution,2) [1] 148.946864 3.289674 Of course, we can use R function to fit that model library(quantreg) rq(rent_euro~area, tau=tau, data=base) Coefficients: (Intercept) area 148.946864 3.289674 Here again, it seems to work quite well. We can use a different probability level, of course, and get a plot plot(base$area,base$rent_euro,xlab=expression(paste("surface (",m^2,")")), ylab="rent (euros/month)",col=rgb(0,0,1,.4),cex=.5) sf=0:250 yr=r$solution[2*n+1]+r$solution[2*n+2]*sf lines(sf,yr,lwd=2,col="blue") tau = .9 r = lp("min", c(rep(tau,n), rep(1-tau,n),0,0), rbind(A1, A2), c(rep("&gt;=", 2*n), rep("=", n)), c(rep(0,2*n), y)) tail(r$solution,2) [1] 121.815505 7.865536 yr=r$solution[2*n+1]+r$solution[2*n+2]*sf lines(sf,yr,lwd=2,col="blue")

## Quantile Regression (multiple)

Now that we understand how to run the optimization program with one covariate, why not try with two ? For instance, let us see if we can explain the rent of a flat as a (linear) function of the surface and the age of the building.

require(lpSolve) tau = .3 n=nrow(base) X = cbind( 1, base$area, base$yearc ) y = base$rent_euro A1 = cbind(diag(2*n), 0,0,0) A2 = cbind(diag(n), -diag(n), X) r = lp("min", c(rep(tau,n), rep(1-tau,n),0,0,0), rbind(A1, A2), c(rep("&gt;=", 2*n), rep("=", n)), c(rep(0,2*n), y)) tail(r$solution,3) [1] 0.000000 3.257562 0.077501

Unfortunately, this time, it is not working well…

library(quantreg) rq(rent_euro~area+yearc, tau=tau, data=base) Coefficients: (Intercept) area yearc -5542.503252 3.978135 2.887234

Results are quite different. And actually, another technique can confirm the later (IRLS – Iteratively Reweighted Least Squares)

eps = residuals(lm(rent_euro~area+yearc, data=base)) for(s in 1:500){ reg = lm(rent_euro~area+yearc, data=base, weights=(tau*(eps&gt;0)+(1-tau)*(eps&lt;0))/abs(eps)) eps = residuals(reg) } reg$coefficients (Intercept) area yearc -5484.443043 3.955134 2.857943 I could not figure out what went wrong with the linear program. Not only coefficients are very different, but also predictions… yr = r$solution[2*n+1]+r$solution[2*n+2]*base$area+r$solution[2*n+3]*base$yearc plot(predict(reg),yr) abline(a=0,b=1,lty=2,col="red")

It’s now time to investigate….

# Stupid Economics

Mardi soir, je discutais avec Arnaud (Stupid Economics, @Stupid_Eco), pendant plus d’une heure et demi…. et c’était en live. Et fort gentillement, Accropolis l’a mis en ligne…

# Discrete or continuous modeling ?

Tuesday, we got our conference “Insurance, Actuarial Science, Data & Models” and Dylan Possamaï gave a very interesting concluding talk. In the introduction, he came back briefly on a nice discussion we usually have in economics on the kind of model we should consider. It was about optimal control. In many applications, we start with a one period economy, then a two period economy, and pretend that we can extend it to $n$ period economy. And then, the continuous case can also be considered. A few years ago, I was working on sports game as an optimal effort startegy (within in a game – fixed time). It was with a discrete model, I was running simulations to get an efficient frontier, where coaches might say “ok, now we have enough (positive) difference, and we get closer to the end of the game, so we can ‘lower the effort’ i.e. top players can relax a little bit” (it was on basket-ball games). I asked a good friend of mine, Romuald, to help me on some technical parts of proofs, but he did not like so much my discrete-time model, and wanted to move to continuous time. And for now six years, we keep saying that someday we should get back to that paper….

My initial thoughts were that the difference was really “cultural”: you are either a continuous-time sort of guy, or a discrete-time one (or maybe none of the two, but that’s another problem). He works with stochastic processes, I work with time series. Of course, we can find connections, but most of the time, the techniques are very different. And tuesday, Dylan mentioned a very nice illustration that it’s not necessarily a cultural difference, and sometimes, it is great to move to continuous time. So I wanted to illustrate that idea.

Consider for instance the following curve.

vu = seq(0,1,length=601) vv = sin(vu*pi) plot(vu,vv,type="l",lwd=2)

The goal is to find the value of the maximum, numerically. And here, there are two (very) different strategies

• the discrete one: we see a (finite) collection of points – for instance, the graph above is a collection of 601 points (connected with a straight line) – and in that case, we need a standard algorithm (in $O(n)$) to get the value of the maximum
• the continuous one: we see a function $x\mapsto \sin(\pi x)$, and in that case, we use optimization routines

In the second case, use for instance

optim(0,function(x) -sin(pi*x)) $par [1] 0.5$value [1] -1

For the first case, we can use the standard R function, and see how long it takes to use simulations to get an approximation of the maximum

library(microbenchmark) max_time = function(n) median(microbenchmark(max(sin(runif(n)*pi)))$time) vn = 10^(seq(1,6,length=21)) vt = Vectorize(max_time)(vn) plot(vn,vt/1e9,col="blue",pch=19,type="b",log="xy") but of course, some home-made code can also be used c_max = function(n=100){ x = sin(runif(n)*pi) y = x[1] for(i in 2:length(x)) { if(x[i] &gt; y) { y = x[i] }} return(y)} max_time=function(n) median(microbenchmark(c_max(n))$time) lines(vn,vt/1e9,type="b")

We can add that horizontal red line using

## Bagging logistic regression #2

Another technique that can be used to generate a bootstrap sample is to keep all $\mathbf{x}_i$‘s, but for each of them, to draw (randomly) a value for $y$, with$$Y_{i,b}\sim\mathcal{B}(\widehat{m}_{S}(\mathbf{x}_i))$$since$$\widehat{m}(\mathbf{x})=\mathbb{P}[Y=1|\mathbf{X}=\mathbf{x}].$$Thus, the code for the b part of bagging algorithm is now

L_logit = list() n = nrow(df) reg = glm(y~x1+x2, df, family=binomial) for(s in 1:100){ df_s = df df_s$y = factor(rbinom(n,size=1,prob=predict(reg,type="response")),labels=0:1) L_logit[[s]] = glm(y~., df_s, family=binomial) } The agg part of bagging algorithm remains unchanged. Here we obtain vu = seq(0,1,length=101) vv = outer(vu,vu,Vectorize(function(x,y) mean(p(c(x,y))))) image(vu,vu,vv,xlab="Variable 1",ylab="Variable 2",col=clr10,breaks=(0:10)/10) points(df$x1,df$x2,pch=19,cex=1.5,col="white") points(df$x1,df$x2,pch=c(1,19)[1+(df$y=="1")],cex=1.5) contour(vu,vu,vv,levels = .5,add=TRUE)

Of course, we can use that code we check the prediction obtain on the observations we have in our sample. Just to change, consider here the myocarde data. The entiere code is here

L_logit = list() reg = glm(as.factor(PRONO)~., myocarde, family=binomial) for(s in 1:1000){ myocarde_s = myocarde myocarde_s$PRONO = 1*rbinom(n,size=1,prob=predict(reg,type="response")) L_logit[[s]] = glm(as.factor(PRONO)~., myocarde_s, family=binomial) } p = function(x){ nd=data.frame(FRCAR=x[1], INCAR=x[2], INSYS=x[3], PRDIA=x[4], PAPUL=x[4], PVENT=x[5], REPUL=x[6]) unlist(lapply(1:1000,function(z) predict(L_logit[[z]],newdata=nd,type="response")))} For the first observation, with our 1000 simulated datasets, and our 1000 models, we obtained the following estimation for the probability to die. histo = function(i){ x = as.numeric(myocarde[i,1:7]) v_x = p(x) hist(v_x,proba=TRUE,breaks=seq(0,1,by=.05),xlab="",main="", col=rep(c(rgb(0,0,1,.4),rgb(1,0,0,.4)),each=10),ylim=c(0,5)) segments(mean(v_x),0,mean(v_x),5,col="red",lty=2) points(myocarde$PRONO[i],0,pch=19,cex=2) xi = round(mean(v_x.5)*1000)/10 text(.75,-.1,paste(xi,"%",sep=""),col=rgb(1,0,0,.6))} histo(1) histo(4)

Hence, for the first observation, in 77.8% of the models, the predicted probability was higher than 50%, and the average probability was actually close to 75%.

or, for observation 22, predictions very close to the first one (except that the first one died, while the 22nd survived)

histo(23) histo(11)

and, we observe here

## Bagging trees

Let’s now get back on our trees, mentioned in the previous post. Bagging was introduced in 1994 by Leo Breiman in Bagging Predictors. If the first section describes the procedure, the second one introduces “Bagging Classification Trees”. Trees are nice for interpretation, but most of the time, they are rather poor predictors. The idea of bagging was to improve the accuracy of classification trees.

The idea of bagging to to generate a lot of trees

clr12 = c("#8dd3c7","#ffffb3","#bebada","#fb8072","#80b1d3","#fdb462","#b3de69","#fccde5","#d9d9d9","#bc80bd","#ccebc5","#ffed6f") n = nrow(myocarde) par(mfrow=c(4,3)) sed=c(1,2,4,5,6,10,11,21,22,24,27,28,30) for(i in 1:12){ set.seed(sed[i]) idx = sample(1:n, size=n, replace=TRUE) cart = rpart(PRONO~., myocarde[idx,]) prp(cart,type=2,extra=1,box.col=clr12[i])}

The strategie is actually the same as before. For the bootstrap part, store the tree in a list

L_tree = list() for(s in 1:1000){ idx = sample(1:n, size=n, replace=TRUE) L_tree[[s]] = rpart(as.factor(PRONO)~., myocarde[idx,]) }

and for the aggregation part, just take the average of predicted probabilities

p = function(x){ nd=data.frame(FRCAR=x[1], INCAR=x[2], INSYS=x[3], PRDIA=x[4], PAPUL=x[4], PVENT=x[5], REPUL=x[6]) unlist(lapply(1:1000,function(z) predict(L_tree[[z]],newdata=nd,type="prob")[,2]))}

Because with this example, we cannot visualize predictions, let us run the same code on the smaller dataset

L_tree = list() n = nrow(df) for(s in 1:1000){ idx = sample(1:n, size=n, replace=TRUE) L_tree[[s]] = rpart(y~x1+x2, df[idx,],control = rpart.control(cp = 0.25, minsplit = 2)) } p = function(x){ nd=data.frame(x1=x[1], x2=x[2]) unlist(lapply(1:1000,function(z) predict(L_tree[[z]],newdata=nd,type="prob")[,2]))} vu=seq(0,1,length=101) vv=outer(vu,vu,Vectorize(function(x,y) mean(p(c(x,y))))) image(vu,vu,vv,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+(df$y=="1")],cex=1.5) contour(vu,vu,vv,levels = .5,add=TRUE) ## Fronm bags to forest Here, we grew a lot of trees, but it is not stricto sensus a random forest algorithm, as introduced in 1995, in Random decision forests. Actually, the difference is in the creation of decision trees. To understand what happens, get back to the previous post on classification trees. As we’ve seen, when we have a node, we look at possible splits : we consider all possible variable, and all possible threshold. The startegy here will be to draw randomly $k$ variables out of $p$ (with of course $k, for instance $k=\sqrt{p}$). That's interesting in high dimension, because at each split, we should look for all variables, all cutoffs, and that can take quite some time (especially with the bootstrap procedure, where the goal will be to grow 1000 trees). To be continued… # Classification from scratch, trees 9/8 Nineth post of our series on classification from scratch. Today, we’ll see the heuristics of the algorithm inside classification trees. And yes, I promised eight posts in that series, but clearly, that was not sufficient… sorry for the poor prediction. ## Decision Tree Decision trees are easy to read. So easy to read that they are everywhere We start from the top, and we go down, with a binary choice, at each stop, each node. Let us see how it works on our dataset library(rpart) cart = rpart(PRONO~.,data=myocarde) library(rpart.plot) prp(cart,type=2,extra=1) We start here with one single leaf. If we have two explanatory variable (the $x$-axis and the $y$-axis if we want to plot it), we will check what happens if we cut the leaf accoring to the value of the first variable (and there will be two subgroups, the one on the left and the one on the right) or if we cut according to the second one (and there will be two subgroups, the one on top and the one below). Why and where do we cut? Let us formalize a little bit. A node (a leaf) constains observations, i.e. $\{y_i,\mathbf{x})i\})$ for some $i\in\mathcal{I}\subset\{1,\cdots,n\}$. Hence, a leaf a caracterized by $\mathcal{I}$. For instance, the first node in the tree is $\mathcal{I}=\{1,\cdots,n\}$. A (binary) split is based on one specific variable – say $x_j$ – and a cutoff, say $s$. Then, there are two options: • either $x_{i,j}\leq s$, then observation $i$ goes on the left, in $\mathcal{I}_L$ • or $x_{i,j}> s$, then observation $i$ goes on the right, in $\mathcal{I}_R$ Thus, $\mathcal{I}=\mathcal{I}_L\cup\mathcal{I}_R$. Now, define some impurity index, in some node. In the context of a classification tree, the most popular index used (the so-called impurity index) is Gini for node $\mathcal{I}$ is defined as $$G(\mathcal{I})=-\sum_{y\in\{0,1\}}p_y(1-p_y)$$where $p_y$ is the proportion of individuals in the leaf of type $y$. I use this notation here because it can be extended to the case of more than one class. Here, we consider only binary classification. Now, why $p_y(1-p_y)$? Because we want leaves that are extremely homogeneous. In our dataset, out of 71 individuals, 42 died, 29 survived. A perfect classification would be obtained if we can split in two, with the 29 survivors on the left, and the 42 dead on the right. In that case, leaves would be perfectly homogneous. So, when $p_0\approx1$ or $p_1\approx1$, we have strong homogenity. If we want an index to maximize, $-p_y(1-p_y)$ might be an interesting candidate. Further more, the worst case would be a leaf with $p_0\approx1/2$, which is exactly what we have here. Note that we can also write$$G(\mathcal{I})=-\sum_{y\in\{0,1\}}\frac{n_{y,\mathcal{I}}}{n_{\mathcal{I}}}\left(1-\frac{n_{y,\mathcal{I}}}{n_{\mathcal{I}}}\right)$$where $n_{y,\mathcal{I}}$ is the number of individuals of type $y$ in the leaf $\mathcal{I}$, and $n_{\mathcal{I}}$ is the number of individuals in the leaf $\mathcal{I}$. If we do not split, we have index$$G(\mathcal{I})=-\sum_{y\in\{0,1\}}\frac{n_{y,\mathcal{I}}}{n_{\mathcal{I}}}\left(1-\frac{n_{y,\mathcal{I}}}{n_{\mathcal{I}}}\right)$$while if we split, define index$$G(\mathcal{I}_L,\mathcal{I}_R)=-\sum_{x\in\{L,R\}}\frac{n_x}{n_{\mathcal{I}_x}}{n_{\mathcal{I}}}\sum_{y\in\{0,1\}}\frac{n_{y,\mathcal{I}_x}}{n_{\mathcal{I}_x}}\left(1-\frac{n_{y,\mathcal{I}_x}}{n_{\mathcal{I}_x}}\right)$$The code to compute is would be gini = function(y,classe){ T. = table(y,classe) nx = apply(T,2,sum) n. = sum(T) pxy = T/matrix(rep(nx,each=2),nrow=2) omega = matrix(rep(nx,each=2),nrow=2)/n g. = -sum(omega*pxy*(1-pxy)) return(g)} Actually, one can consider other indices, like the entropic measure$$E(\mathcal{I})=-\sum_{y\in\{0,1\}}\frac{n_{y,\mathcal{I}}}{n_{\mathcal{I}}}\log\left(\frac{n_{y,\mathcal{I}}}{n_{\mathcal{I}}}\right)$$while if we split, $$E(\mathcal{I}_L,\mathcal{I}_R)=-\sum_{x\in\{L,R\}}\frac{n_x}{n_{\mathcal{I}_x}}{n_{\mathcal{I}}}\sum_{y\in\{0,1\}}\frac{n_{y,\mathcal{I}_x}}{n_{\mathcal{I}_x}}\log\left(\frac{n_{y,\mathcal{I}_x}}{n_{\mathcal{I}_x}}\right)$$ entropy = function(y,classe){ T. = table(y,classe) nx = apply(T,2,sum) n. = sum(T) pxy = T/matrix(rep(nx,each=2),nrow=2) omega = matrix(rep(nx,each=2),nrow=2)/n g = sum(omega*pxy*log(pxy)) return(g)} This index was used originally in C4.5 algorithm. ## Dividing a leaf (or not) For instance, consider the very first split. Assume that we want to split according to the very first variable CLASSE = myocarde[,1] &lt;=100 table(CLASSE) CLASSE FALSE TRUE 13 58 In that case, there will be 13 invididuals on one side (the left, say), and 58 on the other side (the right). gini(y=myocarde$PRONO,classe=CLASSE) [1] -0.4640415

Initially, without any split, it was

-2*mean(myocarde$PRONO)*(1-mean(myocarde$PRONO)) [1] -0.4832375

which can actually also be obtained with

CLASSE = myocarde[,1] gini(y=myocarde$PRONO,classe=CLASSE) [1] -0.4832375 There is a net gain in spliting of gini(y=myocarde$PRONO,classe=(myocarde[,1]&lt;=100))- gini(y=myocarde$PRONO,classe=(myocarde[,1]&lt;=Inf)) [1] 0.01919591 Now, how do we split? Which variable and which cutoff? Well… let’s try all possible splits… Here, we have 7 variables. We can consider all possible values, using sort(unique(myocarde[,1])) But in massive datasets, it can be very long. Here, I prefer seq(min(myocarde[,1]),max(myocarde[,1]),length=101) so that we try 101 values of possible cutoff. Overall, the number of computations is rather low, with 707 Gini indices to compute. Again, I won’t get back here on the motivations for such a technique to create partitions, I will keep that for the course in Barcelona, but it is fast. mat_gini = mat_v=matrix(NA,7,101) for(v in 1:7){ variable=myocarde[,v] v_seuil=seq(quantile(myocarde[,v], 6/length(myocarde[,v])), quantile(myocarde[,v],1-6/length( myocarde[,v])),length=101) mat_v[v,]=v_seuil for(i in 1:101){ CLASSE=variable&lt;=v_seuil[i] mat_gini[v,i]= gini(y=myocarde$PRONO,classe=CLASSE)}}

Actually, the range of possible values is slightly different: I do not want cutoff too much on the left or on the right… having a leaf with one or two observations is not the idea, here. Not, if we plot all the functions, we get

par(mfrow=c(3,2)) for(v in 2:7){ plot(mat_v[v,],mat_gini[v,],type="l", ylim=range(mat_gini),xlab="",ylab="", main=names(myocarde)[v]) abline(h=max(mat_gini),col="blue") }

Here, the most homogenous leaves obtained using a cut in two parts is when we use variable ‘INSYS’. And the optimal cutoff variable is close to 19. So far, that’s the only information we use. Well, actually no. If the gain is sufficiently large, we go for a split. Here, the gain is

gini(y=myocarde$PRONO,classe=(myocarde[,3]&lt;19))- gini(y=myocarde$PRONO,classe=(myocarde[,3]&lt;=Inf)) [1] 0.2832801

which is large. Sufficiently large to go for it, and to split in two. Actually, we look at the relative gain

-(gini(y=myocarde$PRONO,classe=(myocarde[,3]&lt;19))- gini(y=myocarde$PRONO,classe=(myocarde[,3]&lt;=Inf)))/ gini(y=myocarde$PRONO,classe=(myocarde[,3]&lt;=Inf)) [1] 0.5862131 If that gain exceed 1% (the default value in R), we split in two. Then, we do it again. Twice. First, on go on the leaf on the left, with 27 observations. And we try to see if we can split it. idx = which(myocarde$INSYS&lt;19) mat_gini = mat_v = matrix(NA,7,101) for(v in 1:7){ variable = myocarde[idx,v] v_seuil = seq(quantile(myocarde[idx,v], 7/length(myocarde[idx,v])), quantile(myocarde[idx,v],1-7/length( myocarde[idx,v])), length=101) mat_v[v,] = v_seuil for(i in 1:101){ CLASSE = variable&lt;=v_seuil[i] mat_gini[v,i]= gini(y=myocarde$PRONO[idx],classe=CLASSE)}} par(mfrow=c(3,2)) for(v in 2:7){ plot(mat_v[v,],mat_gini[v,],type="l", ylim=range(mat_gini),xlab="",ylab="", main=names(myocarde)[v]) abline(h=max(mat_gini),col="blue") } The graph is here the following, and observe that the best split is obtained using ‘REPUL’, with a cutoff around 1585. We check that the (relative) gain is sufficiently large, and then we go for it. And then, we consider the other leaf, and we run the same code idx = which(myocarde$INSYS&gt;=19) mat_gini = mat_v = matrix(NA,7,101) for(v in 1:7){ variable=myocarde[idx,v] v_seuil=seq(quantile(myocarde[idx,v], 6/length(myocarde[idx,v])), quantile(myocarde[idx,v],1-6/length( myocarde[idx,v])), length=101) mat_v[v,]=v_seuil for(i in 1:101){ CLASSE=variable&lt;=v_seuil[i] mat_gini[v,i]= gini(y=myocarde$PRONO[idx], classe=CLASSE)}} par(mfrow=c(3,2)) for(v in 2:7){ plot(mat_v[v,],mat_gini[v,],type="l", ylim=range(mat_gini),xlab="",ylab="", main=names(myocarde)[v]) abline(h=max(mat_gini),col="blue") } Here, we should split according to ‘REPUL’, and the cutoff is about 1094. Here again, we have to make sure that the split is worth it. And we cut. Now we have four leaves. And we should run the same code, again. Actually, not on the very first one, which is homogenous. But we should do the same for the other three. If we do it, we can see that we cannot split them any further. Gains will not be sufficiently interesting. Now guess what… that’s exactly what we have obtained with our initial code Note that the case of categorical explanatory variables has been discussed in a previous post, a few years ago. ## Application on our small dataset On our small dataset, we obtain (after changing the default values since in R, we should not have leaves with less than 10 observations… and here, the dataset is too small). tree = rpart(y ~ x1+x2,data=df, control = rpart.control(cp = 0.25, minsplit = 7)) prp(tree,type=2,extra=1) u = seq(0,1,length=101) p = function(x,y){predict(tree,newdata=data.frame(x1=x,x2=y),type="prob")[,2]} v = outer(u,u,p) image(u,u,v,xlab="Variable 1",ylab="Variable 2",col=clr10,breaks=(0:10)/10) points(df$x1,df$x2,pch=19,cex=1.5,col="white") points(df$x1,df$x2,pch=c(1,19)[1+z],cex=1.5) contour(u,u,v,levels = .5,add=TRUE) We have a nice and simple cut With less observations in the leaves, we can easily get a perfect model here tree = rpart(y ~ x1+x2,data=df, control = rpart.control(cp = 0.25, minsplit = 2)) prp(tree,type=2,extra=1) u = seq(0,1,length=101) p = function(x,y){predict(tree,newdata=data.frame(x1=x,x2=y),type="prob")[,2]} v = outer(u,u,p) image(u,u,v,xlab="Variable 1",ylab="Variable 2",col=clr10,breaks=(0:10)/10) points(df$x1,df$x2,pch=19,cex=1.5,col="white") points(df$x1,df$x2,pch=c(1,19)[1+z],cex=1.5) contour(u,u,v,levels = .5,add=TRUE) Nice, isn’t it? Now, just two little additional comments before growing some more trees… ## Pruning I did not mention pruning here. Because there are two possible strategies when growing trees. Either we keep spliting, until we obtain only homogeneous leaves. Once we have a big, deep tree, we go for pruning. Or we use the stategy mentionned here : at each step, we check if the split is worth it. If not, we stop. ## Variable Importance An interesting tool is the variable importance function. The heuristic idea is that if we use variable ‘INSYS’ to split, it is an important variable. And its importance is related to the gain in Gini index. If we get back to the visualization of the tree, it seems that two variables are interesting here: ‘INSYS’ and ‘REPUL’. And we should get back to previous computation to quantify how important both are. This will be used in our next post, on random forests. But actually it is not the case here, with one single tree. Let us get back to the graph on the initial node. Indeed, ‘INSYS’ is important, since we decided to use it. But what about ‘INCAR’ or ‘REPUL’? They were very close… And actually, in R, those surrogate splits are considered in the computation, as briefly explained in the vignette. Let us look more carefully at the output of the R function cart = rpart(PRONO~., myocarde) split = summary(cart)$splits

If we look at the first part of that object, we get

split count ncat improve index adj INSYS 71 -1 0.58621312 18.850 0.0000000 REPUL 71 1 0.55440034 1094.500 0.0000000 INCAR 71 -1 0.54257020 1.690 0.0000000 PRDIA 71 1 0.27284114 17.000 0.0000000 PAPUL 71 1 0.20466714 23.250 0.0000000

So indeed, ‘INSYS’ was the most important variable, but surrogate splits can also be considered, and ‘INCAR’ and ‘REPUL’ are indeed very important. The gain was 58% (as we obtained) using ‘INSYS’ but there were gains of 55% (nothing to be ashamed of). So it would be unfair to claim that they have no importance, at all. And it is the same for the other leaves that we split,

REPUL 27 1 0.18181818 1585.000 0.0000000 PVENT 27 -1 0.10803571 14.500 0.0000000 PRDIA 27 1 0.10803571 18.500 0.0000000 PAPUL 27 1 0.10803571 22.500 0.0000000 INCAR 27 1 0.04705882 1.195 0.0000000

On the left, we did use ‘REPUL’ (with 18% gain), but ‘PVENT’, ‘PRDIA’ and ‘PAPUL’ were not that bad, with (almost) 11% gain… We can obtain variable importance by summing all those values, and we have

cart$variable.importance INSYS REPUL INCAR PAPUL PRDIA FRCAR PVENT 10.3649847 10.0510872 8.2121267 3.2441501 2.8276121 1.8623046 0.3373771 that we can visualize using barplot(t(cart$variable.importance),horiz=TRUE)

To be continued with more trees…

# 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 rule$$m^\star(\mathbf{x})=\text{argmin}_y\{\mathbb{P}[Y=y\vert\mathbf{X}=\mathbf{x}]\}$$or$$m^\star(\mathbf{x})=\text{argmin}_y\left\{\frac{\mathbb{P}[\mathbf{X}=\mathbf{x}\vert Y=y]}{\mathbb{P}[\mathbf{X}=\mathbf{x}]}\right\}$$(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 as$$m^\star(\mathbf{x})=\begin{cases}1\text{ if }\mathbb{E}(Y\vert \mathbf{X}=\mathbf{x})>\displaystyle{\frac{1}{2}}\\0\text{ otherwise}\end{cases}$$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 $\delta_y$be 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)

# Paul the octopus is back

This afternoon, Ewen Gallic (aka @3wen) uploaded some recent work on his blog, based on some joint work with Enora Belz, Romain Gaté, Vincent Malardé and Jimmy Merlet (aka @JimmyMerlet), where they used the 2018 FIFA World Cup to play with predictive models (neural nets, SVM, boosting, trees, etc) on a large training dataset. Their joint paper is online, so far only in French.

# 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…)

# Insurance, Actuarial Science, Data and Models

Next week, we will have our two-day conference in Paris, on Insurance, Actuarial Science, Data and Models.

Enora (who started her PhD under my supervision last September) will present a poster on agregated data (in the context of anonymity, GDPR, etc) with two focus: econometric regression on compositional data (when an explanatory variable is defined on the simplex, and classical Euclidean geometry cannot be used) and ecological inference.

Ewen (who is enjoying a postdoc position with me) will present a poster also, on our recent work on genealogical data