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Classification from scratch, logistic with splines 2/8

Today, second post of our series on classification from scratch, following the brief introduction on the logistic regression.

Piecewise linear splines

To illustrate what’s going on, let us start with a “simple” regression (with only one explanatory variable). The underlying idea is natura non facit saltus, for “nature does not make jumps”, i.e. process governing equations for natural things are continuous. That seems to be a rather strong assumption, because we can assume that there is a fixed threshold to explain death. For instance, if patients die (for sure) if the “stroke index” exceeds a threshold, we might expect some discontinuity. Exceept that if that threshold is an heterogeneous (non-observable continuous) variable, then we get back to the continuity assumption.

The most simple model we can think of to extend the linear model we’ve seen in the previous post is to consider a piecewise linear function, with two parts : small values of x, and larger values of x. The most convenient way to do so is to use the positive part function (x-s)_+ which is the difference between x and s if that difference is positive, and 0 otherwise. For instance \beta_1 x+\beta_2(x-s)_+ is the following piecewise linear function, continuous, with a “rupture” at knot s.

Observe also the following interpretation: for small values of x, there is a linear increase, with slope \beta_1, and for lager values of x, there is a linear decrease, with slope \beta_1+\beta_2. Hence, \beta_2 is interpreted as a change of the slope.

And of course, it is possible to consider more than one knot. The function to get the positive value is the following

pos = function(x,s) (x-s)*(x>=s)

then we can use it direcly in our regression model

reg = glm(PRONO~INSYS+pos(INSYS,15)+
pos(INSYS,25),data=myocarde,family=binomial)

The output of the regression is here

summary(reg)
 
Coefficients:
               Estimate Std. Error z value Pr(>|z|)  
(Intercept)     -0.1109     3.2783  -0.034   0.9730  
INSYS           -0.1751     0.2526  -0.693   0.4883  
pos(INSYS, 15)   0.7900     0.3745   2.109   0.0349 *
pos(INSYS, 25)  -0.5797     0.2903  -1.997   0.0458 *

Hence, the original slope, for very small values is not significant, but then, above 15, it become significantly positive. And above 25, there is a significant change again. We can plot it to see what’s going on

u = seq(5,55,length=201)
v = predict(reg,newdata=data.frame(INSYS=u),type="response")
plot(u,v,type="l")
points(myocarde$INSYS,myocarde$PRONO,pch=19)
abline(v=c(5,15,25,55),lty=2)

Using bs() linear splines

Using the GAM function, things are slightly different. We will use here so called b-splines,

library(splines)

We can define spline functions with support (5,55) and with knots \{15,25\}

clr6 = c("#1b9e77","#d95f02","#7570b3","#e7298a","#66a61e","#e6ab02")
x = seq(0,60,by=.25)
B = bs(x,knots=c(15,25),Boundary.knots=c(5,55),degre=1)
matplot(x,B,type="l",lty=1,lwd=2,col=clr6)


as we can see, the functions defined here are different from the one before, but we still have (piecewise) linear functions on each segment (5,15), (15,25) and (25,55). But linear combinations of those functions (the two sets of functions) will generate the same space. Said differently, if the interpretation of the output will be different, predictions should be the same

reg = glm(PRONO~bs(INSYS,knots=c(15,25),
Boundary.knots=c(5,55),degre=1),
data=myocarde,family=binomial)
summary(reg)
 
Coefficients:
              Estimate Std. Error z value Pr(>|z|)  
(Intercept)    -0.9863     2.0555  -0.480   0.6314  
bs(INSYS,..)1  -1.7507     2.5262  -0.693   0.4883  
bs(INSYS,..)2   4.3989     2.0619   2.133   0.0329 *
bs(INSYS,..)3   5.4572     5.4146   1.008   0.3135

Observe that there are three coefficients, as before, but again, the interpretation is here more complicated…

v=predict(reg,newdata=data.frame(INSYS=u),type="response")
plot(u,v,ylim=0:1,type="l",col="red")
points(myocarde$INSYS,myocarde$PRONO,pch=19)
abline(v=c(5,15,25,55),lty=2)


Nevertheless, the prediction is the same… and that’s nice.

Piecewise quadratic splines

Let us go one step further… Can we have also the continuity of the derivative ? Yes, and that’s easy actually, considering parabolic functions. Instead of using a decomposition on x,(x-s_1)_+ and (x-s_2)_+ consider now a decomposition on x,x^{\color{red}{2}},(x-s_1)^{\color{red}{2}}_+ and (x-s_2)^{\color{red}{2}}_+.

 pos2 = function(x,s) (x-s)^2*(x>=s)
reg = glm(PRONO~poly(INSYS,2)+pos2(INSYS,15)+pos2(INSYS,25),
data=myocarde,family=binomial)
summary(reg)
 
Coefficients:
                Estimate Std. Error z value Pr(>|z|)  
(Intercept)      29.9842    15.2368   1.968   0.0491 *
poly(INSYS, 2)1 408.7851   202.4194   2.019   0.0434 *
poly(INSYS, 2)2 199.1628   101.5892   1.960   0.0499 *
pos2(INSYS, 15)  -0.2281     0.1264  -1.805   0.0712 .
pos2(INSYS, 25)   0.0439     0.0805   0.545   0.5855

As expected, there are here five coefficients: the intercept and two for the part on the left (three parameters for the parabolic function), and then two additional terms for the part in the center – here (15,25) – and for the part on the right. Of course, for each portion, there is only one degree of freedom since we have a parabolic function (three coefficients) but two constraints (continuity, and continuity of the first order derivative).

On a graph, we get the following

v = predict(reg,newdata=data.frame(INSYS=u),type="response")
plot(u,v,ylim=0:1,type="l",col="red",lwd=2,xlab="INSYS",ylab="")
points(myocarde$INSYS,myocarde$PRONO,pch=19)
abline(v=c(5,15,25,55),lty=2)

Using bs() quadratic splines

Of course, we can do the same with our R function. But as before, the basis of function is expressed here differently

 x = seq(0,60,by=.25)
B=bs(x,knots=c(15,25),Boundary.knots=c(5,55),degre=2)
matplot(x,B,type="l",xlab="INSYS",col=clr6)


If we run R code, we get

reg = glm(PRONO~bs(INSYS,knots=c(15,25),
Boundary.knots=c(5,55),degre=2),data=myocarde,
family=binomial)
summary(reg)
 
Coefficients:
               Estimate Std. Error z value Pr(>|z|)  
(Intercept)       7.186      5.261   1.366   0.1720  
bs(INSYS, ..)1  -14.656      7.923  -1.850   0.0643 .
bs(INSYS, ..)2   -5.692      4.638  -1.227   0.2198  
bs(INSYS, ..)3   -2.454      8.780  -0.279   0.7799  
bs(INSYS, ..)4    6.429     41.675   0.154   0.8774

But that’s not really a big deal since the prediction is exactly the same

v = predict(reg,newdata=data.frame(INSYS=u),type="response")
plot(u,v,ylim=0:1,type="l",col="red")
points(myocarde$INSYS,myocarde$PRONO,pch=19)
abline(v=c(5,15,25,55),lty=2)

Cubic splines

Last, but not least, we can reach the cubic splines. With our previous notions, we would consider a decomposition on (guess what) x,x^2,x^{\color{red}{3}},(x-s_1)^{\color{red}{3}}_+,(x-s_2)^{\color{red}{3}}_+, to get this time continuity, as well as continuity of the first two derivatives (and to get a very smooth function, since even variations will be smooth). If we use the bs function, the basis is the followin

B=bs(x,knots=c(15,25),Boundary.knots=c(5,55),degre=3)
matplot(x,B,type="l",lwd=2,col=clr6,lty=1,ylim=c(-.2,1.2))
abline(v=c(5,15,25,55),lty=2)

and the prediction will now be

reg = glm(PRONO~bs(INSYS,knots=c(15,25),
Boundary.knots=c(5,55),degre=3),
data=myocarde,family=binomial)
u = seq(5,55,length=201)
v = predict(reg,newdata=data.frame(INSYS=u),type="response")
plot(u,v,ylim=0:1,type="l",col="red",lwd=2)
points(myocarde$INSYS,myocarde$PRONO,pch=19)
abline(v=c(5,15,25,55),lty=2)


Two last things before concluding (for today), the location of the knots, and the extension to additive models.

Location of knots

In many applications, we do not want to specify the location of the knots. We just want – say – three (intermediary) knots. This can be done using

reg = glm(PRONO~1+bs(INSYS,degree=1,df=4),data=myocarde,family=binomial)

We can actually get the locations of the knots by looking at

attr(reg$terms, "predvars")[[3]]
bs(INSYS, degree = 1L, knots = c(15.8, 21.4, 27.15), 
Boundary.knots = c(8.7, 54), intercept = FALSE)

which provides us with the location of the boundary knots (the minumun and the maximum from from our sample) but also the three intermediary knots. Observe that actually, those five values are just (empirical) quantiles

quantile(myocarde$INSYS,(0:4)/4)
   0%   25%   50%   75%  100% 
 8.70 15.80 21.40 27.15 54.00

If we plot the prediction, we get

v = predict(reg,newdata=data.frame(INSYS=u),type="response")
plot(u,v,ylim=0:1,type="l",col="red",lwd=2)
points(myocarde$INSYS,myocarde$PRONO,pch=19)
abline(v=quantile(myocarde$INSYS,(0:4)/4),lty=2)


If we get back on what was computed before the logit transformation, we clealy see ruptures are the different quantiles

B = bs(x,degree=1,df=4)
B = cbind(1,B)
y = B%*%coefficients(reg)
plot(x,y,type="l",col="red",lwd=2)
abline(v=quantile(myocarde$INSYS,(0:4)/4),lty=2)


Note that if we do specify anything about knots (number or location), we get no knots…

reg = glm(PRONO~1+bs(INSYS,degree=2),data=myocarde,family=binomial)
attr(reg$terms, "predvars")[[3]]
bs(INSYS, degree = 2L, knots = numeric(0), 
Boundary.knots = c(8.7,54), intercept = FALSE)

and if we look at the prediction

u = seq(5,55,length=201)
v = predict(reg,newdata=data.frame(INSYS=u),type="response")
plot(u,v,ylim=0:1,type="l",col="red",lwd=2)
points(myocarde$INSYS,myocarde$PRONO,pch=19)


actually, it is the same as a quadratic regression (as expected actually)

reg = glm(PRONO~1+poly(INSYS,degree=2),data=myocarde,family=binomial)
v = predict(reg,newdata=data.frame(INSYS=u),type="response")
plot(u,v,ylim=0:1,type="l",col="red",lwd=2)
points(myocarde$INSYS,myocarde$PRONO,pch=19)

Additive models

Consider now the second dataset, with two variables. Consider here a model like
\mathbb{P}[Y|X_1=x_1,X_2=x_2]=\frac{\exp[\eta(x_1,x_2)]}{1+\exp[\eta(x_1,x_2)]}
where
\exp[\eta(x_1,x_2)]=\beta_0+\color{red}{s_1(x_1)}+\color{blue}{s_2(x_2)}
\color{red}{s_1(x_1)}=\beta_{1,0}x_1+\beta_{1,1}(x_1-s_{11})_++\beta_{1,2}(x_1-s_{12})_+
and
\color{blue}{s_2(x_2)}=\beta_{2,0}x_2+\beta_{2,1}(x_2-s_{21})_++\beta_{2,2}(x_2-s_{22})_+
It might seem a little bit restrictive, but that’s actually the idea of additive models.

reg = glm(y~bs(x1,degree=1,df=3)+bs(x2,degree=1,df=3),data=df,family=binomial(link = "logit"))
u = seq(0,1,length=101)
p = function(x,y) predict.glm(reg,newdata=data.frame(x1=x,x2=y),type="response")
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+(df$y=="1")],cex=1.5)
contour(u,u,v,levels = .5,add=TRUE)


Now, if think about is, we’ve been able to get a “perfect” model, so, somehow, it seems no longer continuous…

persp(u,u,v,theta=20,phi=40,col="green"


Of course, it is… it is piecewise linear, with hyperplane, some being almost vertical.

And one can also consider piecewise quadratic functions

reg = glm(y~bs(x1,degree=2,df=3)+bs(x2,degree=2,df=3),data=df,family=binomial(link = "logit"))
u = seq(0,1,length=101)
p = function(x,y) predict.glm(reg,newdata=data.frame(x1=x,x2=y),type="response")
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+(df$y=="1")],cex=1.5)
contour(u,u,v,levels = .5,add=TRUE)


Funny thing, we now have two “perfect” models, with different areas for the white and the black dots… Don’t ask me how to choose on that one.

In R, it is possible to use the mgcv package to run a gam regression. It is used for generalized additive models, but here, we have only one variable, so it is difficult to see the “additive” part, actually. And to be more specific, mgcv is using penalized quasi-likelihood from the nlme package (but we’ll get back on penalized routines later on).

But maybe I should also mention another smoothing tool before, kernels (and maybe also k-nearest neighbors). To be continued

Classification from scratch, logistic regression 1/8

Let us start today our series on classification from scratch

The logistic regression is based on the assumption that given covariates \mathbf{x}, Y has a Bernoulli distribution,Y|\mathbf{X}=\mathbf{x}\sim\mathcal{B}(p_{\mathbf{x}}),~~~~p_\mathbf{x}=\frac{\exp[\mathbf{x}^T\mathbf{\beta}]}{1+\exp[\mathbf{x}^T\mathbf{\beta}]}The goal is to estimate parameter \mathbf{\beta}.

Recall that the heuristics for the use of that function for the probability is that\log[\text{odds}(Y=1)]=\log\frac{\mathbb{P}[Y=1]}{\mathbb{P}[Y=0]}=\mathbf{x}^T\mathbf{\beta}

Maximimum of the (log)-likelihood function

The log-likelihood is here\log\mathcal{L} = \sum_{i=1}^n y_i\log p_i+(1-y_i)\log (1-p_i) where p_{i}=(1+\exp[-\mathbf{x}_i^T\mathbf{\beta}])^{-1}. Numerical techniques are based on (numerical) gradient descent to compute the maximum of the likelihood function. The (negative) log-likelihood is the following function

y = myocarde$PRONO
X = cbind(1,as.matrix(myocarde[,1:7]))
negLogLik = function(beta){
 -sum(-y*log(1 + exp(-(X%*%beta))) - (1-y)*log(1 + exp(X%*%beta)))
 }

We use the minus sign since standard optimization routines compute minima, not maxima. Now, to find the minimum of that function, we need a starting point to initiate the algorithm

beta_init = lm(PRONO~.,data=myocarde)$coefficients

Why not start with the parameter of the OLS. Somehow, we might think that at least, sign should be ok for instance. Anyway, we need a starting point, and let us use that one.

logistic_opt = optim(par = beta_init, negLogLik, hessian=TRUE, method = "BFGS", control=list(abstol=1e-9))

Here, we obtain

 logistic_opt$par
 (Intercept)        FRCAR        INCAR        INSYS    
 1.656926397  0.045234029 -2.119441743  0.204023835 
       PRDIA        PAPUL        PVENT        REPUL 
-0.102420095  0.165823647 -0.081047525 -0.005992238

Let us verify here that this output is valid. For instance, what if we change the value of the starting point (randomly)

simu = function(i){
logistic_opt_i = optim(par = rnorm(8,0,3)*beta_init, 
negLogLik, hessian=TRUE, method = "BFGS", 
control=list(abstol=1e-9))
logistic_opt_i$par[2:3]
}
v_beta = t(Vectorize(simu)(1:1000))
plot(v_beta)
par(mfrow=c(1,2))
hist(v_beta[,1],xlab=names(myocarde)[1])
hist(v_beta[,2],xlab=names(myocarde)[2])

Ooops. There is a problem here. Clearly, we cannot rely on numerical optimization here. We can think about using another optimization routine

library(optimx)
logit = function(mX, vBeta) {
  exp(mX %*% vBeta)/(1+ exp(mX %*% vBeta)) 
}
logLikelihoodLogitStable = function(vBeta, mX, vY) {
  -sum(vY*(mX %*% vBeta - log(1+exp(mX %*% vBeta))) + 
(1-vY)*(-log(1 + exp(mX %*% vBeta)))) 
}
likelihoodScore = function(vBeta, mX, vY) {
  return(t(mX) %*% (logit(mX, vBeta) - vY) )
}
optimLogitLBFGS = optimx(beta_init, logLikelihoodLogitStable, 
method = 'L-BFGS-B', gr = likelihoodScore, 
mX = X, vY = y, hessian=TRUE)

The optimum is here

attr(optimLogitLBFGS, "details")[[2]]
              [,1]
       0.066680272
FRCAR  0.003080542
INCAR  0.079031364
INSYS -0.001586194
PRDIA  0.040500697
PAPUL -0.041870705
PVENT -0.014162756
REPUL  0.195632244

Let’s be honest here, I do not feel confortable with those techniques. So, what happened here ?

Here, the technique we use is based on the following idea,\mathbf{\beta}_{new}=\mathbf{\beta}_{old} -\left(\frac{\partial^2\log\mathcal{L}(\mathbf{\beta}_{old})}{\partial\mathbf{\beta}\partial\mathbf{\beta}^T}\right)^{-1}\cdot \frac{\partial\log\mathcal{L}(\mathbf{\beta}_{old})}{\partial\mathbf{\beta}}The problem is that my computer does not know this first and second derivatives. So it will compute them using approximation techniques.

Actually, it is possible to use functions dedicated to such computation

library(numDeriv)
library(MASS)
logit = function(x){1/(1+exp(-x))}
logLik = function(beta, X, y){
 -sum(y*log(logit(X%*%beta)) + 
(1-y)*log(1-logit(X%*%beta)))
}
optim_second = function(beta, num_iter){
  LL = vector()
  for(i in 1:num_iter){
    grad = (t(X)%*%(logit(X%*%beta) - y)) 
    H = hessian(logLik, beta, method = "complex", X = X, y = y)
    beta = beta - ginv(H)%*%grad
    LL[i] = logLik(beta, X, y)
  }
  result = list(beta, H)
return(result)
}

With our OLS starting point, we obtain

opt0 = optim_second(beta_init,500)
opt0[[1]]
             [,1]
[1,]  0.951074420
[2,]  0.018860280
[3,]  0.275428978
[4,]  0.144803636
[5,] -0.058535606
[6,]  0.001182178
[7,] -0.108651776
[8,] -0.002940315

But if we try with another starting point

opt1 = optim_second(beta_init*runif(8),500)
opt1[[1]]
             [,1]
[1,]  0.052894794
[2,]  0.024718435
[3,]  0.167953661
[4,]  0.171662947
[5,] -0.057458066
[6,] -0.011361034
[7,] -0.107532114
[8,] -0.002679064

Clearly, some coefficients are rather close. But other aren’t. From my point of viezw, that is a major problem (keep in mind that we do not deal here with massive data ! There are only 7 explanatory variables, and only 71 observations).

Why not try to be clever, and use the analytical values of those derivatives ? Even if some people claim the oppositive, sometimes, it can actually be usefull to do the maths, instead of considering only numerical values.

Newton (or Fisher) Algorithm

If you open any Econometrics textbooks (one can also try to derive it), you will get \frac{\partial\log\mathcal{L}(\mathbf{\beta}_{old})}{\partial\mathbf{\beta}}=\mathbf{X}^T(\mathbf{y}-\mathbf{p}_{old})
while\frac{\partial^2\log\mathcal{L}(\mathbf{\beta}_{old})}{\partial\mathbf{\beta}\partial\mathbf{\beta}^T}=-\mathbf{X}^T\mathbf{\Delta}_{old}\mathbf{X}

Y=myocarde$PRONO
X=cbind(1,as.matrix(myocarde[,1:7]))
colnames(X)=c("Inter",names(myocarde[,1:7]))
 beta=as.matrix(lm(Y~0+X)$coefficients,ncol=1)
 for(s in 1:9){
   pi=exp(X%*%beta[,s])/(1+exp(X%*%beta[,s]))
   gradient=t(X)%*%(Y-pi)
   omega=matrix(0,nrow(X),nrow(X));diag(omega)=(pi*(1-pi))
   Hessian=-t(X)%*%omega%*%X
   beta=cbind(beta,beta[,s]-solve(Hessian)%*%gradient)}

Observe that here, I use only ten iterations of the algorithm !

 beta[,8:10]
                [,1]          [,2]          [,3]
XInter -10.187641685 -10.187641696 -10.187641696
XFRCAR   0.138178119   0.138178119   0.138178119
XINCAR  -5.862429035  -5.862429037  -5.862429037
XINSYS   0.717084018   0.717084018   0.717084018
XPRDIA  -0.073668171  -0.073668171  -0.073668171
XPAPUL   0.016756506   0.016756506   0.016756506
XPVENT  -0.106776012  -0.106776012  -0.106776012
XREPUL  -0.003154187  -0.003154187  -0.003154187

The thing is that is seems to converge extremely fast. And it is rather robust ! Look at what we get if we change our starting point

beta=as.matrix(lm(Y~0+X)$coefficients,ncol=1)*runif(8)
 for(s in 1:9){
   pi=exp(X%*%beta[,s])/(1+exp(X%*%beta[,s]))
   gradient=t(X)%*%(Y-pi)
   omega=matrix(0,nrow(X),nrow(X));diag(omega)=(pi*(1-pi))
   Hessian=-t(X)%*%omega%*%X
   beta=cbind(beta,beta[,s]-solve(Hessian)%*%gradient)}
 beta[,8:10]
                [,1]          [,2]          [,3]
XInter -10.187641586 -10.187641696 -10.187641696
XFRCAR   0.138178118   0.138178119   0.138178119
XINCAR  -5.862429017  -5.862429037  -5.862429037
XINSYS   0.717084013   0.717084018   0.717084018
XPRDIA  -0.073668172  -0.073668171  -0.073668171
XPAPUL   0.016756508   0.016756506   0.016756506
XPVENT  -0.106776012  -0.106776012  -0.106776012
XREPUL  -0.003154187  -0.003154187  -0.003154187

Nice, isn’t it? Looks like we got our winner, don’t we? And one can use the inverse of the Hessian matrix to get standard deviations.

Weighted Least-Squares

Let us go one step further. We’ve seen that we want to compute something like\mathbf{\beta}_{new} =(\mathbf{X}^T\mathbf{\Delta}_{old}\mathbf{X})^{-1}\mathbf{X}^T\mathbf{\Delta}_{old}\mathbf{z}(if we do substitute matrices in the analytical expressions) where \mathbf{z}=\mathbf{X}\mathbf{\beta}_{old}+\mathbf{\Delta}_{old}^{-1}[\mathbf{y}-\mathbf{p}_{old}]. But actually, that’s simply a standard least-square problem\mathbf{\beta}_{new} = \text{argmin}\left\lbrace(\mathbf{z}-\mathbf{X}\mathbf{\beta})^T\mathbf{\Delta}_{old}^{-1}(\mathbf{z}-\mathbf{X}\mathbf{\beta})\right\rbraceThe only problem here is that weights \mathbf{\Delta}_{old} are functions of unknown \mathbf{\beta}_{old}. But actually, if we keep iterating, we should be able to solve it : given the \mathbf{\beta} we got the weights, and with the weights, we can use weighted OLS to get an updated \mathbf{\beta}. That’s the idea of iteratively reweighted least squares.

The algorithm will be

df = myocarde
beta_init = lm(PRONO~.,data=df)$coefficients
X = cbind(1,as.matrix(myocarde[,1:7]))
beta = beta_init
for(s in 1:1000){
p = exp(X %*% beta) / (1+exp(X %*% beta))
omega = diag(nrow(df))
diag(omega) = (p*(1-p))
df$Z = X %*% beta + solve(omega) %*% (df$PRONO - p)
beta = lm(Z~.,data=df[,-8], weights=diag(omega))$coefficients
}

and the output is here

 beta
  (Intercept)         FRCAR         INCAR         INSYS         PRDIA 
-10.187641696   0.138178119  -5.862429037   0.717084018  -0.073668171 
        PAPUL         PVENT         REPUL 
  0.016756506  -0.106776012  -0.003154187

which is almost what we’ve obtained before. Nice isn’t it ? Actually, here we also have standard deviations of estimators

summary( lm(Z~.,data=df[,-8], weights=diag(omega)))
 
Coefficients:
              Estimate Std. Error t value Pr(>|t|)
(Intercept) -10.187642  10.668138  -0.955    0.343
FRCAR         0.138178   0.102340   1.350    0.182
INCAR        -5.862429   6.052560  -0.969    0.336
INSYS         0.717084   0.503527   1.424    0.159
PRDIA        -0.073668   0.261549  -0.282    0.779
PAPUL         0.016757   0.306666   0.055    0.957
PVENT        -0.106776   0.099145  -1.077    0.286
REPUL        -0.003154   0.004386  -0.719    0.475

The standard glm function

Of course, it is possible to use an R built-in function to get our estimate

summary(glm(PRONO~.,data=myocarde,family=binomial(link = "logit")))
 
Coefficients:
              Estimate Std. Error z value Pr(>|z|)
(Intercept) -10.187642  11.895227  -0.856    0.392
FRCAR         0.138178   0.114112   1.211    0.226
INCAR        -5.862429   6.748785  -0.869    0.385
INSYS         0.717084   0.561445   1.277    0.202
PRDIA        -0.073668   0.291636  -0.253    0.801
PAPUL         0.016757   0.341942   0.049    0.961
PVENT        -0.106776   0.110550  -0.966    0.334
REPUL        -0.003154   0.004891  -0.645    0.519

Application and visualisation

Let us visualize the prediction obtained from the logistic regression, on our second dataset

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))
reg = glm(y~x1+x2,data=df,family=binomial(link = "logit"))
u = seq(0,1,length=101)
p = function(x,y) predict.glm(reg,newdata=data.frame(x1=x,x2=y),type="response")
v = outer(u,u,p)
image(u,u,v,xlab="Variable 1",ylab="Variable 2",col=clr10,breaks=(0:10)/10)
points(x,y,pch=19,cex=1.5,col="white")
points(x,y,pch=c(1,19)[1+z],cex=1.5)
contour(u,u,v,levels = .5,add=TRUE)


Here level curves – or iso-probabilities – are linear, so the space is divided in two (0 and 1, survival and death, white and black) by a straight line (or an hyperplane in higher dimension). Furthermore, since we have a linear model, if we change the cutoff (the threshold used to create the two classes), we obtain another straight line (or hyperplane) parallel to the first one.

Next time, we will introduce splines to smooth those continuous covariates… to be continued.

Classification from scratch, overview 0/8

Before my course on « big data and economics » at the university of Barcelona in July, I wanted to upload a series of posts on classification techniques, to get an insight on machine learning tools.

According to some common idea, machine learning algorithms are black boxes. I wanted to get back on that saying. First of all, isn’t it the case also for regression models, like generalized additive models (with splines) ? Do you really know what the algorithm is doing ? Even the logistic regression. In textbooks, we can easily find math formulas. But what is really done when I run it, in R ?

When I started working on academia, someone told me something like « if you really want to understand a theory, teach it ». And that has been my moto for more than 15 years. I wanted to add a second part to that statement: « if you really want to understand an algorithm, recode it ». So let’s try this… My ambition is to recode (more or less) most of the standard algorithms used in predictive modeling, from scratch, in R. What I plan to mention, within the next two weeks, will be

I will use two datasets to illustrate. The first one is inspired by the cover of « Foundations of Machine Learning » by Mehryar Mohri, Afshin Rostamizadeh and Ameet Talwalkar. At least, with this dataset, it will be possible to plot predictions (since there are only two – continuous – features)

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))
plot(x,y,pch=c(1,19)[1+z])

Here is some code to get a visualization of the prediction (here the probability to be a black point)

rmatrix_model = function(model){
u = seq(0,1,length=101)
p = function(x,y) predict(model,newdata=data.frame(x1=x,x2=y),type="response")
v = outer(u,u,p)
return(v)}
nice_graph=function(v){
u = seq(0,1,length=101)
image(u,u,v,xlab="Variable 1",ylab="Variable 2",col=clr10[c(1,10)],breaks=c(0,5,10)/10)
points(x,y,pch=19,cex=1.5,col="white")
points(x,y,pch=c(1,19)[1+z],cex=1.5)
contour(u,u,v,levels = .5,add=TRUE)
}
reg = glm(y~x1+x2,data=df,family=binomial)
nice_graph(rmatrix_model(reg))

Note that colors are defined here as

clr10= c("#ffffff","#f7fcfd","#e5f5f9","#ccece6","#99d8c9","#66c2a4","#41ae76","#238b45","#006d2c","#00441b")

or with some nonlinear model

The second one is a dataset I got from Gilbert Saporta, about heart attacks and decease (our binary variable).

myocarde = read.table("http://freakonometrics.free.fr/myocarde.csv",head=TRUE, sep=";")
myocarde$PRONO = (myocarde$PRONO=="SURVIE")*1
y = myocarde$PRONO
X = as.matrix(cbind(1,myocarde[,1:7]))

So far, I do not plan to talk (too much) on the choice of tunning parameters (and cross-validation), on comparing models, etc. The goal here is simply to understand what’s going on when we call either glm, glmnet, gam, random forest, svm, xgboost, or any function to get a predict model.

Some sort of Otto Neurath (isotype picture) map

Yesterday evening, I was walking in Budapest, and I saw some nice map that was some sort of Otto Neurath style. It was hand-made but I thought it should be possible to do it in R, automatically.

A few years ago, Baptiste Coulmont published a nice blog post on the package osmar, that can be used to import OpenStreetMap objects (polygons, lines, etc) in R. We can start from there. More precisely, consider the city of Douai, in France,

The code to read information from OpenStreetMap is the following

library(osmar)
src = osmsource_api()
bb = center_bbox(3.07758808135,50.37404355, 1000, 1000)
ua = get_osm(bb, source = src)

We can extract a lot of things, like buildings, parks, churches, roads, etc. There are two kinds of objects so we will use two functions

listek = function(vc,type="polygons"){
nat_ids = find(ua, way(tags(k %in% vc)))
nat_ids = find_down(ua, way(nat_ids))
nat = subset(ua, ids = nat_ids)
nat_poly = as_sp(nat, type)}
 
listev = function(vc,type="polygons"){
  nat_ids = find(ua, way(tags(v %in% vc)))
  nat_ids = find_down(ua, way(nat_ids))
  nat = subset(ua, ids = nat_ids)
  nat_poly  as_sp(nat, type)}

For instance to get rivers, use

W=listek(c("waterway"))

and to get buildings

M=listek(c("building"))

We can also get churches

C=listev(c("church","chapel"))

but also train stations, airports, universities, hospitals, etc. It is also possible to get streets, or roads

H1=listek(c("highway"),"lines")
H2=listev(c("residential","pedestrian","secondary","tertiary"),"lines")

but it will be more difficult to use afterwards, so let’s forget about those.

We can check that we have everything we need

plot(M)
plot(W,add=TRUE,col="blue")
plot(P,add=TRUE,col="green")
if(!is.null(B)) plot(B,add=TRUE,col="red")
if(!is.null(C)) plot(C,add=TRUE,col="purple")
if(!is.null(T)) plot(T,add=TRUE,col="red")

Now, let us consider a rectangular grid. If there is a river in a cell, I want a river. If there is a church, I want a church, etc. Since there will be one (and only one) picture per cell, there will be priorities. But first we have to check intersections with polygons, between our grid, and the OpenStreetMap polygons.

library(sp)
library(raster)
library(rgdal)
library(rgeos)
library(maptools)
identification = function(xy,h,PLG){
  b=data.frame(x=rep(c(xy[1]-h,xy[1]+h),each=2),
               y=c(c(xy[2]-h,xy[2]+h,xy[2]+h,xy[2]-h)))
  pb1=Polygon(b)    
  Pb1 = list(Polygons(list(pb1), ID=1))
  SPb1 = SpatialPolygons(Pb1, proj4string = CRS("+proj=longlat +ellps=WGS84 +datum=WGS84 +no_defs +towgs84=0,0,0"))
  UC=gUnionCascaded(PLG)
  return(gIntersection(SPb1,UC))
}

and then, we identify, as follows

whichidtf = function(xy,h){
  h=.7*h
  label="EMPTY"
if(!is.null(identification(xy,h,M))) label="HOUSE"
if(!is.null(identification(xy,h,P))) label="PARK"
if(!is.null(identification(xy,h,W))) label="WATER"
if(!is.null(identification(xy,h,U))) label="UNIVERSITY"
if(!is.null(identification(xy,h,C))) label="CHURCH"
return(label)
}

Let is use colored rectangle to make sure it works

nx=length(vx)
vx=as.numeric((vx[2:nx]+vx[1:(nx-1)])/2)
ny=length(vy)
vy=as.numeric((vy[2:ny]+vy[1:(ny-1)])/2)
 plot(M,border="white")
 for(i in 1:(nx-1)){
     for(j in 1:(ny-1)){
         lb=whichidtf(c(vx[i],vy[j]),h)
         if(lb=="HOUSE") rect(vx[i]-h,vy[j]-h,vx[i]+h,vy[j]+h,col="grey")
         if(lb=="PARK") rect(vx[i]-h,vy[j]-h,vx[i]+h,vy[j]+h,col="green")
         if(lb=="WATER") rect(vx[i]-h,vy[j]-h,vx[i]+h,vy[j]+h,col="blue")
         if(lb=="CHURCH") rect(vx[i]-h,vy[j]-h,vx[i]+h,vy[j]+h,col="purple")      
     }}

As a first start, we us agree that it works. To use pics, I did borrow them from https://fontawesome.com/. For instance, we can have a tree

 library(png)
 library(grid)
 download.file("http://freakonometrics.hypotheses.org/files/2018/05/tree.png","tree.png")
 tree = readPNG("tree.png")

Unfortunatly, the color is not good (it is black), but that’s easy to fix using the RGB decomposition of that package

 rev_tree=tree
 rev_tree[,,2]=tree[,,4]

We can do the same for houses, churches and water actually

 download.file("http://freakonometrics.hypotheses.org/files/2018/05/angle-double-up.png","angle-double-up.png")
 download.file("http://freakonometrics.hypotheses.org/files/2018/05/home.png","home.png")
 download.file("http://freakonometrics.hypotheses.org/files/2018/05/church.png","curch.png")
water = readPNG("angle-double-up.png")
 rev_water=water
 rev_water[,,3]=water[,,4]
 home = readPNG("home.png")
 rev_home=home
 rev_home[,,4]=home[,,4]*.5
 church = readPNG("church.png")
 rev_church=church
 rev_church[,,1]=church[,,4]*.5
 rev_church[,,3]=church[,,4]*.5

and that’s almost it. We can then add it on the map

 plot(M,border="white")
 for(i in 1:(nx-1)){
   for(j in 1:(ny-1)){
     lb=whichidtf(c(vx[i],vy[j]),h)
     if(lb=="HOUSE")  rasterImage(rev_home,vx[i]-h*.8,vy[j]-h*.8,vx[i]+h*.8,vy[j]+h*.8)
     if(lb=="PARK") rasterImage(rev_tree,vx[i]-h*.9,vy[j]-h*.8,vx[i]+h*.9,vy[j]+h*.8)
     if(lb=="WATER") rasterImage(rev_water,vx[i]-h*.8,vy[j]-h*.8,vx[i]+h*.8,vy[j]+h*.8)
     if(lb=="CHURCH") rasterImage(rev_church,vx[i]-h*.8,vy[j]-h*.8,vx[i]+h*.8,vy[j]+h*.8)     
   }}

Nice, isn’t it? (as least as a first draft, done during the lunch break of the R conference in Budapest, today).

 

Graduate Course on Advanced Tools for Econometrics (2)

This Tuesday, I will be giving the second part of the (crash) graduate course on advanced tools for econometrics. It will take place in Rennes, IMAPP room, and I have been told that there will be a visio with Nantes and Angers. Slides for the morning are online, as well as slides for the afternoon.

In the morning, we will talk about variable section and penalization, and in the afternoon, it will be on changing the loss function (quantile regression).

When “learning Python” becomes “practicing R” (spoiler)

15 years ago, a student of mine told me that I should start learning Python, that it was really a great language. Students started to learn it, but I kept postponing. A few years ago, I started also Python for Kids, which is really nice actually, with my son. That was nice, but not really challenging. A few weeks ago, I also started a crash course in Python, taught by Pierre. The truth is I think I will probably give up. I keep telling myself (1) I can do anything much faster in R (2) Python is not intuitive, especially when you’re used to practice R for almost 20 years… Last week, I also had to link Python and R for our pricing game : Ali wrote some template codes in Python, and I had to translate them in R. And it was difficult…

Anyway, since it was a school break this week, I said to my son that we should try to practice together, with a nice challenge. For those willing to try it, you’d better stop here, because I will spoil it.

Continue reading When “learning Python” becomes “practicing R” (spoiler)

Using convolutions (S3) vs distributions (S4)

Usually, to illustrate the difference between S3 and S4 classes in R, I mention glm (from base) and vglm (from VGAM) that provide similar outputs, but one is based on S3 codes, while the second one is based on S4 codes. Another way to illustrate is to manipulate distributions.

Consider the case where we want to sum (independent) random variables. For instance two lognormal distribution. Let us try to compute the median of the sum.

The distribution function of the sum of two independent (positive) random variables is F_{S_2}(x)=\int_0^x F_{X_1}(x-y)dF_{X_2}(x)

pSum2 = function(x) integrate(function(y) 
plnorm(x-y,1,2)*dlnorm(y,2,1),0,x)$value

Let us visualize that cumulative distribution function

vx=seq(0.1,50,by=.1)
vy=Vectorize(pSum2)(vx)
plot(vx,vy,type="l",ylim=c(0,1))
abline(h=.5,lty=2)

Let us find an upper bound to compute (in a decent time) quantiles

pSum2(350)
[1] 0.99195

and then use the uniroot function to inverse that function

qSum = function(u) uniroot(function(x) 
Vectorize(pSum2)(x)-u, interval=c(0,350))$root
vu=seq(.01,.99,by=.01)
vv=Vectorize(qSum)(vu)

The median is here

qSum(.5)
[1] 14.155

Why not consider the sum of three (independent) distributions ? Its cumulative distribution function can be writen using our previous function F_{S_3}(x)=\int_0^x F_{S_2}(x-y)dF_{X_3}(x)

pSum3 = function(x) integrate(function(y) 
pSum2(x-y)*dlnorm(y,2,2),0,x)$value

If we look at some values we good

pSum3(4)
[1] 0.015624
pSum3(5)
Error in integrate(function(y) plnorm(x - y, 1, 2) * 
dlnorm(y, 2, 1),  : 
  maximum number of subdivisions reached

So obviously, there are computational issues here.

Let us consider the following alternative expression F_{S_3}(x)=\int_0^x F_{X_3}(x-y)dF_{S_2}(x). Of course, it is necessary here to compute the density of the sum of two variables

dSum2 = function(x) integrate(function(y) 
dlnorm(x-y,1,2)*dlnorm(y,2,1),0,x)$value
pSum3 = function(x) integrate(function(y) 
dlnorm(x-y,2,2)*dSum2(y),0,x)$value

Again, let us compute some values

pSum3(4)
[1] 0.0090285
pSum3(5)
[1] 0.01186

This one seems to work quite well. But it is just an illusion.

pSum3(9)
Error in integrate(function(y) dlnorm(x - y, 1, 2) *
 dlnorm(y, 2, 1),  : 
  maximum number of subdivisions reached

Clearly, with those S3-type functions, it wlll be complicated to run computations with 3 variables, or more.

Let us consider distributions in the S4-type format of the following package

library(distr)
X1 = Lnorm(mean=1,sd=2)
X2 = Lnorm(mean=2,sd=1)
S2 = X1+X2

To compute the median, we simply have to use

distr::q(S2)(.5)
[1] 14.719

We can also visualize it easily

plot(q(S2))

which looks (very) close to what we got, manually.  But here, it is also possible to work with the sum of 3 (independent) random variables

X3 = Lnorm(mean=2,sd=2)
S3 = X1+X2+X3

To compute the median, use

distr::q(S3)(.5)
[1] 33.208

The function is here

plot(q(S3))

(Advanced) R Crash Course, for Actuaries

The fourth year of the Data Science for Actuaries program started this morning. I will be there for the introduction to R. The slides are available online (created with slidify, the .Rmd file is also available)

A (standard) markdown is also available (as well as the .Rmd file). I have to thank Ewen for his help on slidify (especially for the online quizz, and the integration of leaflet maps or the rgl animated graph….)

Visualizing effects of a categorical explanatory variable in a regression

Recently, I’ve been working on two problems that might be related to semiotic issues in predictive modeling (i.e. instead of a standard regression table, how can we plot coefficient values in a regression model). To be more specific, I have a variable of interest Y that is observed for several individuals i, with explanatory variables \mathbf{x}_i, year t, in a specific region z_i\in\{A,B,C,D,E\}. Suppose that we have a simple (standard) linear model (forget about time here) y_i=\beta_0+\beta_1x_{1,i}+\cdots+\beta_kx_{k,i}+\sum_j \alpha_j \mathbf{1}(z_i\in j)+\varepsilon_i

Let us forget the temporal effect to focus on the spatial effect today. And consider some simulated dataset. There will be only one (continuous) explanatory variable. And I will generate correlated covariates, just to be more realistic.

n=1000
library(mnormt)
r=.5
Sigma=matrix(c(1,r,r,1), 2, 2)
set.seed(1)
X=rmnorm(n,c(0,0),Sigma)
X1=cut(X[,1],c(-100,quantile(X[,1],c(.1,.4,.7,.85)),
100),labels=LETTERS[1:5])
X2=X[,2]
Y=5+X[,1]-X[,2]+rnorm(n)/2
db=data.frame(Y,X1,X2)

Here we have y_i=\beta_0+\beta_1x_{1,i}+\sum_{j\in\{A,B,C,D,E\}} \alpha_j \mathbf{1}(z_i\in j)+\varepsilon_i The goal here is to get to graph to visualize the vector \hat\alpha=(\hat\alpha_A,\cdots,\hat\alpha_E). Let us run the linear regression

reg1=lm(Y~X1+X2,data=db)
idx=which(substr(names(reg1$coefficients), 1,2)=="X1")
v1=reg1$coefficients[idx]
names(v1)=LETTERS[2:5]
barplot(v1,col=rgb(0,0,1,.4))

Note that it is possible to add some sort of “confidence interval” to discuss significance (or to avoid to spend hours discussing differences in bar heights that are not significantly different)

library(Hmisc)
sv1=summary(reg1)$coefficients[idx,2]
(bp1=barplot(v1,ylim=range(c(0,v1+2*sv1))))
errbar(bp1[,1],v1,v1-2*sv1,v1+2*sv1,add=TRUE)

My main concern here is the “reference” that is considered. Should A be the reference? Why not B

db$X1=relevel(db$X1,"B")
reg1=lm(Y~X1+X2,data=db)
idx=which(substr(names(reg1$coefficients),1,2)=="X1")
v1=reg1$coefficients[idx]
names(v1)=LETTERS[c(1,3:5)]
library(Hmisc)
sv1=summary(reg1)$coefficients[idx,2]
(bp1=barplot(v1)
errbar(bp1[,1],v1,v1-2*sv1,v1+2*sv1,add=TRUE)

Why not the smallest one? Why not the largest one?… What if there is no simple way to choose. Furthermore, let us get back to the original point, which is that there might be some temporal aspects. More precisely, we can have \hat\alpha^{(t)}=(\hat\alpha_A^{(t)},\cdots,\hat\alpha_E^{(t)}). If we have also \hat\alpha^{(t+1)} and we get another plot, how do we interpret it. If for E the bar is taller, it means that relative to A, the difference has increased. I have the feeling that the interpretation is more complicated because we do not see, on that graph, changes in \hat\alpha^{(t)}_A.

Let us try something else. First, let us get back to the original setting

db$X1=relevel(db$X1,"A")

Consider here the regression without the intercept, so that all values remain

reg1=lm(Y~0+X1+X2,data=db)
idx=which(substr(names(reg1$coefficients),1,2)=="X1")
v1=reg2$coefficients[idx]
names(v1)=LETTERS[1:5]
barplot(v1)

It can be hard to read, especially if Y takes (very) large values, and you think that barplots should start at 0. But still, having those 5 values is nice. Why not rescale that graph?

A natural idea my be to consider the case where no spatial component is considered, and to look at the difference with that reference.

reg1=lm(Y~1+X2,data=db)
reg2=lm(Y~0+X1+X2,data=db)
idx=which(substr(names(reg2$coefficients),1,2)=="X1")
v1=reg2$coefficients[idx]
v2=v1-reg1$coefficients["(Intercept)"]
barplot(v2,col=rgb(0,0,1,.4))
sv2=summary(reg2)$coefficients[idx,2]
(bp2=barplot(v2,ylim=range(c(v2-2*sv2,v2+2*sv2))))
errbar(bp2[,1],v2,v2-2*sv2,v2+2*sv2,add=TRUE)

I like that graph, I should admit it. Now, I still have some remaining questions. For instance, can we insure that when only the intercept is considered, the value of \hat\beta_0 is somewhere between \hat\beta_A,\cdots,\hat\beta_E? Is it possible that \hat\beta_A-\hat\beta_0,\cdots,\hat\beta_E-\hat\beta_0 are all positive? In that case, I would find that hard to interpret.

Actually, if I really want values that can be seen as compared to some average, why not consider a (weighted) average of \hat\beta_A,\cdots,\hat\beta_E? (weights being here proportion in each class, in each region)

w=table(db$X1)
v3=v1-sum(w*v1)/sum(w)
(bp3=barplot(v3,ylim=range(c(v3-2*sv3,v3+2*sv3))))
errbar(bp3[,1],v3,v3-2*sv3,v3+2*sv3,add=TRUE)

I like that one. But what if, instead of normalizing at the end, we normalize the original dependent variable. By “normalize”, I mean “rescale”, to have a centered variable.

db$Y0=db$Y-mean(db$Y)
reg3=lm(Y0~0+X1+X2,data=db)
sv3=summary(reg3)$coefficients[idx,2]
(bp3=barplot(v3,ylim=range(c(v3-2*sv3,v3+2*sv3))))
errbar(bp3[,1],v3,v3-2*sv3,v3+2*sv3,add=TRUE)

This one is nice, because it is extremely simple to explain. But what if instead of a linear regression, we add a logistic one (with Y\in\{0,1\})? or a Poisson regression…

So maybe it cannot be the best solution here. Let us try something else… In insurance ratemaking, people like to use “zonier“. It is a two-stage regression. The idea is to run a regression without any spatial components, first. Then, consider the regression of residuals on spatial variables. Here, it would be something like

reg1=lm(Y~1+X2,data=db)
reg4=lm(residuals(reg1)~0+db$X1)

Since we focus on residuals, those are centered, and we have an easy interpretation of respective values

sv4=summary(reg4)$coefficients[idx,2]
v4=reg4$coefficients
(bp4=barplot(v4,names.arg=LETTERS[1:5])))
errbar(bp4[,1],v4,v4-2*sv4,v4+2*sv4,add=TRUE)

I guess that it can also be use in generalized linear models, with Pearson (or deviance) residuals.

Another possible idea can be the following. Again, the goal is not to have the true values, but to visualize on a graph how regions can be different. Here, all of them are significantly different. And in region A, Y is smaller, ceteris paribus (other things equal in the sense that we have taken into account x_1). And in region E it is larger. Here, the graph helps to “see” those differences.

Why not consider a completely different graph. What if we plot vector a instead of \alpha, where a_A can be interpreted as the value of the coefficient if we consider region A against “not region A“. What if we consider 5 regressions where dichotomous versions of Z are considered : Z_j=\mathbf{1}_{Z=j}.

v5=sv5=rep(NA,5)
names(v5)=LETTERS[1:5]
for(k in 1:5){
reg=lm(Y~I(X1==LETTERS[k])+X2,data=db)
v5[k]=reg$coefficients[2]
sv5[k]=summary(reg)$coefficients[2,2]}

We can plot that sequence of values, including some confidence intervals (that would be related to significance with respect to all other regions)

(bp5=barplot(v5,ylim=range(c(v5-2*sv5,v5+2*sv5))))
errbar(bp5[,1],v5,v5-2*sv5,v5+2*sv5,add=TRUE)

Looking at values does not give intuitive results, but I have the feeling that it is easy to explain what we plot (we compare each region to “the rest of the world”), and the ordering of a seems to be consistent with \alpha (but I could not prove it).

Here are some ideas I got. I should be able to provide other graphs, but I would love to discuss with anyone on that topics, to find a proper and nice way to visualize effects of a categorical explanatory variable in a regression model (that can be a logistic one). Comments are open…

Holt-Winters with a Quantile Loss Function

Exponential Smoothing is an old technique, but it can perform extremely well on real time series, as discussed in Hyndman, Koehler, Ord & Snyder (2008)),

when Gardner (2005) appeared, many believed that exponential smoothing should be disregarded because it was either a special case of ARIMA modeling or an ad hoc procedure with no statistical rationale. As McKenzie (1985) observed, this opinion was expressed in numerous references to my paper. Since 1985, the special case argument has been turned on its head, and today we know that exponential smoothing methods are optimal for a very general class of state-space models that is in fact broader than the ARIMA class.

Furthermore, I like it because I think it has nice pedagogical features. Consider simple exponential smoothing, L_{t}=\alpha Y_{t}+(1-\alpha)L_{t-1} where \alpha\in(0,1) is the smoothing weight. It is locally constant, in the sense that {}_{t}\hat Y_{t+h} = L_{t}

 library(datasets)
 X=as.numeric(Nile)
 SimpleSmooth = function(a){
  T=length(X)
  L=rep(NA,T)
  L[1]=X[1]
  for(t in 2:T){L[t]=a*X[t]+(1-a)*L[t-1]}
  return(L)
 }
 plot(X,type="b",cex=.6)
 lines(SimpleSmooth(.2),col="red")

When using the standard R function, we get

hw=HoltWinters(X,beta=FALSE,gamma=FALSE, l.start=X[1])
hw$alpha
[1] 0.2465579

Of course, one can replicate that optimal value

V=function(a){
     T=length(X)
     L=erreur=rep(NA,T)
     erreur[1]=0
     L[1]=X[1]
     for(t in 2:T){
         L[t]=a*X[t]+(1-a)*L[t-1]
         erreur[t]=X[t]-L[t-1] }
     return(sum(erreur^2))
}
optim(.5,V)$par
[1] 0.2464844

Here, the optimal value for \alpha is the one that minimizes the one-step prediction, for the \ell_2 loss function, i.e. \sum_{t=2}^n(Y_t-{}_{t-1}\hat Y_t)^2 where here {}_{t-1}\hat Y_t = L_{t-1}. But one can consider another loss function, for instance the quantile loss function, \ell_{\tau}(\varepsilon)=\varepsilon(\tau-\mathbb{I}_{\varepsilon\leq 0}). The optimal coefficient is then obtained using

HWtau=function(tau){
loss=function(e) e*(tau-(e<=0)*1)
 V=function(a){
  T=length(X)
  L=erreur=rep(NA,T)
  erreur[1]=0
  L[1]=X[1]
  for(t in 2:T){
  L[t]=a*X[t]+(1-a)*L[t-1]
  erreur[t]=X[t]-L[t-1] }
 return(sum(loss(erreur)))
 }
 optim(.5,V)$par
}

Here is the evolution of \alpha^\star_\tau as a function of \tau (the level of the quantile considered).

T=(1:49)/50
HW=Vectorize(HWtau)(T)
plot(T,HW,type="l")
abline(h= hw$alpha,lty=2,col="red")

Note that the optimal \alpha is decreasing with \tau. I wonder how general this result can be…

Of course, one can consider more general exponential smoothing, for instance the double one, with L_t=\alpha Y_t+(1-\alpha)[L_{t-1}+B_{t-1}]andB_t=\beta[L_t-L_{t-1}]+(1-\beta)B_{t-1}so that the prediction is now {}_{t}\hat Y_{t+h} = L_{t}+hB_t (it is now locally linear – and no longer constant).

hw=HoltWinters(X,gamma=FALSE,l.start=X[1])
hw$alpha
    alpha 
0.4200241 
hw$beta
      beta 
0.05973389

The code to compute the smoothed series is the following

DoubleSmooth = function(a,b){
  T=length(X)
  L=B=rep(NA,T)
  L[1]=X[1]; B[1]=0
  for(t in 2:T){
  L[t]=a*X[t]+(1-a)*(L[t-1]+B[t-1])
  B[t]=b*(L[t]-L[t-1])+(1-b)*B[t-1] }
 return(L+B)
 }

Here also it is possible to replicate R using the \ell_2 loss function

V=function(A){
     a=A[1]
     b=A[2]
     T=length(X)
     L=B=erreur=rep(NA,T)
     erreur[1]=0
     L[1]=X[1]; B[1]=X[2]-X[1]
     for(t in 2:T){
         L[t]=a*X[t]+(1-a)*(L[t-1]+B[t-1])
         B[t]=b*(L[t]-L[t-1])+(1-b)*B[t-1] 
         erreur[t]=X[t]-(L[t-1]+B[t-1]) }
     return(sum(erreur^2))
}
optim(c(.5,.05),V)$par
[1] 0.41904510 0.05988304

(up to numerical optimization approximation, I guess). But here also, a quantile loss function can be considered

HWtau=function(tau){
loss=function(e) e*(tau-(e<=0)*1)
 V=function(A){
  a=A[1]
  b=A[2]
  T=length(X)
  L=B=erreur=rep(NA,T)
  erreur[1]=0
  L[1]=X[1]; B[1]=X[2]-X[1]
  for(t in 2:T){
   L[t]=a*X[t]+(1-a)*(L[t-1]+B[t-1])
   B[t]=b*(L[t]-L[t-1])+(1-b)*B[t-1] 
   erreur[t]=X[t]-(L[t-1]+B[t-1]) }
  return(sum(loss(erreur)))
  }
     optim(c(.5,.05),V)$par
}

and we can plot those values on a graph

T=(1:49)/50
HW=Vectorize(HWtau)(T)
plot(HW[1,],HW[2,],type="l")
abline(v= hw$alpha,lwd=.4,lty=2,col="red")
abline(h= hw$beta,lwd=.4,lty=2,col="red")
points(hw$alpha,hw$beta,pch=19,col="red")

(with \alpha on the x-axis, and \beta on the y-axis). So here, it is extremely simple to change the loss function, but so far, it should be done manually. Of course, one do it also for the seasonal exponential smoothing model.

The myth of interpretability of econometric models

There are important discussions nowadays about data modeling, to choose between the “two cultures” (as mentioned in Breiman (2001)), i.e. either econometrics models or machine/statistical learning models. We did discuss this issue recently in Econométrie et Machine Learning (so far only in French) with Emmanuel Flachaire and Antoine Ly. One argument often used by econometricians is the interpretability of econometric models. Or at least the attempt to get an interpretable model.

We also have this discussion in actuarial science, for instance in ratemaking (or insurance pricing). Machine learning based models usually perform better (for some a priori chosen metric), but actuaries claim that econometric models are more easily interpretable. In actuarial literature, we assume that claim frequency Y is driven by some non-observable risk factor \Theta, and therefore, we do have heterogeneous risks in our portfolio. And, it can be seen as legitimate to differentiate prices. Assume that this risk factor \Theta is strongly correlated with X_1, the age of the driver. Because in our portfolio, old drivers tend to have more accidents. Here, we could pretend to have a “causal story” (as defined in Freedman (2009)) because of a possible interpretation of the model. So it is natural here to consider a regression model of Y on X_1 to derive our actuarial pricing model. But assume that, possibly, risk factor \Theta is also strongly correlated with X_2, that can be related to spatial features (say latitude, which denoted a north/south position). Because in our portfolio, drivers living in the south tend to have more accidents (reads are known to be more dangerous there). Here, we could pretend to have a second “causal story”.

Of course, since \Theta is strongly correlated with X_1 and X_2, it means that X_1 and X_2 are strongly correlated. Here also, this correlation can be interpreted (not in a causal way as previously, but still), since we know that old people like to live in southern regions. So, what should we do here ? Let us run some simulations to  illustrate.

 set.seed(123)
 n=1e5
 Theta=rnorm(n)
 X1=Theta+rnorm(n)/8
 X2=Theta+rnorm(n)/8
 L=exp(-3+Theta)
 Y=rpois(n,L)
 B=data.frame(Y,X1,X2)

Our first idea was to consider a model where Y is “explained” by the first variable X_1,

 g1=glm(Y~X1,data=B,family=poisson)
 summary(g1)
 
Coefficients:
         Estimate Std. Error z value Pr(>|z|)    
(Inter.) -2.97778    0.01544 -192.88   <2e-16 ***
X1        0.97926    0.01092   89.64   <2e-16 ***

As expected, our variable is “significant”, but also, probably more interesting, X_2, has no impact on the residuals

 B$e1=residuals(g1,type="pearson")
 g1e=lm(e1~X2,data=B)
 summary(g1e)
 
Coefficients:
          Estimate Std. Error t value Pr(>|t|)
(Inter.) 0.0003618  0.0031696   0.114    0.909
X2       0.0028601  0.0031467   0.909    0.363

The interpretation is that once we corrected claim frequency for the age of the drivers, there is no spatial effect here. So, a good model should be based only on the age of the drivers.

But we can also consider the other story. We can consider a model where Y is “explained” by the second variable X_2,

 g2=glm(Y~X2,data=B,family=poisson)
summary(g2)
 
Coefficients:
         Estimate Std. Error z value Pr(>|z|)    
(Inter.) -2.97724    0.01544 -192.81   <2e-16 ***
X2        0.97915    0.01093   89.56   <2e-16 ***

Here also we have a valid model, that can be interpreted, and here also X_1, has no impact on the residuals

 B$e2=residuals(g2,type="pearson")
 g2e=lm(e2~X1,data=B)
 summary(g2e)
 
Coefficients:
          Estimate Std. Error t value Pr(>|t|)
(Inter.) 0.0004863  0.0031733   0.153    0.878
X1       0.0027979  0.0031504   0.888    0.374

The story is similar here. If we correct from the spatial pattern, claims frequency does not depend on the age of the driver.

So, what should we do now? We do have two models, and each of them is as interpretable as the other one. Note that we can not use any statistical tool to distinguish the two: they are comparable

 AIC(g1)
[1] 51013.39
 AIC(g2)
[1] 51013.15

Why not incorporate the two explanatory variables X_1 and X_2, at the same time, in our regression model, and let “the model” decide what to do…?

 g=glm(Y~X1+X2,data=B,family=poisson)
 summary(g)
 
Coefficients:
         Estimate Std. Error  z value Pr(>|z|)    
(Inter.) -2.98132    0.01547 -192.723    2e-16 ***
X1        0.49310    0.06226    7.920 2.38e-15 ***
X2        0.49375    0.06225    7.931 2.17e-15 ***

It looks like we completely lost the interpretability of the model, since our two explanatory variables are (strongly) correlated. Actually, instead of saying “use one, and drop the other one (since it brings no further information)”, it says “use both, each one will explain half of the variable”. Strange interpretation, isn’t it?  So why not try some LASSO here?

library(glmnet)
fit=glmnet(x=as.matrix(B[,c("X1","X2")]), 
    y=B$Y,family="poisson")
plot(fit,xvar="lambda")

Here also, it says that we either keep both, or none. So it cannot be used for variable selection (which is an important motivation to use LASSO technique). So, what should be do if we several interpretable models, but no way to choose? Because usually, we claim that we prefer to use a model with an interpretation. But what should be done here?

Optimal Portfolios, or sort of…

Last week, we got our first class on portfolio optimization. We’ve seen Markowitz’s theory where expected returns and the covariance matrix are given,

> download.file(url="http://freakonometrics.free.fr/portfolio.r",destfile = "portfolio.r")
> source("portfolio.r")
> library(zoo)
> library(FRAPO)
> library(IntroCompFinR)
> library(rrcov)
> data( StockIndex )
> pzoo = zoo ( StockIndex , order.by = rownames ( StockIndex ) )
> rzoo = ( pzoo / lag ( pzoo , k = -1) - 1 ) * 100
> Moments <- function ( x , method = c ( "CovClassic" , "CovMcd" , "CovMest" , "CovMMest" , "CovMve" , "CovOgk" , "CovSde" , "CovSest" ) , ... ) {
method <- match.arg ( method )
ans <- do.call ( method , list ( x = x , ... ) ) + return ( getCov ( ans ) )} > covmat=Moments(as.matrix(rzoo),"CovClassic")
> (covmat=round(covmat,1))
SP500 N225 FTSE100 CAC40 GDAX HSI
SP500   17.8 12.7 13.8 17.8 19.5 18.9
N225    12.7 36.6 10.8 15.0 16.2 16.7
FTSE100 13.8 10.8 17.3 18.8 19.4 19.1
CAC40   17.8 15.0 18.8 30.9 29.9 22.8
GDAX    19.5 16.2 19.4 29.9 38.0 26.1
HSI     18.9 16.7 19.1 22.8 26.1 58.1
> er=apply(as.matrix(rzoo),2,mean)
> (er=round(er,1))
SP500 N225 FTSE100 CAC40 GDAX HSI
0.6 -0.2 0.4 0.5 0.8 1.0
> ef <- efficient.frontier(er, covmat, alpha.min=-2.5, alpha.max=2.5, nport=50)

We can now visualize the efficient frontier (and admissible portfolios) below

> u=c(12,ef$sd,12,12)
> v=c(5,ef$er,-1,5)
> plot(ef$sd,ef$er,type="l",xlab="Standard Deviation",ylab="Expected Return", xlim=c(3.5,11),ylim=c(0,2.5),col="red",lwd=1.5)
> points(sqrt(diag(covmat)),er,pch=19,col="blue")
> text(sqrt(diag(covmat)),er,names(er),pos=4, col="blue",cex=.6)
> polygon(u,v,border=NA,col=rgb(0,0,1,.3))

https://freakonometrics.hypotheses.org/files/2017/11/image-voronoi-post-026-1.png

That was the starting point of our class. We did also mention that something important was actually hard to visualize on that graph : the correlation between returns. It is not in the points (which are univariate, with expected return and standard deviation), but in the efficient frontier. For instance, here is our correlation matrix

> (cormat=covmat/(sqrt(diag(covmat) %*% t(diag(covmat)))))
SP500 N225 FTSE100 CAC40 GDAX HSI
SP500   1.00 0.50 0.79 0.76 0.75 0.59
N225    0.50 1.00 0.43 0.45 0.43 0.36
FTSE100 0.79 0.43 1.00 0.81 0.76 0.60
CAC40   0.76 0.45 0.81 1.00 0.87 0.54
GDAX    0.75 0.43 0.76 0.87 1.00 0.56
HSI     0.59 0.36 0.60 0.54 0.56 1.00

We can actually change the correlation between FT500 and FTSE100 (which is here .786)

courbe=function(r=.786){
R=cormat
R[1,3]=R[3,1]=r
covmat2=(sqrt(diag(covmat) %*% t(diag(covmat))))*R
ef <- efficient.frontier(er, covmat2, alpha.min=-2.5, alpha.max=2.5, nport=50)
plot(ef$sd,ef$er,type="l",xlab="Standard Deviation",ylab="Expected Return",
xlim=c(3.5,11),ylim=c(0,2.5),col="red",lwd=1.5)
points(sqrt(diag(covmat)),er,pch=19,col=c("blue","red")[c(2,1,2,1,1,1)])
text(sqrt(diag(covmat)),er,names(er),pos=4,col=c("blue","red")[c(2,1,2,1,1,1)],cex=.6)
polygon(u,v,border=NA,col=rgb(0,0,1,.3))
}

for instance, with a correlation of 0.6, we get the following efficient frontier

> courbe(.6)

and with a stronger correlation

> courbe(.9)

So clearly, correlation does matter. A lot. But more important, one should keep in mind that expected returns and covariances are not given, but estimated. Previously, we did use the standard estimator for the variance matrix. But another (more robust) estimator can be considered

covmat=Moments(as.matrix(rzoo),"CovSde")
er=apply(as.matrix(rzoo),2,mean)
ef <- efficient.frontier(er, covmat, alpha.min=-2.5, alpha.max=2.5, nport=50)
plot(ef$sd,ef$er,type="l",xlab="Standard Deviation",ylab="Expected Return",xlim=c(3.5,11),ylim=c(0,2.5),col="red",lwd=1.5)
points(sqrt(diag(covmat)),er,pch=19,col="blue")
text(sqrt(diag(covmat)),er,names(er),pos=4,col="blue",cex=.6)
polygon(u,v,border=NA,col=rgb(0,0,1,.3))

It did influence (horizontal) position of points, since variances are now different, as well as the efficient frontier, with clearly much lower variances that can be reached.

And to illustrate a last point, to illustrate the fact that we do have estimators based on observed returns, what if we had observed different ones? A way to get an idea of what might happened is to use bootstrap, e.g. of daily returns.

> covmat=Moments(as.matrix(rzoo),"CovClassic")
> er=apply(as.matrix(rzoo),2,mean)
> ef <- efficient.frontier(er, covmat, alpha.min=-2.5, alpha.max=2.5, nport=50) > a=sqrt(diag(covmat))
> b=er
> k=1
> plot(ef$sd,ef$er,type="l",xlab="Standard Deviation",ylab="Expected Return", xlim=c(3.5,11),ylim=c(0,2.5),col="white",lwd=1.5)
> polygon(u,v,border=NA,col=rgb(0,0,1,.3))
> for(i in 1:100){
+ id=sample(nrow(rzoo),replace=TRUE)
+ covmat=Moments(as.matrix(rzoo)[id,],"CovClassic")
+ er=apply(as.matrix(rzoo)[id,],2,mean)
+ points(sqrt(diag(covmat))[k],er[k],cex=.5)
+ }

or for another asset

Here is what we got on the (estimated) efficient frontier

> covmat=Moments(as.matrix(rzoo),"CovClassic")
> er=apply(as.matrix(rzoo),2,mean)
> ef <- efficient.frontier(er, covmat, alpha.min=-2.5, alpha.max=2.5, nport=50) > plot(ef$sd,ef$er,type="l",xlab="Standard Deviation",ylab="Expected Return", xlim=c(3.5,11),ylim=c(0,2.5),col="white",lwd=1.5)
> points(sqrt(diag(covmat)),er,pch=19,col="blue")
> text(sqrt(diag(covmat)),er,names(er),pos=4, col="blue",cex=.6)
> polygon(u,v,border=NA,col=rgb(0,0,1,.3))
> for(i in 1:100){
+ id=sample(nrow(rzoo),replace=TRUE)
+ covmat=Moments(as.matrix(rzoo)[id,],"CovClassic")
+ er=apply(as.matrix(rzoo)[id,],2,mean)
+ ef <- efficient.frontier(er, covmat, alpha.min=-2.5, alpha.max=2.5, nport=50)
+ lines(ef$sd,ef$er,col="red")
+ }

Thus, it is somehow rather difficult to assess wheter a portfolio is optimal, or not… At least from a statistical perspective….

Traffic Flow of Kota Kinabalu (with R)

This morning, we had our first practicals on network flows, using  an example mentioned in some papers published by Noraini Abdullah and Ting Kien Hua, max flow min cut theorem to minimize traffic congestion in Kota Kinabalu and application of the Shortest Path and Maximum Flow with Bottleneck in Traffic Flow of Kota Kinabalu. From the roads mentioned in the articles, I did try my best to locate the nodes on a map,

m=matrix(c(0,5.995910, 116.105520,
1,5.992737, 116.093718,
2,5.992066, 116.109883,
3,5.976947, 116.095760,
4,5.985766, 116.091580,
5,5.988940, 116.080112,
6,5.968318, 116.080764,
7,5.977454, 116.075460,
8,5.974226, 116.073604,
9,5.969651, 116.073753,
10,5.972341, 116.069270,
11,5.978818, 116.072880),3,12)

we can be visualized below

library(OpenStreetMap)
map = openmap(c(lat= 6.000, lon= 116.06),
c(lat= 5.960, lon= 116.12))
map=openproj(map)
plot(map)
points(t(m[3:2,]),col="black", pch=19, cex=3 )
text(t(m[3:2,]),c("s",1:10,"t"),col="white")

If the source is realistic (up north), I do not feel very confortable with the location of the sink (on the west). But let’s pretend it’s find (to do the maths, at least).

To extract information about edge capacity, on that network use the following code that will extract the three tables from the paper

library(devtools)
install_github("ropensci/tabulizer")
library(tabulizer)
location <- 'http://www.jistm.com/PDF/JISTM-2017-04-06-02.pdf'
out <- extract_tables(location)

with Windows, it seems to be necessary to download another package first

library(devtools)
install_github("ropensci/tabulizerjars")
install_github("ropensci/tabulizer")
library(tabulizer)
location <- 'http://www.jistm.com/PDF/JISTM-2017-04-06-02.pdf'
out <- extract_tables(location)

Now we can get out data frame with capacities

B1=as.data.frame(out[[2]])
B2=as.data.frame(out[[3]])
E=data.frame(from=B1[3:20,"V3"],
to=B1[3:20,"V4"])
E=E[-c(6,8),]
capacity=as.character(B2$V3[-1])
capacity[6]="843"
capacity[4]="2913"
E$capacity=as.numeric(capacity)

We can add those edges on our map (without the arrows to indicate the direction, it would be to heavy to read)

plot(map)
points(t(m[3:2,]),col="black", pch=19, cex=3 )
B=data.frame(i=as.character(c("s",paste("V",1:10,sep=""),"t")),
x=m[3,],y=m[2,])
for(i in 1:nrow(E)){
i1=which(B$i==as.character(E$from[i]))
i2=which(B$i==as.character(E$to[i]))
segments(B[i1,"x"],B[i1,"y"],B[i2,"x"],B[i2,"y"],lwd=3)
}
text(t(m[3:2,]),c("s",1:10,"t"),col="white")

To get the graph with capacities, an alternative is to use

library(igraph)
g=graph_from_data_frame(E)
E(g)$label=E$capacity
plot(g)

but it does not respect geographical locations of nodes. It can actually be done using

plot(g, layout=as.matrix(B[,c("x","y")]))

To get a better understanding of the capacities of the road, use

plot(g, layout=as.matrix(B[,c("x","y")]),
edge.width=E$capacity/200)

From that network with capacities, the goal is to determine maximum flow on that network, from the source to the sink. This can be done with R using

> (m=max_flow(graph=g, source="s", target="t"))
$value
[1] 2571

$flow
[1] 1191 1380 1422 1380 231 0 231 0 1149 1422 1149 0 0 1149 1422
[16] 1149

Our maximum flow is here 2571, which is different from was is actually claimed both in the two papers  max flow min cut theorem to… and application of the Shortest Path… (“the maximum flow for the capacitated network with 12 nodes and 16 edges of the selected scope in this study was 2598 vehicles per hour“) where there are clearly typos since values in the table and on the graph are different. Here I did use the ones from the tables.

E$flux1=m$flow
E(g)$label=E$flux1
plot(g, layout=as.matrix(B[,c("x","y")]),
edge.width=E$flux1/200)

That is nice, but rather odd. Actually, a much simpler flow can be considered, but the same global value

E$flux2=c(1422,1149,1422,1149,0,0,0,0,
1149,1422,1149,0,0,1149,1422,1149)
E(g)$label=E$flux2
plot(g, layout=as.matrix(B[,c("x","y")]),
edge.width=E$flux2/200)

Nice, isn’t it. It is actually possible to do exactly the same on another paper they have, on the same city, traffic congestion problem of road networks in Kota Kinabalu.

location <- 'http://www.worldresearchlibrary.org/up_proc/pdf/999-150486366625-30.pdf'
out <- extract_tables(location)
dim(out[[3]])
B1=as.data.frame(out[[3]])
E=data.frame(from=B1[2:61,"V2"],
to=B1[2:61,"V3"],
capacity=B1[2:61,"V4"])
E$capacity=as.numeric(
as.character(E$capacity))
library(igraph)
g=graph_from_data_frame(E)
m=max_flow(graph=g,
source="S",
target="T")
E$flux1=m$flow
E(g)$label=E$flux1
plot(g,
edge.width=E$flux1/200,
edge.arrow.size=0.15)

Here the value of the maximal flow is 4017, just as they found in the original paper

Multinomial Logit as an Iterated Logit Regression

For the second section of the course at ENSAE, yesterday, we’ve seen how to run a multinomial logistic regression model. It is simply an extension of the binomial logistic regression. But actually, it is also possible to consider iterative binomial regressions.

Consider here a response variable Y with a multinomial distribution (3 factors to have something more general than the binomial), taking values \{A,B,C\}, with respective probabilities \mathbf{p}=(p_A,p_B,p_C). Here is a code to generate some multinomial variables

msample=function(A,B,C){
Y=rep(NA,B)
for(i in 1:B){Y[i]=sample(A,size=1,prob=C[i,])}
return(Y)
}

and here is a code to generate a dataset with n rows,

generate3=function(n,x,pb=c(-2,0)){
set.seed(x)
X1=runif(n)
X2=runif(n)
X3=runif(n)
s1=pb[1]+X1+X2
s2=pb[2]-X1+X2
P1=exp(s1)/(1+exp(s1)+exp(s2))
P2=exp(s2)/(1+exp(s1)+exp(s2))
Y=msample(0:2,n,cbind(1-P1-P2,P1,P2))
df=data.frame(Y=Y,X1=X1,X2=X2,X3=X3)
return(df)
}

Let us generate a training dataset and a validation one

pb=c(.31,.42)
DF1=generate3(1000,1,pb=pb)
DF2=generate3(500,2,pb=pb)

With a multivariate logistic regression
\mathbb{P}[Y=A|\mathbf{x}]=\frac{\exp[\mathbf{x}^{\text{T}}\mathbf{\alpha}]}{1+\exp[\mathbf{x}^{\text{T}}\mathbf{\alpha}]+\exp[\mathbf{x}^{\text{T}}\mathbf{\beta}]}
\mathbb{P}[Y=B|\mathbf{x}]=\frac{\exp[\mathbf{x}^{\text{T}}\mathbf{\beta}]}{1+\exp[\mathbf{x}^{\text{T}}\mathbf{\alpha}]+\exp[\mathbf{x}^{\text{T}}\mathbf{\beta}]}
\mathbb{P}[Y=B|\mathbf{x}]=\frac{1}{1+\exp[\mathbf{x}^{\text{T}}\mathbf{\alpha}]+\exp[\mathbf{x}^{\text{T}}\mathbf{\beta}]}

For convenience, consider the most popular factor in our training dataset

modalite=names(sort(table(DF1$Y),decreasing = TRUE))

Consider a regression model on the simulated dataset (with several covariates), let us estimate it, and let us get predictions.

library(nnet)
reg=multinom(as.factor(Y) ~ ., data = DF1)
mp1=predict (reg, DF1, "probs")
mp2=predict (reg, DF2, "probs")

An alternative can be the following.
consider a first regression model on the Bernoulli variable Y_A=\mathbf{1}(Y=A). Actually, we will consider the most important factor, but for convenience, assume that it is A.
\mathbb{P}[Y_A=A|\mathbf{x}]=\frac{\exp[\mathbf{x}^{\text{T}}\mathbf{a}]}{1+\exp[\mathbf{x}^{\text{T}}\mathbf{a}]}
On our dataset, estimate that model, and get predictions. In the case where Y\neq A, define another Bernoulli variable Y_B=\mathbf{1}(Y=B|Y\neq A). We can estimate that model and derive two probabilities, \mathbb{P}(Y=B|Y\neq A) and \mathbb{P}(Y=C|Y\neq A) (the sum of the two being equal to 1). Based on those two models, it is possible to compute the three probabilities we are looking for. \mathbb{P}[Y=A] is obtained from the first model, and we can derive the other two from \mathbb{P}[Y=B|Y\neq A]\cdot\mathbb{P}[Y\neq A] and \mathbb{P}[Y=C|Y\neq A]\cdot\mathbb{P}[Y\neq A].

reg1=glm((Y==modalite[1])~.,data=DF1,family=binomial)
reg2=glm((Y==modalite[2])~.,data=DF1[-which(DF1$Y==modalite[1]),],family=binomial)
p11=predict (reg1, newdata=DF1, type="response")
p12=predict (reg2, newdata=DF1, type="response")
p21=predict (reg1, newdata=DF2, type="response")
p22=predict (reg2, newdata=DF2, type="response")
mmp1=cbind(p11,(1-p11)*p12,(1-p11)*(1-p12))
mmp2=cbind(p21,(1-p21)*p22,(1-p21)*(1-p22))
colnames(mmp1)=colnames(mmp2)=modalite

Let us compare the predicted probabilites, on the same dataset (here the training dataset)

> mmp1[1:9,c("0","1","2")]
0 1 2
1 0.19728737 0.4991805 0.3035321
2 0.17244580 0.5648537 0.2627005
3 0.19291753 0.5971058 0.2099767
4 0.09087176 0.7787304 0.1303978
5 0.23400225 0.4083022 0.3576955
6 0.18063647 0.6637352 0.1556283
7 0.13188881 0.7402710 0.1278401
8 0.13776970 0.6524959 0.2097344
9 0.12325864 0.6790336 0.1977078
> mp1[1:9,c("0","1","2")]
0 1 2
1 0.19691036 0.5022692 0.3008205
2 0.17123189 0.5680647 0.2607034
3 0.19293066 0.5984402 0.2086291
4 0.08821851 0.7813318 0.1304497
5 0.23470739 0.4109990 0.3542936
6 0.18249687 0.6602168 0.1572863
7 0.13128711 0.7400898 0.1286231
8 0.13525341 0.6553618 0.2093848
9 0.12090016 0.6815915 0.1975084

The two are very close. So yes, it is possible to see the multinomial regression as some sequential binomial regressions.