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.


One thought on “Classification from scratch, logistic regression 1/8”

  1. Hi, great post thank’s for sharing.
    My comment (or question) may be a little out scope here but still… I was wondering about colinearity problems and their diagnostic for logistic regression . For linear models, Variance Inflation Factor (VIF) can be used and they are well studied. I have seen here and there people using them in logistic regression estimation contexts, but I was rather unsatisfied because I couldn’t find any single convincing academic study on the properties of VIF in this context. Are you aware of any reference on this matter ?
    Best regards

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