# Quantile Regression (home made, part 2)

**R-english – Freakonometrics**, and kindly contributed to R-bloggers]. (You can report issue about the content on this page here)

Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.

A few months ago, I posted a note with some home made codes for quantile regression… there was something odd on the output, but it was because there was a (small) mathematical problem in my equation. So since I should teach those tomorrow, let me fix them.

## 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 \(y_i-\mu=a_i-b_i\), \(\forall i=1,\cdots,n\). Heuristically, the idea is to write \(y_i=\mu+\varepsilon_i\), and then define \(a_i\)‘s and \(b_i\)‘s so that \(\varepsilon_i=a_i-b_i\) and \(|\varepsilon_i|=a_i+b_i\), i.e. \(a_i=(\varepsilon_i)_+=\max\lbrace0,\varepsilon_i\rbrace=|\varepsilon|\cdot\boldsymbol{1}_{\varepsilon_i>0}\)and\(b_i=(-\varepsilon_i)_+=\max\lbrace0,-\varepsilon_i\rbrace=|\varepsilon|\cdot\boldsymbol{1}_{\varepsilon_i<0}[/latex]denote respectively the positive and the negative parts.

Unfortunately (that was the error in my previous post), the expression of linear programs is[latex display="true"]\min_{\mathbf{z}}\left\lbrace\boldsymbol{c}^\top\mathbf{z}\right\rbrace\text{ s.t. }\boldsymbol{A}\mathbf{z}=\boldsymbol{b},\mathbf{z}\geq\boldsymbol{0}\)In the equation above, with the \(a_i\)‘s and \(b_i\)‘s, we’re not far away. Except that we have \(\mu\in\mathbb{R}\), while it should be positive. So similarly, set \(\mu=\mu^+-\mu^-\) where \(\mu^+=(\mu)_+\) and \(\mu^-=(-\mu)_+\).

Thus, let\(\mathbf{z}=\big(\mu^+;\mu^-;\boldsymbol{a},\boldsymbol{b}\big)^\top\in\mathbb{R}_+^{2n+2}\)and then write the constraint as \(\boldsymbol{A}\mathbf{z}=\boldsymbol{b}\) with \(\boldsymbol{b}=\boldsymbol{y}\) and \(\boldsymbol{A}=\big[\boldsymbol{1}_n;-\boldsymbol{1}_n;\mathbb{I}_n;-\mathbb{I}_n\big]\)And for the objective function\(\boldsymbol{c}=\big(\boldsymbol{0},\boldsymbol{1}_n,-\boldsymbol{1}_n\big)^\top\in\mathbb{R}_+^{2n+2}\)

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,

library(lpSolve) X = rep(1,n) A = cbind(X, -X, diag(n), -diag(n)) b = y c = c(rep(0,2), rep(1,n),rep(1,n)) equal_type = rep("=", n) r = lp("min", c,A,equal_type,b) head(r$solution,1) [1] 1.077415

It looks like it’s working well…

## Quantile

Of course, we can adapt our previous code for quantiles

tau = .3 quantile(y,tau) 30% 0.6741586

The linear program is now\(\min_{q^+,q^-,\mathbf{a},\mathbf{b}}\left\lbrace\sum_{i=1}^n\tau a_i+(1-\tau)b_i\right\rbrace\)with \(a_i,b_i,q^+,q^-\geq 0\) and \(y_i=q^+-q^-+a_i-b_i\), \(\forall i=1,\cdots,n\). The R code is now

c = c(rep(0,2), tau*rep(1,n),(1-tau)*rep(1,n)) r = lp("min", c,A,equal_type,b) head(r$solution,1) [1] 0.6741586

So far so good…

## Quantile Regression

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\(\min_{\boldsymbol{\beta}^+,\boldsymbol{\beta}^-,\mathbf{a},\mathbf{b}}\left\lbrace\sum_{i=1}^n\tau a_i+(1-\tau)b_i\right\rbrace\)with \(a_i,b_i\geq 0\) and \(y_i=\boldsymbol{x}^\top[\boldsymbol{\beta}^+-\boldsymbol{\beta}^-]+a_i-b_i\)\(\forall i=1,\cdots,n\) and \(\beta_j^+,\beta_j^-\geq 0\) \(\forall j=0,\cdots,k\). So use here

require(lpSolve) tau = .3 n=nrow(base) X = cbind( 1, base$area) y = base$rent_euro K = ncol(X) N = nrow(X) A = cbind(X,-X,diag(N),-diag(N)) c = c(rep(0,2*ncol(X)),tau*rep(1,N),(1-tau)*rep(1,N)) b = base$rent_euro const_type = rep("=",N) r = lp("min",c,A,const_type,b) beta = r$sol[1:K] - r$sol[(1:K+K)] beta [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,A,const_type,b) 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")

And we can adapt the later to multiple regressions, of course,

X = cbind(1,base$area,base$yearc) K = ncol(X) N = nrow(X) A = cbind(X,-X,diag(N),-diag(N)) c = c(rep(0,2*ncol(X)),tau*rep(1,N),(1-tau)*rep(1,N)) b = base$rent_euro const_type = rep("=",N) r = lp("min",c,A,const_type,b) beta = r$sol[1:K] - r$sol[(1:K+K)] beta [1] -5542.503252 3.978135 2.887234

to be compared with

library(quantreg) rq(rent_euro~ area + yearc, tau=tau, data=base) Coefficients: (Intercept) area yearc -5542.503252 3.978135 2.887234 Degrees of freedom: 4571 total; 4568 residual

**leave a comment**for the author, please follow the link and comment on their blog:

**R-english – Freakonometrics**.

R-bloggers.com offers

**daily e-mail updates**about R news and tutorials about learning R and many other topics. Click here if you're looking to post or find an R/data-science job.

Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.