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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\rbracewhich 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\rbracewith 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,

1 2 3 4 5 | 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,

1 2 3 4 5 6 7 | library(lpSolve) A1 = cbind(diag(2*n),0) A2 = cbind(diag(n), -diag(n), 1) r = lp("min", c(rep(1,2*n),0), rbind(A1, A2),c(rep(">=", 2*n), rep("=", n)), c(rep(0,2*n), y)) tail(r$solution,1) [1] 1.077415 |

It looks like it’s working well…

## Quantile

Of course, we can adapt our previous code for quantiles

1 2 3 4 | tau = .3 quantile(x,tau) 30% 0.6741586 |

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

1 2 3 4 5 6 | A1 = cbind(diag(2*n),0) A2 = cbind(diag(n), -diag(n), 1) r = lp("min", c(rep(tau,n),rep(1-tau,n),0), rbind(A1, A2),c(rep(">=", 2*n), rep("=", n)), c(rep(0,2*n), y)) tail(r$solution,1) [1] 0.6741586 |

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.

1 | 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\rbracewith 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

1 2 3 4 5 6 7 8 9 10 11 12 | 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(">=", 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

1 2 3 4 5 | 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

1 2 3 4 5 6 7 8 9 10 11 12 13 | 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(">=", 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.

1 2 3 4 5 6 7 8 9 10 11 12 | require(lpSolve) tau = .3 n=nrow(base) X = cbind( 1, base$area, base$yearc ) y = base$rent_euro A1 = cbind(diag(2*n), 0,0,0) A2 = cbind(diag(n), -diag(n), X) r = lp("min", c(rep(tau,n), rep(1-tau,n),0,0,0), rbind(A1, A2), c(rep(">=", 2*n), rep("=", n)), c(rep(0,2*n), y)) tail(r$solution,3) [1] 0.000000 3.257562 0.077501 |

Unfortunately, this time, it is not working well…

1 2 3 4 5 | 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)

1 2 3 4 5 6 7 8 | 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>0)+(1-tau)*(eps<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…

1 2 3 | 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….

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