When LAD is more efficient than OLS

* There are many potential cases when LAD (least absolute deviations) is a more e


fficient estimators of the linear model than OLS (ordinary least squares).

* These cases are characterized by data in which tails of the errors tend to be wide.

* With errors distributed normally we know that OLS is the most efficient estimator.

* Let's define a small command to show us the relative estimator efficiencies
cap program drop OLSLAD
program define OLSLAD, rclass

  clear
  set obs `1'  // set the observations equal to the first argument 

  if "`2'"=="norm" gen u = rnormal()*5
  
  * Since Stata does not have a built in LaPlace distribution
  * I will simulate one by solving for the probability
  * and generating the error term from an inversion of a uniform draw.
  if "`2'"=="laplace" {
    gen u0 = runiform()
 gen u1 = rbinomial(1,.5)
    gen u = (-1)^u1 * ln(2*u0) * 6
  }

  gen x = rnormal()

  gen y = x*1 + u

  reg y x
  return scalar OLSb = _b[x]
  
  qreg y x
  return scalar LADb = _b[x]
  
end

OLSLAD 10000 "norm"
hist u, title("Normal Distribution")


OLSLAD 10000 "laplace"
hist u, title("LaPlace Distribution")


* Seems to be working well, let's simulate it 1,000 times.

simulate OLSb=r(OLSb) LADb=r(LADb), rep(1000): OLSLAD 1000 "norm"

sum
* We can see that both models produce unbiased estimators though OLS
* has a standard deviation which is about 3/4 the size of LAD

simulate OLSb=r(OLSb) LADb=r(LADb), rep(1000): OLSLAD 1000 "laplace"
sum
* Again, both models are unbiased yet now LAD is slightly more efficient
* than OLS. As sample size increases LAD will gain efficiency.  This
* is the result of LAD being the MLE estimator of a linear model when
* the distribution of errors follows a LaPlace.

* It is easy to show that a LaPlace distribution is maximized when
* the estimator choice is LAD.

Formatted By Econometrics by Simulation