```
* 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

## Tuesday, September 3, 2013

### When LAD is more efficient than OLS

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