## Tuesday, September 3, 2013

### 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

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

end

hist u, title("Normal Distribution") hist u, title("LaPlace Distribution") * Seems to be working well, let's simulate it 1,000 times.

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

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
```