Minimize parameter for Least Absolute Deviation LAD

AI Thread Summary
The discussion focuses on computing the parameter β for the Least Absolute Deviation (LAD) method, highlighting that there is no closed-form solution unlike least squares. Various software tools like R and SAS employ different methods for this calculation, and users are encouraged to consult the documentation for specific guidance. A common point of confusion is clarified: while minimizing the absolute differences from a constant yields the sample median, this does not apply to regression scenarios. The estimates in LAD are not simply the medians of the x values, which would allow for a closed-form solution. Understanding these distinctions is crucial for correctly applying LAD in regression analysis.
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How to compute \beta = arg min_\beta \sum_{i=1}^N {|y_i - x_i^T \beta|
 
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There is no closed-form solution for this (contrary to the situation with least squares). The software you use (R, SAS, etc) use a variety of methods. check the relevant documentation for those programs.
 
statdad said:
There is no closed-form solution for this (contrary to the situation with least squares). The software you use (R, SAS, etc) use a variety of methods. check the relevant documentation for those programs.

I was just interested in the calculation not in applying it to real data.
The estimate is the median of the data x1,...,xn and I wanted to see how they derived that result.
 
No, the estimates are not the medians of the x values - if that were the case we would have a "closed form" method of calculation.

Where you may be confused is this: if you want to minimize

<br /> \sum_{i=1}^n |x_i - a|<br />

as a function of a, the solution is the sample median. This does not generalize to regression.
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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