# Different error weighting for positive and negative residuals for OLS?

For OLS-estimators in multivariate regression analysis, it logically doesn't matter whether an error is positive or negative. I was wondering if in some situations it might make sense to weight a positive error less than a negative error (or maybe the other way around). So something like a intentionally optimistic/pessimistic estimation. So I tried to develop something like this myself, but since I'm not a mathematician or statistician, I didn't quite manage it and can't judge very well if something like this really makes sense, or maybe already exists. This is what i got so far:

Instead of the error being $$\begin{equation} SQR_1(\beta) = ||y-X\beta||^2 \end{equation}$$ I was thinking of something like the function: $$\begin{equation} f(x) = ax^2\cdot s(x) + bx^2\cdot s(-x) \end{equation}$$ where: $$\begin{equation} s(x) = \frac{1}{2}\cdot( sgn(x)+1) \end{equation}$$ and $$a$$ and $$b$$ are parameters weighting the positive/negative errors. $$f(x)$$ with for example $$a=0.2$$ and $$b=1$$ looks like this: Instead of $$SQR_1$$ we can now define: $$\begin{equation} SQR_2(\beta) = f(y-X\beta) \end{equation}$$

I know that I now have to solve $$\frac{\partial f(y-X\beta)}{\partial \beta} = 0$$ for $$\beta$$. And at this point, I'm not getting any further. And even if I solve this equation, wouldn't there be the next problem to prove that the extrema is a minima because the second derivative of my function is not defined for $$x=0$$?

So my questions: Do you know how to solve this equation? Do you think that anything I was doing here was useful? Does something like this already exist?

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• Yes, this makes sense. Such a weighting is routinely done in quantile regression. Note that global minima can occur where the second derivative (or even the derivative itself) is undefined: that's not an inherent problem. Checking the second derivative is merely a tool to use in some cases. For your loss functions, which are globally strictly convex, there is no problem at all: there will be a unique local minimum and it therefore is the global minimum. – whuber May 19 at 14:34