Initially I modeled my objective function as

$$\arg \min \operatorname{Var}(f(x),g(x)) + \operatorname{Var}(c(x),d(x)) + \cdots$$

where $f$, $g$, $c$, $x$ are linear functions.

To be able to solve the problem with linear solvers I have change the problem as follows ($\operatorname{abs}$ stands for absolute value):

$$\arg \min\; \operatorname{abs}(f(x)-g(x)) + \operatorname{abs}(c(x)-d(x)) + \cdots$$

Is this form correct and has the same meaning as the first model?

  • 3
    $\begingroup$ You haven't defined what you mean by $\mbox{Var}(f(x),g(x))$, and there aren't any random variables or data mentioned in your question, so it's very hard to make any sense of the question. What do you mean by $\mbox{Var}$? $\endgroup$ Aug 31, 2015 at 3:19
  • $\begingroup$ I imagine you meant "absolute value" and not "abstract value". $\endgroup$
    – Yair Daon
    Aug 31, 2015 at 15:11
  • 1
    $\begingroup$ Cross-posted: Stack Overflow and Mathematics SE. Cross-posting is discouraged, since you are not respecting the time of people that answer to you in each site. $\endgroup$
    – nicoguaro
    Sep 3, 2015 at 3:42

1 Answer 1


Without knowing much about your question it is hard to answer more specifically. So first and foremost - no. These are not equivalent. Minimizing variangce will generally make you converge to some kind of sample mean, whereas minimizing absolute deviation would (generally) make you find some kind of sample median.

A generally favourable property of sample means is fast convergence (at least, when compared to median). A good property of sample medians is robustness - insensitivity to outliers.


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