# Can Variance be replaced by absolute value in this optimization problem

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?

• 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 $\mbox{Var}$? Aug 31 '15 at 3:19
• I imagine you meant "absolute value" and not "abstract value". Aug 31 '15 at 15:11
• 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. Sep 3 '15 at 3:42