Absolute variable as dependent variable

I have the following model: $$|X| = B_0 +B_1 \cdot y + B_2 \cdot z ,$$ where $z$ and $y$ are normally distributed random variables, and $B_1$ and $B_2$ denote the coefficients.

My dependent variable contains either the positive or negative values of $X$. The positive and negative values of $X$ together are normally distributed, with a mean and median of around zero. For statistical reasons I would like to separately test the positive and negative values. This splits the left and right side of the otherwise normally distributed variable $X$. This clearly violates the normality assumption used for linear regression models. If I want to compare groups I would use a nonparametric type of test, but in this case I want to run the equivalent of a linear regression. Is this possible?

• The normal distribution is symmetric, so you only need a solution for one case. – emcor Jun 25 '14 at 22:56
• I presume you mean with one case it is either positive or negative. – Mr. Radical Jun 25 '14 at 23:00
• It is difficult, if not impossible, to formulate a legitimate statistical model based on this information. Could you please explain the context of this question and the basis for your suppositions about $X$? – whuber Jun 26 '14 at 1:42
• "For statistical reasons I would like to separately test the positive and negative values" - what are these reasons? I'm afraid you are trying to formulate the problem that you dont need to solve – Aksakal Oct 13 '15 at 15:10
• Is it possible to restate the model? As it stands it doesn't make a lot of sense since the left side is nonnegative but the right side can be negative. – dsaxton Mar 1 '16 at 13:57

I think you can run a regression with $X_+$ and $X_-$ as independents, long as you assume a linear relationship to $Y$.
• By $X_+$ you mean a dummy variable for when $X$ is positive correct? Given the OP's comments, I would suggest a paramterization that includes the interactions with one dummy (you can't include both as the are collinear unless you do not include the intercept). – Andy W Jun 25 '14 at 23:31