3
$\begingroup$

The problem I encounter is the following:

Imagine a (perfect) inverted U-shaped relation between an independent variable and a dependent variable. When you look at the curve estimation there is indeed a perfect inverted U-shaped relation.

If you conduct a regression analysis you have the normal independent variable and the squared term of the independent variable. In such a case, usually the normal independent variable gives a positive beta and the independent variable squared gives a negative beta. If these are both significant, this indicates that there is an inverted U-shaped relationship. However, it is well-known that in this way you have a very high VIF value (multicolliniearity), because there is an almost perfectly positive correlation between the normal independent variable and independent variable squared (because you square it of course).

The solution for reducing the multicolliniearity is to mean center the independent variable, and after then taking the squared term of the mean centered independent variable. If i perform a regression now (with both of the two predictors mean centered) I do not have problems anymore with multicollinearity, but I have two negative beta's. How is this possible? How can i have two negative beta's while i have a perfect inverted U shaped relationship according to the curve estimation and data? Is it still possible to prove the inverted U shaped relation with two negative beta's?

Improve this question
$\endgroup$
5
  • 2
    $\begingroup$ Is y =ax$^2$ with a>0 the onlt way to get a perfect inverted U? I don't think so. If the inverted U corresponds to a different function then there will not be an a that gives a perfect fit to y=ax$^2$. Then it would be possible for some other model to fit better. $\endgroup$
    – Michael R. Chernick
    Commented Sep 23, 2012 at 16:40
  • 1
    $\begingroup$ I suspect that you squared the mean-centered x instead of mean centering the squared x. If you mean center before you square, you get a u-shaped predictor. $\endgroup$
    – Fojtasek
    Commented Sep 23, 2012 at 17:03
  • $\begingroup$ I'm not sure there's a problem here, but if you can provide your data, people may be able to say more. $\endgroup$
    – gung - Reinstate Monica
    Commented Sep 23, 2012 at 19:39
  • $\begingroup$ @Fojtasek I also tried to mean centering the squared X, then I have a positive beta for the normal independent variable and a negative beta for the squared one. That is exactly the case when there is an inverted U-shaped relation, but in that case there is a multicolliniearity problem because the VIF value is higher than 10. $\endgroup$
    – user14316
    Commented Sep 23, 2012 at 19:56
  • 1
    $\begingroup$ orm.sagepub.com/content/15/3/339.abstract To all who come across this in the future, I recommend reading Dalal & Zickar (2012) -- The "solution" of using mean centering is not as straightforward as one would think. $\endgroup$
    – user39659
    Commented Feb 7, 2014 at 0:11

1 Answer 1

Reset to default
5
$\begingroup$

Let's make up some data that has a nearly perfect inverted U:

x <- rnorm(100, 10, 2)
x2 <- x^2
centx <- x - mean(x)
centx2 <- centx^2
y <- (x-10)^2*-1 + rnorm(100)
plot(x,y)

Now, let's check the lm on the centered x and centered $x^2$

m1 <- lm(y~centx+centx2)
summary(m1)

And both have negative coefficients. So, why is there a problem?

The U shape is implied by the large coefficient on the quadratic term; its inverse shape by the negative sign.

Improve this answer
$\endgroup$
3
  • 2
    $\begingroup$ When I ran @Peter's code several times I sometimes got a pos. coeff. for centx and sometimes a neg. The fact that y ranged from about -25 to 5, whereas x was always pos., seems to help explain why the centx coeff. was sometimes neg. $\endgroup$
    – rolando2
    Commented Sep 23, 2012 at 22:38
  • 1
    $\begingroup$ @Peter Thank you very much for your answer. The confusing point was that in every academic article that I read they elaborated that there was an inverted U-shaped relation, because the normal predictor was positive and significant and the squared term was negative and significant. $\endgroup$
    – user14322
    Commented Sep 23, 2012 at 23:53
  • 1
    $\begingroup$ That is weird, and reveals that people don't know elementary algebra. $\endgroup$
    – Peter Flom
    Commented Sep 23, 2012 at 23:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.