Quadratic term when the level variable has both positive and negative values

[bhobdari]

Hi All,

We are estimating a regression of Y on X where X takes both positive and negative values. To account for nonlinearities in the relationship, we include X-squared as right hand side variable. By construction X-squared will have only positive values. My dilema is whether this changes the type of the relationship between Y and X we are trying to model?

Best regards,
Bersant

[Joseph Coveney]

Well, let's see . . .

version 15.0

clear *
set seed 1410833
quietly set obs 25
generate double x = runiform() - 0.5
generate double y = x + x * x + rnormal()

summarize x, meanonly
generate double x_pos = x - r(min) + 1

graph twoway qfit y x || qfit y x_pos

exit

[Nick Cox]

I don't understand the question. Clearly you can fit a quadratic or a linear term alone regardless of the sign of the predictor. As Joseph implies, you need to plot the data and the fit to see whether the quadratic is a good idea.

[River Huang]

The fact that X takes both positive and negative values does not justify the use of quadratic specification. I doubt that bhobdari mis-understand the ideas of linear vs quadratic specifications.

[Nick Cox]

River: I see no problem in principle in fitting a quadratic here. Much depends on the definition of the x variable and whether y should behave smoothly for negative, zero and positive x.

[Maarten Buis]

Rather then it being a problem, it can actually be desirable to change X such that it has both negative and positive numbers. Take year of birth, which typically takes quite high values (1976 to take a not very random example). If you take squares of those fairly large numbers, then x and its square will be near perfectly correlated. Say our year of birth variable ranges from 1950 to 2000, then you will get twoway function y = x^2, range(1950 2000). The correlation will be so strong that it may cause problems estimating the individual effects of x and x squared. We could (and should) center our variable at something like 1975, then the new x and its square will be almost uncorrelated (twoway function y = x^2, range(-25 25)), and their separate effects will be much easier to estimate.

[Nick Cox]

Maarten's point is taken in various directions within

SJ-15-2 st0394  . . . . . . . . . . . . . .  Speaking Stata: Species of origin
        . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  N. J. Cox
        Q2/15   SJ 15(2):574--587                                (no commands)
        explores origins (some fixed and natural, others just
        conventional and sometimes not even convenient) and considers
        how best to model simple trends and seasonal periodicities as
        well as defining noncalendar years

[bhobdari]

Thank you all for the input provided so far. I should have been more clear what I am modeling as my concern is more conceptual than statistical. I am modeling Cummulative Abnormal Returns (CARs) as a function of changes in size of blockholding by several groups (families, banks, mutual funds). Over time blockholding increases or decreases, leading to X variable take both positive and negative values. When taking the square of X the negative values become positive. So, I am concerned of two things: (1) whether the relationship I am modeling is still the same, and (2) whether this implies anything about statistical significance (the coefficient of X-squared is insignificant although the scatter plot of Y on X is by far strictly linear).

[Maarten Buis]

(1) whether the relationship I am modeling is still the same

The sign of an effect is not the same as the sign of a variable. If you add a square of a variable \(x\), then the sign of the coefficient of \(x\) will tell you the direction of the effect when \(x\) is 0. The sign of the coefficient of \x^2\) tells you whether that effect increases or decreases as \(x\) increases. For example, when the coefficient of \(x\) is positive, then \(y\) will increase when \(x\) increases when \(x\) is 0. When the coefficient of \(x^2\) is negative, then the effect of \(x\) will decrease when \(x\) increases, and eventually that effect will become negative. So in this case the effect will be concave. So it is irrelevant for the direction of the effect that squaring turns all values of the variable positive.

(2) whether this implies anything about statistical significance (the coefficient of X-squared is insignificant although the scatter plot of Y on X is by far strictly linear).

Again this has nothing to do with the sign of the values on the squared term. A quadratic is not the only way in which a relationship can be non-linear, so it is perfectly possible for a relationship to be non-linear and the quadratic term to add nothing to the model. You may want to look at help fp

[bhobdari]

Thank you Maarten for the thoughtful response and the suggestion to look at fp.