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re: Re: st: Not Quite Quadratic Regression

From   "Ariel Linden, DrPH" <>
To   <>
Subject   re: Re: st: Not Quite Quadratic Regression
Date   Sun, 5 Aug 2012 15:45:15 -0400

I agree with David's general assessment of the situation here. I would point
you to two possible options to model your data here: -mfp- and -mvrs- are
user written programs by Patrick Royston (both available through ssc). The
first program fits multivariable fractional polynomial models, and the
second fits multivariable regression spline models. 


Date: Sat, 4 Aug 2012 08:05:34 -0400
From: David Hoaglin <>
Subject: Re: st: Not Quite Quadratic Regression

If theory predicts a u-shaped relation, it may be useful to transform
y, but it probably will not make sense to transform x. Does theory
provide any guidance on the scale for y or the scale for x?  Common
transformations include log and square root.

I suppose theory does not predict the functional form of the relation.
 Since the relation is u-shaped, you may want to center x at the value
corresponding to the bottom of the U (either based on theory or based
on the data).

You may be able to use lowess to trace a fairly smooth summary curve
through your noisy scatterplot.  That may suggest a functional form.

Many nonlinear relations are not well approximated by a quadratic.
When the amount of data is large enough, I have sometimes been able to
let the data guide the choice of functional form by replacing x by a
piecewise-constant function of x with fairly narrow intervals.  In
your model, the piecewise-constant function would replace ax + bx^2.
This approach works when the model contains predictors other than x.
It may lead you to a suitable (linear) spline.

David Hoaglin

On Sat, Aug 4, 2012 at 6:56 AM, A. Shaul <> wrote:
> Hello Statalisters,
> Theory predict an u-shaped relation between two variables, y and x.
> When I perform a quadratic linear regression with a model like
>      y = ax + bx^2 + constant + error,
> the coefficients a and b are not significant. However, if I change the
> exponent to something less than 2, e.g. 1.5, I obtain significance. In
> other words a model like
>      y = ax + bx^1.5 + constant + error,
> yields significant estimates of a and b. The curvature is still quite
> marked using the exponent of 1.5. I can even use an exponent of 1.1
> and obtain significance and a nice shape. But I don't think I can
> simply choose the exponent based on whatever yields significance. Or
> can I? This is my question.
> I have tried to run a non-linear regression where the exponent was a
> free parameter. Although it tend to yield an exponent around 1 to 2,
> everything turns out highly insignificant. If I plug the estimated
> exponent into an OLS model, like the ones above, I get significance. I
> have also tried to use splines as well as a piecewise constant
> formulation. Again the results are less than ideal (although I get the
> same overall picture).
> The non-linearity is rather apparant in a scatterplot (although
> extremely noisy), and the problem shows up when controlling for other
> covariates where a simple graphical/nonparametric approach is
> unfeasible.

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