```Date: Nov 26, 2012 3:20 AM
Author: Ray Koopman
Subject: Re: Interpretation of coefficients in multiple regressions which<br> model linear dependence on an IV

On Nov 21, 8:13 pm, djh <halitsk...@att.net> wrote:> In a different thread, Ray Koopman explained that if one> suspects these regressions to be dependent on the IV ?u?:>> c on u> c on e> c on (e,u)>> then under the usual initial assumption that the dependence> is linear, these three regressions should be modified to:>> c on (u, u^2)            instead of c on u> c on (e, u, u*e)         instead of c on e> c on (e, u, u*e, u^2)    instead of c on (e,u)In this post I want to talk about only your first case, where thed.v. is a quadratic function of the i.v.: y = a0 + a1*x + a2*x^2.(I use the usual generic variable names x and y so as not to getcaught up in any peculiarities of your particular variables.)First, be aware that this is called "linear regression" by somepeople, and "nonlinear regression" by others. Both are right:y is linear in the coefficients a0,a1,a2 but quadratic in x. Inmy experience, statisticians will usually say "linear" (presumablybecause estimating the coefficients is simple when y is linear inthe coefficients but complicated otherwise), but users/consumersof the results will usually say "nonlinear" (presumably because theyare looking at, or at least thinking of, a plot of the regressionfunction).I suggested the quadratic function as a replacement for thepiecewise linear function:    |b0 + b1*x, x <= x'y = |                   , for some pre-specified value x'.    |c0 + c1*x, x >  x'Note that that function is not realistic, in the sense that itallows a discontinuity at x'. A more realistic function would be    |d0 + d1*(x-x'), x <= x'y = |                        .    |d0 + d2*(x-x'), x >  x'But that function, too, is unrealistic because its gradient can bediscontinuous. The quadratic has neither of those problems but isusually criticized because its slope is not bounded, so choosingbetween the two reduces to deciding which is less unrealistic, withneither being completely satisfactory. (There are, of course, otherpossible functions, but they are all more complicated.)The general question is whether the slope of the regression isdifferent for low and high x-values, and (more importantly?) howthat difference varies as a function of other factors. That is,you have many regressions to do, and the results must be comparedto one another.Your current analyses look at the slope differences, c1-b1.(The intercepts and their differences are generally meaninglesswhen the slopes differ.)a2 in the quadratic model carries the same general meaning as c1-b1.To see this, think of a linear function y = A0 + A1*x  in which boththe intercept (A0) and the slope (A1) are themselves linear functionsof x: A0 = a00 + a01*x, A1 = a10 + a11*x. Clearly, a11 specifies howthe slope A1 changes as a function of x. Substituting for A0 and A1gives the quadratic model, with a2 = a11:y = (    A0     ) + (    A1     )*x  = (a00 + a01*x) + (a10 + a11*x)*x  = a00 + (a01 + a10)*x + a11*x^2  = a0  + (    a1   )*x + a2 *x^2.As with the piecewise-linear model, the intercept (a0) in thequadratic model is generally meaningless.Now for what to many people seems surprising and counterintuitive,even just plain wrong: in the quadratic model, a1 is generallymeaningless. This is because a1 is only the slope of the function atx = 0, whereas many people think of a1 as some sort of conceptual,if not literal, "average" slope. If a single number is wanted thatrepresents some sort of average slope then it can be computed, butfirst that "average" must be defined.
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