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I have a linear mixed effect model and I add the quadratic term of time in my model and it was significant and improve the AIC & BIC of the model, but the problem is that the linear term of time become non significant.My questions are that

  1. Should I add the linear term into the model?
  2. If not, how to interpret the coefficient of I(time^2)? I have already read the Does it make sense to add a quadratic term but not the linear term to a model? but I don`t understand how to check that the global extremum occurs at x=0 (time=0) or not?
Saba S s
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  • Looks as if this may be a problem about throwing a ball into the air or dropping a ball from a balcony. Too sketchy as it is. Please edit your question to give some details of the situation you're modeling. Maybe, this Q & A is relevant. If you're just modeling acceleration and there's no linear term, then the coefficient of $t^2$ might have to do with determining acceleration due to gravity. – BruceET Jun 08 '19 at 20:29

1 Answers1

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Without knowing more about your data and seeing some regression output, it is difficult to say whether you should drop the linear term, retaining only the quadratic term. However, there is no reason to deprecate this idea immediately.

If you have predictor variables $x, x^2,$ and $x^3,$ then you may have co-linearity problems. In the following extremely simple example Minitab output is as follows:

Model Summary

       S     R-sq  R-sq(adj)  R-sq(pred)
0.236001  100.00%    100.00%     100.00%

Coefficients

Term         Coef  SE Coef  T-Value  P-Value      VIF
Constant   12.335    0.510    24.17    0.000
x1         -2.269    0.418    -5.42    0.003   188.59
x2         0.5383   0.0947     5.68    0.002  1016.06
x3        0.36447  0.00625    58.31    0.000   366.71

Regression Equation

y = 12.335 - 2.269 x1 + 0.5383 x2 + 0.36447 x3

A matrix plot of all four variables is as follows:

enter image description here

One may conclude it is best to use only $x^3$ as a predictor variable:

Regression Analysis: y versus x3 

Analysis of Variance

Source      DF   Adj SS   Adj MS    F-Value  P-Value
Regression   1  83744.0  83744.0  278128.45    0.000
  x3         1  83744.0  83744.0  278128.45    0.000
Error        7      2.1      0.3
Total        8  83746.1

Model Summary

       S     R-sq  R-sq(adj)  R-sq(pred)
0.548724  100.00%    100.00%      99.99%

Coefficients

Term          Coef   SE Coef  T-Value  P-Value   VIF
Constant     9.984     0.250    39.90    0.000
x3        0.400237  0.000759   527.38    0.000  1.00

Regression Equation

y = 9.984 + 0.400237 x3

In view of these results, it seems difficult to argue with using $x^3$ alone in a linear regression. So, this is an indication that there are situations in polynomial regression, in which it makes sense to drop lower powers of the $x_i.$

(Confession: These are fake data generated according to $Y_i = 10 + 0.4x_i^3 + e_i,$ where $e_i$ are normal with 0 mean and small variance. The rough idea is that $Y_i$ is the weight of a full cubical container of a certain kind and $x_i$ is the length of one of its sides.)

BruceET
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