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Say one wishes - and believes - to fit a mixed effects model where time ($x_{ij}$ below) is included in a linear and quadratic term in the fixed part, but only linear in the random part (hence random intercept and slope).

$$ y_{ij} = \beta_1 + \beta_2 x_{ij} + \beta_3 x_{ij}^2 + \zeta_{1j} + \zeta_{2j}x_{ij} + \epsilon_{ij} $$

Question

Is the model mis-specified? Is it the case that the quadratic term has to be included also in the random part?

References

Looking a bit around, looks like there are mixed views:

Against: Link 1

In favor: Link 1 and also Link 3

(I'm sure there will be much more for one or the other)

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    $\begingroup$ No, it doesn't have to be. Whether it should be or not is going to be application-specific. $\endgroup$ – jbowman Jan 14 '18 at 22:18
  • $\begingroup$ I would appreciate a complete answer that justifies one or the other $\endgroup$ – Steve Jan 16 '18 at 14:06
  • $\begingroup$ If we had to include everything in the random part which we had included in the fixed part we could never have a pure random intercept model which included any covariates at all. $\endgroup$ – mdewey Jan 16 '18 at 14:31
  • $\begingroup$ I understand that. But given that we have a variable in the fixed and random part (e.g. time) and all the other covariates stay in the fixed part. Is it correct to put a quadratic term of time in the fixed part, but not also in the random part? $\endgroup$ – Steve Jan 16 '18 at 15:07
  • $\begingroup$ Sorry, Steve, but there really is no one right answer to this. It's a model, and the model can't be based on an abstract one-rule-fits-all situations when it comes down to details such as these. It's not hard to construct an example where the "best" model has the random effect term only linear and another where the "best" model has the random effect term both for the linear and quadratic part, while both have linear and quadratic fixed effects. $\endgroup$ – jbowman Jan 16 '18 at 16:13

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