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I am running a model selection analysis with a continuous dependent variable and a variety of continuous and categorical explanatory variables. For two of my continuous explanatory variables I am fitting curvature terms as it looks like there is a quadratic relationship between them and the dependent variable.

When I run the model selection analysis using MuMIn in R, I get a variety of models out, some of which contain only the quadratic term, and not the lower order associated linear term, in them. In my head this seems mathematically incorrect - is the linear term not essential when fitting a higher order polynomial (unless that linear term = 0...)?

Is there anyway to get around this issue other than carrying out the model selection by hand (pretty impossible for me since I am trying to fit 24 parameters)? Can I tell R not to include any quadratic term in a model without its associated linear term?

Sarah
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  • Why do you believe a quadratic polynomial has to include the linear term? I don't see a reason for this. – Roland Mar 11 '13 at 16:33
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    I was basing it on this thread: http://stats.stackexchange.com/questions/28730/does-it-make-sense-to-add-a-quadratic-term-but-not-the-linear-term-to-a-model but perhaps this is not relevant to my models here? I may be getting confused - apologies – Sarah Mar 11 '13 at 16:47
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    Generally speaking you're correct - you normally want to include lower order terms if higher order terms are in a model. I can't speak to what MuMIn might be doing though. – Glen_b Mar 11 '13 at 23:17

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No, there isn't really a way to do it. Of course any model which contains a quadratic and not the linear term is statistically invalid. Also, any model which includes any interactions and not the individual linear terms of the interactions is also invalid. If you do an AIC or adjusted R2, R will print out dozens of models as the 'best' model. You'll have to go down and find the first one that is also statistically valid. Perhaps a better method would be to run a 'full model' regression to see which variables

Eric Peterson
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  • Thanks Eric. This being the case, I thought I might also run a stepwise deletion analysis to provide a second check of what might be the important variables - then hopefully I won't miss anything crucial. – Sarah Mar 11 '13 at 19:11
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    (1) Did part of your answer get cut off? (2) I believe the thread Sarah references disputes your "of course," Eric. (We should also wonder what it means to be "statistically invalid" in the first place.) Exceptions to your "also" have also been discussed. – whuber Mar 11 '13 at 19:21
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    Wether a model containing a quadratic term, but not the linear term is valid depends on what you are modelling. I can easily imagine data (e.g., from kinematic experiments) where you know that the coefficient of the linear term is zero. – Roland Mar 12 '13 at 08:07
  • If the linear term is zero then sure, I wouldn't include it, but for my data I don't think this is the case (it's messy biological data ;-) ). – Sarah Mar 12 '13 at 10:23
  • Then how do you know that a quadratic relationship is appropriate and not, e.g., an exponential or Michaelis-Menten relationship, which are more common in biological systems? – Roland Mar 12 '13 at 10:28
  • Because I looked at the distriibution of the data plotted against the dependent variable and also fitted a gam with a non-parametric smoother to get a feel for the fit. Fr two of the continuous explanatory variables they look distinctly quadratic (one is a negative quadratic and one positive). I could try other fits also I suppose, though I don't think Michaelis Menten is applicable as the fit wouldn't be right in this case... – Sarah Mar 12 '13 at 10:47