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A conjecture that I have heard is this:

When you add a quadratic or a cubic term to a linear latent growth curve model, the fit will always improve.

Is this correct? Why/why not?

As I have had the idea explained to me, adding additional terms will capture a additional portion of the variance and should therefore increase the amount of variance explained, therefore increasing the goodness of fit.

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  • $\begingroup$ I don't have experience with Latent Growth Models but the Wikipedia article says that Models with higher order components, e.g., quadratic, cubic, do not predict ever-increasing variance, but require more than two occasions of measurement. $\endgroup$
    – steadyfish
    Feb 27, 2013 at 6:25
  • $\begingroup$ This refers to something different, meaning that models with higher order components do not assume ever-increasing variance. $\endgroup$
    – histelheim
    Feb 27, 2013 at 13:35

1 Answer 1

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It would be possible to create a dataset in such a way that the residuals from the linear model are orthogonal to the quadratic and cubic terms and in that case adding the terms would not change the fit to the model. However the probability of getting such a dataset in the real world is close enough to 0 that you will not likely ever see this happen with a real dataset.

Adding additional terms (whether polynomial, additional variables, randomly generated noise) will never cause the raw measures of fit (such as SSE or R-squared) to decrease because you could always get the same residuals and measure of fit by assigning a coefficient of 0 to the added terms. And if there is any possible value of the coefficient that gives a better fit (which will happen by chance most of the time) then that better value will be chosen.

So for real data (and even most simulated datasets) then adding any additional terms to the model will improve raw measures of fit. Adjusted/Penalized measures of fit such as adjusted R-squared, AIC, and BIC were developed to counter this idea. They try to penalize adding complexity to a model so that if the raw fit increases only due to chance the penalty counters that and the overall measure does not increase (and could possibly decrease).

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  • $\begingroup$ This is a good answer, but it still skirts around the core reason as to why fit always improves - is it because of the purported reason I stated in the end of my answer? $\endgroup$
    – histelheim
    Feb 27, 2013 at 1:03

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