If $\text{RMSEA} = 0$, it means $\chi^2 < df$.

Does it disqualify RMSEA as a criterion to evaluate the model fit, or is it just the explanation why it is zero?


1 Answer 1


$\chi^2_p < p$ is not such an unusual phenomenon: it should be happening about half of the time with perfectly specified models (including the multivariate normality distributional assumptions). RMSEA is a monotone transformation of non-centrality:

$$ {\rm RMSEA} = \sqrt{\max\biggl(\frac T{(N-1)p},0\biggr)}=\sqrt{\hat\lambda/p} $$

where $\hat\lambda$ is the estimated normalized (per observation) non-centrality. I take the latter to be a more fundamental property of an SEM than RMSEA (the latter is prone to weird inconsistencies, see Chen, Curran, Bollen, Kirby and Paxton (2008), SocMethodRes). For these perfectly specified models, the population non-centrality is zero, and thus the estimate coincides with the population value in the samples in which chi-square is less than the degrees of freedom (which, if you think of it, is a pretty rare property: when you estimate a mean or a regression coefficient, the probability that the estimate is equal to the population quantity is zero).

BTW, Kris Preacher has a neat RMSEA calculator on this website, see http://www.quantpsy.org/rmsea/rmsea.htm.


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