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I am performing a generalized least squares regression based on a design matrix $X$, a response vector $Y$ and a (non-diagonal) covariance matrix $C$, assuming Gaussian errors. I'm not sure what goodness-of-fit tests are applicable. As a first step I could go for a simple chi-squared approach, using the usual formula $\chi^2 = r^T C^{-1} r$, where $r$ is the vector of residuals, but:

  1. Is this formula applicable to the case of a non-diagonal matrix $C$?
  2. Can the generated $\chi^2$ statistic be used in the same way as if $C$ were diagonal (e.g., the weighted least squares case)
  3. Are there better goodness-of-fit statistics that are more frequently used for a problem like this one?
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  • $\begingroup$ How do you know $C$? $\endgroup$ Nov 12 '20 at 16:31
  • $\begingroup$ Known analytical uncertainties + classical error propagation rules. But I was asking in the general case. $\endgroup$
    – Mathieu
    Jul 18 at 16:50
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If the residuals, $r$ have a known population covariance matrix $C$ then $r^TC^{-1}r\sim \chi_n^2$ as long as $r\sim \mathcal{N}(0, C)$, regardless of the structure of $C$. For a proof, Google "quadratic forms of random variables". So it has the exact same meaning as goodness-of-fit for weighted least squares and I don't see anything reason why you couldn't use it in a similar way. However, in my experience statistical packages will give you coefficient of determination as a goodness-of-fit statistic.

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  • $\begingroup$ Won't the coefficient of determination will be the same if I multiply $C$ by an arbitrary factor? My initial choice of $\chi^2$ was motivated by the need to estimate whether my assigned uncertainties ($C$) are sufficient to explain the fit residuals $r$. It seems to me that I'm losing that piece of information if I use $R^2$. $\endgroup$
    – Mathieu
    Mar 23 '18 at 11:05
  • $\begingroup$ Yes, but note that the GLS estimator is unchanged by scaling of C. The relative structure of C is what provides improved efficiency to GLS, not the absolute scale, just as the magnitude of the variance in LLS is unimportant as long as it is approximately constant. $\endgroup$ Mar 23 '18 at 13:03
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Use directed tests/graphical diagnostics:

  • For linearity, expand into a spline function
  • For additivity, test interaction terms
  • For correlation structure, construct a semivariogram and compare it to the theoretical semivariogram you are assuming

These are exemplified in the longitudinal modeling chapter of RMS for GLS. In the RMS course notes you'll also see a flexible Markov model which allows for a richer correlation structure.

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