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deasmhumnha
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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.

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".

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.

Source Link
deasmhumnha
  • 1.1k
  • 5
  • 10

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".