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As per the heading, is it possible to add AIC to some previously computed models based on the stats I have (which include $r^2$, its p-value, $\sigma$ for each variable individually)?

They are all bivariate models, though with some calibration parameters (I know the number of parameters in each model to feed into AIC).

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(If that's not enough I also have bivariate linear regression results (parameters for best fit line, including $F$ statistic, and $\sigma$ and $t$ for the gradient and intercept); though I don't think, for a bivariate model in which I don't care about the means, that fundamentally tells me anything different does it?)

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  • $\begingroup$ Oh dear, I'd better paste that again (can't find how to edit it!) $\endgroup$ – Sideshow Bob Jun 23 '14 at 14:22
  • $\begingroup$ The current answers do not contain enough detail. Putting a bounty on this as I'd like to express AIC directly in terms of n, r2 and p(the number of parameters). Frank has given a good initial answer and what I want to know should be a simple variable substitution from that, except that it′s not clear what he means by 'degrees of freedom' − could be p or n-p. $\endgroup$ – Sideshow Bob Jun 23 '14 at 14:23
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    $\begingroup$ $AIC = n \cdot ln \left( RSS/n \right) + 2 \cdot k$ where $n$ is number of data points, $RSS$ is residual sum of squares, and $K$ is the number of parameters in the model. This assumes that the inequality $n/K < 40 $ is true, otherwise you want to use corrected AIC, or $AIC_c$. You can get RSS from ($R^2$) if you know the residual sum of squares to the mean, as well as to your fit function. $R^2=1-\frac {RSS_{fit}} {RSS_{mean}}$. Many fit programs provide RSS directly. $\endgroup$ – EngrStudent Jun 23 '14 at 14:32
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I'm not sure how to exactly get AIC (i.e., which term to add) but you can use the likelihood ratio $\chi^2$ statistic minus twice the degrees of freedom as a substitute in many cases, and $\chi^{2} = - n \times \log(1 - R^{2})$ for a Gaussian model.

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  • $\begingroup$ So you are saying $AIC \approx -n log (1-R^2) - 2(DF)$ where $DF = n - p$ and $p$ is the number of parameters in the model? $\endgroup$ – Sideshow Bob Jun 18 '14 at 12:27
  • $\begingroup$ Can this be used to compare goodness of fit across two models which use different data points? $\endgroup$ – Sideshow Bob Jun 18 '14 at 13:01
  • $\begingroup$ Hang on - doesn't low $R^2$ therefore imply low $\chi^2$ and hence low AIC? Shouldn't low $R^2$ mean a worse (higher) AIC not better? $\endgroup$ – Sideshow Bob Jun 18 '14 at 13:29
  • $\begingroup$ And do you mean $log$ or $ln$? $\endgroup$ – Sideshow Bob Jun 18 '14 at 13:31
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    $\begingroup$ You want numerator d.f. p not error d.f. n-p-1 $\endgroup$ – Frank Harrell Jun 23 '14 at 14:35

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