2
$\begingroup$

It is well known that the AIC can be used to compare nested models.

Additionally, I believe I am correct in saying that you can also use the AIC to compare non-nested models on the same dataset (please correct me if I am in fact wrong). However, it is not correct to use the AIC to compare between different data sets.

In my scenario, I have 5 measurements on individuals over time plus an outcome variable. If I was to regress the outcome variable linearly on all the measurements over time, I would be able to obtain an AIC for this model which uses the entire dataset.

Now I want to consider only 2 of the measurements of all individuals plus the outcome variable. Technically, I am now using a subset of the original dataset as I have lost information on the other three measurements. However, isn't this the same as fitting a nested model since I kept all individuals but lost 3 explanatory variables? So is it justified to compare the AIC I get from this model with that obtained from the full model?

$\endgroup$
1
  • 1
    $\begingroup$ Check out AICc, which is probably relevant. $\endgroup$
    – Zhubarb
    Nov 6, 2013 at 11:23

2 Answers 2

3
$\begingroup$

To you, all of the predictors in the model are data so, from that perspective, data is changing. But the data that needs to be consistent across models for AIC is the outcome or response variable. As long as that is the same data in each comparison you can use AIC. Search AIC model selection on this site. This has been discussed several times.

$\endgroup$
1
$\begingroup$

@John is correct but I think not quite explicit enough, perhaps leading to confusion (maybe not now, but if someone else reads this thread later).

AIC needs to be on the same data set: That is, it has to have the same subjects. And it needs to have the same dependent variable (John's point). To be nested, one set of independent variables must be a subset of the other, but AIC can work on non-nested subsets (at least, according to some people see this thread).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.