I have a simple model in INLA (the regressor is a single model=iid term), which reports the precision of the hyperparameter. How to convert it to variance?

Question #1: Is my understanding correct that the CI can be simply converted by taking the reciprocal of the endpoints? (As reciprocal is an invertible monotone transformation.)

Question #2: What to do with the point estimator? This is not an MLE, thus no invariance principle applies, so, if my understanding is correct, I can't simple take its reciprocal.

  • $\begingroup$ I'm guessing that INLA is doing enough approximate things (finding a MAP estimator, Laplace approximation, Wald CIs, etc.) that you should just invert. If you want you could assume that the posterior of precision is Gaussian and try to calculate the mean (or mode) of $1/\tau$ (by delta method, or ...) $\endgroup$ – Ben Bolker Jun 8 '18 at 14:40
  • $\begingroup$ @BenBolker Thank you very much! Intuitively I totally understand your logic (of simply taking the reciprocal), but do you have anything written to substantiate this, that I could cite in a paper...? $\endgroup$ – Tamas Ferenci Jun 9 '18 at 8:34
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    $\begingroup$ Unfortunately not. Maybe you can ask on the INLA mailing list (available via the INLA web page ? $\endgroup$ – Ben Bolker Jun 9 '18 at 12:09
  • $\begingroup$ @BenBolker Thanks for the suggestion, I received an extremely quick answer from them: r-inla.org/… . (I don't know what is the Stackexchange policy in this case; is it enough if I leave the link here among the comments...?) $\endgroup$ – Tamas Ferenci Jul 24 '18 at 17:56
  • $\begingroup$ If you got a useful answer from them, you should post it (or at least the highlights) here, with appropriate attribution, as an answer. Comments are considered ephemeral, off-site links are semi-ephemeral. $\endgroup$ – Ben Bolker Jul 25 '18 at 1:56

Following the suggestion of @BenBolker, I've asked this on the R-INLA mailing list, where I received an extremely rapid and detailed response from Håvard Rue.

Briefly, log precision should be extracted and then transformed:

 m <- fit$internal.marginals.hyperpar[[1]]
 m.var <- inla.tmarginal(function(x) 1/exp(x), m)

where fit is the fitted model.

Again, credit goes to Håvard Rue.


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