Not resolved as far as I know. And there is another issue I should have noticed before: the log likelihood is wrong also. The "mean" of minus the log likelihood is used (i.e., the log of the likelihood for the data and proposed model is divided by the sample size). The ranking of models isn't affected but if one attempts to apply the common threshold of "2 AIC units" suggesting a very different model, well, that won't work with the AIC values produced by FindDistribution.
I'll write up a summary and send to Wolfram, Inc., and report back. Below working example:
(* Generate data and find the 5 best fitting distributions *)
n = 100; (* Sample size *)
SeedRandom[12345];
data = RandomVariate[ExponentialDistribution[1], n];
nbest = 5;
TableForm[(fd = FindDistribution[data, nbest, {"LogLikelihood", "AIC"}]),
TableHeadings -> {("Distribution " <> # &) /@ (ToString[#] & /@ Range[nbest]), {"Distribution", "log(L) and AIC"}}]
(* Log of the likelihood as calculated by FindDistribution *)
Column[{Style["Log of the likelihood\n", 18, Bold],
TableForm[Table[{fd[[i, 2, 1]], LogLikelihood[fd[[i, 1]], data]/n,
LogLikelihood[fd[[i, 1]], data]}, {i, nbest}],
TableHeadings -> {("Distribution " <> # &) /@ (ToString[#] & /@ Range[nbest]),
{"\nFrom\nFindDistribution", "Duplicating what\nFindDistribution\ndoes",
"What the\nlog likelihood\nshould be"}}]}]
(* AIC as calculated by FindDistribution *)
Column[{Style["AIC\n", 18, Bold], TableForm[Table[{fd[[i, 2, 2]],
2 LogLikelihood[fd[[i, 1]], data]/n - 2 k[[i]]/(n - k[[i]] - 1),
-2 LogLikelihood[fd[[i, 1]], data] + 2 k[[i]]}, {i, nbest}],
TableHeadings -> {("Distribution " <> # &) /@ (ToString[#] & /@ Range[nbest]),
{"\nFrom\nFindDistribution", "Duplicating what\nFindDistribution\ndoes",
"\nWhat AIC\nshould be"}}]}]
(Note the sign of AIC is arbitrary. A common approach is to choose the "smaller is better" approach which is what I used for the "What AIC should be" column.)
