I am posting this, hoping that it will also be useful to others. see also Fitting t-distribution in R: scaling parameter .
my data series x is fat-tailed, 1063 obs. it seems straightforward to determine whether I can reject a cauchy:
> library(fitdistrplus)
> ca <- fitdistr(x, "cauchy")
> summary(ca)
Fitting of the distribution ' cauchy ' by maximum likelihood
Parameters :
estimate Std. Error
location 0.01140 0.001277
scale 0.02612 0.001031
Loglikelihood: 1588 AIC: -3172 BIC: -3162
Goodness-of-fit statistics
1-mle-cauchy
Kolmogorov-Smirnov statistic 0.07703
Cramer-von Mises statistic 1.21856
Anderson-Darling statistic 12.35285
Goodness-of-fit criteria
1-mle-cauchy
Aikake's Information Criterion -3172
I would like to know whether the data can reject the null hypothesis that x was drawn from the cauchy. It would be nice if fitdist gave an interpretation of the statistics itself, but I think I can explain it. I just need to compare, e.g., my K-S of 0.077 to the K-S critical value of 1.63/sqrt(length(x))=0.05. we can reject the null that the data is from a cauchy.
now, the cauchy and normal are both special cases of the Student-T. I would like to see where in the family of student-t distributions x would be, and how well freeing this parameter helps compared with the cauchy. alas, fitdistr first needs a df provided as a parameter. I can match the kurtosis of x to determine the best match. a student-T has a kurtosis of 6/(df-4), so for my x with kurtosis of 10.7, my df should be about 4.5. and now,
t4 <- fitdistr(x, "t", df=4.5)
which for me gives a t4$loklik of about 1699.
the coup de grace should be to test how good this freed-up df performs vs. the restricted cauchy with something like
library(lmtest)
lrtest( ca, t4 )
i.e., something like 2*log-LR compared to the Chi-sq. with a difference of 111, and we are way above the 10.8 critical level at 0.1% for a dimensionality difference of 1. is this built into R fitdistr somehow, but I just overlooked it? right approach?