[R] Help : stablereg parameter interpretation

[ripley@stats.ox.ac.uk]

In which package is the stablereg function?

I could not find it in any of the CRAN packages, and vaguely recall it
may be in one of Jim Lindsey's.  In any case, the best thing to do is to 
ask the author.

On Tue, 25 Mar 2003 pfm401 at lineone.net wrote:

> Dear all,
> 
> I am having difficulty interpreting the parameter estimates from the stablereg
> function. Specifically I am trying to keep things simple to start with by
> using stablereg to fit a normal distribution to a simulated data set from
> that distribution (in order to understand the way that stablereg reports
> parameter estimates). I cannot work out the scale on which the dispersion
> parameter (some function of the standard deviation if the data are normal)
> is reported. The location parameter i.e. the mean is obvious.
> 
> For example:
> 
> set.seed(1234)
> # Simulate normal data with mean 5 and sd 2 using rstable> normal5.2<-rstable(10000,loc=5,disp=2/sqrt(2),skew=0,tail=2)
> mean(normal5.2)
> [1] 5.010073
> sd(normal5.2)
> [1] 2.005362
> 
> # summary(lm(normal5.2~1)) yields similar results
> 
> 
> 
> fittest<-stablereg(y=normal5.2,loc=~1,disp=~1,iloc=5,idisp=2/sqrt(2),iskew=0,itail=2,oskew=F,otail=F)
> # Normal model, as skew=0 and tail=2
> 
> yields :
> 
> Call:
> stablereg(y = normal5.2, loc = ~1, disp = ~1, iloc = 5, idisp = 2/sqrt(2),
> 
>     iskew = 0, itail = 2, oskew = F, otail = F)
> 
> -Log likelihood              21498.69
> No. of obs                   10000
> No. of estimated parameters  2
> No. of parameters            4
> Degrees of freedom           9998
> AIC                          21500.69
> Iterations                   6
> 
> Location parameters
> ~1
>              estimate       se
> (Intercept)     5.015  0.02011
> 
> Dispersion parameters
>    estimate        se
> 1    0.2888  0.008753
> 
> Correlations:
>           1         2
> 1  1.000000 -0.001761
> 2 -0.001761  1.000000
> 
> 
> so that 5.015 is indeed the (approx.) mean. As dispersion uses a log link
> function and disp=sd/sqrt(2) for the normal (I think!) I tried
> sqrt(2)*exp(0.2888)
> which gives 1.887727
> and 95% confidence limits at sqrt(2)*exp(0.2888+-1.96*0.008753)
> i.e. (1.854968,1.921065)
> 
> Without going into more detail I have tried this for different parameter
> values and simulations and cannot resolve the dispersion parameter to the
> standard deviation. Clearly I am missing something!!
> 
> Any help is much appreciated.
> 
> Thanks in advance,
> Paul.
> 
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> 

-- 
Brian D. Ripley,                  ripley at stats.ox.ac.uk
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford,             Tel:  +44 1865 272861 (self)
1 South Parks Road,                     +44 1865 272866 (PA)
Oxford OX1 3TG, UK                Fax:  +44 1865 272595