R help - Mar 2003 - Help : stablereg parameter interpretation

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.

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.
> 
> ______________________________________________
> R-help at stat.math.ethz.ch mailing list
> https://www.stat.math.ethz.ch/mailman/listinfo/r-help
> 
-- 
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