responses to my question on non-central t

[Robert Kinley <KINLEY_ROBERT@LILLY.COM>]

The original question is attached at the end of this message
...

In the world of R the excellent  mvtnorm  package
answers my original 
question and many more that I didn't know I needed to
ask.

In the world of Splus ( version 6 , Windows XP )  this
link :-
        http://www.biostat.wustl.edu/archives/html/s-news/2002-11/msg00079.html
which was provided by Dimitris Rizopoulos, is useful.

Sadly in Splus there seems to be no library equivalent
to the R mvtnorm package...

... unless someone out there has ported it, or has something
similar up their sleeve ??

             
  cheers            
   Bob


----- Forwarded by Robert
Kinley/EMA/LLY on 28/09/2005 17:35 -----

Robert Kinley <KINLEY_ROBERT@lilly.com>
Sent by: r-help-bounces@stat.math.ethz.ch

14/09/2005 14:24

To	s-news@lists.biostat.wustl.edu, r-help@stat.math.ethz.ch
cc
Subject	[R] non-central t   : R v.Splus

Hi

For bureaucratic reasons beyond my control I need to rewrite an R function
(for producing operating characteristic curves) as an Splus function  (
version 6 , windows XP ).

The R function makes extensive use of the fact that the student's t 
distribution function  pt()  has a non-centrality parameter built
in ... 
sadly that parameter is not present in the Splus pt() function .

However, the Splus f distribution function pf() does have such a parameter
, so I have tried to write my own non-central version of pt() based around
 pf() , using the relationship between the t and F distributions.

Unfortunately my success has been limited  ... I can only get correct
probabilities for part of the range of the quantile space , failing when
the quantile becomes small ... and I'm beginning to wonder whether it's
actually possible to do what I want at all , given that the range of x
in 
F(x) is [ 0:Inf ] while that in t(x) is [-Inf , Inf ] , and the 
non-central t  distribution is not symmetric ...

Do any wiser heads than mine have any experience or advice to offer  
... 
?

        thanks          Bob
Kinley

                
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