Re: [boost] [math/staticstics/design] How besttonamestatisticalfunctions?

From: Paul A Bristow (pbristow_at_[hidden])
Date: 2006-07-12 12:08:01

|  -----Original Message-----
|  From: boost-bounces_at_[hidden] 
|  [mailto:boost-bounces_at_[hidden]] On Behalf Of Topher Cooper
|  Sent: 12 July 2006 14:11
|  To: boost_at_[hidden]
|  Subject: Re: [boost] [math/staticstics/design] How 
|  besttonamestatisticalfunctions?
  
|  This brings to mind another function that, though easily derived 
|  would be good to have to allow internal computations less subject to 
|  round-off error.  This is a two parameter function that is the 
|  cumulative probability between a lower and an upper 
|  bound.  Mathematically this can always be computed as "CDF(x[ub]) - 
|  CDF(x[lb])" (read the square brackets as mathematical subscript 
|  notation) but numerically with very small intervals, you can easily 
|  end up with 0 when you want something close to 
|  "PDF((x[ub]+x[lb])/2)*(x[ub]-x[lb])". 

An interesting suggestion.

John Maddock has been muttering about using Boost.Interval with these
functions.
It's on his TODO list allegedly ;-)

Would this help with the "CDF(x[ub]) - CDF(x[lb])"?

And/or allow one to produce "PDF((x[ub]+x[lb])/2)*(x[ub]-x[lb])"
using the density/mass/distribution?

|  You don't need to make any 
|  general guarantee about precision and so could do initial 
|  implementations as the difference of the cumulative functions, but 
|  then go back and do better for individual distributions.
|  
|  I don't know any standard term for this off the top of my head.  I 
|  would suggest just using a two argument version of whatever is 
|  decided on for the cumulative distribution.  So, using my suggested 
|  function name:
|  
|            standard_normal.cdf(-1.0, 1.0)
|  
|  would return the probability that a random variate with a normal 
|  distribution is within one standard deviation of the mean.
|  
|  The only problem I have with this is that if we look at the one 
|  parameter version as being the two parameter version with one 
|  parameter defaulted its the *first* parameter that is 
|  defaulted since:
|  
|            dist.cdf(x) = dist.cdf(-INFINITY, x)
|  
|  That would suggest using the complementary cdf instead, but that 
|  seems a lot less natural.

The language doesn't make this natural :-(

Paul

---
Paul A Bristow
Prizet Farmhouse, Kendal, Cumbria UK LA8 8AB
[phone] & SMS, Mobile [phone] & SMS
pbristow_at_[hidden]