URL:
<http://savannah.gnu.org/bugs/?25039>
Summary: gsl_ran_exponential range should be [0,inf]
Project: GNU Scientific Library
Submitted by: bjg
Submitted on: Mon 08 Dec 2008 07:26:47 PM GMT
Category: Accuracy problem
Severity: 2 - Minor
Operating System:
Status: Confirmed
Assigned to: None
Open/Closed: Open
Release: 1.12
Discussion Lock: Any
_______________________________________________________
Details:
From: "Adam Stinchcombe" <[EMAIL PROTECTED]>
To: [email protected]
Subject: [Bug-gsl] bug: 0 not possible result in gsl_ran_exponential
Date: Sat, 6 Dec 2008 19:04:35 -0500
Hello,
Section 19.5 The Exponential Distribution
(http://www.gnu.org/software/gsl/manual/gsl-ref.html#The-Exponential-Distribution)
states that the distribution produces x >= 0, meaning that x=0 is a
possible result. The standard definition of the exponential
distribution also gives that x=0 is a possible value from the
distribution. The implementation in randist/exponential.c of gsl-1.11
is
double
gsl_ran_exponential (const gsl_rng * r, const double mu)
{
double u = gsl_rng_uniform_pos (r);
return -mu * log (u);
}
gsl_rng_uniform_pos returns a floating point number uniformly
distributed number on (0,1) , excluding both 0.0 and 1.0
(http://www.gnu.org/software/gsl/manual/gsl-ref.html#index-gsl_005frng_005funiform_005fpos-1486
). log maps (0,1) one-to-one and onto (-inf,0), excluding 0.0, which
means that the function above returns values on (0,inf),where inf is
the largest double. The point being that 0.0 is excluded as a possible
return value of the function.
I suggest the following implementation:
double
gsl_ran_exponential (const gsl_rng * r, const double mu)
{
double u = gsl_rng_uniform (r);
return -mu * log(1-u)
}
gsl_rng_uniform gives a value on [0,1) which is mapped to (0,1] by 1-u
and then to [0,inf) by log.
Since this distribution looks like the exponential everywhere except a
single point, I suspect that the statistics produced in the past were
not significantly incorrect. Please let me know if this is in fact a
bug or enlighten me as to my misunderstanding.
Cheers,
Adam Stinchcombe
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