[Octave-bug-tracker] [bug #46238] normrnd() produces NaN for 0 in standard deviation vector

[anonymous]

URL:
  <http://savannah.gnu.org/bugs/?46238>

                 Summary: normrnd() produces NaN for 0 in standard deviation
vector
                 Project: GNU Octave
            Submitted by: None
            Submitted on: Sat 17 Oct 2015 03:36:02 PM UTC
                Category: Octave Function
                Severity: 3 - Normal
                Priority: 5 - Normal
              Item Group: Incorrect Result
                  Status: None
             Assigned to: None
         Originator Name: Robert
        Originator Email: address@hidden
             Open/Closed: Open
         Discussion Lock: Any
                 Release: 4.0.0
        Operating System: Any

    _______________________________________________________

Details:

When calling normrnd() with 0 standard deviation on scalar arguments, it
(correctly) returns the given mean, e.g.

>> normrnd(10, 0)
ans =  10


However, when called with vector arguments, the elements where the standard
deviation is 0 are replaced by NaN, e.g.

>> normrnd([10 10 10 10], [1 0 1 0])
ans =

   9.5101      NaN   7.9095      NaN


The issue is on line 93 or normrnd.m:

k = ! isfinite (mu) | !(sigma > 0) | !(sigma < Inf);

which should be 

k = ! isfinite (mu) | !(sigma >= 0) | !(sigma < Inf);

similar to line 86, which deals with scalar arguments:

if (isfinite (mu) && (sigma >= 0) && (sigma < Inf))


Correcting this would make behaviour between scalar and vector operations
consistent, and also make normrnd() behave the same way as Matlab.




    _______________________________________________________

Reply to this item at:

  <http://savannah.gnu.org/bugs/?46238>

_______________________________________________
  Message sent via/by Savannah
  http://savannah.gnu.org/