# bug #46238: normrnd() produces NaN for 0 in standard deviation vector

 Submitter: None Submitted: Sat 17 Oct 2015 03:36:02 PM UTC Category: Octave Function Severity: 3 - Normal Priority: 5 - Normal Item Group: Incorrect Result Status: Fixed Assigned to: None Originator Name: Robert Originator Email: -email is unavailable- Open/Closed:  * Closed Release:  * 4.0.0 Operating System:  * Any Fixed Release: None Planned Release: None
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## Discussion

Sat 17 Oct 2015 04:14:28 PM UTC, comment #1:

Thanks for catching this.  Obviously the scalar and vector cases were meant to be the same, but a probable typo let in the error.

I've fixed this on the stable branch of Octave and it will be part of the next 4.0.1 bug fix release.  You can either wait until then, or use the m-file directly from the Mercurial stable repository.

Rik <rik5>
Sat 17 Oct 2015 03:36:02 PM UTC, original submission:

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.

Anonymous

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