I meant, of course, that large samples of =FLOOR(10^RAND();1;1) should satisfy the chi-squared
distribution for conformance to the Benford Distribution.
-----Original Message-----
From: Dennis E. Hamilton [mailto:dennis.hamilton@acm.org]
Sent: Sunday, July 14, 2013 02:55 PM
To: 'Toki Kantoor'
Cc: 'users@global.libreoffice.org'
Subject: RE: [libreoffice-users] Benford's Law
Uniform random number generators do not conform to Benford's law.
To get uniform digits in the range 1 to 10, try =FLOOR(10*RAND();1;1)
However, Benford's law is about the *first* digit of a wide variety of numbers.
See <http://en.wikipedia.org/wiki/Benford%27s_law#Mathematical_statement>.
To get the Benford distribution of digits 1 to 9,
I think you want =FLOOR(10^RAND();1;1)
What makes you think these do not have the Benford distribution? How are you testing that.
You should be able to create a histogram for the frequencies of values of 1, 2, 3, ..., 9 and show
that it approaches the Benford distribution as you increase the number of samples.
While the RNG used for RAND() may not be cryptographically wonderful, I expect it would pass a
reasonable test (say chi-squared) for correspondence to the Benford distribution.
- Dennis
-----Original Message-----
From: Toki Kantoor [mailto:toki.kantoor@gmail.com]
Sent: Sunday, July 14, 2013 02:01 PM
Cc: users@global.libreoffice.org
Subject: Re: [libreoffice-users] Benford's Law
On 07/13/2013 11:01 PM, Brian Barker wrote:
Unless I misunderstand,the formula =10^RAND() should create random variates in the range (1,10)
following the law.
10^RAND generates a set of random numbers that does _not_ adhere to
Benford's Law. I need a random number generator whose output does
adhere to Benford's Law.
jonathon
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