Re: [libreoffice-users] Benford's Law

[Brian Barker]

At 21:03 18/07/2013 +, Toki "Jonathan" Kantoor wrote:

On 07/15/2013 12:42 PM, Brian Barker wrote:
... don't stop being what they are after some 
arbitrary number of significant figures - 
whether it be one, three, or any other.


At the fourth significant digit, 0 and 9 occur 
slightly (¿1:10,000?) more frequently than 1, 2, 
3, 4, 5, 6, 7, and 8. For most practical 
purposes, the fourth digit can be treated as a 
uniformly random number. At the fifth, and 
subsequent digits, the numbers are randomly, and uniformly distributed.


It's surely intuitively obvious that this cannot 
be so.  The first digits are very non-uniformly 
distributed, the second ones less so, and so 
on.  What your source is telling you is that the 
fourth digit is very, very nearly uniformly 
distributed and that subsequent digits are so 
nearly so that they may be considered so for all 
practical purposes - not that they really are.  (That would be wrong.)


In any case, if the formula I suggested works 
(and we've seen plenty of evidence that it does 
and none that it doesn't - but I'm still open to 
correction), then according to your theory it 
*will* provide such uniformly distributed digits 
after the fourth.  You've already got what you 
want: the problem appears to be that you cannot 
believe the distribution will turn out the way 
that you say it will!  It's irrational of you to 
suggest removing one set of digits that you claim 
are already uniformly distributed and replacing 
them with another also uniformly distributed 
set!  And if there were any difference, how many 
variates would you have to call upon before any 
difference would be noticeable?  Billions of 
billions of billions?!  More than you are going to use, at any rate.


If you want values that follow Benford's Law up 
to three digits, you can easily take the true 
values from my suggested formula, truncate (or 
round?) them after three digits, and add 
further random digits selected from a uniform distribution.


And my original post was asking what happened to 
the macro that automatically did that.


And you now have an even simpler solution: a 
formula that does it.  But you are very welcome 
not to use it if you don't like it.  Even if you 
wanted to make the irrational change, you could 
easily construct a formula to do this.


Brian Barker


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Re: [libreoffice-users] Benford's Law

[Toki Kantoor]

On 07/15/2013 12:42 PM, Brian Barker wrote:

> don't stop being what they are after some arbitrary number of
> significant figures - whether it be one, three, or any other.  

At the fourth significant digit, 0 and 9 occur slightly (¿1:10,000?)
more frequently than 1, 2, 3, 4, 5, 6, 7, and 8. For most practical
purposes, the fourth digit can be treated as a uniformly random number.
At the fifth, and subsequent digits, the numbers are randomly, and
uniformly distributed.

> If you want values that follow Benford's Law up to three digits, you can
> easily take the true values from my suggested formula, truncate (or
> round?) them after three digits, and add further random digits selected
> from a uniform distribution.

And my original post was asking what happened to the macro that
automatically did that.

jonathon
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Re: [libreoffice-users] Benford's Law

[Brian Barker]

At 12:26 15/07/2013 +, Toki "Jonathan" Kantoor wrote:

On 07/14/2013 09:54 PM, Dennis E. Hamilton wrote:
However, Benford's law is about the *first* digit of a wide variety 
of numbers.


First three digits, not first digit. The fourth and subsequent 
digits should be uniformly distributed.


The striking thing about variates that follow Benford's Law is indeed 
that the initial digits are not equally distributed.  But such 
variates don't stop being what they are after some arbitrary number 
of significant figures - whether it be one, three, or any other.  The 
values come from the distribution - so all their decimal digits are 
part of the story.


If you want values that follow Benford's Law up to three digits, you 
can easily take the true values from my suggested formula, truncate 
(or round?) them after three digits, and add further random digits 
selected from a uniform distribution.


But I'm not at all sure why you would want to do this.  Does the 
source of this suggestion mean merely that, although the early digits 
are not uniformly distributed, later ones are more nearly so - and 
that there is little point in worrying about the difference after, 
say, three digits?  If so, there is equally no point in worrying that 
these later digits might be too right!


Brian Barker


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Re: [libreoffice-users] Benford's Law

[Toki Kantoor]

On 07/14/2013 09:54 PM, Dennis E. Hamilton wrote:

> However, Benford's law is about the *first* digit of a wide variety of 
> numbers.  

First three digits, not first digit.
The fourth and subsequent digits should be uniformly distributed.

jonathon
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Re: [libreoffice-users] Benford's Law

[Mark LaPierre]

On 07/14/2013 05:01 PM, Toki Kantoor wrote:

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



Hey Jonathon,

This is the distribution I got after 100 million cycles:
10^RAND

9 - 0.045
8 - 0.051
7 - 0.057
6 - 0.066
5 - 0.079
4 - 0.096
3 - 0.124
2 - 0.176
1 - 0.300

I don't think you are going to get any closer to Benford's law 
distribution than that.


Here is the code I used:

use strict;
use warnings;

my $num;
my $lpcnt;
my @Distribution;
$Distribution[0]=0;

srand;

$lpcnt = 1;
while ($lpcnt) {
$num=substr(10**rand(),0,1);
++$Distribution[$num];
--$lpcnt;
}

$lpcnt = 9;
while ($lpcnt) {
print "$lpcnt - ".substr($Distribution[$lpcnt]/1,0,5)."\n";
--$lpcnt;
}


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https://linuxcounter.net/


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RE: [libreoffice-users] Benford's Law

[Dennis E. Hamilton]

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.hamil...@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.kant...@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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Re: [libreoffice-users] Benford's Law

[Brian Barker]

At 21:01 14/07/2013 +, Toki "Jonathan" Kantoor wrote:
10^RAND generates a set of random numbers that does _not_ adhere to 
Benford's Law.


Incidentally, if you would like your random numbers in binary instead 
of decimal, I can provide an even easier formula for the initial 
digit of Benford's Law variates:

=1

Brian Barker


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RE: [libreoffice-users] Benford's Law

[Dennis E. Hamilton]

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.kant...@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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Re: [libreoffice-users] Benford's Law

[Steve Edmonds]

On 2013-07-15 09:01, Toki Kantoor wrote:

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

Hi
From http://en.wikipedia.org/wiki/Benford%27s_law
Therefore, this is the distribution expected if the mantissae 
 of the /logarithms/ of the 
numbers (but not the numbers themselves) are uniformly and randomly 
distributed 
.


In a logarithm the part after the decimal point is the mantissa. Convert 
a logarithm to decimal by number=10^(logarithm).
For numbers between 1 and 10 logarithm will be between 0 and 1. 
Therefore 10^(rand()) should produce numbers randomly between 1 and 10 
conforming with Benford's law.


=10^RAND() is the same as 10^(rand())

Steve



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Re: [libreoffice-users] Benford's Law

[Brian Barker]

At 21:01 14/07/2013 +, Toki "Jonathan" Kantoor wrote:

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.


Well, I must say you have not exactly offered much evidence for this 
assertion!  If we knew why you thought this was so, we might be able to help.


Probabilities of initial digits according to Benford's Law and 
proportions from 10 000 trials of my formula:

  1   2   3  4   5  6   7  8   9
.301  .176  .125  .097  .079  .067  .058  .051  .046
.310  .171  .126  .093  .080  .066  .059  .053  .043

How much closer would you expect these trials to get?  Have you done 
a chi-squared test?



I need a random number generator whose output does adhere to Benford's Law.


And I wish you good luck in persuading someone to help!

Brian Barker


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Re: [libreoffice-users] Benford's Law

[Toki Kantoor]

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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Re: [libreoffice-users] Benford's Law

[Brian Barker]

At 18:59 13/07/2013 +, Toki "Jonathan" Kantoor wrote:
Once upon a time I had an extension that generated random numbers 
that adhered to Benford's Law.


Do you need one?  Unless I misunderstand, the formula
=10^RAND()
should create random variates in the range (1,10) following the law.

(Here's hoping you will not use this to create plausible fake scientific data!)

I trust this helps.

Brian Barker


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