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Benford: Why and Why Not (non-Math-heads: SNOOZE ALERT)
Howdy,
Here is an explanation of why more numbers start with '1' than with '9',
how to make a system where that isn't true, and why you don't need to. If
you don't like math, you've already extracted everything from this message
that you could possibly need.
-----
For any integral magnitude x in the range 0..N, there are infinitely many
polynomials of the form
c_1 b_1^e_1 + c_2 b_2^e_2 + ... + c_L b_L^e_L = x
c_i, b_i, e_i integers. This is trivially provable given the single term
(x b_1^0), in combination with all the possible terms where c_i = 0.
Let us distinguish polynomials of a form where, for terms i and j, i != j:
1. b_i = j_i
2. e_i = L-i
3. b_i^e_i <= x
4. 0 <= c_i < b_i
Condition 1 allows only polynomials where b is the same for every term.
Condition 2 allows only polynomials whose terms are in strictly decreasing
order by the exponent, and contain every integral exponent from L-1 downto
0. Condition 3 ensures that the c_1 != 0.
These four conditions, restrict the set of polynomials that can represent
x, for any B such that b_i=B, to 1. These conditions are, in fact, the
'normalization' rules for representing numbers as strings of symbols, in
any base B. Since B can be assumed, and e_i adduced, any x can be
represented by concatenating symbols drawn from an alphabet of size B,
representing the magnitudes 0..B-1 according to c_i. When B=10, and the
alphabet is { '0'=0, '1'=1, '2'=2, ... '9'=9 }, we call this the decimal
system.
Consider the journey of x as it goes from 0 to B^(i+1)-1. It has to
progress from 0 to B^i-1 to 2B^i-1 to B^(i+1)-1.
Graph(?) A:
B^(i-1)-1 (B^i)-1 (2B^i)-1 (B^(i+1))-1
|..........|..........|........................................|
R r S s T t R'
Distance r = s = B^i. Distance t = (B-2)B^i. These three 'legs' of x's
journey represent, starting with s, a contiguous range of representations
beginning with '1', t, a contiguous range of representations _not_ starting
with '1', and r, a range 1/B the size we are disecting, which is itself
divided in the same proportions. Thus, from S to T, the fraction of
strings beginning with '1' increases to (about) 1/2, and from T to R' it
decreases to 1/B. And since r is a minor copy of rst, at R and S, the
fraction must be 1/B. So, for any period, the fraction rises to 1/2 in
time s, and falls to 1/B in time (B-2)s.
Consider the normalized decimal representations of integers drawn from the
range [0,N]. When N=1, the set in question is {'0', '1'} of which 1/2
begin with '1'. As N goes to 9, eight strings are added to the set, none
of which begin with '1', thus, at N=9, only 1/10 of the sets elements begin
with '1'. As N goes to 19, the fraction swings quickly back up to 1/2. As
N goes to 99, the fraction drops, again, more slowly (8 times more slowly),
to 1/10. On a log graph, this behavior is a saw-tooth pattern:
Graph B:
. . .
. . .
0.5 ....*.........*.........*..........
. * . * . *
p . * *. * *. *
. * . * . *
0.1 ............*.........*.........*..
* . . .
. . N .
0 1 19 199 1999
Note that this graph describes properties of a representation for numbers.
That is, this is the graph of a distribution of strings, not of integers.
A different representation would yield a different behavior.
Consider the graph of log_B(N). Here the x-axis is N, the y-axis log_B(N),
and it is graphed on log_B, paper as '*'s:
Graph C:
3 - +---------*
| *
2 - +---------*
| *
1 - +---------*
| *
0 - *
| | | |
B^0 B^1 B^2 B^3
Ceiling(log_B(N)), shown in the graph with lines, is the length of the base
B representation of N. This shows that even if the 'unfair first digit'
problem were avoided by a different representation, any log based
distribution would still exhibit a similar problem in string _lengths_.
i.e., only 1/B of all strings from 0..N are shorter than log_b(N): an
'unfair distribution of lengths'.
For a flat distribution of strings, you require a representation scheme
where, for any N, 1/B strings begin with a_i (a symbol from B's alphabet),
and for range of allowable lengths of strings in the scheme (1..L), for any
l, the number of strings for numbers in N is the same.
Here are two such schemes.
'Base-1' representation: the alphabet has one symbol, any magnitude x is
represented by x concatenated instances of the symbol. The first property
is satisfied because the alphabet has 1 symbol, and 1/1 of the strings from
1..N begin with that symbol. The second property is satisfied because the
allowable lengths vary from 1..N, and for each length, only one string
exists of that length, thus every possible length has the name number of
strings.
Fixed length representation: All numbers from 0..N, N = B^L, are
represented by strings of length L comprising symbols from B's alphabet,
constructed according to the normalization rules but replacint condition 3
with the rule e_i=L-(i+1) (forcing all strings to be the same length). The
second property is satisfied because there is only one allowable length.
The first property is satisfied, because, since N=B^L, every distinct
string of length L of symbols from B's alphabet is a valid number in this
representation, and of those strings, for any position p in the string,
including p=0 (the first symbol), 1/B of the strings will symbol a_i in
that position. An example of this scheme is a byte of computer memory used
to hold magnitudes from 0..255.
But, since the 'Base-1' scheme doesn't exhibit the predictability problem
of our normalized decimal system, and is a dramatically 'real'
representation of numbers (although inconvenient) one might guess that even
if one can predict features of representations, it does not necessarily
follow that such a prediction can be exploited into guessing the number
itself. Something near half of all decimal representations begin with a
symbol from { '1', '2', '3' }, but that doesn't actually help me guess the
next number, just make some predictions about its representation.
This is, in fact the case. Our normalized log-based representation schemes
are 'unfair' in the distribution of strings, essentially using all the
'good' ones first (for oft' used small numbers). The further out we go,
the longer strings we need, and the log distrubition problem surfaces. But
for any range, the actual (formless) magnitudes themselves are evenly
distributed, in spite of what you might guess from our biased strings.
Thus, Benford distribution is the kernel of a great bar bet, but an
unlikely alley in predicting the underlying magnitudes.
-----
Fractals? Well, r is a 'little copy' of rst. Chaos theory? No. How
about just some good basic algebra (OK, a little calculus if you want to
calculate the area under under the curve in Graph B, and from that, the
amount to bet at the bar).
Hope you enjoyed this,
Scott Collins | "Few people realize what tremendous power there
| is in one of these things." -- Willy Wonka
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