Why the IRS cares about the Riemann Zeta Function and Number … · 2019. 9. 9. · Intro General...

Intro General Theory Why Benford? Apps Benford Good Stick Refs Products F Chains

Why the IRS cares about the RiemannZeta Function and Number Theory

(and why you should too!)

Steven J. [email protected],

[email protected]

http://web.williams.edu/Mathematics/sjmiller/public_html/

Stresa, Italy, July 11, 1, 20191




Introduction

A. Berger and T. P. Hill, An Introduction to Benford’s Law,Princeton University Press, Princeton, 2015. See also http://www.benfordonline.net/.

A. E. Kossovsky, Benford’s Law: Theory, the General Law ofRelative Quantities, and Forensic Fraud Detection Applications,WSPC, 2014.

S. J. Miller (editor), Theory and Applications of Benford’s Law,Princeton University Press, 2015.

M. Nigrini, Benford’s Law: Applications for Forensic Accounting,Auditing, and Fraud Detection, 1st Edition, Wiley, 2014.

2

http://www.benfordonline.net/

http://www.benfordonline.net/


Interesting Question

Motivating Question: For a nice data set, such as theFibonacci numbers, stock prices, street addresses ofcollege employees and students, ..., what percent of theleading digits are 1?

Natural guess: 10% (but immediately correct to 11%!).

3


Interesting Question

Motivating Question: For a nice data set, such as theFibonacci numbers, stock prices, street addresses ofcollege employees and students, ..., what percent of theleading digits are 1?

Answer: Benford’s law!

4


Examples with First Digit Bias

Fibonacci numbers

Most common iPhone passcodes

Twitter users by # followers

Distance of stars from Earth

5


Summary

Explain Benford’s Law.

Discuss examples and applications.

Sketch proofs.

Describe open problems.

6


Caveats!

A math test indicating fraud is not proof of fraud:unlikely events, alternate reasons.

7


Caveats!

A math test indicating fraud is not proof of fraud:unlikely events, alternate reasons.

8


Examples

recurrence relations

special functions (such as n!)

iterates of power, exponential, rational maps

products of random variables

L-functions, characteristic polynomials

iterates of the 3x + 1 map

differences of order statistics

hydrology and financial data

many hierarchical Bayesian models

9


Applications

Analyzing round-off errors.

Determining the optimal way to storenumbers.

Detecting tax and image fraud, and dataintegrity.

10


General Theory

11


Benford’s Law: Newcomb (1881), Benford (1938)

StatementFor many data sets, probability of observing afirst digit of d base B is logB

(d+1

d

); base 10

about 30% are 1s.

Benford’s Law (probabilities)12


Background Material

Modulo: a = b mod c if a − b is an integer times c; thus17 = 5 mod 12, and 4.5 = .5 mod1.

13


Background Material


Significand: x = S10(x) · 10k , k integer, 1 ≤ S10(x) < 10.

14


Background Material



S10(x) = S10(x) if and only if x and x have the sameleading digits. Note log10 x = log10 S10(x) + k .

15


Background Material



S10(x) = S10(x) if and only if x and x have the sameleading digits. Note log10 x = log10 S10(x) + k .

Key observation: log10(x) = log10(x) mod 1 if and only if xand x have the same leading digits.

Thus often study y = log10 x mod 1.Advanced: e2πiu = e2πi(u mod 1).

16


Equidistribution and Benford’s Law

Equidistribution{yn}∞n=1 is equidistributed modulo 1 if probabilityyn mod 1 ∈ [a, b] tends to b − a:

#{n ≤ N : yn mod 1 ∈ [a, b]}N

→ b − a.

17




#{n ≤ N : yn mod 1 ∈ [a, b]}N

→ b − a.

Thm: β 6∈ Q, nβ is equidistributed mod 1.

18




#{n ≤ N : yn mod 1 ∈ [a, b]}N

→ b − a.


Examples: log10 2, log10

(1+

√5

2

)6∈ Q.

19




#{n ≤ N : yn mod 1 ∈ [a, b]}N

→ b − a.



(1+

√5

2

)6∈ Q.

Proof: if rational: 2 = 10p/q.

20




#{n ≤ N : yn mod 1 ∈ [a, b]}N

→ b − a.



(1+

√5

2

)6∈ Q.

Proof: if rational: 2 = 10p/q.Thus 2q = 10p or 2q−p = 5p, impossible.

21


Example of Equidistribution: n√π mod 1

0.2 0.4 0.6 0.8 1

0.5

1.0

1.5

2.0

n√π mod 1 for n ≤ 10

22



0.2 0.4 0.6 0.8 1

0.2

0.4

0.6

0.8

1.0


23



0.2 0.4 0.6 0.8 1

0.2

0.4

0.6

0.8

1.0


24



0.2 0.4 0.6 0.8 1

0.2

0.4

0.6

0.8

1.0

n√π mod 1 for n ≤ 10, 000

25


Logarithms and Benford’s Law

Fundamental EquivalenceData set {xi} is Benford base B if {yi} isequidistributed mod 1, where yi = logB xi .

26




x = S10(x) · 10k then

log10 x = log10 S10(x) + k = log10 S10x mod 1.

27




x = S10(x) · 10k then

log10 x = log10 S10(x) + k = log10 S10x mod 1.

28



Prob(leading digit d)= log10(d +1)− log10(d)= log10

(d+1

d

)

= log10

(1 + 1

d

).

Have Benford’s law ↔mantissa of logarithmsof data are uniformlydistributed

29


The Power of the Right Perspective

30


Examples

2n is Benford base 10 as log10 2 6∈ Q.

31


Examples

Fibonacci numbers are Benford base 10.

32


Examples

Fibonacci numbers are Benford base 10.an+1 = an + an−1.

33


Examples

Fibonacci numbers are Benford base 10.an+1 = an + an−1.Guess an = rn: rn+1 = rn + rn−1 or r2 = r + 1.

34


Examples

Fibonacci numbers are Benford base 10.an+1 = an + an−1.Guess an = rn: rn+1 = rn + rn−1 or r2 = r + 1.Roots r = (1 ±

√5)/2.

35


Examples


√5)/2.

General solution: an = c1rn1 + c2rn

2 .

36


Examples


√5)/2.


2 .

Binet: an = 1√5

(1+

√5

2

)n− 1√

5

(1−

√5

2

)n.

37


Examples


√5)/2.


2 .

Binet: an = 1√5

(1+

√5

2

)n− 1√

5

(1−

√5

2

)n.

Most linear recurrence relations Benford:

38


Examples


√5)/2.


2 .

Binet: an = 1√5

(1+

√5

2

)n− 1√

5

(1−

√5

2

)n.

Most linear recurrence relations Benford:⋄ an+1 = 2an

39


Examples


√5)/2.


2 .

Binet: an = 1√5

(1+

√5

2

)n− 1√

5

(1−

√5

2

)n.

Most linear recurrence relations Benford:⋄ an+1 = 2an − an−1

40


Examples


√5)/2.


2 .

Binet: an = 1√5

(1+

√5

2

)n− 1√

5

(1−

√5

2

)n.

Most linear recurrence relations Benford:⋄ an+1 = 2an − an−1

⋄ take a0 = a1 = 1 or a0 = 0, a1 = 1.

41


Digits of 2n

First 60 values of 2n (only displaying 30)1 1024 1048576 digit # Obs Prob Benf Prob2 2048 2097152 1 18 .300 .3014 4096 4194304 2 12 .200 .1768 8192 8388608 3 6 .100 .125

16 16384 16777216 4 6 .100 .09732 32768 33554432 5 6 .100 .07964 65536 67108864 6 4 .067 .067

128 131072 134217728 7 2 .033 .058256 262144 268435456 8 5 .083 .051512 524288 536870912 9 1 .017 .046

42


Digits of 2n

First 60 values of 2n (only displaying 30)1 1024 1048576 digit # Obs Prob Benf Prob2 2048 2097152 1 18 .300 .3014 4096 4194304 2 12 .200 .1768 8192 8388608 3 6 .100 .125

16 16384 16777216 4 6 .100 .09732 32768 33554432 5 6 .100 .07964 65536 67108864 6 4 .067 .067

128 131072 134217728 7 2 .033 .058256 262144 268435456 8 5 .083 .051512 524288 536870912 9 1 .017 .046

43


Digits of 2n

First 60 values of 2n (only displaying 30): 210 = 1024 ≈ 103.1 1024 1048576 digit # Obs Prob Benf Prob2 2048 2097152 1 18 .300 .3014 4096 4194304 2 12 .200 .1768 8192 8388608 3 6 .100 .125

16 16384 16777216 4 6 .100 .09732 32768 33554432 5 6 .100 .07964 65536 67108864 6 4 .067 .067

128 131072 134217728 7 2 .033 .058256 262144 268435456 8 5 .083 .051512 524288 536870912 9 1 .017 .046

44



χ2 values for αn, 1 ≤ n ≤ N (5% 15.5).N χ2(γ) χ2(e) χ2(π)

100 0.72 0.30 46.65200 0.24 0.30 8.58400 0.14 0.10 10.55500 0.08 0.07 2.69700 0.19 0.04 0.05800 0.04 0.03 6.19900 0.09 0.09 1.71

1000 0.02 0.06 2.90

45


Logarithms and Benford’s Law: Base 10 (5%: log(χ2) ≈ 2.74)

log(χ2) vs N for πn (red) and en (blue),n ∈ {1, . . . ,N}. Note π175 ≈ 1.0028 · 1087.

200 400 600 800 1000

-1.5

-1.0

-0.5

0.5

1.0

1.5

2.0

46


Logarithms and Benford’s Law: Base 10 (5%: log(χ2) ≈ 2.74)

log(χ2) vs N for πn (red) and en (blue),n ∈ {1, . . . ,N}. Note π175 ≈ 1.0028 · 1087.

200 400 600 800 1000

-1.5

-1.0

-0.5

0.5

1.0

1.5

2.0

47


New Result: Linear Recurrence Relations of Degree 2

an+1 = f (n)an + g(n)an−1 with non-constantcoefficients f (n) and g(n).

Explore conditions on f and g such that thesequence generated obeys Benford’s Lawfor all initial values.

First solve the closed form of the sequence(an), then analyze its main term.

48


Main idea: reduce the degree of recurrence.

an+1 = (λ(n) + µ(n))an − µ(n)λ(n − 1)an−1,and compare the coefficients:

f (n) = λ(n) + µ(n)

g(n) = −λ(n − 1)µ(n).

We show that for any given pair of f and g,such λ and µ always exist.

49


Linear Recurrence Relations of Degree 2

Recurrence relations of degree 1:

an+1 = λ(n)an + bn

bn = µ(n)bn−1.

an+1 = r(n)(

1 +n∑

k=3

n∏i=k

λ(i)µ(i) +

a2

b1

n∏i=2

λ(i)µ(i)

),

where r(n) := b1

n∏i=2

µ(i).

Find conditions on µ, λ such that main termdominates; Benford if

∏µ(i) is.

50


Examples when f and g are functions

If µ(k) = k , then r(n) = n!.

If µ(k) = kα where α ∈ R, then r(n) = (n!)α.

If µ(k) = exp(αh(k)) where α is irrational andh(k) is a monic polynomial, then

log r(n) = αn∑

k=1h(k).

LemmaThe sequence {αp(n)} is equidistributed mod 1if α /∈ Q and p(n) a monic polynomial.

51


Examples when f and g are random variables

Take µ(n) ∼ h(n)Un where the Un’s areindependent uniform distributions on [0, 1],and h(n) is a deterministic function in n such

thatn∏

i=1h(i) is Benford.

Then r(n) =n∏

i=1h(i)

n∏i=1

Ui is Benford.

Take µ(n) ∼ exp(Un) where the Un’s are i.i.d.random variables. Then take logarithm andsum up log(µ(n)). Apply Central LimitTheorem and get a Gaussian distributionwith increasing variance.52


Linear Recurrences of Higher Degree

Use recurrence relation of degree 3 as anexample. Similar main idea: reduce thedegree.

Define the sequence {an}∞n=1 byan+1 = f1(n)an + f2(n)an−1 + f3(n)an−2.

Define an auxiliary sequence (bn)∞n=1 by

bn = an+1 − λ(n)an. Then (bn) is degree 2.

53


Why Benford’s Law?

54


Streets

Not all data sets satisfy Benford’s Law.Long street [1,L]: L = 199 versus L = 999.Oscillates b/w 1/9 and 5/9 with first digit 1.

Probability first digit 1 versus street length L.What if we have many streets of different lengths?

55


Streets


Probability first digit 1 versus street length L.What if we have many streets of different lengths?

56


Streets


Probability first digit 1 versus log(street length L).What if we have many streets of different lengths?

57


Streets


Probability first digit 1 versus log(street length L).What if we have many streets of different lengths?

58


Amalgamating Streets

All houses: 1000 Streets,

each from 1 to 10000.

First digit and first two digits vs Benford.Conclusion: More processes, closer to Benford.

59




each from 1 to rand(10000).


60




each 1 to rand(rand(10000)).


61




each 1 to rand(rand(rand(10000))).


62


Probability Review

Let X be random variable with density p(x):⋄ p(x) ≥ 0;

∫∞

−∞p(x)dx = 1;

⋄ Prob (a ≤ X ≤ b) =∫ b

a p(x)dx .

Mean µ =∫∞

−∞xp(x)dx .

Variance σ2 =∫∞

−∞(x − µ)2p(x)dx .

Independence: knowledge of one random variablegives no knowledge of the other.

63


Central Limit Theorem

Normal N(µ, σ2) : p(x) = e−(x−µ)2/2σ2/√

2πσ2.

TheoremIf X1,X2, . . . independent, identically distributed randomvariables (mean µ, variance σ2, finite moments) then

SN :=X1 + · · ·+ XN − Nµ

σ√

Nconverges to N(0, 1).

64


Central Limit Theorem: Sums of Uniform Random VariablesXi ∼ Unif(−1/2,1/2) (adjusted to mean 0, variance 1)

Y1 = X1/σX1 vs N(0, 1).

Density of Y1 versus N(0, 1).65



Y2 = (X1 + X2)/σX1+X2 vs N(0, 1).




Y4 = (X1 +X2 +X3 +X4)/σX1+X2+X3+X4 vs N(0, 1).




Y8 = (X1 + · · ·+ X8)/σX1+···+X8 vs N(0, 1).




Density of Y4 = (X1 + · · ·+ X4)/σX1+···+X4.

(Don’t even think of asking to see Y8’s!)69


Normal Distributions Mod 1

As σ → ∞, N(0, σ2) mod 1 → Unif(0, 1).

Variance is .01.70



As σ → ∞, N(0, σ2) mod 1 → Unif(0, 1).

Variance is .1.71



As σ → ∞, N(0, σ2) mod 1 → Unif(0, 1).

Variance is .5.72


Products and Benford’s Law

Pavlovian Response: See a product, take a logarithm.

73




X1,X2, . . . nice, WN = X1 · X2 · · ·XN .

74




X1,X2, . . . nice, WN = X1 · X2 · · ·XN .

Yi = log10 Xi , VN := log10 WN .

75




X1,X2, . . . nice, WN = X1 · X2 · · ·XN .


VN = log10(X1 · X2 · · ·XN)

76




X1,X2, . . . nice, WN = X1 · X2 · · ·XN .


VN = log10(X1 · X2 · · ·XN)

= log10 X1 + log10 X2 + · · ·+ log10 XN

77




X1,X2, . . . nice, WN = X1 · X2 · · ·XN .


VN = log10(X1 · X2 · · ·XN)

= log10 X1 + log10 X2 + · · ·+ log10 XN

= Y1 + Y2 + · · ·+ YN .

Need distribution of VN mod 1, which by CLT becomes uniform,

implying Benfordness!

78


Applications

79


Applications for the IRS: Detecting Fraud

A Tale of Two Steve Millers....80


Detecting Fraud

Bank FraudAudit of a bank revealed huge spike ofnumbers starting with 48 and 49, most dueto one person.

Write-off limit of $5,000. Officer had friendsapplying for credit cards, ran up balancesjust under $5,000 then he would write thedebts off.

81


Can you see the cat in the tree?

82


Transmitting Images

How to transmit an image?

Have an L × W grid with LW pixels.

Each pixel a triple: (Red, Green, Blue).

Often each value in {0, 1, 2, 3, . . . , 2n − 1}.

n = 8 gives 256 choices for each, or16,777,216 possibilities.

83


Steganography

Steganography: Concealing a message inanother message: https://en.wikipedia.org/wiki/Steganography.

84

https://en.wikipedia.org/wiki/Steganography



Steganography

Steganography: Concealing a message inanother message: https://en.wikipedia.org/wiki/Steganography.

Take one of the colors, say red, a number from 0to 255.

Write in binary: r727 + r626 + · · ·+ r12 + r0.

If change just the last or last two digits, veryminor change to image.

Can hide an image in another.85





86



87


Benford Good Processes

A. Kontorovich and S. J. Miller, Benford’sLaw, values of L-functions and the 3x + 1problem, Acta Arithmetica 120 (2005), no. 3,269–297.

88


Poisson Summation and Benford’s Law: Definitions

Feller, Pinkham (often exact processes)

89



Feller, Pinkham (often exact processes)data YT ,B = logB

−→X T (discrete/continuous):

P(A) = limT→∞

#{n ∈ A : n ≤ T}T

90



Feller, Pinkham (often exact processes)data YT ,B = logB

−→X T (discrete/continuous):

P(A) = limT→∞

#{n ∈ A : n ≤ T}T

Poisson Summation Formula: f nice:∞∑

ℓ=−∞f (ℓ) =

∞∑

ℓ=−∞f (ℓ),

Fourier transform f (ξ) =

∫ ∞

−∞f (x)e−2πixξdx .

91


Benford Good Process

XT is Benford Good if there is a nice f st

CDF−→Y T ,B

(y) =

∫ y

−∞

1T

f(

tT

)dt+ET (y) := GT (y)

and monotonically increasing h (h(|T |) → ∞):

92




CDF−→Y T ,B

(y) =

∫ y

−∞

1T

f(

tT


and monotonically increasing h (h(|T |) → ∞):Small tails: GT (∞)− GT (Th(T )) = o(1),GT (−Th(T ))− GT (−∞) = o(1).

93




CDF−→Y T ,B

(y) =

∫ y

−∞

1T

f(

tT


and monotonically increasing h (h(|T |) → ∞):Small tails: GT (∞)− GT (Th(T )) = o(1),GT (−Th(T ))− GT (−∞) = o(1).Decay of the Fourier Transform:∑

ℓ 6=0

∣∣∣ f (T ℓ)ℓ

∣∣∣ = o(1).

94




CDF−→Y T ,B

(y) =

∫ y

−∞

1T

f(

tT


and monotonically increasing h (h(|T |) → ∞):Small tails: GT (∞)− GT (Th(T )) = o(1),GT (−Th(T ))− GT (−∞) = o(1).Decay of the Fourier Transform:∑

ℓ 6=0

∣∣∣ f (T ℓ)ℓ

∣∣∣ = o(1).

Small translated error: E(a, b,T )) =∑|ℓ|≤Th(T ) [ET (b + ℓ)− ET (a + ℓ)] = o(1).

95


Main Theorem

Theorem (Kontorovich and M–, 2005)XT converging to X as T → ∞ (think spreadingGaussian). If XT is Benford good, then X isBenford.

96


Main Theorem

Theorem (Kontorovich and M–, 2005)XT converging to X as T → ∞ (think spreadingGaussian). If XT is Benford good, then X isBenford.

Examples⋄ L-functions⋄ characteristic polynomials (RMT)⋄ 3x + 1 problem⋄ geometric Brownian motion.

97


Sketch of the proof

Structure Theorem:⋄ main term is something nice spreading out⋄ apply Poisson summation

98


Sketch of the proof

Structure Theorem:⋄ main term is something nice spreading out⋄ apply Poisson summation

Control translated errors:⋄ hardest step⋄ techniques problem specific

99


Sketch of the proof (continued)

∞∑

ℓ=−∞P

(a + ℓ ≤ −→

Y T ,B ≤ b + ℓ)

100



∞∑

ℓ=−∞P

(a + ℓ ≤ −→

Y T ,B ≤ b + ℓ)

=∑

|ℓ|≤Th(T )

[GT (b + ℓ)− GT (a + ℓ)] + o(1)

101



∞∑

ℓ=−∞P

(a + ℓ ≤ −→

Y T ,B ≤ b + ℓ)

=∑

|ℓ|≤Th(T )

[GT (b + ℓ)− GT (a + ℓ)] + o(1)

=

∫ b

a

∑

|ℓ|≤Th(T )

1T

f(

t + ℓ

T

)dt + E(a, b,T ) + o(1)

102



∞∑

ℓ=−∞P

(a + ℓ ≤ −→

Y T ,B ≤ b + ℓ)

=∑

|ℓ|≤Th(T )

[GT (b + ℓ)− GT (a + ℓ)] + o(1)

=

∫ b

a

∑

|ℓ|≤Th(T )

1T

f(

t + ℓ

T

)dt + E(a, b,T ) + o(1)

= f (0) · (b − a) +∑

ℓ 6=0

f (T ℓ)e2πibℓ − e2πiaℓ

2πiℓ+ o(1).

103


Riemann Zeta Function (for real part of s greater than 1)

ζ(s) =∞∑

n=1

1ns

=∏

p prime

(1 − 1

ps

)−1

, Re(s) > 1.

Geometric Series Formula: (1 − x)−1 = 1 + x + x2 + · · · .Unique Factorization: n = pr1

1 · · · prmm .

∏

p

(1 − 1

ps

)−1

=

[1 +

12s +

(12s

)2

+ · · ·][

1 +13s +

(13s

)2

+ · · ·]· · ·

=∑

n

1ns .

104


Riemann Zeta Function

∣∣ζ(

12 + ik

4

)∣∣, k ∈ {0, 1, . . . , 65535}.

2 4 6 8

0.05

0.1

0.15

0.2

0.25

0.3

105


3x + 1 Problem

Kakutani (conspiracy), Erdös (not ready).

x odd, T (x) = 3x+12k , 2k ||3x + 1.

Conjecture: for some n = n(x), T n(x) = 1.

106


3x + 1 Problem


x odd, T (x) = 3x+12k , 2k ||3x + 1.


7

107


3x + 1 Problem


x odd, T (x) = 3x+12k , 2k ||3x + 1.


7 →1 11

108


3x + 1 Problem


x odd, T (x) = 3x+12k , 2k ||3x + 1.


7 →1 11 →1 17

109


3x + 1 Problem


x odd, T (x) = 3x+12k , 2k ||3x + 1.


7 →1 11 →1 17 →2 13

110


3x + 1 Problem


x odd, T (x) = 3x+12k , 2k ||3x + 1.


7 →1 11 →1 17 →2 13 →3 5

111


3x + 1 Problem


x odd, T (x) = 3x+12k , 2k ||3x + 1.


7 →1 11 →1 17 →2 13 →3 5 →4 1

112


3x + 1 Problem


x odd, T (x) = 3x+12k , 2k ||3x + 1.


7 →1 11 →1 17 →2 13 →3 5 →4 1 →2 1,

113


3x + 1 Problem


x odd, T (x) = 3x+12k , 2k ||3x + 1.


7 →1 11 →1 17 →2 13 →3 5 →4 1 →2 1,2-path (1, 1), 5-path (1, 1, 2, 3, 4).m-path: (k1, . . . , km).

114


3x + 1 and Benford

Theorem (Kontorovich and M–, 2005)As m → ∞, xm/(3/4)mx0 is Benford.

Theorem (Lagarias-Soundararajan, 2006)

X ≥ 2N , for all but at most c(B)N−1/36X initialseeds the distribution of the first N iterates ofthe 3x + 1 map are within 2N−1/36 of theBenford probabilities.

115


3x + 1 Data: random 10,000 digit number, 2k ||3x + 1

80,514 iterations ((4/3)n = a0 predicts 80,319);χ2 = 13.5 (5% 15.5).

Digit Number Observed Benford1 24251 0.301 0.3012 14156 0.176 0.1763 10227 0.127 0.1254 7931 0.099 0.0975 6359 0.079 0.0796 5372 0.067 0.0677 4476 0.056 0.0588 4092 0.051 0.0519 3650 0.045 0.046

116


3x + 1 Data: random 10,000 digit number, 2|3x + 1

241,344 iterations, χ2 = 11.4 (5% 15.5).

Digit Number Observed Benford1 72924 0.302 0.3012 42357 0.176 0.1763 30201 0.125 0.1254 23507 0.097 0.0975 18928 0.078 0.0796 16296 0.068 0.0677 13702 0.057 0.0588 12356 0.051 0.0519 11073 0.046 0.046

117


Stick Decomposition

T. Becker, D. Burt, T. C. Corcoran, A. Greaves-Tunnell, J. R.Iafrate, J. Jing, S. J. Miller, J. D. Porfilio, R. Ronan, J.Samranvedhya, F. W. Strauch and B. Talbut, Benford’s Law andContinuous Dependent Random Variables, Annals of Physics388 (2018), 350–381.

J. Iafrate, S. J. Miller and F. W. Strauch, Equipartitions and adistribution for numbers: A statistical model for Benford’s law,Physical Review E 91 (2015), no. 6, 062138 (6 pages).

118


Fixed Proportion Decomposition Process

Decomposition Process

1 Consider a stick of length L.

119





2 Uniformly choose a proportion p ∈ (0, 1).

120






3 Break the stick into two pieces—lengths pLand (1 − p)L.

121






3 Break the stick into two pieces—lengths pLand (1 − p)L.

4 Repeat N times (using the same proportion).

122



123


Fixed Proportion Conjecture (Joy Jing ’13)

Conjecture: The above decomposition processis Benford as N → ∞ for any p ∈ (0, 1), p 6= 1

2 .

124


Fixed Proportion Conjecture (Joy Jing ’13)

Conjecture: The above decomposition processis Benford as N → ∞ for any p ∈ (0, 1), p 6= 1

2 .

Counterexample (SMALL REU ’13): p = 111 , 1 − p = 10

11 .

125


Benford Analysis

At N th level,

2N sticks

N + 1 distinct lengths: write pN−j(1 − p)j as

pN

(1 − p

p

)j

, j ∈ {0, . . . ,N}, have

(Nj

)times.

126


Benford Analysis

At N th level,

2N sticks


pN

(1 − p

p

)j

, j ∈ {0, . . . ,N}, have

(Nj

)times.

(Weighted) Geometric with ratio 1−pp = 10y ;

behavior depends on irrationality of y !

127


Benford Analysis

At N th level,

2N sticks


pN

(1 − p

p

)j

, j ∈ {0, . . . ,N}, have

(Nj

)times.

(Weighted) Geometric with ratio 1−pp = 10y ;

behavior depends on irrationality of y !Theorem: Benford if and only if y irrational.

128


Benford Analysis (cont)

Say 1−pp = 10r/q for r , q integers.

All terms with index j mod q have same leading digit; probability index j mod q is

1

2N

[(

N

j

)

+

(

N

j + q

)

+

(

N

j + 2q

)

+ · · ·

]

=1

q

q−1∑

s=0

(

cosπs

q

)Ncos

π(N − 2j)s

q

=1

q

1 +

q−1∑

s=1

(

cosπs

q

)Ncos

π(N − 2j)s

q

=1

q

(

1 + Err

[

(q − 1)(

cosπ

q

)N])

,

where Err[X ] indicates an absolute error of size at most X

129


Examples

p = 3/11, 1000 levels; y = log10(8/3) 6∈ Q

(irrational)

130


Examples

p = 1/11, 1000 levels; y = 1 ∈ Q

(rational)

131


Examples

p = 1/(1 + 1033/10), 1000 levels; y = 33/10 ∈ Q

(rational)

132


Random Cuts

L

L K1 LH1-K1L

L K1K2 L K1H1-K2L LH1-K1LK3 LH1-K1LH1-K3L

L K1K2K4 L K1K2H1-K4L L K1H1-K2LK5 L K1H1-K2LH1-K45L LH1-K1LK3K6 LH1-K1LK3H1-K6L LH1-K1LH1-K3LK7 LH1-K1LH1-K3LH1-K7L

Figure: Unrestricted Decomposition: Breaking L into pieces, N = 3.133


Conclusions and References

134


Conclusions and Future Investigations

See many different systems exhibit Benfordbehavior.

Ingredients of proofs (logarithms,equidistribution).

Applications to fraud detection / dataintegrity.

135


A. K. Adhikari, Some results on the distribution of the mostsignificant digit, Sankhya: The Indian Journal of Statistics, SeriesB 31 (1969), 413–420.

A. K. Adhikari and B. P. Sarkar, Distribution of most significantdigit in certain functions whose arguments are random variables,Sankhya: The Indian J. of Statistics, Series B 30 (1968), 47–58.

R. N. Bhattacharya, Speed of convergence of the n-foldconvolution of a probability measure ona compact group, Z.Wahrscheinlichkeitstheorie verw. Geb. 25 (1972), 1–10.

F. Benford, The law of anomalous numbers, Proceedings of theAmerican Philosophical Society 78 (1938), 551–572. http://www.jstor.org/stable/984802?seq=1#page_scan_tab_contents.

136


A. Berger, L. A. Bunimovich and T. Hill, One-dimensionaldynamical systems and Benford’s Law, Trans. AMS 357 (2005),no. 1, 197–219. http://www.ams.org/journals/tran/2005-357-01/S0002-9947-04-03455-5/.

A. Berger and T. Hill, Newton’s method obeys Benford’s law, TheAmer. Math. Monthly 114 (2007), no. 7, 588-601. http://digitalcommons.calpoly.edu/cgi/viewcontent.cgi?article=1058&context=rgp_rsr.

A. Berger and T. Hill, Benford on-line bibliography, http://www.benfordonline.net/.

J. Boyle, An application of Fourier series to the most significantdigit problem Amer. Math. Monthly 101 (1994), 879–886.http://www.jstor.org/stable/2975136?seq=1#page_scan_tab_contents.

J. Brown and R. Duncan, Modulo one uniform distribution of thesequence of logarithms of certain recursive sequences,Fibonacci Quarterly 8 (1970) 482–486.

137


P. Diaconis, The distribution of leading digits and uniformdistribution mod 1, Ann. Probab. 5 (1979), 72–81. http://statweb.stanford.edu/~cgates/PERSI/papers/digits.pdf.

W. Feller, An Introduction to Probability Theory and itsApplications, Vol. II, second edition, John Wiley & Sons, Inc.,1971.

R. W. Hamming, On the distribution of numbers, Bell Syst. Tech.J. 49 (1970), 1609-1625. https://archive.org/details/bstj49-8-1609.

T. Hill, The first-digit phenomenon, American Scientist 86 (1996),358–363. http://www.americanscientist.org/issues/feature/1998/4/the-first-digit-phenomenon/99999.

T. Hill, A statistical derivation of the significant-digit law,Statistical Science 10 (1996), 354–363. https://projecteuclid.org/euclid.ss/1177009869.

138


P. J. Holewijn, On the uniform distribuiton of sequences ofrandom variables, Z. Wahrscheinlichkeitstheorie verw. Geb. 14(1969), 89–92.

W. Hurlimann, Benford’s Law from 1881 to 2006: a bibliography,http://arxiv.org/abs/math/0607168.

D. Jang, J. Kang, A. Kruckman, J. Kudo & S. J. Miller, Chains ofdistributions, hierarchical Bayesian models and Benford’s Law,Journal of Algebra, Number Theory: Advances and Applications,volume 1, number 1 (March 2009), 37–60. http://arxiv.org/abs/0805.4226.

E. Janvresse and T. de la Rue, From uniform distribution toBenford’s law, Journal of Applied Probability 41 (2004) no. 4,1203–1210. http://www.jstor.org/stable/4141393?seq=1#page_scan_tab_contents.

A. Kontorovich and S. J. Miller, Benford’s Law, Values ofL-functions and the 3x + 1 Problem, Acta Arith. 120 (2005),269–297. http://arxiv.org/pdf/math/0412003.pdf.

139


D. Knuth, The Art of Computer Programming, Volume 2:Seminumerical Algorithms, Addison-Wesley, third edition, 1997.

J. Lagarias and K. Soundararajan, Benford’s Law for the 3x + 1Function, J. London Math. Soc. (2) 74 (2006), no. 2, 289–303.http://arxiv.org/pdf/math/0509175.pdf.

S. Lang, Undergraduate Analysis, 2nd edition, Springer-Verlag,New York, 1997.

P. Levy, L’addition des variables aleatoires definies sur unecirconference, Bull. de la S. M. F. 67 (1939), 1–41.

E. Ley, On the peculiar distribution of the U.S. Stock IndicesDigits, The American Statistician 50 (1996), no. 4, 311–313.http://www.jstor.org/stable/2684926?seq=1#page_scan_tab_contents.

140


R. M. Loynes, Some results in the probabilistic theory ofasympototic uniform distributions modulo 1, Z.Wahrscheinlichkeitstheorie verw. Geb. 26 (1973), 33–41.

S. J. Miller, Benford’s Law: Theory and Applications, PrincetonUniversity Press, 2015. http://web.williams.edu/Mathematics/sjmiller/public_html/benford/.

S. J. Miller and M. Nigrini, The Modulo 1 Central Limit Theoremand Benford’s Law for Products, International Journal of Algebra2 (2008), no. 3, 119–130. http://arxiv.org/pdf/math/0607686v2.

S. J. Miller and M. Nigrini, Order Statistics and Benford’s law,International Journal of Mathematics and MathematicalSciences, Volume 2008 (2008), Article ID 382948, 19 pages.http://arxiv.org/pdf/math/0601344v5.pdf.

141


S. J. Miller and R. Takloo-Bighash, An Invitation to ModernNumber Theory, Princeton University Press, Princeton, NJ, 2006.http://web.williams.edu/Mathematics/sjmiller/public_html/book/index.html.

S. Newcomb, Note on the frequency of use of the different digitsin natural numbers, Amer. J. Math. 4 (1881), 39-40. http://www.jstor.org/stable/2369148?seq=1#page_scan_tab_contents.

M. Nigrini, Digital Analysis and the Reduction of Auditor LitigationRisk. Pages 69–81 in Proceedings of the 1996 Deloitte & Touche/ University of Kansas Symposium on Auditing Problems, ed. M.Ettredge, University of Kansas, Lawrence, KS, 1996.

M. Nigrini, The Use of Benford’s Law as an Aid in AnalyticalProcedures, Auditing: A Journal of Practice & Theory, 16 (1997),no. 2, 52–67.

142


M. Nigrini and S. J. Miller, Benford’s Law applied to hydrologydata – results and relevance to other geophysical data,Mathematical Geology 39 (2007), no. 5, 469–490. http://link.springer.com/article/10.1007%2Fs11004-007-9109-5?LI=true.

M. Nigrini and S. J. Miller, Data diagnostics using second ordertests of Benford’s Law, Auditing: A Journal of Practice andTheory 28 (2009), no. 2, 305–324. http://accounting.uwaterloo.ca/uwcisa/symposiums/symposium_2007/AdvancedBenfordsLaw7.pdf.

R. Pinkham, On the Distribution of First Significant Digits, TheAnnals of Mathematical Statistics 32, no. 4 (1961), 1223-1230.

R. A. Raimi, The first digit problem, Amer. Math. Monthly 83(1976), no. 7, 521–538.

143


H. Robbins, On the equidistribution of sums of independentrandom variables, Proc. Amer. Math. Soc. 4 (1953), 786–799.http://projecteuclid.org/download/pdf_1/euclid.aoms/1177704862.

H. Sakamoto, On the distributions of the product and the quotientof the independent and uniformly distributed random variables,Tohoku Math. J. 49 (1943), 243–260.

P. Schatte, On sums modulo 2π of independent randomvariables, Math. Nachr. 110 (1983), 243–261.

P. Schatte, On the asymptotic uniform distribution of sumsreduced mod 1, Math. Nachr. 115 (1984), 275–281.

144


P. Schatte, On the asymptotic logarithmic distribution of thefloating-point mantissas of sums, Math. Nachr. 127 (1986), 7–20.

E. Stein and R. Shakarchi, Fourier Analysis: An Introduction,Princeton University Press, 2003.

M. D. Springer and W. E. Thompson, The distribution of productsof independent random variables, SIAM J. Appl. Math. 14 (1966)511–526. http://www.jstor.org/stable/2946226?seq=1#page_scan_tab_contents.

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P. R. Turner, The distribution of leading significant digits, IMA J.Numer. Anal. 2 (1982), no. 4, 407–412.

145


Productsof

Random Variables

S. J. Miller and M. Nigrini, The Modulo 1Central Limit Theorem and Benford’s Law forProducts, International Journal of Algebra 2(2008), no. 3, 119–130.

146


Preliminaries

X1 · · ·Xn ⇔ Y1 + · · ·+ Yn mod 1, Yi = logB Xi

Density Yi is gi , density Yi + Yj is

(gi ∗ gj)(y) =

∫ 1

0gi(t)gj(y − t)dt .

hn = g1 ∗ · · · ∗ gn, hn(ξ) = g1(ξ) · · · gn(ξ).

147


Modulo 1 Central Limit Theorem

Theorem (M– and Nigrini 2007){Ym} independent continuous random variableson [0, 1) (not necc. i.i.d.), densities {gm}.Y1 + · · ·+ YM mod 1 converges to the uniformdistribution as M → ∞ in L1([0, 1]) if and only iffor all n 6= 0, limM→∞ g1(n) · · · gM(n) = 0.

⋄ Gives info on rate of convergence.

148


Generalizations

Levy proved for i.i.d.r.v. just one year afterBenford’s paper.

Generalized to other compact groups, withestimates on the rate of convergence.⋄ Stromberg: n-fold convolution of a regularprobability measure on a compact Hausdorffgroup G converges to normalized Haarmeasure in weak-star topology iff support ofthe distribution not contained in a coset of aproper normal closed subgroup of G.

149


Distribution of digits (base 10) of 1000 productsX1 · · ·X1000, where g10,m = φ11m.φm(x) = m if |x − 1/8| ≤ 1/2m (0 otherwise).

2 4 6 8 10

0.1

0.2

0.3

0.4

0.5

150


Proof under stronger conditions

Use standard CLT to show Y1 + · · ·+ YM

tends to a Gaussian.

Use Poisson Summation to show theGaussian tends to the uniform modulo 1.

151



-4 -2 2 4

0.1

0.2

0.3

0.4

Figure: Plot of normal (mean 0, stdev 1).

152



0.2 0.4 0.6 0.8 1.0

1

2

3

4

Figure: Plot of normal (mean 0, stdev .1) modulo 1.

153



0.2 0.4 0.6 0.8 1.0

0.995

1.000

1.005

1.010

1.015

Figure: Plot of normal (mean 0, stdev .5) modulo 1.

154


Inputs

Poisson Summation Formulaf nice: ∞∑

ℓ=−∞f (ℓ) =

∞∑

ℓ=−∞f (ℓ),

Fourier transform f (ξ) =

∫ ∞

−∞f (x)e−2πixξdx .

Lemma2√

2πσ2

∫∞σ1+δ e−x2/2σ2

dx ≪ e−σ2δ/2.

155


Proof Under Weaker Conditions

Lemma

As N → ∞, pN(x) = e−πx2/N√N

becomesequidistributed modulo 1.

∫∞x=−∞

x mod 1∈[a,b]pN(x)dx =

1√N

∑n∈Z∫ b

x=a e−π(x+n)2/Ndx .

e−π(x+n)2/N = e−πn2/N + O(max(1,|n|)

N e−n2/N).

Can restrict sum to |n| ≤ N5/4.1√N

∑n∈Z e−πn2/N =

∑n∈Z e−πn2N .

156



1√N

∑

|n|≤N5/4

∫ b

x=ae−π(x+n)2/Ndx

=1√N

∑

|n|≤N5/4

∫ b

x=a

[e−πn2/N + O

(max(1, |n|)

Ne−n2/N

)]dx

=b − a√

N

∑

|n|≤N5/4

e−πn2/N + O

1

N

N5/4∑

n=0

n + 1√N

e−π(n/√

N)2

=b − a√

N

∑

|n|≤N5/4

e−πn2/N + O

(1N

∫ N3/4

w=0(w + 1)e−πw2√

Ndw

)

=b − a√

N

∑

|n|≤N5/4

e−πn2/N + O(

N−1/2).

157



Extend sums to n ∈ Z, apply PoissonSummation:

1√N

∑

n∈Z

∫ b

x=ae−π(x+n)2/Ndx ≈ (b−a) ·

∑

n∈Ze−πn2N .

For n = 0 the right hand side is b − a.For all other n, we trivially estimate the sum:∑

n 6=0

e−πn2N ≤ 2∑

n≥1

e−πnN ≤ 2e−πN

1 − e−πN,

which is less than 4e−πN for N sufficiently large.158


Proof in General Case: Fourier input

Fejér kernel:

FN(x) =N∑

n=−N

(1 − |n|

N

)e2πinx .

Fejér series TN f (x) equals

(f ∗ FN)(x) =N∑

n=−N

(1 − |n|

N

)f (n)e2πinx .

Lebesgue’s Theorem: f ∈ L1([0, 1]). AsN → ∞, TN f converges to f in L1([0, 1]).TN(f ∗ g) = (TN f ) ∗ g: convolution assoc.

159


Proof of Modulo 1 CLT

Density of sum is hℓ = g1 ∗ · · · ∗ gℓ.

Suffices show ∀ǫ: limM→∞∫ 1

0 |hM(x)− 1|dx < ǫ.

Lebesgue’s Theorem: N large,

||h1 − TNh1||1 =

∫ 1

0|h1(x)− TNh1(x)|dx <

ǫ

2.

Claim: above holds for hM for all M.

160


Proof of Modulo 1 CLT : Proof of Claim

TNhM+1 = TN(hM ∗ gM+1) = (TNhM) ∗ gM+1

||hM+1 − TNhM+1||1 =

∫ 1

0|hM+1(x)− TNhM+1(x)|dx

=

∫ 1

0|(hM ∗ gM+1)(x)− (TNhM) ∗ gM+1(x)|dx

=

∫ 1

0

∣∣∣∣∣

∫ 1

0(hM(y)− TNhM(y))gM+1(x − y)

∣∣∣∣∣ dydx

≤∫ 1

0

∫ 1

0|hM(y)− TNhM(y)|gM+1(x − y)dxdy

=

∫ 1

0|hM(y)− TNhM(y)|dy · 1 <

ǫ

2.

161



Show limM→∞ ||hM − 1||1 = 0.Triangle inequality:

||hM − 1||1 ≤ ||hM − TNhM ||1 + ||TNhM − 1||1.

Choices of N and ǫ:

||hM − TNhM ||1 < ǫ/2.

Show ||TNhM − 1||1 < ǫ/2.

162



||TNhM − 1||1 =

∫ 1

0

∣∣∣∣∣∣∣

N∑

n=−Nn 6=0

(1 − |n|

N

)hM(n)e2πinx

∣∣∣∣∣∣∣dx

≤N∑

n=−Nn 6=0

(1 − |n|

N

)|hM(n)|

hM(n) = g1(n) · · · gM(n) −→M→∞ 0.For fixed N and ǫ, choose M large so that |hM(n)| < ǫ/4N whenevern 6= 0 and |n| ≤ N.

163


Productsand

Chains of Random Variables

D. Jang, J. U. Kang, A. Kruckman, J. Kudoand S. J. Miller, Chains of distributions,hierarchical Bayesian models and Benford’sLaw, Journal of Algebra, Number Theory:Advances and Applications, volume 1,number 1 (March 2009), 37–60.

164


Key Ingredients

Mellin transform and Fourier transformrelated by logarithmic change of variable.

Poisson summation from collapsing tomodulo 1 random variables.

165


Preliminaries

Ξ1, . . . ,Ξn nice independent r.v.’s on [0,∞).Density Ξ1 · Ξ2:

∫ ∞

0f2(x

t

)f1(t)

dtt

166


Preliminaries

Ξ1, . . . ,Ξn nice independent r.v.’s on [0,∞).Density Ξ1 · Ξ2:

∫ ∞

0f2(x

t

)f1(t)

dtt

⋄ Proof: Prob(Ξ1 · Ξ2 ∈ [0, x ]):∫ ∞

t=0Prob

(Ξ2 ∈

[0,

xt

])f1(t)dt

=

∫ ∞

t=0F2

(xt

)f1(t)dt ,

differentiate.167


Mellin Transform

(Mf )(s) =

∫ ∞

0f (x)xs dx

x

(M−1g)(x) =1

2πi

∫ c+i∞

c−i∞g(s)x−sds

g(s) = (Mf )(s), f (x) = (M−1g)(x).

(f1 ⋆ f2)(x) =

∫ ∞

0f2(x

t

)f1(t)

dtt

(M(f1 ⋆ f2))(s) = (Mf1)(s) · (Mf2)(s).168


Mellin Transform Formulation: Products Random Variables

TheoremXi ’s independent, densities fi . Ξn = X1 · · ·Xn,

hn(xn) = (f1 ⋆ · · · ⋆ fn)(xn)

(Mhn)(s) =n∏

m=1

(Mfm)(s).

As n → ∞, Ξn becomes Benford: Yn = logB Ξn,|Prob(Yn mod 1 ∈ [a, b])− (b − a)| ≤

(b − a) ·∞∑

ℓ 6=0,ℓ=−∞

n∏

m=1

(Mfi)(

1 − 2πiℓlogB

).

169


Proof of Kossovsky’s Chain Conjecture for certain densitie s

Conditions{Di(θ)}i∈I: one-parameter distributions,densities fDi(θ) on [0,∞).p : N → I, X1 ∼ Dp(1)(1), Xm ∼ Dp(m)(Xm−1).m ≥ 2,

fm(xm) =

∫ ∞

0fDp(m)(1)

(xm

xm−1

)fm−1(xm−1)

dxm−1

xm−1

limn→∞

∞∑

ℓ=−∞ℓ 6=0

n∏

m=1

(MfDp(m)(1))

(1 − 2πiℓ

logB

)= 0

170


Chains of Random Variables

Return to street problem: chain of uniforms.

Let Dunif(θ) be the density of a uniform randomvariable on [0, θ].

Let X1 ∼ Dunif(1) and Xn+1 ∼ Dunif(Xn).

171


Proof of Kossovsky’s Chain Conjecture for certain densitie s

Theorem (JKKKM)If conditions hold, as n → ∞ the distributionof leading digits of Xn tends to Benford’s law.

The error is a nice function of the Mellintransforms: if Yn = logB Xn, then

|Prob(Yn mod 1 ∈ [a, b])− (b + a)| ≤∣∣∣∣∣∣∣(b − a) ·

∞∑

ℓ=−∞ℓ 6=0

n∏

m=1

(MfDp(m)(1))

(1 − 2πiℓ

logB

)∣∣∣∣∣∣∣

172


Example: All Xi ∼ Exp(1)

Xi ∼ Exp(1), Yn = logB Ξn.

Needed ingredients:⋄∫∞

0 exp(−x)xs−1dx = Γ(s).⋄ |Γ(1 + ix)| =

√πx/ sinh(πx), x ∈ R.

|Pn(s)− log10(s)| ≤

logB s∞∑

ℓ=1

(2π2ℓ/ logB

sinh(2π2ℓ/ logB)

)n/2

.

173


Example: All Xi ∼ Exp(1)

Bounds on the error|Pn(s)− log10 s| ≤⋄ 3.3 · 10−3 logB s if n = 2,⋄ 1.9 · 10−4 logB s if n = 3,⋄ 1.1 · 10−5 logB s if n = 5, and⋄ 3.6 · 10−13 logB s if n = 10.

Error at most

log10 s∞∑

ℓ=1

(17.148ℓ

exp(8.5726ℓ)

)n/2

≤ .057n log10 s

174

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