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Math 341: ProbabilityTwenty-first Lecture (11/24/09)
Steven J MillerWilliams College
[email protected]://www.williams.edu/go/math/sjmiller/
public html/341/
Bronfman Science CenterWilliams College, November 24, 2009
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Summary for the Day
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Summary for the day
Benford’s Law:⋄ Review.⋄ Inputs (equidistribution).⋄ Clicker question.⋄ Difference equations.⋄ Products.
More Sum Than Difference Sets:⋄ Definition.⋄ Inputs (Chebyshev’s Theorem).
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Introduction
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Caveats!
Not all fraud can be detected by Benford’s Law.
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Caveats!
Not all fraud can be detected by Benford’s Law.
A math test indicating fraud is not proof of fraud:unlikely events, alternate reasons.
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Daily Summary Introduction Clicker Questions Theory Applications Mod 1 CLT Products ℱ
Caveats!
Not all fraud can be detected by Benford’s Law.
A math test indicating fraud is not proof of fraud:unlikely events, alternate reasons.
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Notation
Logarithms: logB x = y means x = By .⋄ Example: log10 100 = 2 as 100 = 102.⋄ logB(uv) = logB u + logB v .⋄ log10(100 ⋅ 1000) = log10(100) + log10(1000).Set Theory:⋄ ℚ = rational numbers = {p/q : p, q integers}.⋄ x ∈ S means x belongs to S.⋄ [a, b] = {x : a ≤ x ≤ b}.Modulo 1:⋄ Any x can be written as integer + fraction.⋄ x mod 1 means just the fractional part.⋄ Example: � mod 1 is about .14159.
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Benford’s Law: Newcomb (1881), Benford (1938)
StatementFor many data sets, probability of observing a first digit ofd base B is logB
(d+1
d
); base 10 about 30% are 1s.
Not all data sets satisfy Benford’s Law.⋄ Long street [1, L]: L = 199 versus L = 999.⋄ Oscillates between 1/9 and 5/9 with first digit 1.⋄ Many streets of different sizes: close to Benford.
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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
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Applications
analyzing round-off errors
determining the optimal way to storenumbers
detecting tax and image fraud, and dataintegrity
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ClickerQuestion
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First 60 numbers of the form 2n
digit # Obs # Pred Obs Prob Benf Prob1 18 18.1 .300 .3012 12 10.6 .200 .1763 6 7.5 .100 .1254 6 5.8 .100 .0975 6 4.8 .100 .0796 4 4.0 .067 .0677 2 3.5 .033 .0588 5 3.1 .083 .0519 1 2.7 .017 .046
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First 60 numbers of the form 2n
digit # Obs # Pred Obs Prob Benf Prob1 18 18.1 .300 .3012 12 10.6 .200 .1763 6 7.5 .100 .1254 6 5.8 .100 .0975 6 4.8 .100 .0796 4 4.0 .067 .0677 2 3.5 .033 .0588 5 3.1 .083 .0519 1 2.7 .017 .046
As N → ∞, is {2n}Nn=0 Benford? (a) yes (b) no (c) open.
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First 60 numbers of the form 2n
digit # Obs # Pred Obs Prob Benf Prob1 18 18.1 .300 .3012 12 10.6 .200 .1763 6 7.5 .100 .1254 6 5.8 .100 .0975 6 4.8 .100 .0796 4 4.0 .067 .0677 2 3.5 .033 .0588 5 3.1 .083 .0519 1 2.7 .017 .046
As N → ∞, is {2n}Nn=0 Benford? (a) yes (b) no (c) open.
Are the 9s low in limit? (a) yes (b) no (c) open.15
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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
).
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
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Data Analysis
�2-Tests: Test if theory describes data⋄ Expected probability: pd = log10
(d+1
d
).
⋄ Expect about Npd will have first digit d .⋄ Observe Obs(d) with first digit d .
⋄ �2 =∑9
d=1(Obs(d)−Npd )
2
Npd.
⋄ Smaller �2, more likely correct model.
Will study n, en, �n.
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Logarithms and Benford’s Law
�2 values for �n, 1 ≤ n ≤ N (5% 15.5).N �2( ) �2(e) �2(�)
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
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Logarithms and Benford’s Law: Base 10
log(�2) vs N for �n (red) and en (blue),n ∈ {1, . . . ,N}. Note �175 ≈ 1.0028 ⋅ 1087, (5%,log(�2) ≈ 2.74).
200 400 600 800 1000
-1.5
-1.0
-0.5
0.5
1.0
1.5
2.0
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Logarithms and Benford’s Law: Base 20
log(�2) vs N for �n (red) and en (blue),n ∈ {1, . . . ,N}. Note e3 ≈ 20.0855, (5%,log(�2) ≈ 2.74).
5000 10 000 15 000 20 000
-2
-1
1
2
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General Theory
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Mantissas
Mantissa: x = M10(x) ⋅ 10k , k integer.
M10(x) = M10(x) if and only if x and x have thesame leading digits.
Key observation: log10(x) = log10(x) mod 1 ifand only if x and x have the same leading digits.Thus often study y = log10 x .
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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.
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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.
Thm: � ∕∈ ℚ, n� is equidistributed mod 1.
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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.
Thm: � ∕∈ ℚ, n� is equidistributed mod 1.
Examples: log10 2, log10
(1+
√5
2
)∕∈ ℚ.
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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.
Thm: � ∕∈ ℚ, n� is equidistributed mod 1.
Examples: log10 2, log10
(1+
√5
2
)∕∈ ℚ.
Proof: if rational: 2 = 10p/q.
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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.
Thm: � ∕∈ ℚ, n� is equidistributed mod 1.
Examples: log10 2, log10
(1+
√5
2
)∕∈ ℚ.
Proof: if rational: 2 = 10p/q.Thus 2q = 10p or 2q−p = 5p, impossible.
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Example of Equidistribution: n√� mod 1
0.2 0.4 0.6 0.8 1
0.5
1.0
1.5
2.0
n√� mod 1 for n ≤ 10
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Example of Equidistribution: n√� mod 1
0.2 0.4 0.6 0.8 1
0.2
0.4
0.6
0.8
1.0
n√� mod 1 for n ≤ 100
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Example of Equidistribution: n√� mod 1
0.2 0.4 0.6 0.8 1
0.2
0.4
0.6
0.8
1.0
n√� mod 1 for n ≤ 1000
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Example of Equidistribution: n√� mod 1
0.2 0.4 0.6 0.8 1
0.2
0.4
0.6
0.8
1.0
n√� mod 1 for n ≤ 10, 000
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Denseness
DenseA sequence {zn}∞n=1 of numbers in [0, 1] isdense if for any interval [a, b] there are infinitelymany zn in [a, b].
Dirichlet’s Box (or Pigeonhole) Principle:If n + 1 objects are placed in n boxes, atleast one box has two objects.
Denseness of n�:Thm: If � ∕∈ ℚ then zn = n� mod 1 is dense.
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Proof n� mod 1 dense if � ∕∈ ℚ
Enough to show in [0, b] infinitely often forany b.
Choose any integer Q > 1/b.
Q bins:[0, 1
Q
],[
1Q ,
2Q
], . . . ,
[Q−1
Q ,Q].
Q + 1 objects:{� mod 1, 2� mod 1, . . . , (Q + 1)� mod 1}.
Two in same bin, say q1� mod 1 andq2� mod 1.
Exists integer p with 0 < q2�− q1�− p < 1Q .
Get (q2 − q1)� mod 1 ∈ [0, b].33
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Logarithms and Benford’s Law
Fundamental EquivalenceData set {xi} is Benford base B if {yi} isequidistributed mod 1, where yi = logB xi .
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Logarithms and Benford’s Law
Fundamental EquivalenceData set {xi} is Benford base B if {yi} isequidistributed mod 1, where yi = logB xi .
0 1log 2 � log 10
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Logarithms and Benford’s Law
Fundamental EquivalenceData set {xi} is Benford base B if {yi} isequidistributed mod 1, where yi = logB xi .
0 1
1 102
log 2 � log 10
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Logarithms and Benford’s Law
Fundamental EquivalenceData set {xi} is Benford base B if {yi} isequidistributed mod 1, where yi = logB xi .
Proof:x = MB(x) ⋅ Bk for some k ∈ ℤ.FDB(x) = d iff d ≤ MB(x) < d + 1.logB d ≤ y < logB(d + 1), y = logB x mod 1.If Y ∼ Unif(0, 1) then above probability islogB
(d+1
d
).
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Examples
2n is Benford base 10 as log10 2 ∕∈ ℚ.
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Daily Summary Introduction Clicker Questions Theory Applications Mod 1 CLT Products ℱ
Examples
Fibonacci numbers are Benford base 10.
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Daily Summary Introduction Clicker Questions Theory Applications Mod 1 CLT Products ℱ
Examples
Fibonacci numbers are Benford base 10.an+1 = an + an−1.
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Daily Summary Introduction Clicker Questions Theory Applications Mod 1 CLT Products ℱ
Examples
Fibonacci numbers are Benford base 10.an+1 = an + an−1.Guess an = nr : rn+1 = rn + rn−1 or r2 = r +1.
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Daily Summary Introduction Clicker Questions Theory Applications Mod 1 CLT Products ℱ
Examples
Fibonacci numbers are Benford base 10.an+1 = an + an−1.Guess an = nr : rn+1 = rn + rn−1 or r2 = r +1.Roots r = (1 ±
√5)/2.
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Daily Summary Introduction Clicker Questions Theory Applications Mod 1 CLT Products ℱ
Examples
Fibonacci numbers are Benford base 10.an+1 = an + an−1.Guess an = nr : rn+1 = rn + rn−1 or r2 = r +1.Roots r = (1 ±
√5)/2.
General solution: an = c1rn1 + c2rn
2 .
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Daily Summary Introduction Clicker Questions Theory Applications Mod 1 CLT Products ℱ
Examples
Fibonacci numbers are Benford base 10.an+1 = an + an−1.Guess an = nr : rn+1 = rn + rn−1 or r2 = r +1.Roots r = (1 ±
√5)/2.
General solution: an = c1rn1 + c2rn
2 .
Binet: an = 1√5
(1+
√5
2
)n− 1√
5
(1−
√5
2
)n.
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Daily Summary Introduction Clicker Questions Theory Applications Mod 1 CLT Products ℱ
Examples
Fibonacci numbers are Benford base 10.an+1 = an + an−1.Guess an = nr : rn+1 = rn + rn−1 or r2 = r +1.Roots r = (1 ±
√5)/2.
General solution: an = c1rn1 + c2rn
2 .
Binet: an = 1√5
(1+
√5
2
)n− 1√
5
(1−
√5
2
)n.
Most linear recurrence relations Benford:
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Daily Summary Introduction Clicker Questions Theory Applications Mod 1 CLT Products ℱ
Examples
Fibonacci numbers are Benford base 10.an+1 = an + an−1.Guess an = nr : rn+1 = rn + rn−1 or r2 = r +1.Roots r = (1 ±
√5)/2.
General solution: an = c1rn1 + c2rn
2 .
Binet: an = 1√5
(1+
√5
2
)n− 1√
5
(1−
√5
2
)n.
Most linear recurrence relations Benford:⋄ an+1 = 2an
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Daily Summary Introduction Clicker Questions Theory Applications Mod 1 CLT Products ℱ
Examples
Fibonacci numbers are Benford base 10.an+1 = an + an−1.Guess an = nr : rn+1 = rn + rn−1 or r2 = r +1.Roots r = (1 ±
√5)/2.
General solution: an = c1rn1 + c2rn
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
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Daily Summary Introduction Clicker Questions Theory Applications Mod 1 CLT Products ℱ
Examples
Fibonacci numbers are Benford base 10.an+1 = an + an−1.Guess an = nr : rn+1 = rn + rn−1 or r2 = r +1.Roots r = (1 ±
√5)/2.
General solution: an = c1rn1 + c2rn
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.
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Daily Summary Introduction Clicker Questions Theory Applications Mod 1 CLT Products ℱ
Applications
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Daily Summary Introduction Clicker Questions Theory Applications Mod 1 CLT Products ℱ
Stock Market
Milestone Date Effective Rate from last milestone108.35 Jan 12, 1906500.24 Mar 12, 1956 3.0%
1003.16 Nov 14, 1972 4.2%2002.25 Jan 8, 1987 4.9%3004.46 Apr 17, 1991 9.5%4003.33 Feb 23, 1995 7.4%5023.55 Nov 21, 1995 30.6%6010.00 Oct 14, 1996 20.0%7022.44 Feb 13, 1997 46.6%8038.88 Jul 16, 1997 32.3%9033.23 Apr 6, 1998 16.1%
10006.78 Mar 29, 1999 10.5%11209.84 Jul 16, 1999 38.0%12011.73 Oct 19, 2006 1.0%13089.89 Apr 25, 2007 16.7%14000.41 Jul 19, 2007 28.9%
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Daily Summary Introduction Clicker Questions Theory Applications Mod 1 CLT Products ℱ
Applications for the IRS: Detecting Fraud
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Daily Summary Introduction Clicker Questions Theory Applications Mod 1 CLT Products ℱ
Applications for the IRS: Detecting Fraud
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Daily Summary Introduction Clicker Questions Theory Applications Mod 1 CLT Products ℱ
Applications for the IRS: Detecting Fraud
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Daily Summary Introduction Clicker Questions Theory Applications Mod 1 CLT Products ℱ
Applications for the IRS: Detecting Fraud (cont)
Embezzler started small and then increased dollaramounts.
Most amounts below $100,000 (critical threshold fordata requiring additional scrutiny).
Over 90% had first digit of 7, 8 or 9.
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Daily Summary Introduction Clicker Questions Theory Applications Mod 1 CLT Products ℱ
Detecting Fraud
Bank FraudAudit of a bank revealed huge spike of numbersstarting with 48 and 49, most due to one person.
Write-off limit of $5,000. Officer had friends applyingfor credit cards, ran up balances just under $5,000then he would write the debts off.
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Daily Summary Introduction Clicker Questions Theory Applications Mod 1 CLT Products ℱ
Detecting Fraud
EnronBenford’s Law detected manipulation of revenuenumbers.
Results showed a tendency towards round EarningsPer Share (0.10, 0.20, etc.).Consistent with a small but noticeable increase inearnings management in 2002.
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Data Integrity: Stream Flow Statistics: 130 years, 457,440 records
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Election Fraud: Iran 2009
Numerous protests and complaints over Iran’s 2009elections.Lot of analysis done; data is moderately suspicious.Tests done include
First and second leading digits;Last two digits (should almost be uniform);Last two digits differing by at least 2.
Warning: do enough tests, even if nothing is wrong willfind a suspicious result, but when all tests are on theboundary....
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Daily Summary Introduction Clicker Questions Theory Applications Mod 1 CLT Products ℱ
The Modulo 1Central Limit Theorem
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Daily Summary Introduction Clicker Questions Theory Applications Mod 1 CLT Products ℱ
Needed Input: Poisson Summation Formula
Poisson Summation Formulaf nice:
∞∑
ℓ=−∞
f (ℓ) =
∞∑
ℓ=−∞
f (ℓ),
Fourier transform f (�) =
∫∞
−∞
f (x)e−2�ix�dx .
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Daily Summary Introduction Clicker Questions Theory Applications Mod 1 CLT Products ℱ
Needed Input: Poisson Summation Formula
Poisson Summation Formulaf nice:
∞∑
ℓ=−∞
f (ℓ) =
∞∑
ℓ=−∞
f (ℓ),
Fourier transform f (�) =
∫∞
−∞
f (x)e−2�ix�dx .
What is ‘nice’?f Schwartz more than enough.
f twice continuously differentiable & f , f ′, f ′′ decay likex−(1+�) for an � > 0 (g decays like x−a if ∃x0,C st∣x ∣ > x0, ∣g(x)∣ ≤ C/∣x ∣a).
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Daily Summary Introduction Clicker Questions Theory Applications Mod 1 CLT Products ℱ
Modulo 1 Central Limit Theorem
The Modulo 1 Central Limit Theorem for Independent
Let {Ym} be independent continuous random variables on[0, 1), not necessarily identically distributed, with densities{gm}. A necessary and sufficient condition forY1 + ⋅ ⋅ ⋅+ YM modulo 1 to converge to the uniformdistribution as M → ∞ (in L1([0, 1]) is that for each n ∕= 0we have limM→∞ g1(n) ⋅ ⋅ ⋅ gM(n) = 0.
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Daily Summary Introduction Clicker Questions Theory Applications Mod 1 CLT Products ℱ
Modulo 1 Central Limit Theorem
The Modulo 1 Central Limit Theorem for Independent
Let {Ym} be independent continuous random variables on[0, 1), not necessarily identically distributed, with densities{gm}. A necessary and sufficient condition forY1 + ⋅ ⋅ ⋅+ YM modulo 1 to converge to the uniformdistribution as M → ∞ (in L1([0, 1]) is that for each n ∕= 0we have limM→∞ g1(n) ⋅ ⋅ ⋅ gM(n) = 0.
Application to Benford’s law: If X = X1 ⋅ ⋅ ⋅XM then
log10 X = log10 X1 + ⋅ ⋅ ⋅+ log10 XM := Y1 + ⋅ ⋅ ⋅+ YM .
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Products of Random Variablesand the Fourier Transform
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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, g(�) = g1(�) ⋅ ⋅ ⋅ gn(�).Dirac delta functional:
∫��(y)g(y)dy = g(�).
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Daily Summary Introduction Clicker Questions Theory Applications Mod 1 CLT Products ℱ
Fourier input
Fejér kernel:
FN(x) =N∑
n=−N
(1 − ∣n∣
N
)e2�inx .
Fejér series:
TN f (x) = (f ∗ FN)(x) =N∑
n=−N
(1 − ∣n∣
N
)f (n)e2�inx .
Lebesgue’s Theorem: f ∈ L1([0, 1]). As N → ∞, TN fconverges to f in L1([0, 1]).TN(f ∗ g) = (TN f ) ∗ g: convolution assoc.
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Daily Summary Introduction Clicker Questions Theory Applications Mod 1 CLT Products ℱ
Modulo 1 Central Limit Theorem
Theorem (M– and Nigrini 2007)
{Ym} independent continuous random variables on [0, 1)(not necc. i.i.d.), densities {gm}. Y1 + ⋅ ⋅ ⋅+ YM mod 1converges to the uniform distribution as M → ∞ inL1([0, 1]) iff ∀n ∕= 0, limM→∞ g1(n) ⋅ ⋅ ⋅ gM(n) = 0.
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Daily Summary Introduction Clicker Questions Theory Applications Mod 1 CLT Products ℱ
Generalizations
Levy proved for i.i.d.r.v. just one year after Benford’spaper.
Generalized to other compact groups, with estimateson the rate of convergence.⋄ Stromberg: n-fold convolution of a regularprobability measure on a compact Hausdorff group Gconverges to normalized Haar measure in weak-startopology iff support of the distribution not contained ina coset of a proper normal closed subgroup of G.
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Daily Summary Introduction Clicker Questions Theory Applications Mod 1 CLT Products ℱ
Theorem (M– and Nigrini 2007)
{Ym} indep. discrete random variables on [0, 1), not necc.identically distributed, densities
gm(x) =rm∑
k=1
wk ,m��k,m(x),wk ,m > 0,rm∑
k=1
wk ,m = 1.
Assume that there is a finite set A ⊂ [0, 1) such that all�k ,m ∈ A. Y1 + ⋅ ⋅ ⋅+ YM mod 1 converges weakly to theuniform distribution as M → ∞ iff ∀n ∕= 0,limM→∞ g1(n) ⋅ ⋅ ⋅ gM(n) = 0.
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Daily Summary Introduction Clicker Questions Theory Applications Mod 1 CLT Products ℱ
Distribution of digits (base 10) of 1000 productsX1 ⋅ ⋅ ⋅X1000, where g10,m = �11m .�m(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
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Daily Summary Introduction Clicker Questions Theory Applications Mod 1 CLT Products ℱ
Proof of Modulo 1 CLT
Density of sum is hℓ = g1 ∗ ⋅ ⋅ ⋅ ∗ gℓ.
Suffices show ∀�: limM→∞
∫ 10 ∣hM(x)− 1∣dx < �.
Lebesgue’s Theorem: N large,
∣∣h1 − TNh1∣∣1 =
∫ 1
0∣h1(x)− TNh1(x)∣dx <
�
2.
Claim: above holds for hM for all M.
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Daily Summary Introduction Clicker Questions Theory Applications Mod 1 CLT Products ℱ
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 <
�
2.
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Daily Summary Introduction Clicker Questions Theory Applications Mod 1 CLT Products ℱ
Proof of Modulo 1 CLT (continued)
Show limM→∞ ∣∣hM − 1∣∣1 = 0.Triangle inequality:
∣∣hM − 1∣∣1 ≤ ∣∣hM − TNhM ∣∣1 + ∣∣TNhM − 1∣∣1.
Choices of N and �:
∣∣hM − TNhM ∣∣1 < �/2.
Show ∣∣TNhM − 1∣∣1 < �/2.
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Daily Summary Introduction Clicker Questions Theory Applications Mod 1 CLT Products ℱ
∣∣TNhM − 1∣∣1 =
∫ 1
0
∣∣∣∣∣∣∣
N∑
n=−Nn ∕=0
(1 − ∣n∣
N
)hM(n)e2�inx
∣∣∣∣∣∣∣dx
≤N∑
n=−Nn ∕=0
(1 − ∣n∣
N
)∣hM(n)∣
hM(n) = g1(n) ⋅ ⋅ ⋅ gM(n) −→M→∞ 0.For fixed N and �, choose M large so that ∣hM(n)∣ < �/4Nwhenever n ∕= 0 and ∣n∣ ≤ N.
74