Question

2. Suppose that you read through this years issues of the New York Times and record each number that appears in a news article: the income of a CEO, the number of cases of wine produced by a winery, the total charitable contribution of a politician during the previous tax year, the age of a celebrity, and so on. Now focus on the leading digit of cach number, which could be 1,2,,8, or 9. Your first thought might be that the leading digit X of a randomly selected mber would be eay likely to be one of the nine possibilities (a discrete uniforin distribution). However, much empirical evidence as well as some theoretical argunents suggest an alternative probability distribution called Benfords law: Let X be the leading digit in a naturally occurring observed number p(z) = P(First digit ish-log10件1 r + (a) Show that the probability mass function sums to one for 1,,9, without computing indi vidual probabilities (b) Compute the individual probabilities and display them in a table. How does this probability digits wre eqally likely? Your distribution compare to the distribution of first digitsf answer may be visual or simply descriptive (c) Give the form for the CDF of X (d) Using this CDF, what is the probability that the leading digit of a natural number is less than 4? At most 4? More than 6?

Answer 1

Here X is a discreet random variable that can take up values 1,2,3…upto 9

We are given the probability mass function of X

p(x) = P(first digit is x) = log₁₀((x+1)/x)      x=1,2,3….,9

past. (a) To prove "Taking Left Hand Side We have, lo l t 10 eg togmithmic Propert -lo,a/c 7(可 で10 l D lo C log (2)4 19 104.19) + Opening both the brackets

LogTTI) t-log102) ーーLe g 13) -...log (g) all the other ogs get cancelled out except-Por-log (10) and logTO(1) tơ t o 느 log (10) _ Log (1) TO z Right Hand Side ot the problem Hente proved

PART(b)

Calculating individual probabilities using the given probability mass function

P(X=1) = log₁₀((x+1)/x) = log₁₀((1+1)/1) =  log₁₀2 = 0.301

P(X=2) = log₁₀((x+1)/x) = log₁₀((2+1)/2) =  log₁₀3/2 = 0.176

P(X=3) = log₁₀((x+1)/x) = log₁₀((3+1)/3) =  log₁₀4/3 = 0.125

P(X=4) = log₁₀((x+1)/x) = log₁₀((4+1)/4) =  log₁₀5/4 = 0.097

P(X=5) = log₁₀((x+1)/x) = log₁₀((5+1)/5) =  log₁₀6/5 = 0.079

P(X=6) = log₁₀((x+1)/x) = log₁₀((6+1)/6) =  log₁₀7/6 = 0.067

P(X=7) = log₁₀((x+1)/x) = log₁₀((7+1)/7) =  log₁₀8/7 = 0.058

P(X=8) = log₁₀((x+1)/x) = log₁₀((8+1)/8) =  log₁₀9/8 = 0.051

P(X=9) = log₁₀((x+1)/x) = log₁₀((9+1)/9) =  log₁₀10/9 =0.046

if all the digits were equally likely the probabilty would have been same for all the digits from 1,2,3...,9.

equal to 1/9 or 0.11

Random Variable X (first digit)	Probability Calculated from PMF (prob. mass function)	Probability P(X=x) if all the digits were equally likely
1	0.301	0.111
2	0.176	0.111
3	0.125	0.111
4	0.097	0.111
5	0.079	0.111
6	0.067	0.111
7	0.058	0.111
8	0.051	0.111
9	0.046	0.111

from the above table we observe that

the probability(first digit is x) calculated from the probability mass function keeps on decreasing, the digit 1 having the highest probability i.e., 0.301 and the digit 9 having the lowest probability i.e., 0.046.

while the probability if all the didgits were equally likely remains the same i.e., 0.111.

PART(c)

Since X is a discreet random variable

The form for the CDF of X is:

%;ール

where F(x) or P(X<=x) is the CDF of X

and p(x_i) or P(X=x_i) is the PMF of X

PART(d)

we have to find the probability that the leading digit of a natural number is less than 4

P(X<4) = ΣP(X=x_i) , for xi<4 i.e., x=1,2,3

P(X<4) = P(X=1) + P(X=2) + P(X=3)

since we have already calculated these values in PART(b) hence writing the values

we get,

P(X<4) = 0.301 +0.176 +0.125

P(X<4) = 0.602

we have to find the probability that the leading digit of a natural number is atmost 4

P(X<=4) = ΣP(X=x_i) , for xi<=4 i.e., x=1,2,3,4

P(X<=4) = P(X=1) + P(X=2) + P(X=3) + P(X=4)

since we have already calculated these values in PART(b) hence writing the values

we get,

P(X<=4) = 0.301 +0.176 +0.125 +0.097

P(X<=4) = 0.699

we have to find the probability that the leading digit of a natural number is more than 6

since by the formula for cdf we can only calculate the probabilities of the form P(X<x) or P(X<=x)

therefore, in order to calculate P(X>6), we will calculate P(X<=6) and subtract it from 1

i.e.,

P(X>6) = 1- ΣP(X=x_i) , for xi<=6 i.e., x=1,2,3,4,5,6

P(X>6) = 1 - [P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6)]

since we have already calculated these values in PART(b) hence writing the values

we get,

P(X>6) = 1 - [0.301 + 0.176 + 0.125 + 0.097 + 0.079 + 0.067]

P(X>6) = 1-0.845

P(X>6) = 0.155