Benford's Law
en.wikipedia.org
en.wikipedia.org
Being evenly distributed over multiplication is natural for many measures. I think it is considered the most even, "maximum entropy" distribution for things that can't be negative.
The most even, maximum entropy distribution for an unbounded (-infinity to +infinity) prior is an unbounded uniform distribution but you can't apply it, for example, in the case of the size of objects since it would imply that negative sizes are as likely as positive sizes (and objects don't have negative sizes).
In general position parameters tend to be uniformly distributed over addition/subtraction whereas scale parameters tend to be evenly distributed over multiplication (uniform over log(x)). It's the reason some graphs use log axis. For Bayesians, it's a fairly straightforwards concept. See also improper priors (https://en.wikipedia.org/wiki/Prior_probability#Improper_pri...)
A simple uniform distribution does not obey Benford's law because each digit from 1 to 9 is equally likely to be the leading significant digit. Thus each digit occurs as the leading digit with 11.1% probability. See https://news.ycombinator.com/item?id=22340018 for an example.
So the probability that the number starts with a 5 (in decimal representation) is sort of like the probability of the sum over all N of:
(Maximum > N) (Number starts with 5 given that Maximum is N)
The thing is that when N is anything less than 9,999 etc. the second probability IS NOT 1/10. It is skewed towards the lower numbers due to the most significant digit being less than 9 more frequently!!!
So this is the real explanation. And I just came up with it on the spot while writing this. Yay!
Hehe
Look at: https://news.ycombinator.com/item?id=22340018 from elsewhere in this thread or https://news.ycombinator.com/item?id=21541264 of mine from a couple of months ago.
http://magarshak.com/blog/?p=318
Your script only tests samples which are from ranges where the max is a power of 10.
I’m sorry to tell you this, but you inadvertently misled people with that empirical test. This just goes to show that we have to check our assumptions, as scientists or mathematicians trying to prove a statement. (Even with empirical tests :)
Hopefully this message will fix that at least for those people that are reading this thread! The rest will be confused. But that’s what happens in science all the time.
PS: I edited the original Wikipedia page with the explanation :)
But let us accept your very limited range for a moment and go along with it. Then you say that the numbers in this range follow Benford's law. But clearly, it doesn't. None of the probabilities in this range obey the probabilities in Benford's law.
Someone needs to revert the dubious edit (https://en.wikipedia.org/w/index.php?title=Benford%27s_law&d...) you have made in Wikipedia.
It is trivial to see that literally any range with min = 0 and max = any number other than a power of 10 makes it LESS likely that a 9 will come up as the first digit. For example the range 0-300 has 1 and 2 come up as the first digit way more than the rest. Don’t you think the same is true of 0-30000 and 0-300000000000000000000000? The size of the range doesn’t make your assertion any more true, that for large ranges every leading digit begins to have an equal chance of appearing.
My point is that, given a uniform distribution from 0 to a max, it has to have a max somewhere. If we assume that max itself is uniformly distributed then we derive the proportions you find in Benford’s law.
Look to put it another way, Benford’s law comes from the numbers which are the same number of digits as the max. The rest are evenly distributed but those numbers are the most numerous at that point and they contribute the phenomenon. Ok?
Are you convinced?
PS: There has got to be someone who figured this out before 2020. Come on. Someone post a link to this derivation.
It is impossible on Hacker News for a new green account with less than 500 points to downvote someone else.
With large ranges, even if you exclude a power of 10 in the upper bound, it does not change the 11.11% chance of each digit being the first digit.
The empirical test is cherrypicked also.
If you don’t admit this then there is really no point to continue.
I wrote: > The leading digits of a uniform distribution does not follow Benford's law.
And @EGreg wrote: > I’m sorry to tell you this, but you inadvertently misled people with that empirical test. This just goes to show that we have to check our assumptions, as scientists or mathematicians trying to prove a statement. (Even with empirical tests :)
So, what specific range of the uniform distribution yields leading digits that follows Benford's law?
For example 0-300
One third of numbers are evenly distributed: 0-100
One third starts with 1: 100-200
One third starts with 2: 200-300
Do you understand?
Bensford's law is 1:30.1%, 2:17.6% 3:12.5% etc.
When the max is uniformly distributed then Benford’s law emerges. I mean, all you have to do is read the link - where I derive it.
What exactly is the law — please don’t handwave. If the law is those exact point values mentioned in the article then I just showed you how we arrived at them.
What does "max is uniformly distributed" even mean? If you think that the Benford's law holds good for a set of uniformly distributed numbers, why not simply provide that set? It would be so easy to prove your claim if you just provide an example set of numbers that obeys Benford's law.
All sets of numbers you have presented so far (0-300, 0-30000, 0-300000000000000000000000) do not follow Benford's law. It is very simple to show. In all these sets, the probability of first digit as 1 is equal to the probability of first digit as 2 which contradicts Benford's law.
With large ranges, even if you exclude a power of 10 in the upper bound, it does not change the 11.11% chance of each digit being the first digit.*
That is JUST FALSE ok? For for pretty much any distribution you choose for the max, other than 100% chance it is a power of 10 and 0% chance other numbers, you’ll get that the digit 1 comes up way more than 2, which would come up more than 3, etc. How much more? This comes from the fact that there are just as many numbers 100-200 as there are 0-100. Ok? And that’s all 1s. Then you hit the 2s, and so on.
If the max happens to be anywhere in the range 100-1000 with equal probability, you get that result. Benford’s law. If the max is distributed as some sort of continuous distribution — and not that ridiculous distribution of ONLY ever being powers of 10 — then you likely get something similar.
What are you arguing about?? If you are saying it’s mysterious why the lower digits come up more than higher ones, well the mystery is over. If you want an EXACT fit to the numbers in the article then I think they come out whenever the max is uniformly distributed between 10^n and 10^(n+1). But they may also have a sort of “law of large numbers” thing where pretty much any continuous distribution of the max leads to this law. That part I can’t tell you. What I can tell you is OBVIOUSLY the lower digits will come out more frequently.
(Btw it is NOT true for eg dictionary words for example, an initial A doesnt appear more than B. That should tell you something!)
And to understand the reason we just have to look at the family of uniform distributions, and see that for almost all of them, this proportion holds. Sure, for some of them, the 1,2,3 may be even MORE prevalent relative to 4-9 because the maximum value was 400 or 4000 or 40000. Ok? You can see this. For a uniformly distributed process that happens to have that as the maximum, Benford’s law will have the same proportions between 1,2,3 but then drop for 4-9 since they didn’t get that “boost”.
But if you keep sampling and this maximum keeps growing by some continuous distribution that’s not perfectly synced with the metric system, then it’s as likely to be in the range 100-200 as it is to be in 200-300. And then as likely to be in 1000-2000 as in 2000-3000. Given that, we get something like Benford’s law.
Now, perhaps it is ALSO TRUE for other distributions. I just explained why it’s true for uniform ones.
Benford’s law appears when the max of your range is uniformly distributed
If we are going to argue like this, I might as well present a set of two numbers S = {1, 2} and claim that when we choose numbers from uniform distribution, the probability of 3 occurring as the first digit is 0. Other commenters are not assuming weird distributions like this because this kind of discussion does not provide any new insights and is just a waste of time.
The law states that in many naturally occurring collections of numbers, the leading significant digit is likely to be small.
I have explained why that happens for the vast majority of UNIFORMLY DISTRIBUTED VARIABLES.
The vast majority. That implies that there is a collection of all possible uniformly distributed variables, and in particular those that are sampled from real world processes.
As long as they are uniformly distributed, with 0 as the minimum and M as the maximum, the first digit will appear more commonly.
I explained it several times. Why are you still insisting that statements about MAJORITY of uniform distributions are weird?
Yes statements about collections of uniform distributions are not statements about ONE SPECIFIC uniform distribution. And?
Pretty much all of them.
If you redefine the law like that, then sure, I agree that there are many uniform distributions too where the 1st digit is likely to be small. Here is another simple example: Consider the distribution of positive integers from 1 to 2. If we pick a number at random from {1, 2} then the 1st digit is likely to be small. This kind of analysis is boring.
But (fortunately!) that's not what Benford's law says. Benford's law provides a specific formula. Check https://en.wikipedia.org/wiki/Benford%27s_law#Definition to see the specific formula that must hold good for a set of numbers to be said to obey Benford's law. That's what makes Benford's law so interesting whereas your example ranges are degenerative cases where nothing new, surprising, or interesting is going on.
And once again, this is because of a simple analysis. Let me state it a DIFFERENT way, maybe this is something that you will take note of: the only time all possible digits have equal probability of being leading digits is when we have a uniform distribution with a max that is a power of 10. Literally every other distribution starts to exhibit that phenomenon. Now you can quibble as to what distributions lead to that EXACT curve fit. And there can be explanations for why power law distributions do. But other distributions exhibit this same PHENOMENON, while not necessarily converging to that exact proportion. As I said before, any other uniform distribution would not have those exact proportions, but would exhibit the phenomenon.
Basically, every continuous distribution is highly unlikely to have a cliff at a power of 10. It is going to go down gradually, and therefore if it includes the range 8000-9000 then it will probably include numbers above 10000. And even a discontinuous one with a uniform distribution (with a cliff at the end) will exhibit the phenomenon. OK? So if you have the range 8000-9000 in there, that means 1s and 2s were a lot more prevalent, and if you have a continuous distribution then 10,000+ numbers will be there, but perhaps not numbers 90,000+.
Do you at least get the intuition behind this? As soon as you get numbers close to a power of 10, your distribution probably includes numbers in the next order of magnitude, i.e. a lot of leading 1s. The more numbers you get starting with 9, the more it is highly unlikely you're right at the max of your distribution, with a cliff. Unless that happens to be the contrived "empirical test" that was linked to as "proof" that uniform distributions lead to equal changes for every digit to be leading.
The intuition is what matters. Now, maybe for UNIFORM distributions, or POWER distributions, that exact curve fit can be worked out. Perhaps you can show there is a "large" family of distributions for which the curve fits. Kind of like the law of large numbers. But I didn't do anything quite as ambitious. I simply showed why it's not just true for normally distributed processes, but others which you would think are uniformly distributed. Because chances are, that process has more 1s than 2s, and 2s than 3s, in exactly the proportion that a uniform distribution with some max would have, and the chances are the max isn't exactly a power of 10.
In practice, all this means is that the distribution may look like Benford's law for 1, 2, 3 and then drop to equally small for 4, 5, 6, 7, 8, 9. The 1 is going to be 10x more prevalent than the 9. The 2 would 5x more prevalent or whatever, UNLESS the distribution had a cutoff right before 200 or 2000. Understand? And this DOES HAPPEN.
There is a distribution of leading digits that looks like:
d P(d)
1 30.1%
2 17.6%
3 12.5%
4 9.7%
5 7.9%
6 6.7%
7 5.8%
8 5.1%
9 4.6%
Neat! For each of those data sets you get the same distribution. Now, someone (I won't say who), says that it also is true for the uniform distribution.
But it isn't.
It simply isn't.
And I said as much when I said, "The leading digits of a uniform distribution does not follow Benford's law."
And your counter example is if you take a uniform distribution from 0-300, the leading digits go to something like:
d P(d)
1 36.7%
2 36.7%
3 3.7%
4 3.7%
5 3.7%
6 3.7%
7 3.7%
8 3.7%
9 3.7%
> "The leading digits of a uniform distribution does not follow Benford's law." -- me
And you, directly disagreeing with that correct statement:
> This just goes to show that we have to check our assumptions, as scientists or mathematicians trying to prove a statement. -- EGreg
Indeed.
In practice though you can use a wide Gaussian if it is more mathematically convenient and get about the same results.
To illustrate Benford's Law, Dr. Mark J. Nigrini offered this example: "If we think of the Dow Jones stock average as 1,000, our first digit would be 1.
"To get to a Dow Jones average with a first digit of 2, the average must increase to 2,000, and getting from 1,000 to 2,000 is a 100 percent increase.
"Let's say that the Dow goes up at a rate of about 20 percent a year. That means that it would take five years to get from 1 to 2 as a first digit.
"But suppose we start with a first digit 5. It only requires a 20 percent increase to get from 5,000 to 6,000, and that is achieved in one year.
"When the Dow reaches 9,000, it takes only an 11 percent increase and just seven months to reach the 10,000 mark, which starts with the number 1. At that point you start over with the first digit a 1, once again. Once again, you must double the number -- 10,000 -- to 20,000 before reaching 2 as the first digit.
"As you can see, the number 1 predominates at every step of the progression, as it does in logarithmic sequences."
So the plan was to make the low order bits of the adder slower and more power efficient. Make the high order bits faster and less power efficient. Meet the same overall clock rate target given by the worst case but use less power in the average case.
Sadly memory operations, which look a lot like pure entropy, were sufficiently common that the gains just weren't worth the design effort.
IIRC at one point his numbers showed on a month to month basis losses only 4% of the time.
https://en.wikipedia.org/wiki/Bernie_Madoff#Investment_scand...
Though it doesn't indicate how many previous discussions there were, or what discussion occurred. Dang's comments (or occasionally, those by others, I've been known to add similar myself) manifest that information into the thread in a more immediately tangible way.
That's ... an interesting property of UI/UX and simple perception that itself leads to some interesting questions. I've been looking at the notion of manifest vs. latent functions (Bronislaw Malinowski and Robert K. Merton), which implies the related notion of manifest vs. latent perceptions or awareness (I'm not finding any ... manifest ... discussion of this, though I'm sure it exists).
And there's a strong psychological element here.
Dang's moderation often hinges on this -- retitling of posts and detaching of threads all change the degree of manifestation of certain aspects of a topic. Dang's guidance is generally toward promoting productive discussion, and of avoiding derailment.
Which might start approaching a latent answer to your question ;-)
Clearly even the mere suggestion touched the community’s proverbial nerve. I wonder if Dan G has life aspirations beyond HN moderator and how the community shall survive without his touch.
Of dang's future aspirations I know not.
{'1': 0.6896551724137931, '2': 0.13793103448275862, '3': 0.06896551724137931, '4': 0.06896551724137931, '5': 0.0, '6': 0.0, '7': 0.0, '8': 0.034482758620689655, '9': 0.0}
It's not exactly Benford's distribution. It doesn't span more than two orders of magnitude, but it's somewhat Benfordian, and it is what you might expect from a distribution of comments.
Also, please don't report 16 digits when your data set only has 1 sigfig.
i: 10 min: 0.000000 max: 0.689655 range: 0.689655 width: 70 scale: 101.500000
1 0.69 **********************************************************************
2 0.14 **************
3 0.07 *******
4 0.07 *******
5 0.00
6 0.00
7 0.00
8 0.03 ***
9 0.00However, it's an interesting question whether the dimensionless fundamental constants have a deeper explanation, since they don't depend on unit choices. The fermion masses are roughly evenly distributed logarithmically, for example (which leads to Benford's law). There isn't a clear reason why.
Here's what I did:
Open IDLE and type:
from random import seed
from random import random
seed(1)
>>>random()
0.13436424411240122
>>>random()
0.8474337369372327
>>>random()
0.763774618976614
So with the above basis, I want to look at the distribution of the first nonzero digit after the decimal point.
So next put the following function in. (basically it just gets a random as above, but keeps multiplying by 10 as long as the first digit of the string conversion is a "0". this returns the first non-zero digit at the left.)
def getone():
floaty = random()
while(str(floaty)[0] == '0'):
floaty *= 10
return int(str(floaty)[0])
>>>getone()
3
>>>getone()
2
>>>getone()
4
>>>getone()
2
buckets = [0,0,0,0,0,0,0,0,0,0] # buckets for: 0,1,2,3,4,5,6,7,8,9
for tries in range(10000000):
buckets[getone()] += 1
for result in range(10):
print (result, ":", buckets[result])
0 : 0
1 : 1112927
2 : 1110821
3 : 1109741
4 : 1110793
5 : 1111604
6 : 1112178
7 : 1111272
8 : 1110402
9 : 1110262
Why not?
That's why for base 10, the probability of the leading significant digit to be 1 is log_10(1 + 1/1) = 0.301. In base 2, the probability of the leading significant digit to be 1 is, quite trivially, log_2(1 + 1/1) = 1.0. Of course, all numbers in binary must begin with the digit 1.
# uniform.py
import random
n = 10**4
samples = 10**6
counts = [0] * 10
for i in range(samples):
random_number = random.randint(1, n)
first_digit = int(str(random_number)[0])
counts[first_digit] += 1
for i, count in enumerate(counts):
print('{} => {:.3f}'.format(i, counts[i] / samples))
$ python3 uniform.py
0 => 0.000
1 => 0.111
2 => 0.111
3 => 0.111
4 => 0.111
5 => 0.111
6 => 0.111
7 => 0.111
8 => 0.111
9 => 0.111
# exponential.py
import random
n = 10**4
samples = 10**6
counts = [0] * 10
for i in range(samples):
random_number = 2 ** random.randint(1, n)
first_digit = int(str(random_number)[0])
counts[first_digit] += 1
for i, count in enumerate(counts):
print('{} => {:.3f}'.format(i, counts[i] / samples))
$ python3 exponential.py
0 => 0.000
1 => 0.301
2 => 0.176
3 => 0.125
4 => 0.097
5 => 0.079
6 => 0.067
7 => 0.058
8 => 0.051
9 => 0.046 0 => 0.000
1 => 0.301
2 => 0.200
3 => 0.100
4 => 0.100
5 => 0.100
6 => 0.100
7 => 0.000
8 => 0.100
9 => 0.000
With n = 100, the experimental results become closer to Benford's law:
0 => 0.000
1 => 0.300
2 => 0.170
3 => 0.130
4 => 0.100
5 => 0.070
6 => 0.070
7 => 0.060
8 => 0.050
9 => 0.050
0 => 0.000
1 => 0.301
2 => 0.176
3 => 0.125
4 => 0.097
5 => 0.079
6 => 0.069
7 => 0.056
8 => 0.052
9 => 0.045Here is the code:
import random
N = 10
counts = [0] * 10
for n in range(N):
first_digit = int(str(2**n)[0])
counts[first_digit] += 1
for i, count in enumerate(counts):
print('{} => {:.3f}'.format(i, counts[i] / N))
0 => 0.000
1 => 0.300
2 => 0.200
3 => 0.100
4 => 0.100
5 => 0.100
6 => 0.100
7 => 0.000
8 => 0.100
9 => 0.000
0 => 0.000
1 => 0.300
2 => 0.170
3 => 0.130
4 => 0.100
5 => 0.070
6 => 0.070
7 => 0.060
8 => 0.050
9 => 0.050
0 => 0.000
1 => 0.301
2 => 0.176
3 => 0.125
4 => 0.097
5 => 0.079
6 => 0.069
7 => 0.056
8 => 0.052
9 => 0.045
0 => 0.000
1 => 0.301
2 => 0.176
3 => 0.125
4 => 0.097
5 => 0.079
6 => 0.067
7 => 0.058
8 => 0.051
9 => 0.046
import random
N = 8192
counts = [0] * 10
for n in range(N):
first_digit = int(str(2**n)[0])
counts[first_digit] += 1
for i, count in enumerate(counts):
print('{} => {:.3f}'.format(i, counts[i] / N))
0 => 0.000
1 => 0.301
2 => 0.176
3 => 0.125
4 => 0.097
5 => 0.079
6 => 0.067
7 => 0.058
8 => 0.051
9 => 0.046
https://news.ycombinator.com/item?id=22341416
Any comments?