7 ms·

As the article explains, benford's law applies to distributions over values that evolve over time with multiplicative fluctuations, not multiplicative distribu

As the article explains, benford's law applies to distributions over values that evolve over time with multiplicative fluctuations, not multiplicative distributions themselves. A simple uniform distribution over [0,n] obeys benford's law, averages over any range of N.

BenoitEssiambre 7y ago
Time doesn't have to be a component. Also data distributed uniformly in a range would not obey Benford's law. Each digit would appear with equal frequency.
susam 7y ago
> A simple uniform distribution over [0,n] obeys benford's law 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 https://news.ycombinator.com/item?id=22340018 for an example.
- dmurray 7y ago
  What if N is 19?
  - susam 7y ago
    The larger the value of N is, the more the experiments agree with the Benford's law.
  - foo101 7y ago
    It does not matter what N you choose. The experimental results are always an approximation of Benford's law. The approximation gets better and better as N becomes larger and larger. The fact that you do not see Benford's law working with N = 19 is not in anyway a flaw in the way the experimental results are set up. It just means N is too small to approximate Benford's law closely.
- EGreg 7y ago
  Of course it does, because there is going to be a maximum of the range A, B. Even if the maximum itself would have a uniform distribution, the fact is that all numbers drawn from the samples will be below it. 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
  - jsweojtj 7y ago
    The leading digits of a uniform distribution does not follow Benford's law. Look at: https://news.ycombinator.com/item?id=22340018 https://news.ycombinator.com/item?id=22340018 from elsewhere in this thread or https://news.ycombinator.com/item?id=21541264 https://news.ycombinator.com/item?id=21541264 of mine from a couple of months ago.
    - EGreg 7y ago
      Here, I wrote it up to explain more clearly: http://magarshak.com/blog/?p=318 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 :)
      - mathduck 7y ago
        You are artificially constructing a very limited range (upper bound and lower bound) in which of course the likelihood of any digit being the first digit won't be equal. This is nothing new. This does not contradict what other commenters have said about uniform distribution. 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. 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&diff=941103496&oldid=940396536 https://en.wikipedia.org/w/index.php?title=Benford%27s_law&d...) you have made in Wikipedia.

BenoitEssiambre 7y ago

Time doesn't have to be a component. Also data distributed uniformly in a range would not obey Benford's law. Each digit would appear with equal frequency.

susam 7y ago

> A simple uniform distribution over [0,n] obeys benford's law 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 https://news.ycombinator.com/item?id=22340018 for an example.

dmurray 7y ago
What if N is 19?
- susam 7y ago
  The larger the value of N is, the more the experiments agree with the Benford's law.
- foo101 7y ago
  It does not matter what N you choose. The experimental results are always an approximation of Benford's law. The approximation gets better and better as N becomes larger and larger. The fact that you do not see Benford's law working with N = 19 is not in anyway a flaw in the way the experimental results are set up. It just means N is too small to approximate Benford's law closely.
EGreg 7y ago
Of course it does, because there is going to be a maximum of the range A, B. Even if the maximum itself would have a uniform distribution, the fact is that all numbers drawn from the samples will be below it. 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
- jsweojtj 7y ago
  The leading digits of a uniform distribution does not follow Benford's law. Look at: https://news.ycombinator.com/item?id=22340018 https://news.ycombinator.com/item?id=22340018 from elsewhere in this thread or https://news.ycombinator.com/item?id=21541264 https://news.ycombinator.com/item?id=21541264 of mine from a couple of months ago.
  - EGreg 7y ago
    Here, I wrote it up to explain more clearly: http://magarshak.com/blog/?p=318 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 :)
    - mathduck 7y ago
      You are artificially constructing a very limited range (upper bound and lower bound) in which of course the likelihood of any digit being the first digit won't be equal. This is nothing new. This does not contradict what other commenters have said about uniform distribution. 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. 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&diff=941103496&oldid=940396536 https://en.wikipedia.org/w/index.php?title=Benford%27s_law&d...) you have made in Wikipedia.