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foo101 | 6 years ago
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.
EGreg|6 years ago
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.