You ignored my point that each independent coin toss trial outcome is
uniformly distributed whereas the Poisson distribution is
exponentially distributed.
That is why I asserted that your and xulescu's analogies are inapplicable. Rare trial outcomes in a Poisson distribution occur less often then less rare ones (look at the area under the distribution curve at the tails). Whereas all trial outcomes in a coin toss occur at the same probability.
Ah *snooze*.
Do you really think we're idiots? The analogy was for an error in your modelling that you yourself accepted as valid, not for the numbers.
Er, no I didn't. I said I made no proof whether Poisson distribution is applicable. But note every research paper I read about Bitcoin claims the block chain is a Poisson process.
To my models it makes no difference whatsoever if they're coin tosses, loaded D20's or, indeed, exponentials or Poisson.
And I posited to you that your regression model is blind because you don't know the distribution of the process. You claim there isn't enough entropy to establish one, but that is your assumption. I don't see any proof from you to validate your assumption that block chains are not approximated by Poisson distributions. Why should I believe you when the research papers claim otherwise?
To address your point directly, a uniform distribution CANNOT have a tail by definition.
Exactly I made that point.
In terms of counts, both Poisson and binary distributions have normal tails.
If you mean the Binomial distribution, the random variable is the # of outcomes in a series of trials. I was referring to the distribution of a single trial. In the p=0.5 Bernoulli trial coin toss, the distribution of a trial is uniformall possible outcomes for each independent trial are equally probable aka uniform. Whereas, for the Poisson trial, the probabilities of the possible outcomes has an exponential tail.
Both also completely fail because they assume complete independence between the samples.
You make two afaik unproven assumptions:
1. Blocks are not significantly independent.
2. Imperfect independence makes the Poisson model a useless approximation.
In the meanwhile you keep wasting your time on this triviality.
I will not let myself be insulted on the math, unless I deserve it (which can happen) in which case I will mea culpa.