I'm trying to arrive at the function that describes the following, but can't quite figure it out for multiple blocks. (there are some useful insights for a single block here.)

assume the Poisson point process that is the arrival of proof-of-work blocks on Monero. the mean of the block times is the target block time t (120 seconds).

also assume n subsequent blocks.

also assume p, the probability of n subsequent blocks having a mean block time less than or equal to T.

given t, n and p, how can I calculate T?

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[–] [email protected] 4 points 3 weeks ago (4 children)

Reply by rucknium:

If I understand your question right, I think you're looking for the inverse cumulative distribution function (a.k.a. quantile function) of the Erlang distribution.

The random length of time to mine the next block has an exponential distribution with rate parameter 1/t. The length of time to mine n blocks has an Erlang distribution with shape parameter n and rate parameter 1/t.

The Erlang distribution is a special case of the Gamma distribution. The Erlang distribution's shape parameter must be an integer, but the Gamma distribution's shape parameter can be any positive real number. We can use the Gamma distribution if Erlang isn't given to us by our calculator.

You would compute T in the R language with:

qgamma(p = p, shape = n, rate = 1/t)/n

The results of this simulation match the closed-form computation:

t <- 120
n <- 15
p <- 0.4

set.seed(314)

mining.times <- matrix(rexp(n * 100000, rate = 1/t), ncol = n)

mining.times <- rowSums(mining.times)

quantile(mining.times/n, probs = p)

qgamma(p = p, shape = n, rate = 1/t)/n
# Divide by n to get the mean instead of the total

[–] [email protected] 1 points 3 weeks ago (2 children)

are we having federation issues? or why did you repost it?

[–] [email protected] 2 points 3 weeks ago (1 children)

Town currently still blocks Tor exit nodes (because it is getting ddosed by them) and rucknium understandably doesn't want to connect without it.

[–] [email protected] 1 points 3 weeks ago

got it. they managed to post it anyway.

does having a hidden service introduce the same issues?

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