3 ms·
I agree that E[r() * r() * ...] = 2^-n but that's not the final word. E[log(r() * r() * ...)] = -n and Pr(r() * r() * ... > 2^-n) is very small. That's the intu
by sobellian 3y ago
I agree that E[r() * r() * ...] = 2^-n but that's not the final word. E[log(r() * r() * ...)] = -n and Pr(r() * r() * ... > 2^-n) is very small. That's the intuition part.
The two distributions are identical, I checked numerically and they have the same summary statistics. Furthermore, r() * n is a uniform distribution from 0 to n so by induction we can prove the two are identical.
- sobellian 3y agoFor anyone who cares and read this far (and if so, I'm honestly surprised) I made a colab that demonstrates all this: https://colab.research.google.com/drive/1xCZ24cLPNOqyJD0kUTOUMKm3jDsDb48-?usp=sharing https://colab.research.google.com/drive/1xCZ24cLPNOqyJD0kUTO... To wrap it up: - X == r() * r() * ... == r(r(...)) - E[X] = 2^-n - median[X] ~ 1/2 exp(1 - n) ?! Anyway, I thought it was interesting - especially how the median never quite "catches up" to the value one might naively compute from the CLT. I think there's some subtlety I'm missing there.