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> I'm sorry if you interpreted it that way. is probably a good marker that it is time for me take my bow in this conversation, however, for an alternative appr

> I'm sorry if you interpreted it that way. is probably a good marker that it is time for me take my bow in this conversation, however, for an alternative approach I recommend SideQuark's comment on iterative random variables vs chains of multiplied random variables, which have different characteristics when defined as a sequence.

sobellian 3y ago
I checked his comment, I think he's incorrect on the approaches being different. Using my function `foo` that does the iterative approach, we can compare the distributions and they're fairly identical. >>> np.mean([foo(100) for _ in range(100000)]) 3.258425093913613e-33 >>> np.mean([np.product(np.random.random(100)) for _ in range(100000)]) 8.814732867008917e-33 (There's quite a bit of variance on a log scale, so 3 vs 8 is not a huge difference. I re-ran this and got varying numbers with both approaches. But the iterated code is very slow...) Note that the mean is actually quite a bit closer to 2^-100 even though the vast majority of numbers from either distribution fall below 2^-100. Even so, the mean for both is approximately a factor of 100 less than 2^-100. Suspicious! Although I think we've both burned enough time on this.
- anderskaseorg 3y ago
  Yeah, the problem with your sampling experiment (as your suspicious observation makes it sound like you’ve started to realize) is that this distribution has a very narrow tail to 1 that you won’t hit by accident but nonetheless contributes very significantly to the computation of the mean. You correctly computed that the mean of log(foo(n)) is exactly −n, and the median of foo(n) indeed shrinks like e^−n (more specifically like e^(1/3 − n), I think), but the mean of foo(n) is exactly 2^−n.
  - tbalsam 3y ago
    If this were stackoverflow, I'd mark this as an answer. Also answers my question up above a bit as well, just from a different approach. Many thanks for clarifying here and elsewhere! :) HN is a lovely, curious, and interesting place.

sobellian 3y ago

I checked his comment, I think he's incorrect on the approaches being different. Using my function `foo` that does the iterative approach, we can compare the distributions and they're fairly identical. >>> np.mean([foo(100) for _ in range(100000)]) 3.258425093913613e-33 >>> np.mean([np.product(np.random.random(100)) for _ in range(100000)]) 8.814732867008917e-33 (There's quite a bit of variance on a log scale, so 3 vs 8 is not a huge difference. I re-ran this and got varying numbers with both approaches. But the iterated code is very slow...) Note that the mean is actually quite a bit closer to 2^-100 even though the vast majority of numbers from either distribution fall below 2^-100. Even so, the mean for both is approximately a factor of 100 less than 2^-100. Suspicious! Although I think we've both burned enough time on this.

anderskaseorg 3y ago
Yeah, the problem with your sampling experiment (as your suspicious observation makes it sound like you’ve started to realize) is that this distribution has a very narrow tail to 1 that you won’t hit by accident but nonetheless contributes very significantly to the computation of the mean. You correctly computed that the mean of log(foo(n)) is exactly −n, and the median of foo(n) indeed shrinks like e^−n (more specifically like e^(1/3 − n), I think), but the mean of foo(n) is exactly 2^−n.
- tbalsam 3y ago
  If this were stackoverflow, I'd mark this as an answer. Also answers my question up above a bit as well, just from a different approach. Many thanks for clarifying here and elsewhere! :) HN is a lovely, curious, and interesting place.