4 ms·

e shows up because we're doing an iterated product (of random variables, but still). If you look at the central limit theorem, sum(log(rand())) tends to N * E[l

e shows up because we're doing an iterated product (of random variables, but still). If you look at the central limit theorem, sum(log(rand())) tends to N * E[log(rand())] for large N. Well, what's E[log(rand())]? -1! Like I said, it's fairly easy to test this in a python terminal. I encourage anyone who doubts me to try it :)

tbalsam 3y ago
Alright, I tried it in the terminal and the numbers confirmed my earlier math. .5->.25->.125 mean as N->infinity. I chained the python uniform function like the above code in the originally linked post does and averaged the results. The snark to me and the other commenters is a bit unnecessary in either case. I'm not really sure where you're getting a natural log or an iterated product of random variables in this instance. Could you perhaps show where you're transforming uniform(uniform(uniform(uniform(0, 1)))) into the math you're showing above? I'm trying to follow along here but am having difficulty connecting your line of reasoning to the problem at hand.
- sobellian 3y ago
  Not trying to be snarky by any means. I'm sorry if you interpreted it that way. I don't think the difference will show up for small N. This is an asymptotes thing. Try it for N = 100, that's what I did. For example: >>> np.product(np.random.random(100)) 5.469939152265385e-43 >>> 1 / np.e**100 3.7200759760208555e-44 >>> 1 / 2**100 7.888609052210118e-31 The underlying thing here is that random(random()) in this case is the same as random() * random(). So random(random(random(...))) is the same as random() * random() * random() and then the analysis goes on. And sure, random() * random() has a mean close to 1/4. But the dynamics change as N becomes large. Edit - and just in case you doubt whether random() * random() * ... is a valid way of doing this, I also just checked the laborious way of doing it: >>> def foo(n): ... result = 1.0 ... for _ in range(n): ... result = np.random.uniform(0, result) ... return result ... >>> foo(100) 1.4267531652344414e-46 >>> foo(100) 7.852496730908136e-49 >>> foo(100) 1.3216070221780724e-41
  - tbalsam 3y ago
    > 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.

tbalsam 3y ago

Alright, I tried it in the terminal and the numbers confirmed my earlier math. .5->.25->.125 mean as N->infinity. I chained the python uniform function like the above code in the originally linked post does and averaged the results. The snark to me and the other commenters is a bit unnecessary in either case. I'm not really sure where you're getting a natural log or an iterated product of random variables in this instance. Could you perhaps show where you're transforming uniform(uniform(uniform(uniform(0, 1)))) into the math you're showing above? I'm trying to follow along here but am having difficulty connecting your line of reasoning to the problem at hand.

sobellian 3y ago
Not trying to be snarky by any means. I'm sorry if you interpreted it that way. I don't think the difference will show up for small N. This is an asymptotes thing. Try it for N = 100, that's what I did. For example: >>> np.product(np.random.random(100)) 5.469939152265385e-43 >>> 1 / np.e**100 3.7200759760208555e-44 >>> 1 / 2**100 7.888609052210118e-31 The underlying thing here is that random(random()) in this case is the same as random() * random(). So random(random(random(...))) is the same as random() * random() * random() and then the analysis goes on. And sure, random() * random() has a mean close to 1/4. But the dynamics change as N becomes large. Edit - and just in case you doubt whether random() * random() * ... is a valid way of doing this, I also just checked the laborious way of doing it: >>> def foo(n): ... result = 1.0 ... for _ in range(n): ... result = np.random.uniform(0, result) ... return result ... >>> foo(100) 1.4267531652344414e-46 >>> foo(100) 7.852496730908136e-49 >>> foo(100) 1.3216070221780724e-41
- tbalsam 3y ago
  > 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.