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I did redo it, and the distributions are identical. It's just a very heavily skewed distribution, so the expected value is not very intuitive. I still think eve

I did redo it, and the distributions are identical. It's just a very heavily skewed distribution, so the expected value is not very intuitive. I still think even E[x] decreases faster than 2^-n though. See my other comments. Edit - Actually, I think E[x] = 2^-n, but showing this numerically for large n is impractical because of just how skewed the distribution is. It is simultaneously true that E[iter_rand(N)] = 2^-N and E[log(iter_rand(N))] = -N. Fun stuff :)

SideQuark 3y ago
It's not about intuition. I computed it directly from the integral definition for expected value. I posted it elsewhere in the thread. It proves it for all N. Also, you keep using the new incorrect claim that r() * r() * ... * r() is the same as r(r(...r())..). They are not.
- maxbond 3y ago
  random() * 5 is equivalent to random(5) (the former being the classic way to implement the latter). So why wouldn't random() * random() be equivalent to random(random())?
  - SideQuark 3y ago
    Multiplying by a constant commutes with expectation. Multiplying by a random variable does not. It may in some cases, but intuition is not the way to check it.
    - deleted 3y ago
      [deleted]
    - anderskaseorg 3y ago
      The claim here is not about expectations. random(X) is identically distributed to random() * X, even if X itself is a random variable, because the randomness in random() is independent from the randomness in X. Indeed, the way you would implement random(X), if you only had random() and X available, would be to compute random() * X. So in particular, random(random()) is identically distributed to random() * random(). (To be clear, this is different from random()^2, as pointed out elsewhere.)
      - jasomill 3y ago
        As a quick sanity check, Mathematica 13.2.1 Kernel for Linux ARM (64-bit) Copyright 1988-2023 Wolfram Research, Inc. In[1]:= DI = Distributed; In[2]:= EV = Expectation; In[3]:= SS = Subscript; In[4]:= UD = UniformDistribution; In[5]:= f[1] := EV[x, DI[x, UD[{0, 1}]]] In[6]:= f[n_Integer] := EV[x, DI[x, UD[{0, f[n - 1]}]]] In[7]:= Block[{n = 42}, EV[Product[SS[x, k], {k, 1, n}], Map[DI[SS[x, #], UD[{0, 1}]] &, Range[1, n]]] == Product[EV[x, DI[x, UD[{0, 1}]]], {k, 1, n}] == f[n]] Out[7]= True where n = 42 ∈ ℕ is arbitrary. Alas, Mathematica doesn't appear to support actual nested distributions of the form UD[{1,UD[{1,...}]}], so, e.g., f[2] == EV[x_2, DI[x_2, UD[{0, EV[x_1, DI[x_1, UD[{0,1}]]]}]]] == EV[x_2, DI[x_2, UD[{0,1/2}] == 1/4 which is trivial.
        jasomill 3y ago
        (As is the expectation of a uniform distribution in the first place. From the PDF for the uniform distribution (1/(b - a) for x ∈ [a,b], zero elsewhere) and the definition of EV as the integral of the product of the identity map and the PDF over the reals, EV[x, DI[x, UD[{a,b}]]] == Integrate[x * Piecewise[{ {1/(b - a), a <= x <= b}, {0, x < a || x > b}}], {x, -∞, ∞}] == Integrate[x * 1/(b - a), {x, a, b}] == ((b^2/2) - (a^2/2)) / (b - a) == (a + b)/2 so for a = 0, == b/2 no matter how many nested integrals it takes to define the real number b.)
      - SideQuark 3y ago
        Agreed, in this case they agree, but it's completely a numerical coincidence and not general. So people with a rough feeling that they compute rand * 5 should equate to rand(rand())=rand * rand should be admonished to check it extremely carefully. And the result is not true at the level of code. In practice it's not true, unless extreme care is taken to ensure perfect uniformity, which is rare due to nonuniformity of IEEE floats. I've not seen a codebase in the wild that makes all the required pieces work. It's doable, I've written articles on it, but it's costly to do. Another error in practice is that the method of taking a rand * 5 is not uniform due to numerical representations. So these are not identical in practice. This the best way to analyze, which is trivially analyzable, is the nested form. It's also why I wrote "may" in the previous comment.
- 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 ago
    For 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.

SideQuark 3y ago

It's not about intuition. I computed it directly from the integral definition for expected value. I posted it elsewhere in the thread. It proves it for all N. Also, you keep using the new incorrect claim that r() * r() * ... * r() is the same as r(r(...r())..). They are not.

maxbond 3y ago
random() * 5 is equivalent to random(5) (the former being the classic way to implement the latter). So why wouldn't random() * random() be equivalent to random(random())?
- SideQuark 3y ago
  Multiplying by a constant commutes with expectation. Multiplying by a random variable does not. It may in some cases, but intuition is not the way to check it.
  - deleted 3y ago
    [deleted]
  - anderskaseorg 3y ago
    The claim here is not about expectations. random(X) is identically distributed to random() * X, even if X itself is a random variable, because the randomness in random() is independent from the randomness in X. Indeed, the way you would implement random(X), if you only had random() and X available, would be to compute random() * X. So in particular, random(random()) is identically distributed to random() * random(). (To be clear, this is different from random()^2, as pointed out elsewhere.)
    - jasomill 3y ago
      As a quick sanity check, Mathematica 13.2.1 Kernel for Linux ARM (64-bit) Copyright 1988-2023 Wolfram Research, Inc. In[1]:= DI = Distributed; In[2]:= EV = Expectation; In[3]:= SS = Subscript; In[4]:= UD = UniformDistribution; In[5]:= f[1] := EV[x, DI[x, UD[{0, 1}]]] In[6]:= f[n_Integer] := EV[x, DI[x, UD[{0, f[n - 1]}]]] In[7]:= Block[{n = 42}, EV[Product[SS[x, k], {k, 1, n}], Map[DI[SS[x, #], UD[{0, 1}]] &, Range[1, n]]] == Product[EV[x, DI[x, UD[{0, 1}]]], {k, 1, n}] == f[n]] Out[7]= True where n = 42 ∈ ℕ is arbitrary. Alas, Mathematica doesn't appear to support actual nested distributions of the form UD[{1,UD[{1,...}]}], so, e.g., f[2] == EV[x_2, DI[x_2, UD[{0, EV[x_1, DI[x_1, UD[{0,1}]]]}]]] == EV[x_2, DI[x_2, UD[{0,1/2}] == 1/4 which is trivial.
      - jasomill 3y ago
        (As is the expectation of a uniform distribution in the first place. From the PDF for the uniform distribution (1/(b - a) for x ∈ [a,b], zero elsewhere) and the definition of EV as the integral of the product of the identity map and the PDF over the reals, EV[x, DI[x, UD[{a,b}]]] == Integrate[x * Piecewise[{ {1/(b - a), a <= x <= b}, {0, x < a || x > b}}], {x, -∞, ∞}] == Integrate[x * 1/(b - a), {x, a, b}] == ((b^2/2) - (a^2/2)) / (b - a) == (a + b)/2 so for a = 0, == b/2 no matter how many nested integrals it takes to define the real number b.)
    - SideQuark 3y ago
      Agreed, in this case they agree, but it's completely a numerical coincidence and not general. So people with a rough feeling that they compute rand * 5 should equate to rand(rand())=rand * rand should be admonished to check it extremely carefully. And the result is not true at the level of code. In practice it's not true, unless extreme care is taken to ensure perfect uniformity, which is rare due to nonuniformity of IEEE floats. I've not seen a codebase in the wild that makes all the required pieces work. It's doable, I've written articles on it, but it's costly to do. Another error in practice is that the method of taking a rand * 5 is not uniform due to numerical representations. So these are not identical in practice. This the best way to analyze, which is trivially analyzable, is the nested form. It's also why I wrote "may" in the previous comment.
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 ago
  For 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.