4 ms·

Essentially similar to “after N throws of a coin, what will the difference be between heads and tails?”. It will most likely be away from 0, and probably large.

thaumasiotes 1y ago
Isn't the expectation sqrt(N)? (The expectation of |H - T|, that is; the expectation of (H - T) is 0.)
- pfortuny 1y ago
  I think it is so (I seem to recall it and the example from Feller's book).
  - madcaptenor 1y ago
    It's definitely on the order of sqrt(N). But it's a bit smaller than sqrt(N) typically: (1) H is asymptotically normally distributed with mean N/2 and variance N/4 (central limit theorem) (2) So H - T = H - (N - H) = 2H - N is ~normally distributed with mean 0 and variance N, or st.dev. sigma = sqrt(N) (3) So |H-T| is "half-normal" (https://en.wikipedia.org/wiki/Half-normal_distribution https://en.wikipedia.org/wiki/Half-normal_distribution). This has mean sigma * sqrt(2/pi) or approximately 0.8 sqrt(N). This can be checked with a quick simulation. For example in R I run 1000 simulations with N = 10000 with the one-liner mean(abs(2*rbinom(1000, 10000, 0.5)-10000)) which returns 79.944. The mean square of H-T will be approximately N, though (basically we can ignore the absolute value when computing the variance). In fact, it's exactly N. H is Binomial(N, 1/2) which has variance N/4. H - T = 2H - N has variance N/4 * 2^2 = N, and mean 0, so its mean square is N.
majormajor 1y ago
If you keep flipping a coin you get heads or tails every time. There's no streak-breaking limit where you stop counting consecutive heads after some point. But here a streak can run into a loss-limiting wall "for this turn, you have $0, so you can't give any more away." That boundary - unless you allow infinite negative balances - breaks the first-glance intuition. As people hit 0, the amount of dollars that are redistributed on the next turn goes down. So the richest will still be losing their dollar a turn, but are less likely to gain a dollar. Which will have the most significant effect on people nearly-broke, since they will still lose, but gain less frequently.
- pfortuny 1y ago
  OK, interesting take which I had not realized: the problem is also different because there is a limited amount of cash, I now realize. However, the idea that “the mean distribution (the uniform, in this case)” needs to be near the N-th step one is what is statistically wrong, that is why I turned to the coin example. But yes, thanks for pointing out the differences!

thaumasiotes 1y ago

Isn't the expectation sqrt(N)? (The expectation of |H - T|, that is; the expectation of (H - T) is 0.)

pfortuny 1y ago
I think it is so (I seem to recall it and the example from Feller's book).
- madcaptenor 1y ago
  It's definitely on the order of sqrt(N). But it's a bit smaller than sqrt(N) typically: (1) H is asymptotically normally distributed with mean N/2 and variance N/4 (central limit theorem) (2) So H - T = H - (N - H) = 2H - N is ~normally distributed with mean 0 and variance N, or st.dev. sigma = sqrt(N) (3) So |H-T| is "half-normal" (https://en.wikipedia.org/wiki/Half-normal_distribution https://en.wikipedia.org/wiki/Half-normal_distribution). This has mean sigma * sqrt(2/pi) or approximately 0.8 sqrt(N). This can be checked with a quick simulation. For example in R I run 1000 simulations with N = 10000 with the one-liner mean(abs(2*rbinom(1000, 10000, 0.5)-10000)) which returns 79.944. The mean square of H-T will be approximately N, though (basically we can ignore the absolute value when computing the variance). In fact, it's exactly N. H is Binomial(N, 1/2) which has variance N/4. H - T = 2H - N has variance N/4 * 2^2 = N, and mean 0, so its mean square is N.

majormajor 1y ago

If you keep flipping a coin you get heads or tails every time. There's no streak-breaking limit where you stop counting consecutive heads after some point. But here a streak can run into a loss-limiting wall "for this turn, you have $0, so you can't give any more away." That boundary - unless you allow infinite negative balances - breaks the first-glance intuition. As people hit 0, the amount of dollars that are redistributed on the next turn goes down. So the richest will still be losing their dollar a turn, but are less likely to gain a dollar. Which will have the most significant effect on people nearly-broke, since they will still lose, but gain less frequently.

pfortuny 1y ago
OK, interesting take which I had not realized: the problem is also different because there is a limited amount of cash, I now realize. However, the idea that “the mean distribution (the uniform, in this case)” needs to be near the N-th step one is what is statistically wrong, that is why I turned to the coin example. But yes, thanks for pointing out the differences!