Anticipating critical transitions

The Perl code that you provide will depend a lot on your random number generator. It's worth remembering that most random-number generators aren't; they're pseudo-random-number generators, which are defined in such a way that they mimic a lot of the properties of truly random numbers. (Some random number generators take advantage of external events to generate random numbers - like the timing of keypresses on the keyboard. This type of generator tends to fall back on pseudorandom numbers if it runs out of keyboard-timing-generated random numbers; and for the Perl program that you gave, this type of generator will run out of keyboard-generated random numbers with very high probability).

One property of genuine random numbers that any pseudo-random number generator won't have is the vanishingly small possibility of a sequence of N heads for large N. A pseudo-random number generator will hold a certain amount of state - say, 64 bits. A well-designed 64-bit generator will, in a sequence of 2^64 coinflips, include all possible reasonably short sequences, and a number of suitably random-looking longer ones, with a believable distribution (which means very close to 2^63 heads and 2^63 tails). After that, it will start to repeat itself. What you will not get is 2^64 heads (or 2^64 tails).

So. While the sequence that the Perl program above simulates is potentially infinite, the Perl program itself is not potentially infinite. There is a maximum number of terms that it will consider; therefore a maximum possible value of $sum. And no matter where that maximum value is, such a truncated series does have a finite expected value. Given the results of your run, and this sequence, I expect that your computer's random number generator allows for a series of more or less 1674 'heads' in a row; that would give a truncated series with an expected sum of near to 8.

So the repeatability that you see there is a function of the limitations of the technology, not a function of the problem itself.

1   2       2
2   4       4
3   8       6.66666666666667
4   16      10.6666666666667
5   32      17.0666666666667
6   64      27.7333333333333
7   128     46.0190476190476
8   256     78.0190476190476
9   512     134.907936507937
10  1024    237.307936507937
11  2048    423.489754689755
12  4096    764.823088023088
13  8192    1394.97693417693
14  16384   2565.26264846265
15  32768   4749.79598179598
16  65536   8845.79598179598
17  131072  16555.9136288548
18  262144  31119.4691844104
19  524288  58713.5744475683
20  1048576 111142.374447568
21  2097152 211006.755399949
22  4194304 401656.937218131
23  8388608 766379.024174653
24  16777216 1465429.69084132

7: 12.3408045473115
19: 13.8475564640298
21: 17.1112080535515
22: 27.9980309134702
23: 11.2690660132211
31: 9.17316020399938
70: 9.71114553027328
74: 13.8303629749467
97: 9.9859069856619
115: 12.6414850499443
127: 11.0081608201826
138: 9.95738337454533
181: 12.7233012408077
230: 10.0657909667346

I read an interesting puzzle on Stephen Landsburg's blog that generated a lot of disagreement. Stephen offered to bet anyone $15,000 that the average results of a computer simulation, run 1 million times, would be close to his solution's prediction of the expected value.

Thoroughly discussed on MathOverflow awhile ago.

Just to make this clear, the problem Stephen Landsburg discusses is pretty much irrelevant for the point being made here. It's only mentioned because his suggestion of using a simulation is a lead-in to main topic.

Landsburg's solution is somewhat head-scratching. It works fine if you're sampling families. But the problem as stated doesn't sample individual families. It asks about the overall population.

This washes out his effect pretty strongly, to the point that the expected excess in the case of an infinite-sized country is the same absolute excess as in the case of one family: You expect 0.3 more boys than girls. zero.

In total. Across the whole country. And it wasn't asking about the families, it was asking about the country. So the ratio is a teeny tiny deviation from 50%, not 30.6%. This is very much an ends in 'gry' situation.

(Note: what I was trying to say above, about the 0.3 more boys than girls, and then got confused:

This is all about the stopping condition. The stopping condition produces a bias by the disproportionate effect of the denominators on either side of 1/2. But this stopping condition only steps in once to the country and to the single family all the same: there's effectively only one stopping condition, that being when N boys have been born. Increasing N from 1 to a million just washes out the effect by adding more unbiased random babies on what for clarity you can imagine as the beginning of the sequence. So you have this biased ratio, and then you weighted-average it with 1/2 with a huge weight on the 1/2)

But many things we might simulate with computers, like the weather, the economy, or existential risk, are full of power law distributions that might not have a convergent expected value.

Some of those power laws have to break down. If you find a storm of 10^30 J, it isn't Terrestrial weather any more. Similarly, if we ever had 10000% growth in one year, that's probably because we hit the singularity and all bets are off.

The models may incorporate those divergences, but only because they're wrong.

This post is about two problems. The first half of the post is about errors caused by using averages. This has a simple solution: don't use averages. The second half of the post is about understanding tail risk. This is a hard problem, but I don't think it has much to do with the first problem.

Just don't use averages.
Why do you care about averages?

In almost all situations, the median is a better single number summary of a random variable. For example, the median sex ratio is more salient than the mean sex ratio. Mean temperature change is irrelevant. Median temperature change is important. More important is the probability of a catastrophically large change. So answer that question. And answer both questions! You can almost always afford more than a single number to summarize your random variable.

Once you know that the question is the size of the tail, it should be clear that the mean is irrelevant. Even if you do care about the mean, as by fiat in the random series, it is better not to just sample from it, but to understand the whole distribution. If you do that, you'll see that it is skewed, which should make you nervous.

"Anticipating critical transitions", Marten Scheffer et al., p. 344

PDF.

Similar weird things happen for the Cauchy distribution (whose probability density function is proportional to 1/(1+x^2)), which is symmetric around 0 but does not have mean 0 because the sum doesn't converge.

Exercise: what do you expect to happen if you try to find the mean of the Cauchy distribution by simulation?

my expectation, written down before seeing more than seing one (naive) answer, is: greater than 50%. But not necessarily by a lot at first. continuing it on indefinitely, if our population is sufficiently large, will eventually result in a population dominated by females, depending heavily on the details of the distribution of certain abnormalities (aka is there even one two-headed or two-tailed coin, and if not, then what is the exact nature of graph in that region, at which point it becomes ridiculously complicated...)

terribly sloppy notation follows. E(geneticconditions):(chanceofgivingfemalebirth>malebirth). by (havingchildren untill maleborn): (amplifies the prevelance of such conditions in birthing populations.).

But then, i've heard this puzzle before and have had a LONG time to muse about it. 50% is wrong if we're talking about beings with a biology resembling our own. :P

50% is the correct expected distribution, however, if they're using perfect quantum coinflips.

aaanndd...

(please forgive the spelling errors; i just looked at the clock.)

Can't we get results for all practical purposes by setting some epsilon and saying "we are disregarding the highest epsilon percent of results"?

For most practical purposes, I would suggest that a reasonable pEpsilon would be around 10.

I think it would be helpful to explicitly write out the what you're calculating mathematically, e.g. \sum{k=1}^{\infty} (\frac{2^k}{k})*P[halting~by~step~k] = \sum{k=1}^{infty} (\frac{2^k}{k})(2^k) = \sum_{k=1}^{infty} \frac{1}{k} just to make it clear that this is the harmonic series. I know it's described in English, but the math is often helpful. (For posting latex, see also: http://lwlatex.appspot.com/. I don't know if that's the most up-to-date method, but it works.)

Why aren't you worried about global warming?