Reputation
1,257
Next privilege 2,000 Rep.
Edit questions and answers
Badges
6 7
Newest
 Yearling
Impact
~18k people reached

  • 0 posts edited
  • 0 helpful flags
  • 33 votes cast
Jun
11
comment Why are IQ test results normally distributed?
@LLuuUsErI132UOutOfMemory: Just modify your answer slightly, and it will be accurate, this site is very annoying, and I don't want to help it along, although it is somewhat less annoying than other sites on the network..
Jun
9
comment Why are IQ test results normally distributed?
What you are missing is that if you don't have a definition for the x axis, you can always parametrize the x axis so whatever distribution is normal. I said it already 4 times here, this answer is incorrect. The IQ test is defined so that the population results are normal with mean 100 and s.d. 15, this replaced the earlier definition of 100 times the ratio "mental age"/"chronological age" which gave age-dependent results, but at least had an objectively defined x axis, and identified prodigies. The modern IQ "g" is just defined to be normally distributed, it's true by definition.
Jun
9
comment Why are IQ test results normally distributed?
@TonyK: IQ is normal by definition, you parametrize the raw score by brute force so that the distribution ends up normal. If you use any natural mathematical metric for difficulty of questions, like "size of search space" in a chess problem, or "number of steps of deduction" in a mathematical problem, basically anything, in the natural metric, the distribution of humans would be a power-law like distribution with a heavy tail and different individuals would perform astronomically better at some tasks than others.
Jun
9
comment Why are IQ test results normally distributed?
@LLuuUsErI132UOutOfMemory: You adjust the distribution by defining the original score I(n) as some increasing function of the number n of correct answers, then find the empirical distribution f(I) of the unnormalized I scores in the population, and then you reparametrize I by defining a new score function g so that f(I(g)) dg/dI is a Gaussian with mean 100 and s.d. 15. You can always do it by relabling the x axis, every 1-d distribution can be reparametrized to any other. They also have baskets of gender biased questions, and they also adjust the test to make sure female and male IQ is equal.
Jun
3
comment Why are IQ test results normally distributed?
IQ is adjusted by fiat so that the distribution of scores is normal with a mean of 100 and a S.D. of 15. You can always map the scores so that any distribution, whatever it is, becomes normal.
Jun
1
comment Advantage of accepting non-measurable sets
@MichaelGreinecker: The notion I am using is not standard, and requires a reasonably rigorous development separate from the development of ZFC. The idea is to make sets countable collections, and make the real numbers a proper class. In order to make this convienient, you need to show that the constructions of maps of real numbers, functions of real numbers, can be constructed just as easily in this different perspective, and that requires a heavier slog than the comments here. I will write and link to a development of such a theory. One also needs to embed all countable models of ZFC.
Jun
1
comment Advantage of accepting non-measurable sets
@MichaelGreinecker: An "arbitrary shape" (my mangling of your phrase "arbitrary object") is any logical predicate defined on R^N which is either true or false for any element of R^N. "belonging to a Vitali set" is a predicate in ZFC, for example. Any such logical predicate, when you allow the real numbers to expand through addition of random elements, always defines a notion of volume through the probability that the new random element has the property defined by the predicate. The Vitali set is not a complete enumeration when you adjoin the new random element, and you need to expand it.
May
31
comment Advantage of accepting non-measurable sets
Downvoted, sorry, there is a simple reason why arbitrary shapes should have volume, because there is a notion of volume separate from partitioning in cubes and counting. The Monte-Carlo definition of volume asks you to consider a random point in [0,1]^N, and ask whether it landed in the set. If you do it countably many times, the volume is the ratio of the number of successes to the number of trials. This notion intuitively only requires that you can imagine asking "is x in S" for arbitrary x and get an answer, and the rejection of measurability means Monte-Carlo doesn't work in the universe.
May
22
comment Advantage of accepting non-measurable sets
@ArthurFischer: I corrected the error as you posted the comment, I didn't do it instantly, because I needed to review Solovay's construction first, now that I understand the point of the stupid collapse (thanks @AsafKaragila). The error is regarding a side issue, involving the details of Solovay's construction. The main point of the answer is correct, and will always be correct, the construction of nonmeasurable sets makes probability difficult. The answer is describing a different non-ZF idea to resolve this, the idea that powersets should not be considered as fixed entities.
May
22
revised Advantage of accepting non-measurable sets
fix mistake
May
22
comment Advantage of accepting non-measurable sets
@ArthurFischer: Asaf Karagila's comments pointed out a mistake in the answer, which has not yet been fixed! You need to keep them here until I fix it.
May
18
answered Advantage of accepting non-measurable sets
Nov
14
awarded  Yearling
Oct
6
comment Level of Rigor in Mathematical Physics
This is precisely what is happening, except that physicists are flying on completely different clouds, and they learn to do it just jumping off airplanes without a parachute and smashing on the ground again and again and again. You produce results when you happen to fly a couple of times. The method of flying is by calculating precise results using computationally well-defined methods, rather than by proving theorems using logically defined axiomatic systems. The two methods are complementary, because axiomatic systems are too stupid about statistics to know about the computational methods.
Sep
24
awarded  Autobiographer
Apr
17
comment Foundation for analysis without axiom of choice?
@Did: Yes, you are right, filtrations are an exception to the rule, but in probability filtrations are an abomination. The process by which you define a stochastic process in a measurable universe is simply by giving a convergent sequence of random approximations to the process, and this natural definition is butchered by the infernal nonmeasurables, so that you end up introducing filtrations. The filtrations are useless in this context, and positively harmful, they are simply barrier to entry, even here the probabilist secretly thinks that all path sets in path-space are measurable.
Apr
17
comment Why is the axiom of choice separated from the other axioms?
@Hurkyl: The "presupposition" is that there are any nonmeasurable sets in the first place! It is obvious that these make it impossible to speak about "randomly choosing a real number", and this is something that makes natural arguments onerous and convoluted, because you need to distinguish between "random variables" (which are not real numbers and cannot be when there is uncountable choice) and "real numbers" which are something else. It's an artificial and stifling distinction.
Apr
17
comment Foundation for analysis without axiom of choice?
@AsafKaragila: Downvoted, because of the statement that "in Solovay's model you can decompose R into more parts than it has elements". This is using the ridiculous idea that just because you can't inject aleph-1 into R that aleph-1 somehow has "more elements" than R. Not true, it doesn't, it is simply incomparable. The ordinal tower is separated conceptually from the powersets in Solovay model, and you can't compare the two in size at all. The reals are enormously large, and the ordinals are much smaller, but to demonstrate this by embedding you have to first take the ordinals to be countable.
Apr
17
comment Implication and Interpretation of Banach Tarski
@AsafKaragila: Borel sets are not dependent on the kind of axiom of choice one is talking about here. They have nothing to do with uncountable continuum choice.
Apr
17
comment Implication and Interpretation of Banach Tarski
Of course the Banach Tarski pieces behave non-paradoxically, because the axiom of choice is consistent! But this is not the same as saying that they are not paradoxical, because they contradict the concept of "randomly chosen real number", and require you to formulate the concept of "random real" differently, as a "random variable", for which you cannot do certain operations. For example, you cannot speak about the probability that a random chosen real lands in a non-measurable set, the concept is incoherent. So this means random reals are forbidden in your universe.