Chill2Macht

So it seems like the limit random variable is exponential, which seems kind of weird considering it is $n - Xn$ for Xn = Op(1)

At least that was my reading of it, but I admit I've had to re-read it numerous times to confirm whether the interpretation was correct.

It says "let x (i.e. \xi) denote the nu'th value from the top", so we want the 1st from the top i.e. the max.

No, $\nu = 1$ for the maximum.

@TomChen Oh OK. Yeah I think that corresponds to $n - \Xi_n$, or something like this.

Oh ****, this has already been asked on SE before, see here: stats.stackexchange.com/questions/314782/… So it seems that the asymptotic distribution is exponential, not constant? Although I'm not sure what the limit random variable will tell us about the rate of growth of the sequence itself.

When you say $Beta(n,1)$, which is the $\alpha$ and which is the $\beta$? According to Wikipedia (assuming I am simplifying the gamma functions correctly), the pdf of $Beta(n,1)$ is either $(1/n)^{-1} x^{n-1} = n x^{n-1}$ or $n (1 - x)^{n-1}$ supported on $[0,1]$. It seems either way that the distribution Cramer gives for $\Xi_n$ is $n$ times a $Beta(n,1)$ or $Beta(1,n)$, so it seems you are correct. If you could provide a reference for this I would be very grateful.

@TomChen Maybe? I didn't recognize the pdf from Cramer, he writes that it is $\left( 1 - \frac{\xi}{n}\right)^{n-1} 1_{[0,n]}$. Is that Beta? And that is known to converge in probability to the constant RV 1, even though the support is increasing?

Theoretical computer science is much more like pure math (and is arguably just a branch of) pure math than it is similar to programming, or the lazy non-rigorous accounts of machine learning which are in vogue nowadays. If she focuses on taking theoretical computer science courses, I don't really see why she as someone interested in pure math should have a problem. (Maybe unless she's interested in geometry or topology, but even then there are things like discrete differential geometry, persistent homology, spectral graph theory, category theory which should keep her somewhat interested.)

Not to mention also, that in my question I also explicitly state "all other things being equal" to radically simplify the factors under consideration for generating such an estimate, and I explicitly state that I am only interested in considering geographic, climactic, and ecological factors in such an estimate, assuming that the technology/economics/everything else is not radically changed compared to today.

@Asher That point is vacuous for getting an estimate. There is no such thing as 100% accuracy in any field, or for any question -- only estimates with varying degrees of precision are possible. Not to mention that the question inherently does not ask for a level of precision which is impossible -- it only asks which one would be bigger. By that logic, literally every single question on this website should be closed. Also, regarding the claim that "we already have the technology to build self-sufficient arcologies in the desert" -- citation needed.

@jamesqf They are well-defined, I defined them clearly in the question. (Sacramento: obviously yes -- Reno: obviously no -- Salt Lake City: obviously no -- Albany, Syracuse, Buffalo: obviously yes)

psychology, international relations, business, commerce, law, plate tectonics, climate change, government, society, meteorology, international trade, all have literally nothing to do with it. In my question I say "all other things being equal". This page says California will have approx 50 mill people in 2050 ppic.org/content/pubs/report/R_116HJ3R.pdf, this one says the USA smithsonianmag.com/40th-anniversary/… will have approx 400 million, so approx only 10% of the population growth is projected to be on the West Coast.

@Asher Half of the things you mention have nothing to do with the land's carrying capacity using current technology. en.wikipedia.org/wiki/Carrying_capacity I'm not asking whether it will be cool to move to California in 50 years, or why they'll do it. This was pretty clear from the question, especially given that I only mention ecology, geography, and climate. If people lived to the maximum possible ecological extent, for whatever reason, on both the West and East Coast, which would have the larger population? Education, architecture, civil infrastructure, anthropology, sociology,

@smurtagh I know very well what Bayesian probability is, but that is completely irrelevant to my question.

@James How is this far too broad? What is the maximum number of people the West Coast can support versus the maximum number the East Coast can support?

@jamesqf Why would it be all the way to the Mississippi? That doesn't make any sense to me.

@kbelder No, not at all. ncatlab.org/nlab/show/principle+of+equivalence $i$ is defined to be such that $i^2=-1$. All constructions of $i$ that satisfy this property lead to exactly the same conclusions. Thus the philosophical interpretation is irrelevant. Likewise the properties of probability are specified completely by the Kolmogorov axioms. The properties of an object are all that is necessary to specify it up to equivalence and therefore is all that matters. Sure interpretations can be a helpful learning aid, but it is dangerous and incorrect to overextend the analogies they imply.

@whuber Can you elaborate on the last point please? ("Many classical statistical procedures are viewed as invalid by Bayesians") Why do Bayesians consider them invalid if they are mathematically correct? Because they don't like to assume that the parameter to be estimated is a constant random variable? I don't really understand.

@whuber I wasn't referring specifically to the debate when I said that questions were ill-posed, I was referring to philosophy in general. However, according to the answers below, in Bayesianism the parameter to be estimated can be modeled as both a constant random variable and also as a non-constant random variable. In contrast, in frequentism the parameter to be estimated can only be modeled as a constant random variable.

I know nothing about statistics, so I don't know if this characterization is correct, but if it is, then mathematically speaking it is self-evident that frequentism is just a special case of Bayesianism, and that the difference can be reduced to one of axioms.

@Kodiologist I agree with you entirely -- the reason why these problems can't be definitely solved is because they are all ill-posed. There's no point in wasting time arguing about an ill-posed problem; there's no substance to such a discussion at all whatsoever. It just consists of people ignoring the obvious fact that they aren't sticking to logically consistent concepts, so of course they can't reach any definitive conclusions.

@Kodiologist The only way to decisively settle a philosophical argument, at least in my view, is to agree to and define unambiguous definitions and evaluate the logical consequences of those axioms (reductio ad absurdum), what some people call mathematics. There is no point in arguing if you never expect to be able to arrive at some resolution or understanding of the argument. And the only way to do so is to be entirely explicit and clear about what you mean when defining terms. That is all that a mathematical axiom is (in my opinion) -- functioning and useful philosophy.

math.mit.edu/~dws/177/prob07.pdf; the first page about PM looks helpful. I may be wrong, but it seems like the result you want should follow from applying Fubini on $[0,t]\times \Omega \times E$ for all $t$.

Maybe this will help: ma.utexas.edu/users/gordanz/notes/nonsense.pdf; basically the difference between joint measurability and progressive measurability is analogous to the difference between being a local martingale bounded in expectation and being a true martingale.

Barely. It just means that for every $t$, the process is jointly measurable on $[0,t] \times \Omega$. Just apply Fubini for every $t$ and you're done. It makes no practical difference. Also you might want to cite the exact book and theorem you are looking at, because it is unclear what your question is or why you are even confused.

Yes that is one way to think about it. Progressive measurability is the same as joint measurability. If you integrate a jointly measurable function with respect to one of the variables, it won't affect the joint measurability at all.

I mean that the integral is pathwise, I.e. given $\omega \in \Omega$, and that for each $\omega$, $F(t,x,\omega)$ is a measurable function, so therefore $\int F(t,x,\omega) \nu (dx)$ is measurable also. It's a real analysis result, has nothing to do with probability.

well then obviously it is progressively measurable -- it's a measurable function of the original process, which is progressively measurable

see theorem 1: almostsure.wordpress.com/2010/03/25/…; the ito integral of any predictable process wrt a local martingale will be a local martingale under mild conditions, and in particular progressively measurable (since any local martingale is progressively measurable)

any stopped progressively measurable process is again progressively measurable math.stackexchange.com/questions/991099/…

Oh and also, is the property unique to continuous local martingales? To local martingales which are not necessarily continuous, but whose quadratic variation processes are absolutely continuous (the distribution function of some measure absolutely continuous with respect to Lebesgue measure)?

Maybe a shorter way to phrase my confusion is: the Brownian motions can be different, but do they have to be different? I.e. are there cases where no choice such that they are the same is possible? Or can we always choose some probability space such that $\xi$ and $\tilde{\xi}$ coincide?

I hope this makes sense, I don't quite know how else to phrase my confusion

Oh OK this makes much more sense to me now. One last question; the answer to (4) being true then implies that $\int_0^t \xi_s^s ds$ and $\int_0^t \tilde{\xi}_s^2ds$ are versions of each other? But the problem is that given a quadratic variation process, we can't find a unique process with that quadratic variation (e.g. Brownian motion doesn't have to be the only process with quadratic variation $t$), so we can't use the quadratic variation to represent the process uniquely?

I.e. what I don't understand is why (4) being true doesn't imply (3) being true -- this is my biggest confusion/misunderstanding -- I really appreciate you being able to formulate it in such a precise manner for me.

See also math.stackexchange.com/questions/1796863/… for more examples of my confusion regarding optimal properties of corresponding quadratic variation processes and suitability as stochastic integrators.

Also also to summarize, we can say that most, if not all, of the sub-claims are true if we add the two extra conditions: (1) we allow enlargements of the probability space, (2) the sample paths of the local martingale are absolutely continuous, (3) the measure generated by the quadratic variation process is absolutely continuous w.r.t. Lebesgue measure. Are (2) and (3) equivalent given (1)? (is this what you are saying in your answer of the 2nd sub-claim?)

And finally, the sample paths of the quadratic variation are always absolutely continuous (but this is always true since the quadratic variation is always increasing and thus of finite variation)? But even if the sample paths of the quadratic variation are always absolutely continuous, that does not imply that the measure they generate is necessarily absolutely continuous w.r.t. Lebesgue measure? (I don't understand how the counterexample to the 2nd sub-claim is a local martingale, although I may be misremembering the theorem characterizing local martingales in terms of q.v. processes.)

Yes you are correct that I meant the natural filtration; I apologize for not being more specific. Also I am a little confused; (1) do you mean local martingale instead of martingale in most of the places above? I am more interested in local martingales than martingales. (2) So the continuity of paths (i.e. the distinction between continuous local martingale and general local martingale) doesn't matter, but the absolute continuity of paths (a strictly weaker condition) (the difference between "absolutely continuous local martingales" and general local martingales) does matter?

Also I want to upvote your post but I don't have enough reputation after posting this bounty.

@MarianoSuárez-Alvarez So then why is axiomatizing "discrete space" such a good idea? Also I'm not saying "don't study non-continuous spaces", I'm just pointing out that topology was invented with the idea of making rigorous the notion, and yet more than a hundred years later, while we use topologies all of the time, still no one has bothered to make rigorous the notion.

@MarianoSuárez-Alvarez The distinction between "continuous spaces" and "discrete spaces" is a fundamental concept. You are disagreeing with both common sense and giants like Riesz and Hausdorff if you truly believe it to be unimportant. In any case, if you don't want to answer the question, then there is no need for you to comment or criticize the question; quite frankly I just want an answer, not snobby opinions.

@MarianoSuárez-Alvarez Sure they aren't, but that WAS what Riesz, Hausdorff, and Kuratoski were trying to do when they stumbled across the notion. Moreover, my question doesn't state that, so your statement is a strawman argument.

@user2357112 I am sorry you feel that way, although I disagree. It was never my intent to attack the notion of topology, just the notion that topological spaces are "the axiomatization of the notion of continuous space". I had forgotten to consider the viewpoints of those who are not interested in studying "continuous spaces" but are interested in topology when I originally wrote the question. I had read the quote about Riesz before writing the question, I thought it was better known that Riesz (and I believe also Hausdorff and Kuratoski) wanted to axiomatize "continuous space".

@jdods Yes because every point in $\mathbb{Z}$ is open. Not the case for the reals or even the rationals. The notion of discrete space has an explicit characterization in terms of topological notions which the notion of continuous space does mot seem to have.

@menag Yes you are correct, thank you for pointing out the error. Perhaps "connected" or "locally compact" would remedy it. In either case your observation that all spaces can be made discrete via the discrete metric further supports the contention that we lack an adequate axiomatization for the notion of continuous space analogous to that which we have for discrete space.

between "continuous" and "discrete" spaces. And discrete spaces clearly do have a clear definition under the framework of topology, but continuous spaces don't? They are just as important to mathematics. And the distinction can't be as simple as being between the natural numbers and the reals, because what about processes on $\mathbb{Z}$ or Riemannian manifolds? For example.

@EricWofsey No I don't believe you are understanding my question. Topological spaces in general clearly are NOT "continuous". Some subset thereof, sure, but not the whole category. I just want a rigorous notion of "continuous space", not just a notion of "topological space". If it is possible to rigorously formulate "discrete space" within the framework of topology, I don't understand why it shouldn't also be possible for "continuous space", yet no one seems to do so.