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Oct 24 at 23:03 vote accept Quertiopler
Oct 24 at 23:03 comment added Quertiopler Got it! So it's not the dimensionality of $\mathhb R^{\mathbb N}$ itself, but rather the topology considered. I will try to figure out whether the functionals that are of interest for me, will be continuous wrt that topology. Thank you very much for being so patient with me!
Oct 24 at 20:33 comment added Iosif Pinelis Previous comment continued: In particular, the only linear functionals of the path $y$ that are continuous wrt that topology are the ones depending only on the values of $y$ on a finite set of time moments -- so that we are just back where we were, that is, back to finite-dimensional distributions.
Oct 24 at 20:23 comment added Iosif Pinelis @Quertiopler : The topology considered in that Example 2.6 (cf. Example 1.2 in Billingsley's book) is the one of pointwise convergence, which is of no use in almost any applications. For instance, the functional $\Bbb R^{\Bbb N}\ni y\mapsto\sup_{p\in\Bbb N}y(p)/a(p)$ in my answer is not continuous wrt the pointwise convergence, and hence the weak convergence of probability measures wrt to this topology will be of no consequence for this functional, as well as for integral-type and other functionals usually of interest.
Oct 24 at 19:37 comment added Quertiopler Thanks for expanding. There was a small typo in my comment: Example 2.6 was meant. In fact, to my understanding Example 2.6 states that we can deduce convergence of a process from its finite-dimensional distributions, which would contradict the example you've added. I would say your example + Example 2.6 can be seen as the root cause of my confusion.
Oct 24 at 18:13 history edited Iosif Pinelis CC BY-SA 4.0
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Oct 24 at 18:13 comment added Iosif Pinelis @Quertiopler : (i) I have added details on '$\mathbb R^{\mathbb N}$ is "too high" dimensional'. (ii) Your understanding of Example 2.7 in Billingsley is almost exactly opposite to its actual point, explicitly stated there: "Weak-convergence theory in $C$ [and other spaces -- I.P.] goes beyond the finite-dimensional case in an essential way". That example actually shows that the tightness condition cannot be dropped. I have also added comments on the tightness to my answer.
Oct 24 at 18:07 history edited Iosif Pinelis CC BY-SA 4.0
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Oct 24 at 17:47 history edited Iosif Pinelis CC BY-SA 4.0
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Oct 24 at 16:00 comment added Quertiopler I think my confusion is causing me to explain this poorly. I edited the question again. I've also added my motivation for the title (and why I didn't change it yet). You mention that "$\mathbb R^{\mathbb N}$ is 'too high' dimensional for the finite-dimensional convergence to work.". How do you mean that? This could possibly resolve my confusion (at least partly) as I don't see why we should have any problems with finite-dimensional convergence to work in $\mathbb R^{\mathbb N}$. Also, concerning your edit, I understand Example 2.7 in the book you link to to mean that tightness is not required.
Oct 24 at 14:32 history edited Iosif Pinelis CC BY-SA 4.0
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Oct 24 at 12:32 comment added Iosif Pinelis @Quertiopler : Like your first comment under my answer, your edits are still incomprehensible to me, as a whole. In particular, I don't know in what sense "$\lim_{p\rightarrow\infty}\mathbb R^p$ is identified as $\mathbb R^{\mathbb N}$" and why "it is natural to assume that $P_n$ is the distribution of [...]", and what $P_n$ actually is. Also, I see no clarification in the edits of your original, title question "Why is $\mathbb R^{\mathbb N}$ not high-dimensional enough?" In fact, I see nothing like a question in your edits.
Oct 24 at 9:24 comment added Quertiopler I apologize for any confusion. I've added more information to my question. I would be grateful if you could review it again. Hopefully, my edits will clarify what I mean. Thank you very much!
Oct 24 at 1:33 comment added Iosif Pinelis @Quertiopler : I hardly understand anything in your comment. (i) What do you mean, exactly, by "study the behavior of $S_{n,p}$ for large $n$ and $p$ in $\mathbb R^{\mathbb N}$"? (ii) In what sense do you mean "convergence of $S_{n,p}$ for any finite $p$"? (iii) What is $S_{n,\infty}$, again exactly? (iv) What exactly do you mean by "finite dimensional convergence implies infinite dimensional convergence"?
Oct 23 at 23:03 comment added Quertiopler But why is it not fruitful to study the behavior of $S_{n,p}$ for large $n$ and $p$ in $\mathbb R^{\mathbb N}$? I may have a fundamental wrong perception, but I would expect that if I have convergence of $S_{n,p}$ for any finite $p$, then it follows that I also have convergence of $S_{n,\infty}$ (to a Gaussian process). The fact that finite dimensional convergence implies infinite dimensional convergence sounds like a nice feature in that regard (as it simplifies things by a lot). In the paper, and in your answer, this fact sounds like I huge disadvantage though.
Oct 23 at 20:16 history edited Iosif Pinelis CC BY-SA 4.0
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Oct 23 at 16:38 history edited Iosif Pinelis CC BY-SA 4.0
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Oct 23 at 16:31 history answered Iosif Pinelis CC BY-SA 4.0