Timeline for Why is $\mathbb R^{\mathbb N}$ not high-dimensional enough?
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| Oct 24 at 16:21 | comment | added | Will Sawin | @Quertiopler I think what you say is right, but the key point is that in the setting of Example 2.6 we haven't fixed even a definition of what we mean by error. In the paper [1] they immediately after the part you quote consider concrete problems of estimating the distribution of some parameter and one can see for these problems whether weak convergence suffices to study them or not. | |
| Oct 24 at 16:13 | comment | added | Will Sawin | @Quertiopler Sure, but that's not the point of contention: This is the step that is the same in both your argument and the paper [1]. The problem is what we can deduce about the case of fixed n and fixed p from the n to infinity limit of the case $p=\infty$ with $p=\infty$ understood to mean that we are in $\mathbb R^\mathbb N$. | |
| Oct 24 at 16:11 | comment | added | Quertiopler | Or are you saying that Example 2.6 suggests an approximation with an error, which is valid for $p$ fixed. However, as $p\rightarrow\infty$ the approximation no longer works as the error term does not vanish to zero fast enough. Hence they suggest a CLT/conditions that ensure fast enough convergence | |
| Oct 24 at 16:04 | comment | added | Quertiopler | Thanks for your reply. Isn't Example 2.6 in cermics.enpc.fr/%7Emonneau/Billingsley-2eme-edition.pdf saying exactly that? That we can deduce the $p\rightarrow\infty$ case (with $p =\infty$ understood to mean that we are in $\mathbb R^{\mathbb N}$) from the $p = \text{finite}$ case? | |
| Oct 24 at 14:04 | history | answered | Will Sawin | CC BY-SA 4.0 |