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The authors did not say anything like "$\mathbb R^{\mathbb N}$ is not high-dimensional enough."

Rather, they said

"finite dimensional convergence [...] is too weak a result to develop genuinely high-dimensional inference methods".

The meaning here is rather the opposite: $\mathbb R^{\mathbb N}$ is "too high" dimensional for the finite-dimensional convergence to work. For instance, if you have a sequence $((Y_1(p))_{p\in\Bbb N},(Y_2(p))_{p\in\Bbb N},\dots)$ of stochastic processes of which you only know that the finite-dimensional distributions of the process $((Y_n(p))_{p\in\Bbb N}$ converge to the finite-dimensional distributions of some process $((Y_\infty(p))_{p\in\Bbb N}$ as ($n\to\infty$), you cannot in general conclude anything about convergence of the distributions of a functional of the entire path of the process $Y_n(\cdot)$ to the distribution of that functional of the entire path of the process $Y_\infty(\cdot)$. An example of such a functional is $\Bbb R^{\Bbb N}\ni y\mapsto\sup_{p\in\Bbb N}y(p)/a(p)$ for some positive function $a$ on $\Bbb N$.

So, if one wants to study the behavior of $S_{n,p}:=S_n$ for large $n$ and $p$, then it is not enough to know the behavior of $S_{n,p}$ for large $n$ but only for a fixed finite set of values $p$. You already "understand that weak convergence (= convergence in distribution) in $\mathbb R^{\mathbb N}$ is equivalent to weak convergence of the finite dimensional marginals." So, the highlighted thesis follows.

On a somewhat positive note: In this comment, the OP wrote:

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.

Of course, I said nothing of this sort. In fact, I did not talk about any advantages or disadvantages at all.

What can actually be said on this matter is the following. The convergence of the finite-dimensional distributions is of course necessary for the convergence of the distributions of the entire processes. Moreover, there are a number of results saying that, with the additional tightness condition, the convergence of the finite-dimensional distributions is also sufficient for the convergence of the distribution of the entire processes -- see e.g. Theorems 7.1 and 13.1 in Billingsley.

Furthermore, the tightness condition cannot be dropped -- cf. Example 2.7 in Billingsley's book.

Yet more, according to Prokhorov%27s_theorem, the tightness condition is necessary if the paths of the processes are in a Polish space.

The authors did not say anything like "$\mathbb R^{\mathbb N}$ is not high-dimensional enough."

Rather, they said

"finite dimensional convergence [...] is too weak a result to develop genuinely high-dimensional inference methods".

The meaning here is rather the opposite: $\mathbb R^{\mathbb N}$ is "too high" dimensional for the finite-dimensional convergence to work. For instance, if you have a sequence $((Y_1(p))_{p\in\Bbb N},(Y_2(p))_{p\in\Bbb N},\dots)$ of stochastic processes of which you only know that the finite-dimensional distributions of the process $((Y_n(p))_{p\in\Bbb N}$ converge to the finite-dimensional distributions of some process $((Y_\infty(p))_{p\in\Bbb N}$ as ($n\to\infty$), you cannot in general conclude anything about convergence of the distributions of a functional of the entire path of the process $Y_n(\cdot)$ to the distribution of that functional of the entire path of the process $Y_\infty(\cdot)$. An example of such a functional is $\Bbb R^{\Bbb N}\ni y\mapsto\sup_{p\in\Bbb N}y(p)/a(p)$ for some positive function $a$ on $\Bbb N$.

So, if one wants to study the behavior of $S_{n,p}:=S_n$ for large $n$ and $p$, then it is not enough to know the behavior of $S_{n,p}$ for large $n$ but only for a fixed finite set of values $p$. You already "understand that weak convergence (= convergence in distribution) in $\mathbb R^{\mathbb N}$ is equivalent to weak convergence of the finite dimensional marginals." So, the highlighted thesis follows.

On a somewhat positive note: In this comment, the OP wrote:

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.

Of course, I said nothing of this sort. In fact, I did not talk about any advantages or disadvantages at all.

What can actually be said on this matter is the following. The convergence of the finite-dimensional distributions is of course necessary for the convergence of the distributions of the entire processes. Moreover, there are a number of results saying that, with the additional tightness condition, the convergence of the finite-dimensional distributions is also sufficient for the convergence of the distribution of the entire processes -- see e.g. Theorems 7.1 and 13.1 in Billingsley.

Furthermore, the tightness condition cannot be dropped -- cf. Example 2.7 in Billingsley's book.

Yet more, according to Prokhorov's theorem, the tightness condition is necessary if the paths of the processes are in a Polish space.

The authors did not say anything like "$\mathbb R^{\mathbb N}$ is not high-dimensional enough."

Rather, they said

"finite dimensional convergence [...] is too weak a result to develop genuinely high-dimensional inference methods".

The meaning here is rather the opposite: $\mathbb R^{\mathbb N}$ is "too high" dimensional for the finite-dimensional convergence to work.

Indeed, if one wants to study the behavior of $S_{n,p}:=S_n$ for large $n$ and $p$, then it is not enough to know the behavior of $S_{n,p}$ for large $n$ but only for a fixed finite set of values $p$. You already "understand that weak convergence (= convergence in distribution) in $\mathbb R^{\mathbb N}$ is equivalent to weak convergence of the finite dimensional marginals." So, the highlighted thesis follows.

On a somewhat positive note: In this comment, the OP wrote:

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.

Of course, I said nothing of this sort. In fact, I did not talk about any advantages or disadvantages at all.

What can actually be said on this matter is the following. The convergence of the finite-dimensional distributions is of course necessary for the convergence of the distributions of the entire processes. Moreover, there are a number of results saying that, with the additional tightness condition, the convergence of the finite-dimensional distributions is also sufficient for the convergence of the distribution of the entire processes -- see e.g. Theorems 7.1 and 13.1 in Billingsley.

Furthermore, the tightness condition cannot be dropped -- cf. Example 2.7 in Billingsley's book.

Yet more, according to Prokhorov%27s_theorem, the tightness condition is necessary if the paths of the processes are in a Polish space.

The authors did not say anything like "$\mathbb R^{\mathbb N}$ is not high-dimensional enough."

Rather, they said

"finite dimensional convergence [...] is too weak a result to develop genuinely high-dimensional inference methods".

The meaning here is rather the opposite: $\mathbb R^{\mathbb N}$ is "too high" dimensional for the finite-dimensional convergence to work. For instance, if you have a sequence $((Y_1(p))_{p\in\Bbb N},(Y_2(p))_{p\in\Bbb N},\dots)$ of stochastic processes of which you only know that the finite-dimensional distributions of the process $((Y_n(p))_{p\in\Bbb N}$ converge to the finite-dimensional distributions of some process $((Y_\infty(p))_{p\in\Bbb N}$ as ($n\to\infty$), you cannot in general conclude anything about convergence of the distributions of a functional of the entire path of the process $Y_n(\cdot)$ to the distribution of that functional of the entire path of the process $Y_\infty(\cdot)$. An example of such a functional is $\Bbb R^{\Bbb N}\ni y\mapsto\sup_{p\in\Bbb N}y(p)/a(p)$ for some positive function $a$ on $\Bbb N$.

So, if one wants to study the behavior of $S_{n,p}:=S_n$ for large $n$ and $p$, then it is not enough to know the behavior of $S_{n,p}$ for large $n$ but only for a fixed finite set of values $p$. You already "understand that weak convergence (= convergence in distribution) in $\mathbb R^{\mathbb N}$ is equivalent to weak convergence of the finite dimensional marginals." So, the highlighted thesis follows.

On a somewhat positive note: In this comment, the OP wrote:

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.

Of course, I said nothing of this sort. In fact, I did not talk about any advantages or disadvantages at all.

What can actually be said on this matter is the following. The convergence of the finite-dimensional distributions is of course necessary for the convergence of the distributions of the entire processes. Moreover, there are a number of results saying that, with the additional tightness condition, the convergence of the finite-dimensional distributions is also sufficient for the convergence of the distribution of the entire processes -- see e.g. Theorems 7.1 and 13.1 in Billingsley.

Furthermore, the tightness condition cannot be dropped -- cf. Example 2.7 in Billingsley's book.

Yet more, according to Prokhorov%27s_theorem, the tightness condition is necessary if the paths of the processes are in a Polish space.

The authors did not say anything like "$\mathbb R^{\mathbb N}$ is not high-dimensional enough."

Rather, they said

"finite dimensional convergence [...] is too weak a result to develop genuinely high-dimensional inference methods".

The meaning here is rather the opposite: $\mathbb R^{\mathbb N}$ is "too high" dimensional for the finite-dimensional convergence to work.

Indeed, if one wants to study the behavior of $S_{n,p}:=S_n$ for large $n$ and $p$, then it is not enough to know the behavior of $S_{n,p}$ for large $n$ but only for a fixed finite set of values $p$. You already "understand that weak convergence (= convergence in distribution) in $\mathbb R^{\mathbb N}$ is equivalent to weak convergence of the finite dimensional marginals." So, the highlighted thesis follows.

On a somewhat positive note: In this comment, the OP wrote:

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.

Of course, I said nothing of this sort. In fact, I did not talk about any advantages or disadvantages at all.

What can actually be said on this matter is the following. The convergence of the finite-dimensional distributions is of course necessary for the convergence of the distributions of the entire processes. Moreover, there are a number of results saying that, with the additional tightness condition, the convergence of the finite-dimensional distributions is also sufficient for the convergence of the distribution of the entire processes -- see e.g. Theorems 7.1 and 13.1.

The authors did not say anything like "$\mathbb R^{\mathbb N}$ is not high-dimensional enough."

Rather, they said

"finite dimensional convergence [...] is too weak a result to develop genuinely high-dimensional inference methods".

The meaning here is rather the opposite: $\mathbb R^{\mathbb N}$ is "too high" dimensional for the finite-dimensional convergence to work.

Indeed, if one wants to study the behavior of $S_{n,p}:=S_n$ for large $n$ and $p$, then it is not enough to know the behavior of $S_{n,p}$ for large $n$ but only for a fixed finite set of values $p$. You already "understand that weak convergence (= convergence in distribution) in $\mathbb R^{\mathbb N}$ is equivalent to weak convergence of the finite dimensional marginals." So, the highlighted thesis follows.

On a somewhat positive note: In this comment, the OP wrote:

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.

Of course, I said nothing of this sort. In fact, I did not talk about any advantages or disadvantages at all.

What can actually be said on this matter is the following. The convergence of the finite-dimensional distributions is of course necessary for the convergence of the distributions of the entire processes. Moreover, there are a number of results saying that, with the additional tightness condition, the convergence of the finite-dimensional distributions is also sufficient for the convergence of the distribution of the entire processes -- see e.g. Theorems 7.1 and 13.1 in Billingsley.

Furthermore, the tightness condition cannot be dropped -- cf. Example 2.7 in Billingsley's book.

Yet more, according to Prokhorov%27s_theorem, the tightness condition is necessary if the paths of the processes are in a Polish space.