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Suppose there is some data $X_{1},X_{2},\ldots,X_{n}$. We further suppose that there is some parameter $\theta$, for which we want to do statistical inference. Assume that there is a asymptotical (weak) consistency result, i.e. there is an estimator $\theta_{n}(X_{1},\ldots,X_{2})$, s.t. $\theta_{n}(X_{1},\ldots,X_{2})\overset{\mathbb{P}}{\longrightarrow}\theta$.

Now, I want to go one step further, calculating a weak limit theorem allowing for testing and confidence on $\theta$. This means, I am looking for a deterministic sequence $\ell_{n}$, s.t. $$\ell_{n}\cdot\left(\theta_{n}(X_{1},...,X_{n}\right)-\theta)\longrightarrow V$$ in distribution, with some r.v. $V$, s.t. $\mathbb{P}_{V}$ is a well known distribution.

Question 1: Is there a way to find the optimal convergence rate $\ell_{n}$?

Question 2: Is it possible to change the limit distribution from $V$ to $V'$ to get a different sequence $\ell_{n}'$? Of course, both distributions, $\mathbb{P}_{V}, \mathbb{P}_{V'}$ are assumed to be non-degenerate.

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$\newcommand{\al}{\alpha} \newcommand{\be}{\beta} \newcommand{\de}{\delta} \newcommand{\De}{\Delta} \newcommand{\ep}{\varepsilon} \newcommand{\ga}{\gamma} \newcommand{\Ga}{\Gamma} \newcommand{\la}{\lambda} \newcommand{\si}{\sigma} \newcommand{\Si}{\Sigma} \newcommand{\thh}{\theta} \newcommand{\om}{\omega} \newcommand{\R}{\mathbb{R}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\F}{\mathcal{F}} \newcommand{\E}{\operatorname{\mathsf E}} \newcommand{\Var}{\operatorname{\mathsf Var}} \renewcommand{\P}{\operatorname{\mathsf P}} \newcommand{\ii}[1]{\operatorname{\mathsf I}\{#1\}} \newcommand{\eD}{\overset{\text{D}}\to} \newcommand{\D}{\overset{\text{D}}=} \newcommand{\tsi}{\tilde\si}$

To answer your Question 1, one needs to know what you mean by the optimal rate.

However, as the following answer to your Question 2 shows, the convergence rate is essentially unique, up to the sign and rescaling, and so, it is hardly meaningful to talk about the optimal rate. Indeed, we have

Proposition. Suppose that $(T_n)$ is a sequence of random variables (r.v.'s) and $(\ell_n)$ and $(\ell'_n)$ are sequences of real numbers such that \begin{equation} \ell_n T_n\eD V\quad\text{and}\quad\ell'_n T_n\eD V' \end{equation} for some non-degenerate r.v.'s $V$ and $V'$, where $\eD$ denotes the convergence in distribution. Then for \begin{equation} r_n:=\frac{\ell'_n}{\ell_n} \end{equation} we have $|r_n|\to c$ for some nonzero real $c$. If, moreover, (the distribution of) $V$ is not symmetric, then $r_n\to c$ for some nonzero real $c$.

Proof. Suppose that $r_n\not\to c$ for any nonzero real $c$. Then (passing to subsequences if needed), we see that without loss of generality (wlog) one of the following three cases must obtain.

Case 1: $r_n\to0$. Then $\ell'_n T_n=r_n(\ell_n T_n)\eD0V=0$, which contradicts the conditions that $\ell'_n T_n\eD V'$ and $V'$ is non-degenerate.

Case 2: $|r_n|\to\infty$. This reduces to Case 1 by switching the roles of $\ell_n$ and $V$ with those of $\ell'_n$ and $V'$.

Case 3: $r_{m_k}\to r$ and $r_{n_k}\to s$, where $r$ and $s$ are distinct nonzero real numbers and $(m_k)$ and $(n_k)$ are strictly increasing sequences of natural numbers. Then $$\ell'_{m_k} T_{m_k}=r_{m_k}(\ell_{m_k} T_{m_k})\eD rV=:W$$ and similarly $$\ell'_{n_k} T_{n_k}\eD sV=\ga W,$$ where $\ga:=s/r$. In view of the condition $\ell'_n T_n\eD V'$, we have $$W\D\ga W,$$ where $\D$ means the equality in distribution. Wlog $|\ga|\le1$ and hence one of the following two subcases of Case 3 must obtain.

Subcase 3.1: $|\ga|=1$. Then $\ga=-1$, because $\ga=s/r$ and $r$ and $s$ are distinct. So, in this subcase $W\D-W$, that is, $W$ is symmetric and hence $V$ is so.

Subcase 3.2: $|\ga|<1$. Then $W\D\ga W\D\ga^2 W\D\dots\D\ga^n W\eD0$, so that $W\D0$ and hence $V\D0$, a contradiction.

Thus, one may have $r_n\not\to c$ for any nonzero real $c$ only if $V$ is symmetric.

Similarly, by applying the absolute value function to $\ell_n$, $\ell'_n$, $r_n$, $T_n$, $V$, and $V'$, it is shown that $|r_n|\to c$ for some nonzero real $c$ in all cases without exceptions. (In this latter consideration, there will be no analogue of the exceptional Subcase~3.1.) The proposition is now proved. $\Box$

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