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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.

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

My 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-degenerated.

Xarrus
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  • Calculate the characteristic function of $l_n$ and see if the algebraic form is simple enough to deduce the rate of convergence to the characteristic function of $V$, – stans Jul 01 '18 at 13:17
  • @stans $\ell_{n}$ is a deterministic sequence. So, if you mean the Fourier transform with "characteristic function", then, afaik, the latter is not defined. – Xarrus Jul 01 '18 at 13:23
  • In general, $l_n$ cannot be deterministic since it is a function of random variable $\theta_n(X_1,...,X_n)$. Also, if you were to choose such a degenerate functional form that would make $l_n$ deterministic, then why would you expect it to converge to a *random* variable? – stans Jul 01 '18 at 13:39
  • Hmm, I don't get your point. in the classical CLT it is $\ell_{n}=\sqrt{n}$. It is a weak convergence result. Not convergence in probability. – Xarrus Jul 01 '18 at 13:43
  • I see. Your notation is somewhat ambiguous. I thought $l_n$ was a function of $\theta_n - \theta_n$ but you meant it to be a simple multiplier. In that case, let me rephrase: calculate the characteristic function of $l_n * \theta_n$ and see if the algebraic form is simple enough to deduce the order of $l_n$ for which there is convergence to another characteristic function. – stans Jul 01 '18 at 13:47
  • Ok, but does this implies, that $\ell_{n}$ is optimal? If yes, why? Maybe you can write it down as an answer. – Xarrus Jul 01 '18 at 14:08
  • You would have to define "asymptotic optimality". It is not the same as finite sample optimality. – stans Jul 01 '18 at 14:10
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    Could you explain what you mean by "optimal"? Up to an arbitrary choice of multiplicative constant, asymptotically the sequence $l_n$ is unique, so exactly what are you hoping to optimize? – whuber Jul 01 '18 at 16:34
  • @whuber How can I find this sequence $\ell_{n}$? Does the consistency proof deliver this sequence? So, if there is a sequence $\ell_{n}$, it is unique? – Xarrus Jul 01 '18 at 16:50
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    At https://stats.stackexchange.com/a/153067/919, I describe a general way to find coefficients $a_n$ and $b_n$ that make the distribution of $a_n\theta_n + b_n$ converge. – whuber Jul 01 '18 at 18:28

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