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the OLS estimators of the short-run parameters are $T^{1/2}$ consistent. What is the$T^{1/2}$ consistent ?

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Let the model be, $y=X\beta+e$ where $e_i\sim i.i.d N(0,\sigma^2<\infty)$ and $i=1,2,\ldots,T$. Then square root consistency means,

$$ \sqrt T (\beta-\hat\beta)\overset{d}{\rightarrow} N(0,(X'X)^{-1} \sigma^2) $$

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  • $\begingroup$ Thank you for your explanations. I have two more questions. First one is that what is asymptotically singular ? and second one what is asymptotically perfectly collinear? Please give an explanation thank you again :) $\endgroup$ – 1190 Oct 2 '16 at 14:11
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    $\begingroup$ You should really ask this as new questions! Comments is not the place for new questions. And, please include some more context than in this question. $\endgroup$ – kjetil b halvorsen Oct 2 '16 at 14:38
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    $\begingroup$ As @kjetilbhalvorsen suggests, you should go for a new question with more details. $\endgroup$ – TPArrow Oct 2 '16 at 14:41
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    $\begingroup$ Actually, the answer is not correct. First the asymptotic variance has too many X and second lim is imprecise in terms of the convergence mode. $\endgroup$ – Christoph Hanck Oct 2 '16 at 14:54
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    $\begingroup$ @ChristophHanck Thanks, I fixed that. What do you mean by the second lim? $\endgroup$ – TPArrow Oct 2 '16 at 15:14
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Let the model be, $y=X\beta+e$ where $e_i\sim i.i.d N(0,\sigma^2<\infty)$ and $i=1,2,\ldots,T$. Then by central limit theorem:

$$ \sqrt T (\beta-\hat\beta)\overset{d}{\rightarrow} N(0,(X'X)^{-1} \sigma^2) $$

This in turn implies that $$\sqrt T (\beta-\hat\beta) = O_P(1)$$ The above equation defines "$\hat{\beta}$ is $\sqrt{n}$-consistent".

What does that definition mean? For some statistic $Z_n$, $Z_n = O_P(1)$ i.e. $Z_n$ is "bounded in probability" means the following: for all $\varepsilon > 0$ there exists a constant $C_{\varepsilon}$ such that: $$ \mathbb{P}(|Z_n| > C_{\varepsilon}) \leq \varepsilon $$

For more details also see: root-n consistent estimator, but root-n doesn't converge?

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