# what is $T^{1/2}$ consistent?

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

## 2 Answers

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,

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

• 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 :) – 1190 Oct 2 '16 at 14:11
• 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. – kjetil b halvorsen Oct 2 '16 at 14:38
• As @kjetilbhalvorsen suggests, you should go for a new question with more details. – TPArrow Oct 2 '16 at 14:41
• 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. – Christoph Hanck Oct 2 '16 at 14:54
• @ChristophHanck Thanks, I fixed that. What do you mean by the second lim? – TPArrow Oct 2 '16 at 15:14

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?