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I've heard the term "root-n" consistent estimator' used many times. From the resources I've been instructed by, I thought that a "root-n" consistent estimator meant that:

  • the estimator converges on the true value (hence the word "consistent")
  • the estimator converges at a rate of $1/\sqrt{n}$

This puzzles me, since $1/\sqrt{n}$ doesn't converge? Am I missing something crucial here?

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    $\begingroup$ It means $\sqrt{n}(\hat\theta-\theta)=O_p(1)$. $\endgroup$ – hejseb Apr 14 '16 at 6:19
  • $\begingroup$ but $\hat\theta$ is a variable, so how would you compute this? $\endgroup$ – Candic3 Apr 14 '16 at 6:44
  • $\begingroup$ @hejseb, I appreciate your response, thank you. Would you please explain in words? It helps me to be able to verbalize, rather than just looking at symbols. $\endgroup$ – Candic3 Apr 14 '16 at 6:58
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    $\begingroup$ Nice question! But I'm confused by the claim that $1/\sqrt n$ doesn't converge, what did you mean by that? $\endgroup$ – Silverfish Apr 14 '16 at 14:27
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    $\begingroup$ You confuse the sequence $1/\sqrt{n} = 1/1, 1/\sqrt{2}, 1/\sqrt{3}, \ldots$ with the series $\sum_{i=1}^n 1/\sqrt{k}$ whose general term is $1/1 + 1/\sqrt{2} + 1/\sqrt{3} + \cdots + 1/\sqrt{n}$. The former converges to $0$ as $n$ grows large whereas the latter diverges. The latter, however, is irrelevant. $\endgroup$ – whuber Apr 14 '16 at 15:56
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What hejseb means is that $\sqrt{n}(\hat\theta-\theta)$ is "bounded in probability", loosely speaking that the probability that $\sqrt{n}(\hat\theta-\theta)$ takes on "extreme" values is "small".

Now, $\sqrt{n}$ evidently diverges to infinity. If the product of $\sqrt{n}$ and $(\hat\theta-\theta)$ is bounded, that must mean that $(\hat\theta-\theta)$ goes to zero in probability, formally $\hat\theta-\theta=o_p(1)$, and in particular at rate $1/\sqrt{n}$ if the product is to be bounded. Formally, $$ \hat\theta-\theta=O_p(n^{-1/2}) $$ $\hat\theta-\theta=o_p(1)$ is just another way of saying we have consistency - the error "vanishes" as $n\to\infty$. Note that $\hat\theta-\theta=O_p(1)$ would not be enough (see the comments) for consistency, as that would only mean that the error $\hat\theta-\theta$ is bounded, but not that it goes to zero.

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  • $\begingroup$ So, for a an estimator to be "consistent", it must have an $O(1)$, constant value, because if it were $O(n)$, then the estimate would diverge as n increases. $\endgroup$ – Candic3 Apr 15 '16 at 4:18
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    $\begingroup$ No, not quite, see my edit. $\endgroup$ – Christoph Hanck Apr 15 '16 at 7:33

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