15

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?

Christoph Hanck
  • 25,948
  • 3
  • 57
  • 106
makansij
  • 1,919
  • 5
  • 27
  • 38
  • 6
    It means $\sqrt{n}(\hat\theta-\theta)=O_p(1)$. – hejseb Apr 14 '16 at 06:19
  • but $\hat\theta$ is a variable, so how would you compute this? – makansij Apr 14 '16 at 06:44
  • @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. – makansij Apr 14 '16 at 06:58
  • 2
    Nice question! But I'm confused by the claim that $1/\sqrt n$ doesn't converge, what did you mean by that? – Silverfish Apr 14 '16 at 14:27
  • Look up harmonic series https://en.wikipedia.org/wiki/Harmonic_series_(mathematics), it never reaches zero even as n goes to infinity, for $1/(n^p)$ when $p\le1$ – makansij Apr 14 '16 at 15:25
  • 5
    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. – whuber Apr 14 '16 at 15:56

1 Answers1

21

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.

Christoph Hanck
  • 25,948
  • 3
  • 57
  • 106