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Consider a linear model $y=X\beta+\varepsilon$, and take $\tilde\beta$ as an estimator of the population parameter $\beta$.

If $\left\|\tilde\beta-\beta\right\|_2\rightarrow0$ in probability, does this imply that $\tilde\beta$ is a $\sqrt{n}$-consistent estimator of $\beta$?

I am not used to work in asymptotic theory so I am not sure If what I did is correct, but this is what I have so far: I know that $\left\|\tilde\beta-\beta\right\|_2\rightarrow0$ in probability means that $P(\left\|\tilde\beta-\beta\right\|_2\geq\epsilon)=0$. So from here,

\begin{eqnarray} P(\left\|\tilde\beta-\beta\right\|_2\geq\epsilon)=0&\Rightarrow&P((\tilde\beta_1-\beta_1)^2+\ldots+(\tilde\beta_p-\beta_p)^2\geq\epsilon^2)=0\nonumber\\ &\Rightarrow&P(p(\tilde\beta_1-\beta_1)^2\geq\epsilon^2)=0\nonumber\\ &\Rightarrow&P(\sqrt{p}|\tilde\beta_1-\beta_1|\geq\epsilon)=0\nonumber \end{eqnarray}

Now the definition for being $\sqrt{n}$-consistent is that $$P(\sqrt{n}|\tilde\beta-\beta|>K)<\epsilon$$

EDTI: Correction based on comments.

So I think that what I got proves consistency, and even a sort of $\sqrt{p}$-consistency. But is there a way to prove the $\sqrt{n}$-consistency?

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  • $\begingroup$ Thank you for your comment. The second $\Rightarrow$ was not correct, it was a $p$ not an $n$ what should appear there. $\endgroup$ – Álvaro Méndez Civieta Mar 28 at 11:54
  • $\begingroup$ I am afraid this does not make sense: $p$ is the fixed dimension of $\beta$ while $n$ is the sample size increasing to infinity. $\endgroup$ – Xi'an Mar 28 at 11:59
  • $\begingroup$ If $\tilde\beta$ converges in probability to $\beta$, then it also converges in distribution. And convergence in distribution implies $\tilde\beta$ is tight or bounded in probability (hints here). But not sure if that implies $\sqrt n$ consistency. $\endgroup$ – StubbornAtom Mar 28 at 15:29

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