# Does convergence in probability imply $\sqrt{n}$-consistency?

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?

• Thank you for your comment. The second $\Rightarrow$ was not correct, it was a $p$ not an $n$ what should appear there. – Álvaro Méndez Civieta Mar 28 at 11:54
• I am afraid this does not make sense: $p$ is the fixed dimension of $\beta$ while $n$ is the sample size increasing to infinity. – Xi'an Mar 28 at 11:59
• 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. – StubbornAtom Mar 28 at 15:29