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How can I formally show that a root n consistent estimator is weakly consistent?

Heuristically its clear that if $\sqrt{n}(\theta_n-\theta_0)$ is bounded in probability then since $\sqrt{n} \to \infty$ as $n \to \infty$ clearly $\theta_n-\theta_0 \in o_p(1)$ that is it goes to zero in probability but every time I try to write it formally I get stuck. Could someone help me out?

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  • $\begingroup$ Is this a question from a course or textbook? If so, please add the [self-study] tag & read its wiki. $\endgroup$ – gung Mar 12 '17 at 15:25
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    $\begingroup$ @gung I was reading some notes and although it seemed obvious, surprisingly I could not prove it , so I sought some help. A hint would be appreciated too $\endgroup$ – user3503589 Mar 12 '17 at 15:33
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I'm going to write your estimator with a hat on it: $\hat{\theta}_n$. The tricks involve playing around with subsets. If $A \subset B$, then $P(A) \le P(B)$. Observe that

\begin{align*} P(|\hat{\theta}_n - \theta_0| > \epsilon) &= P(|\hat{\theta}_n - \theta_0| > \epsilon , \sqrt{n}|\hat{\theta}_n - \theta_0| \le M) + P(|\hat{\theta}_n - \theta_0| > \epsilon , \sqrt{n}|\hat{\theta}_n - \theta_0| > M) \\ &\le P(|\hat{\theta}_n - \theta_0| > \epsilon , \sqrt{n}|\hat{\theta}_n - \theta_0| \le M) + P(\sqrt{n}|\hat{\theta}_n - \theta_0| > M) \\ &\le P(M/\sqrt{n} > \epsilon , \sqrt{n}|\hat{\theta}_n - \theta_0| \le M) + P(\sqrt{n}|\hat{\theta}_n - \theta_0| > M) \\ &\le P(M/\sqrt{n} > \epsilon ) + P(\sqrt{n}|\hat{\theta}_n - \theta_0| > M). \end{align*}

So, more formally, pick $\epsilon,\delta >0$. Then pick $M$ such that $$ P(\sqrt{n}|\hat{\theta}_n - \theta_0| > M) \le \delta, $$ if $n > N_1$, which you can do because it's bounded in probability. Then if $n > \text{max}(N_1, M^2/\epsilon^2)$, you have your result: $$ P(|\hat{\theta}_n - \theta_0| > \epsilon) \le \delta. $$

Note that I am not showing how $\sqrt{n}$- consistency implies boundedness in probability, so I hope that this counts as a hint.

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  • $\begingroup$ I have a stupid question . On rereading your proof I am wondering what is random in the set $\{M/\sqrt{n}> \epsilon\}$. Its clear that the set is either empty or the whole set depending on the value of $n$ and we choose n large enough so that its empty(since we have already fixed $M$ and $\epsilon$). The probability of this set can either be $1$ or $0$ and we chose $n$ in such a way so that its $0$. Am I correct? $\endgroup$ – user3503589 Mar 14 '17 at 14:11
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    $\begingroup$ @user3503589 yes, that's right. Sorry, I didn't see this message because you never tagged me. $\endgroup$ – Taylor Mar 15 '17 at 23:03

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