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

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?

• Is this a question from a course or textbook? If so, please add the [self-study] tag & read its wiki. – gung - Reinstate Monica Mar 12 '17 at 15:25
• @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 – user3503589 Mar 12 '17 at 15:33

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
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.
• 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? – user3503589 Mar 14 '17 at 14:11