Suppose that $n^{1/3} \hat{\theta} = O_{p}(1)$.
I want to find the probability limit of $n^{1/4} \hat{\theta}$ .
I have to use the following definition and lemma:
Definition: A sequence of random variables $\left \{ x_{n} \right \}$ is bounded in probability if and only if for every $\epsilon>0$, there exists a $b_{\epsilon} < \infty$ and integer $N$ such that $$P\left [ \left | x_{n} \right |\geq b_{\epsilon} \right ]<\epsilon$$ for all $N \geq N_{\epsilon}$. Then we write $x_{N}= O_{p}(1)$.
Lemma: If $x_{N}\overset{p}{\rightarrow}a$ then $x_{N}= O_{p}(1)$.
I am quite not sure how to start so any help would be appreciated.
But intuitively, will we also have $n^{1/4} \hat{\theta} = O_{p}(1)$?
Let $X_n:=n^{1/3}\hat{\theta}_n$. Then for any $\epsilon>0$, $$ \mathsf{P}(|n^{-1/12}X_n|>\epsilon)=\mathsf{P}(|X_n|>n^{1/12}\epsilon)<\epsilon $$ for all $n$ large enough ($\because X_n=O_p(1)$). It means that $n^{1/4}\hat{\theta}_n=o_p(1)$.