# If $\hat{a}=O_{p}(\sqrt{\frac{logn}{nh^b}}+h^c)$, what is $\hat{a}^2$ in terms of $O_{p}()$?

If $$\hat{a}=O_{p}(\sqrt{\frac{logn}{nh^b}}+h^c)$$, where $$n$$ is sample size, and $$h$$ is bandwidth that also depends on $$n$$. What is the order of $$\hat{a}^2$$ in terms of $$O_{p}()$$?

More specifically, suppose $$\hat{a}^2=O_{p}(x_n)$$, then I'm not sure whether $$x_n=\frac{logn}{nh^b}+h^{2c}$$, or $$x_n=\frac{logn}{nh^b}+h^{2c}+2\frac{h^{2c}logn}{nh^b}$$.

Thanks!

• What does $O_p$ refer to? Aug 4, 2020 at 15:40
• @StephanKolassa It's the standard notation for stochastic boundedness. $\widehat{a}=O_p(x_n)$ means $\frac{\widehat{a}}{x_n}=O_p(1)$, i.e., $\frac{\widehat{a}}{x_n}$ is stochastically bounded. Aug 4, 2020 at 15:47
• Ah, thanks. My first association was algorithmic complexity. Good I didn't answer... Aug 4, 2020 at 15:48

Remember that $$Y_n=O_P(1)$$ means that, for every $$\epsilon>0$$, there is an $$M>0$$ such that $$P(|Y_n|\geq M)<\epsilon$$; and $$Y_n=O_P(a_n)$$ means that $$Y_n/a_n=O_P(1)$$. Suppose that $$Y_n=O_P(a_n)$$. Then, by definition, for every $$\epsilon>0$$, there is an $$M>0$$ such that $$P(Y_n^2/a_n^2\geq M^2) = P(|Y_n/a_n|\geq M) < \epsilon.$$ It follows that $$Y_n^2=O_P(a_n^2)$$.
Let $$a_n=b_n+c_n$$ and suppose that $$b_n c_n>0$$ eventually. If $$Y_n^2=O_P(b_n^2+c_n^2)$$, then, for every $$\epsilon>0$$, there is an $$M>0$$ such that $$P(Y_n^2/a_n^2\geq M)\leq P(Y_n^2/(b_n^2+c_n^2)\geq M) < \epsilon,$$ eventually, yielding that $$Y_n^2=O_P(a_n^2)$$.
• Thanks! This is very helpful. Another related question is: with your notation, suppose $a_n=b_n+c_n$, then does $Y^2_n=O_p(b^2_n+c^2_n)$ imply $Y^2_n=O_p(a^2_n)$? Or, can you construct a counterexample that shows it doesn't imply? Aug 4, 2020 at 18:20
• If $b_n c_n>0$ eventually, then yes, it does. Take a look at the end of the question.