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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!

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  • $\begingroup$ What does $O_p$ refer to? $\endgroup$ Aug 4, 2020 at 15:40
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    $\begingroup$ @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. $\endgroup$ Aug 4, 2020 at 15:47
  • $\begingroup$ Ah, thanks. My first association was algorithmic complexity. Good I didn't answer... $\endgroup$ Aug 4, 2020 at 15:48

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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 $Y_n^2=O_P(a_n^2)$.

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    $\begingroup$ 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? $\endgroup$ Aug 4, 2020 at 18:20
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    $\begingroup$ If $b_n c_n>0$ eventually, then yes, it does. Take a look at the end of the question. $\endgroup$
    – Zen
    Aug 4, 2020 at 18:47

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