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In which condition can I say that $X_n=O_p(\alpha_n)$ implies $X_n^{-1}=O_p(1/\alpha_n)$, since it holds for $X_n\sim N(0,1)$.(In this case, $X_n=O_p(1)$ and $\frac{1}{X_n}$ is a.s. finite and hence $\frac{1}{X_n}=O_p(1)$.)

From the definition of big O notaion, $X_n=O_p(\alpha_n)$ means, for $\forall \epsilon>0,\exists M_\epsilon,N_\epsilon$, that $P(|X_n|\geq M_\epsilon\alpha_n)<\epsilon$. To prove $\frac{1}{X_n}=O_p(\frac{1}{\alpha_n})$, I must find some $C_\epsilon>0$ to make the following probability small enough,

\begin{align} P(|\frac{1}{X_n}|\geq C_\epsilon/\alpha_n)&=P(|X_n|\leq \frac{\alpha_n}{C_\epsilon})\\ &=1-P(|X_n|> \frac{\alpha_n}{C_\epsilon})\\ &=? \end{align} Is something wrong with my proof?

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  • $\begingroup$ More specifically,what if we also know that $EX_n=0$ and $Var(X_n)=O(\alpha_n^2)$? $\endgroup$ Sep 7 '21 at 1:37
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Is the statement true? Suppose $X_n = n$ and $\alpha_n = n^2$.
Then, you can show that $X_n = O_p(\alpha_n)$ but $1/X_n \not= O_p(1/\alpha_n)$.

The problem is that $O_p$ only requires $X_n$ is "bounded above by" $\alpha_n$, not necessarily they are the "same order".

A sufficient condition for $1/X_n = O_p(1/\alpha_n)$ would be a kind of the opposite: $X_n$ is bounded from below by $\alpha_n$. You can write this formally as $\alpha_n/X_n = O_p(1)$.

If this is true, for all $\epsilon$ there exists $M_\epsilon$ such that: $P(|\alpha_n/X_n| \ge M_\epsilon) \to 0$. This implies $P(|1/X_n| \ge M_\epsilon / \alpha_n) \to 0$. So $1/X_n = O_p(1/\alpha_n)$. Note that I assume $\alpha_n > 0$ here.

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  • $\begingroup$ In general, the statement in the question is not true. I'm not sure what does $\alpha_n=O_p(X_n)$ mean, since $X_n$ is a set of the random variables and $\alpha_n$ is a constant sequence. $\endgroup$ Sep 7 '21 at 0:47
  • $\begingroup$ That's right. See the edit. $\endgroup$
    – Kota Mori
    Sep 7 '21 at 2:46
  • $\begingroup$ I think the sufficient condition $\alpha_n/X_n=O_p(1)$ is trivial since we can always divide both sides by $\alpha_n$. Am I missing something? $\endgroup$ Sep 7 '21 at 3:15
  • $\begingroup$ Yes, it is trivial. You need some ways to get $X_n$ bounded below to say something about $1/X_n$ because $O_p$-notation is only about "bounded above". You may want to narrow down the question about what kind of extra conditions you are looking for. $\endgroup$
    – Kota Mori
    Sep 7 '21 at 3:39

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