# Does $X_n=O_p(\alpha_n)$ implies $X_n^{-1}=O_p(1/\alpha_n)$

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?

• More specifically，what if we also know that $EX_n=0$ and $Var(X_n)=O(\alpha_n^2)$? Sep 7 '21 at 1:37

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.

• 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. Sep 7 '21 at 0:47
• That's right. See the edit. Sep 7 '21 at 2:46
• 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? Sep 7 '21 at 3:15
• 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. Sep 7 '21 at 3:39