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

Question

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?

Answer 1

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.