1
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ More specifically,what if we also know that $EX_n=0$ and $Var(X_n)=O(\alpha_n^2)$? $\endgroup$ Sep 7, 2021 at 1:37

1 Answer 1

0
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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, 2021 at 0:47
  • $\begingroup$ That's right. See the edit. $\endgroup$
    – Kota Mori
    Sep 7, 2021 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, 2021 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, 2021 at 3:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.