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One rule for small op and big Op is

$$ (1+o_p(1))^{-1} = O_p(1) $$

(on page 13 of Vaart, A. W. van der. (1998). Asymptotic statistics. Cambridge University Press.)

I am curious whether it is true to hold

$$ (1+o_p(1))^{-1} \overset{?}{=} 1+o_p(1) $$

My thought is that suppose $X_n\rightarrow_p 0$, then $X_n=o_p(1)$. Consider the function $g(x)=1/(1+x)$, then by the continuous mapping theorem (https://en.wikipedia.org/wiki/Continuous_mapping_theorem), we have

$$ g(X_n)\rightarrow_p g(0) = 1\,. $$

Is that correct, or am I missing something?

If so, why not use the tight result $(1+o_p(1))^{-1} = 1+o_p(1)$?

If not, is there any counterexample that $(1+o_p(1))^{-1} = O_p(1)$ holds but not $(1+o_p(1))^{-1} = 1+o_p(1)$?

PS: I have checked the proof of the rule in another question, Op and op Convergence Property Related Question, and it seems that the answer also applies the continuous mapping theorem. No explanation for the difference between $O_p(1)$ and $1+o_p(1)$.

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  • $\begingroup$ Are you sure that the notation "$X_n = 1 + o_p(1)$" has a defined meaning? The sources I have seen (including the wikipedia article for little o in probability notation) only define $X_n = o_p(a_n)$, not $X_n = o_p(a_n) + f(n)$. Maybe the hypothesis should be rewritten to "$(1 - o_p(1))^{-1} - 1 = o_p(1)$"? $\endgroup$
    – fblundun
    Nov 3 '21 at 23:12
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    $\begingroup$ @fblundun, not sure. Yeah, this point might be necessary for a rigorous definition. Just double-checked the textbook, all right-hand sides are either Op or op. But I am guessing for convenient proof, this notation can still be acceptable (and it seems that I came across lots of similar expressions in the proofs). I mean, if in some proof, after some calculation, we obtain $X_n=1+o_p(1)$, then we can easily conclude that $X_n-1\rightarrow_p 0$. $\endgroup$
    – ya wei
    Nov 3 '21 at 23:54
  • $\begingroup$ Maybe the weaker result was simply all the author needed for his/her proof? $\endgroup$ Nov 4 '21 at 16:22

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