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I am using the standard wikipedia definition of $O_p(1)$ and $o_p(1)$. Let $X_n$ be some sequence of random variables. It is known that $X_n = o_p(1) \implies X_n = O_p(1)$, whereas the reverse implication does not hold.

I was wondering if we could still make an even stronger statement:

$X_n = o_p(1) \implies$ there exists some sequence $\delta_n \to 0$ such that $X_n = O_p(\delta_n)$.

Of course, in many applications, this $\delta_n$ would go very slow to zero. I have not been able to come up yet with a possible counterexample where it does not go to zero. Note that the reverse implication of the above statement is definitely true:

$X_n = O_p(\delta_n) = \delta_n O_p(1) = o(1)O_p(1) = o_p(1)$.

Stan
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1 Answers1

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Recall that a sequence $(X_n)$ converges to $0$ in probability if and only if $\mathbb E\left[\min\{1,\lvert X_n\rvert\}\right]\to 0$.

If a sequence $(X_n)$ converges to $0$ in probability, then taking $\delta_n:=\sqrt{\mathbb E\left[\min\{1,\lvert X_n\rvert\}\right]}$, one has $\delta_n\to 0$. Moreover, for $n$ large enough, $\delta_n<1$ hence $$ \mathbb E\left[\min\left\{1,\frac 1{\delta_n}\lvert X_n\rvert\right\}\right]\leqslant \mathbb E\left[\min\left\{\frac{1}{\delta_n},\frac 1{\delta_n}\lvert X_n\rvert\right\}\right] =\frac 1{\delta_n}\mathbb E\left[\min\left\{1, \lvert X_n\rvert\right\}\right]=\sqrt{\mathbb E\left[\min\{1,\lvert X_n\rvert\}\right]}\to 0, $$ in particular $X_n=O_{\mathbb P}\left(\delta_n\right)$.

Davide Giraudo
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  • Thanks a lot for your reply Davide, this is very useful! Did you mean that $X_n = O_p(\delta_n)$ in your last statement? Moreover, in the display you showed that $X_n / \delta_n = o_p(1)$ using that $E[\min{1, |X_n|/\delta_n }] \to 0$, right? But doesn't this mean that you can similarly obtain that $X_n / \delta_n^2 = o_p(1)$? And also $X_n / \delta_n^{1000} = o_p(1)$? I think I'm missing something. – Stan Nov 06 '24 at 12:30
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    @Stan Thanks for having caught the typo. I do not think that we can get $X_n/\delta_n^2$ because the $\delta_n^2$ would not factorize. If you mean that you apply the argument say inductively, the $\delta_n$ will depend on $(X_n)$. More precisely, at the second step, you will need $\delta'_n=\sqrt{E[\min{1,\lvert X_n\rvert/\delta_n}]}$. – Davide Giraudo Nov 06 '24 at 12:45
  • Wow, thank you for a great response. I'm convinced that the characterization of convergence in probability that you cite is correct, but I'm not sure how to prove it, and I'm not having any luck finding it here or on google. I'm struggling with the direction: weak convergence implies $E \rightarrow 0$. Would you happen to know a reference proving it? (or if the argument is sufficiently small, just proving it). – Guillaume Dehaene Nov 06 '24 at 13:15
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    @GuillaumeDehaene I have added a link. More generally, some links to proofs to fact related to the convergence of random variables can be found here https://math.stackexchange.com/questions/2822452/collection-of-standard-facts-about-convergence-of-random-variables – Davide Giraudo Nov 06 '24 at 13:49
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    Thank you very much. As I expected, the proof is quite easy once you write down the right decomposition. – Guillaume Dehaene Nov 06 '24 at 15:30