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I have a sequence of real random variables $(X_n)_{n \in \mathbb{N}}$. If I know that there exists a sequence of strictly positive real numbers $(\epsilon_n)_{n \in \mathbb{N}}$ which is such that $\lim_{n \rightarrow \infty} \epsilon_n = 0$ and \begin{equation} X_n = o_{P}(\epsilon_n), \qquad \text{as } n \rightarrow \infty, \end{equation} it seems clear to me that it implies $X_n = o_P(1)$ as $n \rightarrow \infty$. My question is about the converse statement: if I know that $X_n = o_P(1)$ as $n \rightarrow \infty$, can I always construct a sequence of strictly positive real numbers $(\epsilon_n)_{n \in \mathbb{N}}$ satisfying the statements mentionned before?

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  • $\begingroup$ I'm not famliar with this notation. What do you mean by $X_n = o_p(\epsilon_n)$? $\endgroup$
    – LNT
    Commented 2 days ago
  • $\begingroup$ This means that $\tfrac{X_n}{\epsilon_n}$ converges to zero in probability, that is, for any $\epsilon >0$, we have $\lim_{n \rightarrow \infty} P\left[|\tfrac{X_n}{\epsilon_n}| < \epsilon\right] =1$ $\endgroup$
    – Akurishen
    Commented 2 days ago
  • $\begingroup$ Intuitively, I'm worried about there being a difference of scale between $o_P(1)$ and $o_P(\epsilon_n)$ with $\epsilon_n \rightarrow 0$. The second one feels smaller to me. though I'm unable to use this to formulate a counter-example. $\endgroup$ Commented 2 days ago
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    $\begingroup$ I don't think so. If the random variable $X_n = 1/n$ with probability one, then it converges to zero in probability but it is not $o_p(1/n^2)$. $\endgroup$
    – Akurishen
    Commented 2 days ago
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    $\begingroup$ Of course! This sounds like the type of result where, if it's true (and it does seem true, I agree with your intuition) then it requires some cool topological argument to pull out, and these have always been a huge weakness of mine. Best of luck! $\endgroup$ Commented 2 days ago

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Here's a partial answer.

Suppose not only $X_n\to 0$ stochastically, but even in $L^1$, i.e. $\mathbb E[\lvert X_n \rvert]\to 0$. Then choosing e.g. $\epsilon_n := \sqrt{\mathbb E[\lvert X_n \rvert]}$ we get via Markov's inequality that \begin{equation} \mathbb P(\lvert X_n\rvert \geq \epsilon\cdot\epsilon_n)\leq \frac{\mathbb E[\lvert X_n\rvert]}{\epsilon\cdot\epsilon_n} = \frac{\sqrt{\mathbb E[\lvert X_n\rvert]}}{\epsilon}. \end{equation} Since the root function is continuous, this of course tends to zero. Note that (with your definitions) you would probably need an additional assumption like $\mathbb P(X_n \neq 0)>0$ for all $n$ (we don't want to divide by $0$).

Takeaway: When looking for counterexamples you should focus on sequences of random variables that converge stochastically but not in mean. There are plenty of examples of such sequences out there. Furthermore, this could be useful to play around with.

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  • $\begingroup$ This is already of great help. Thank you for this answer! $\endgroup$
    – Akurishen
    Commented 2 days ago

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