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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?

Akurishen
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  • I'm not famliar with this notation. What do you mean by $X_n = o_p(\epsilon_n)$? – Psychomath Jul 10 '24 at 09:12
  • This means that $\tfrac{X_n}{\epsilon_n}$ converges to zero in probability, that is, for any $\epsilon >0$, we have $\lim_{n \rightarrow \infty} P\left[|\tfrac{X_n}{\epsilon_n}| < \epsilon\right] =1$ – Akurishen Jul 10 '24 at 09:15
  • 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. – Guillaume Dehaene Jul 10 '24 at 09:37
  • I have exactly the same feeling. Although my impression was that if the sequence is $o_P(1)$, it should converge to zero in probability at a certain rate and I could pick $\epsilon_n$ to be of smaller order than this rate, but I am not able to write something rigorous. – Akurishen Jul 10 '24 at 09:42
  • Actually, after giving it some thought, wouldn't the property hold for any $\epsilon_n$ tending to 0? The argument I'm thinking of is that, for any bound $B$, there comes a point $n_0$ after which all $|\epsilon_n| \leq B$. Proving the limit behavior is then obvious, no? – Guillaume Dehaene Jul 10 '24 at 09:43
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    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 $o_p(1/n^2)$. – Akurishen Jul 10 '24 at 09:47
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    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! – Guillaume Dehaene Jul 10 '24 at 10:15

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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.

Joseph Expo
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