Let $X_n, n \geq 1$ be a sequence of random variables that converges to zero in probability, that is, $\forall \varepsilon >0$, $$\lim_{n \to \infty} P(|X_n| < \varepsilon) = 1$$ Moreover, let $X_n=o_p(n^{-1})$, that is, $\forall \varepsilon >0$, $$\lim_{n \to \infty} P\left(\left|\frac{X_n}{n^{-1}}\right| < \varepsilon\right) = 1,$$ or equivalently, $\forall \varepsilon, \eta >0$, there exists $n_0$ such that for $n\geq n_0$, $$P\left(\left|\frac{X_n}{n^{-1}}\right| < \varepsilon\right) \geq 1-\eta,$$

My question is, what can we say about $E(X_n)$ when $n \to \infty$? For instance, is it true that $E(X_n)=o(n^{-1})$? Or more generally, is it true that $E(o_p(n^{-1}))=o(n^{-1})$? How can I prove so?


1 Answer 1


No, you cannot say anything like that. Consider $(X_n)_n$ defined as $$X_n=\begin{cases}0 &\text{ w.p. } 1- \frac{1}{n}\\ n^2 &\text{ w.p. }\frac{1}{n}\end{cases}$$

Then $\mathbb{E}[X_n] = n\xrightarrow[n\to\infty]{} \infty$ (and tweaking the example above, you can replace the growth rate $n$ by anything you'd like), but, for any $\varepsilon > 0$, $$ \mathbb{P}\{ n\lvert X_n\rvert < \varepsilon \} \geq 1-\frac{1}{n} \xrightarrow[n\to\infty]{} 1. $$

  • $\begingroup$ Can you figure if there is any assumption we can make such that my theory holds? $\endgroup$ Mar 27, 2017 at 22:16
  • $\begingroup$ @SlayingTitans Uniform integrability (UI) should do the trick for you I believe. You want to prevent things from going wrong on small sets and that's the typical condition to do that. Certainly convergence in probability and UI implies convergence in expectation, but I'm not 100% sure that will extend to your exact theorem. $\endgroup$
    – David
    Mar 27, 2017 at 22:21
  • $\begingroup$ Boundedness should be sufficient (quite a strong assumption). $\endgroup$
    – Clement C.
    Mar 27, 2017 at 22:23
  • $\begingroup$ @ClementC. I think boundedness should be much stronger than necessary. Domination by an $Y \in L_1$ is probably stronger than necessary, don't you think? $\endgroup$
    – David
    Mar 27, 2017 at 22:25
  • 1
    $\begingroup$ For the (overkill) boundedness assumption, say $\lvert X_n\rvert\leq A$ a.s. for all $n$: $$\mathbb{E}[\lvert X_n\rvert ] = \int_0^\infty \mathbb{P}\{ \lvert X_n\rvert \geq x\} dx \leq \int_0^\varepsilon 1dx + \int_\varepsilon^A\mathbb{P}\{ \lvert X_n\rvert \geq \varepsilon\} dx = \varepsilon + A\mathbb{P}\{ \lvert X_n\rvert \geq \varepsilon\}$$ which is going to be at most $2\varepsilon$ for all $n\geq n_\varepsilon$. $\endgroup$
    – Clement C.
    Mar 28, 2017 at 2:23

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