1

If I have two functions (actually estimators), say $A_n(x)$ and $B_n(x)$, and I can prove $\sup_{x\in\mathbb{R}} |A_n(x)-B_n(x)|\leq o_p(1/n)$, can I conclude \begin{eqnarray*} \lim_{n\rightarrow\infty} \sup_{x\in\mathbb{R}}|A_n(x)-B_n(x)|=0 \; \; \; a.s. \end{eqnarray*} like that? If I can't, is there any way or lemma to prove something like this (strong conistency)? I have tried GC lemma, but it did not work, because my estimators are not step function. Thank you so much in advance

Rizky Reza Fujisaki
  • 586
  • 2
    What does the $p$ in $o_p$ signify? – Greg Martin Jul 22 '19 at 07:23
  • @GregMartin $p$ as in probability, roughly speaking, $o_p$ is similar to the usual little-$o$, but it is for random variable, so if an rv $X_n=o_p(1/n)$, then the rv will converge in probability to $0$ at the rate of $1/n$ – Rizky Reza Fujisaki Jul 22 '19 at 07:28
  • 1
    Are you asking if $\forall x, [\lim_{n\rightarrow\infty} |A_n(x)-B_n(x)|=0 ; ; ; a.s.]$ ? – Gabriel Romon Jul 22 '19 at 08:18
  • @GabrielRomon ah sorry, i forgot to put $\sup$ there, i edited my question, thank you so much, ok, so, can I conclude the almost surely-ness? – Rizky Reza Fujisaki Jul 22 '19 at 14:03
  • 2
    You're essentially asking if the following implication holds: $$[na_n \xrightarrow{P}0] \implies [a_n \xrightarrow{a.s}0]$$ Consider this question of mine with the hypothesis that the $X_n$ be indepedent. For any sequence of real numbers $u_n$ that goes to $0$ (e.g. $u_n=\frac 1n$), one can prove that $X_n=o_P(\frac 1n)$. However, as proved in the question, $X_n$ does not converge a.s. So you should not expect the implication to hold. – Gabriel Romon Jul 22 '19 at 14:44
  • @GabrielRomon i see, then I have to look for another way to solve it, ok, thank you so much – Rizky Reza Fujisaki Jul 22 '19 at 22:39

0 Answers0