3
$\begingroup$

Suppose $X_i$, $i=1,2,...$, are iid random variables with $EX_i=0$ and $EX_i^2 < \infty$.

For each $x \in (0,1)$, $\frac{1}{\lfloor n x \rfloor} \sum_{i=1}^{\lfloor n x \rfloor} X_i \stackrel{a.s.}{\to} 0$ by the standard strong law of large numbers.

Clearly though $\sup_{x \in (1/n,1)} \left| \frac{1}{\lfloor n x \rfloor} \sum_{i=1}^{\lfloor n x \rfloor} X_i \right| \stackrel{a.s.}{\nrightarrow} 0$.

My question is: on what domain can the partial average argument be restricted to so that the maximally selected partial average does converge almost surely to zero? Namely, does there exist a sequence $a_n \to 0$ so that $\sup_{x \in (a_n,1)} \left| \frac{1}{\lfloor n x \rfloor} \sum_{i=1}^{\lfloor n x \rfloor} X_i \right| \stackrel{a.s.}{\to} 0$?

What is the smallest $a_n$ one can take so that this holds? How does the choice of $a_n$ depend on the distribution of the $X_i's$?

A minimal requirement when $a_n \to 0$ would seem to be $na_n \to \infty$. My only thought was to apply the law of the iterated logarithm, which would seem to imply $a_n\sqrt{n}/\sqrt{\log \log (n)} \to \infty$ is required. Is this required or optimal?

$\endgroup$
3
  • $\begingroup$ idk how this has 2 upvotes. this question should be closed. it's necessary and sufficient for $na_n \to \infty$, with the law of the iterated logarithm being completely irrelevant... unless I am just completely missing something $\endgroup$ Jun 11 '20 at 1:13
  • 1
    $\begingroup$ Could you provide the details then @mathworker21? $\endgroup$ Jun 11 '20 at 2:26
  • 2
    $\begingroup$ The question is equivalent to this: for which $b_n$ does it hold that $\sup_{k\ge b_n} \frac 1k \left|\sum_{i=1}^k X_k\right|\to 0$. $b_n\to \infty$ is clearly necessary (unless $X_i\equiv 0$). But it is also sufficient, since $ z_n\to 0$, $n\to\infty$ implies $\sup_{k\ge m} |z_k| \to 0$, $m\to\infty$. $\endgroup$
    – zhoraster
    Jun 11 '20 at 9:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.