# Uniform almost sure convergence of the partial sum process

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?

• 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 Jun 11 '20 at 1:13
• Could you provide the details then @mathworker21? Jun 11 '20 at 2:26
• 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 $\sup_{k\ge m} |z_k| \to 0$, $m\to\infty$. Jun 11 '20 at 9:16