2
$\begingroup$

By the converse of the strong law of large numbers, we know that, given a sequence of i.i.d random variables $X_1,X_2,\dots$ such that $\mathbb{P}(X_1 \ge 0)=1$ and $\mathbb{E}X_1= \infty$, then I have $$ S_N:=\frac{1}{N}\sum_{i=1}^N X_i \longrightarrow \infty \quad \mathbb{P}\textit{-a.s}. $$

I suppose that, just as in the case of the strong law of large numbers (without further assumptions on the moments of such random variables), we don't have a a-priori bound for the speed of divergence. My question is: given the law $\mathbb{P}_X$ of $X_1$, is there a well defined deterministic and diverging sequence $a_N=a_N(\mathbb{P}_X)$ and two positive but finite random variables $c$ and $C$ such that $$ \mathbb{P}\Big(\limsup_{N\to \infty} [S_N \ge C\cdot a_N] \Big)=0 $$ and $$ \mathbb{P}\Big(\limsup_{N\to \infty} [S_N \le c\cdot a_N] \Big)=0. $$

Is it possible to get bounds for $a_N$ in terms of the law $\mathbb{P}_X$?

$\endgroup$
2
  • $\begingroup$ I’m confused by the notation at the end — are you taking a liminf of an inequality, and is it with respect to $n$ or $N$? Also, can you provide a sequence that works if the variables are standard normals? $\endgroup$
    – user44143
    Commented Jul 15, 2020 at 10:46
  • $\begingroup$ Sorry, just fixed the n/N situation. Also, it was supposed to be limsup, not liminf. But I am referring to the limsup of events, putting in words I want to prove that $[S_N > C\cdot a_N]$ only happen a finite (but random) ammount of times almost surely. $\endgroup$
    – Kernel
    Commented Jul 15, 2020 at 11:04

1 Answer 1

2
$\begingroup$

Such a sequence $a_n$ does not exist even for a well studied example like returns to the origin of simple random walk in one dimension. If $X_i$ denotes the number of steps from the $i-1$ time the walk returned to the origin to the $i$'th time, then $X_i$ are i.i.d. and their sum $S_n$ is the number of steps until the $n$'th return time to the origin. In [1] $S_n$ is denoted by $\rho_n$ and we will follow this. Taking $\epsilon=1$ in Theorem 11.6 on page 119 in [1] we find that $$\rho_n>n^2 \log (n)$$ infinitely often almost surely, yet $$\rho_n<\frac{n^2}{\log \log n}$$ infinitely often almost surely. More precise information is in Theorem 11.5 there. To exclude "wild" sequences $a_n$, just separate cases, using Theorem 9.11 in [1] that yields for all $x \in (0,\infty)$, $$\lim_{k \to \infty} P(\rho_k <k^2 x) = f(x) , $$ where $f(x) \in (0,1)$ is explicitly given.

Case 1: If $a_n \le n^2$ infinitely often, then for any constant $C \in (0,\infty)$ we will have $P(\limsup [\rho_n>Ca_n]) \ge 1-f(C) $ by Fatou's lemma so $P(\limsup [\rho_n>Ca_n]) =1 $ by the Hewitt-Savage zero-one law.

Case 2: If $a_n \le n^2$ finitely often, then a similar argument shows that $P(\limsup [\rho_n \le c a_n]) =1 $ for all $c \in (0,\infty)$.

The situation is more extreme for return times of two dimensional simple random walk, for which see Theorem 20.5 page 218 in [1].

[1] P. Revesz, Random Walk in Random and Non-Random Environments, World Scientific Publ., Second edition (2005). https://www.worldscientific.com/worldscibooks/10.1142/5847

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .