Let $X_1, X_2, \dots$ be independently uniformly distributed random variables in $\{-1, +1\}$ and let $a_1, b_1,a_2,b_2, \ldots \in \mathbb{R}$ be fixed, bounded and of non-zero average. Let $Y_n=a_1X_1 + \cdots + a_nX_n$ and $Z_n=b_1X_1 + \cdots + b_nX_n$.

I'm interested in understanding the joint distribution of $Y_n$ and $Z_n$ as $n\to\infty$ and more specifically in giving an upper bound for the probability $$\mathbb{P}[|Y_n| \leq x \land |Z_n| \leq y],$$ which is uniform in $x, y \in \mathbb{R}$ and $n\in\mathbb N$. A quick computation seems to reveal that the distribution and a bound of the form $$\mathbb{P}[|Y_n| \leq x \land |Z_n| \leq y]=O\!\left(\frac{(1+x)(1+y)}{n}\right)$$ should be attainable if the sequences $a_1,a_2,\dots$ and $b_1, b_2, \dots$ are sufficiently "independent" from each other.

I haven't been able to find anything in the literature about this, but I guess that this kind of problems should have been widely studied, so I'm looking for references about it. Thank you

  • $\begingroup$ I don't think you can get $n$ in the denominator unless you assume that $Y_n,Z_n$ are independent or very close to that. $\endgroup$ – Iosif Pinelis May 1 at 14:15
  • $\begingroup$ @IosifPinelis and how would you guarantee such independence? $\endgroup$ – Penchez May 4 at 9:52
  • $\begingroup$ I did not say that I could guarantee independence. However, if $a_jb_j=0$ for all $j$, then $Y_n$ and $Z_n$ will be independent. $\endgroup$ – Iosif Pinelis May 5 at 1:25

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