# Joint distribution of two weighted sums of IID random variables

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

• 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. – Iosif Pinelis May 1 '19 at 14:15
• @IosifPinelis and how would you guarantee such independence? – Penchez May 4 '19 at 9:52
• 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. – Iosif Pinelis May 5 '19 at 1:25