# Variation of Hoeffding's Inequality for (weakly) dependent random variables

Is there a version, or modification, of this theorem for weakly dependent random variables? Or perhaps at least one for the special case involving Bernoulli random variables (that are now weakly dependent)? I can't seem to find anything formal on the subject.

I've stated Hoeffding's theorem below for when we have iid random variables.

Theorem (Hoeffding's Inequality). For iid random variables $$X_1, \dots, X_n$$ satisfying

$$a_i \leq X_i \leq b_i~\text{a.s.}, \\ \gamma_i = b_i - a_i, \\ \gamma_i \leq \Gamma_i,$$

Hoeffding's inequality says

$$\mathbb{P}\left[|S_n - \mathbb{E}[S_n]| > t\right] < 2\exp\left\{-\frac{2t^2}{\sum_{i=1}^n \gamma_i^2}\right\} < 2\exp\left\{-\frac{2t^2}{n\Gamma^2}\right\},$$ where $$S_n := X_1 + \dots + X_n$$.

Edit:

Weak Dependence. We can think of weakly dependent random variables in a time sense. That is, suppose we are given a time dependent sequence of random variables $$\{X_t\}_{t=1}^{\infty}$$. If we fix a $$t$$ and let $$s \in \mathbb{N}$$, then for any $$X_t$$ and $$X_{t + s}$$ as $$s$$ increases the $$\text{Cov}(X_t, X_{t+s})$$ decreases to $$0$$ asymptotically (e.g. exponential decay).

• just small covariance isn't enough to give a Hoeffding-like inequality; even pairwise independence (zero covariance) is not enough Oct 25, 2020 at 2:33
• Are there any such concentration bounds we can work with when we don’t have strict independence? Oct 26, 2020 at 2:56
• You can get the same bound if your variables are negatively associated, Proposition 5 in Dubhashi-Ranjan paper brics.dk/RS/96/25/BRICS-RS-96-25.pdf
– Max
Oct 26, 2020 at 10:00