Let $X_{1}, \ldots, X_{T} \in \{0, 1\}$ be a sequence of Boolean random variables with $$ \mathbb{E}[X_{t} | X_{1}, \dots, X_{t - 1}] = p_{t}. $$ Consider the sequence $Y_{t} := X_{t} - p_{t}$ (which can be written as a Martingale Difference sequence since $\mathbb{E}[Y_{t} | X_{1}, \dots, X_{t - 1}] = 0$). An application of Freedman's inequality yields that for a fixed $\delta \in [0, 1]$, $$ \bigg|\sum_{t = 1} ^ {T} Y_{t}\bigg| \le \mu V_{Y} + \frac{1}{\mu} \log \frac{2}{\delta} $$ with probability $\ge 1 - \delta$, for any $\mu \in [0, 1]$ (referencing Theorem 1 in this paper; the proof is correct), where $$ V_{Y} := \sum_{t = 1} ^ T \mathbb{E}[Y_{t}^2 | X_{1}, \dots, X_{t - 1}] = \sum_{t = 1} ^ {T} p_{t}(1-p_{t}). $$ However, clearly, this is not tight by simply considering the following construction: for a fixed large $N \in \mathbb{N}$, let $$ p_t = \begin{cases} \frac{1}{N}, &t=1\\ 0 & t \ge 2, \end{cases} $$ so that $Y_{1} = 1 - \frac{1}{N}$ with probability $\frac{1}{N}$ and $\frac{-1}{N}$ with probability $1 - \frac{1}{N}$, and $Y_{t} = 0$ for all $t \ge 2$.
For this sequence, Freedman's inequality above yields that $$ \bigg|\sum_{t = 1} ^ {T} Y_{t}\bigg| \le \mu \cdot \frac{1}{N}\bigg(1 - \frac{1}{N}\bigg) + \frac{1}{\mu} \log \frac{2}{\delta} $$ which is minimized at $\mu = 1$ since $N$ is large, $\delta = \frac{1}{N}$ to yield a bound $$ \Bigg|\sum_{t = 1} ^ {T} Y_{t}\Bigg| \le \frac{1}{N}\bigg(1 - \frac{1}{N}\bigg) + \log \frac{2}{\delta} = \frac{1}{N}\bigg(1 - \frac{1}{N}\bigg) + \log 2N. $$ However, the r.v. $|\sum_{t = 1} ^ T Y_{t}|$ is $\frac{1}{N}$ with probability $1 - \frac{1}{N}$ implying that the concentration obtained via Freedman is not tight (in particular, the $\log 2N$ term is undesirable).
Is there a tighter alternative to Freedman's inequality that holds for the particular form of $Y_{t} = X_{t} - p_{t}$, where $X_{t}$ is boolean? Any pointers would be greatly appreciated.
