My question is related the hitting time of not a random walk, but a more general martingale process.
Suppose we start with an arbitrary $x_0=x$ with $0\leq x\leq 1$. We compute $x_{t+1}$ from $x_t$ as follows:
\begin{equation} x_{t+1}= \begin{cases} x_t + \epsilon_t & \textrm{with probability $\frac{\epsilon'_t}{\epsilon_t+\epsilon'_t}$}\\ x_t - \epsilon'_t & \textrm{with probability $\frac{\epsilon_t}{\epsilon_t+\epsilon'_t}$} \end{cases} \end{equation} Note that this means $\mathbb{E}\left[x_{t+1} \bigm| x_t\right]=x_t$. The values for $\epsilon_t,\epsilon'_t$ are chosen such that $0\leq x_t\leq 1$ for all $t$. There are no other restrictions on $\epsilon_t,\epsilon'_t$, e.g. they can be dependent, etc. We stop when $x_t\in\{0,1\}$ (and say, it is guaranteed that we stop in at most $T$ steps).
I am looking for an upper bound on $\displaystyle\sum_{t=1}^T (x_{t+1}-x_t)^2$ in the form of \begin{align*} \Pr\left[\displaystyle\sum_{t=1}^T (x_{t+1}-x_t)^2 > \alpha\right] \leq \beta. \end{align*}
For instance, I think if $\epsilon_t=\epsilon_t'$ for all $t$, then hitting time of the standard brownian motion can give an upper bound? Thanks for your ideas!