Prove an anti-concentration inequality for a martingale My problem can be described easily:
I have a sequence $(X_l)_{l \in \mathbb{N}}$ of r.v. adapted to some filtration $(\mathcal{F}_l)_{l \in \mathbb{N}}$, such that 


*

*$\left|X_{l+1}-X_l\right|\le R$ a. s.

*$\mathbb{E}[X_{l+1}-X_l| \mathcal{F}_l]\le \delta$  a. s.

*$X_0 = x_0$ a.s. where $x_0 \in \mathbb{R}$ is fixed

*$\operatorname{Var}[X_{l+1}-X_l |\mathcal{F}_l] \ge v$ where $v>0$.


I want to prove $\mathbb{P}[X_l \le x_0-c\sqrt{vl}+\delta l]\ge p $ for some $p>0$ and some constant $c>0$. 
By using 2) one can easily reduce the problem to a martingale 
$$\bar{X}_l:= \left(\sum_{k=1}^l D_k - \mathbb{E}[D_k| \mathcal{F}_{k-1}]\right)+x_0$$ where $D_k:=X_k-X_{k-1}$. The martingale again has bounded increments, $\bar{X}_l=x_0$ and $\operatorname{Var}[\bar{X}_l]\ge vl$. The task then is to prove $\mathbb{P}[\bar{X}_l \le x_0-c\sqrt{vl}]\ge p$.
Sadly, the proof in the original paper (Lemma 6.5) was wrong and as I've found out one can't adapt it after having corrected the wrong inequality (Jensen's inequality used in the wrong direction). 
The only possibility I see so far is using the Martingale Central limit theorem and then try to control the speed of convergence for my original process using the inequalities described in the beginning of this paper. This seems pretty hard to check for my concrete considered process.
May I miss any obvious proof using simple martingale theory? Can anybody prove this result using only the above assumptions (or maybe some small additional assumptions, e.g. $L^{2+\delta}$ moments) ?
 A: Basically, the proof goes along the following lines:
(1) Take a small $\varepsilon>0$ and show that the expected exit time from the interval $[-\varepsilon\sqrt{vl},\varepsilon\sqrt{vl}]$ is less than $\varphi l$ (this is standard, using the fact that your martingale squared becomes a submartingale with uniformly positive drift, see e.g. Example 7.1 of Section 4.7 of Durrett/Probability) with small $\varphi$. Chebyshev's inequality then will show that you martingale will go out of that interval with probability close to 1 until time $l$.
(2) By the Optional Stopping Theorem, with probability bounded away from 0 (in fact, even close to $1/2$) it will exit the above interval through $(-\varepsilon\sqrt{vl})$.
(3) now, you only need to show that the walker will remain to the left of (say) $(-\frac{1}{2}\varepsilon\sqrt{vl})$ till time $l$.
(4) For this, first note that the Optional Stopping Theorem implies that, starting at $(-\varepsilon\sqrt{vl})$, the probability that the walker exits the interval $[-M\sqrt{vl},-\frac{1}{2}\varepsilon\sqrt{vl}]$ through the left side is at least constant (depending on $M$).
(5) Using the Doob's inequality, we then observe that the process is unlikely to reach $(-\frac{1}{2}\varepsilon\sqrt{vl})$ in time $l$.
You may find it interesting to look at the proof of Lemma 2.1 in https://arxiv.org/abs/1201.6089, it contains all these ideas.
